There is a lot more fun to be had with cubed numbers! First of all, you have a four link loop: 33³ = 35937 - thirty five thousand nine hundred and thirty seven - 43 letters (exluding whitespace) 43³ = 79547 - seventy nine thousand five hundred and seven - 38 letters 38³ = 54872 - fifty four thousand eight hundred and seventy two - 42 letters 42³ = 74088 - seventy four thousand and eighty eight - 33 letters! On top of that there are also 4 stand alone end points! 41³ = 68921 - sixty eight thousand nine hundred and twenty one - 41 letters 51³ = 132651 - one hundred and thirty two thousand six hundred and fifty one - 51 letters 54³ = 157464 - one hundred and fifty seven thousand four hundred and sixty four - 54 letters 55³ = 166375 - one hundred and sixty six thousand three hundred and seventy five - 55 letters
And every number ends up at one of these fixed points or gets caught into this cycle with f(x) = number of letters in x^3 I wonder what happens in my language, Slovene...
@Ian O'Keefe I wonder what the longest chain is... Also, after "squared" and "cubed", do you use "TO THE n-th POWER"? 4096: TWO TO THE TWELFTH POWER so 20
Spanish has a loop of two at the numbers four and sick ("cuatro" and "seis"), but it also has a fixed point at five ("Cinco"). I find this interesting since it's never easy to tell if a number in Spanish will resolve to a fixed point at five or get trapped forever in the loop between six and four.
Using only numbers fifteen or smaller, Latin has an eight-length chain before the lasso: Quindecim (XV) --> novem (IX) --> quinque (V) --> septem (VII) --> sex (VI) --> tres (III) --> quattuor (IV) --> octo (VIII), and then it loops between quattuor and octo.
Matt, Having searched a couple of hundred languages, I finally found the first 5 cycle! Also, you were very lucky to have one of the few you checked have a four cycle (French), since of the ones I checked, I only found one other: Semelai, the cycle is 3 (hmpe) --> 4(hmpom) --> 5(mesong) --> 6(pru) --> 3(hmpe). This doesn't really count but also in Icelandic if the things you are counting are feminine, then 3(prjár) --> 5(fimm) --> 4(fjórar) --> 6(sex) --> 3(prjár). The five cycle which I finally found is in Sardinian, where: 3(tres) --> 4(battoro) --> 7(sette) --> 5(chimbe) --> 6(ses) --> 3(tres) Hope you are interested, and thank you for the video
I found another five loop in the Guanche language, which is the aboriginal language of the peoples from the Canary Islands. 5(simusetti) -> 9(alda-marava) -> 10(marava) -> 6(sesetti) -> 7(satti) -> 5 again
Quick note depends on how you spell them, because some conventions use k instead of ch to represent the word chimbe; plus, the word battoro, according to Sardinian orthographic rules, can also be spelt battor and this won't affect at all it's pronunciation
If you do this with Pokemon they all end at Blastoise. For example: Charmander (10) 10. Caterpie (8) 8. Wartortle (9) 9. Blastoise (9) I think the longest chain possible is: Fletchinder (11) 11. Metapod (7) 7. Squirtle (8) 8. Wartortle (9) 9. Blastoise (9)
In Danish the numbers 2(to), 3(tre), 4(fire) are all stopping points. After this minefield every number has more letters than its value so there are no loops.
Swedish has quite a few 3-letter numerals in single digits and 2 stop points, tre (3) and fyra (4). But since the majority of single digits have 3 letters, I would imagine 3 is where most sequences would stop.
Assuming no "and"s and no counting spaces Lowest 7 Chain: 323 Lowest 8 Chain: 1,103,323,373,373,373,373,373,373,373,373 The lowest number with a 9 chain would be super huge, somewhere around 10^100,000,000,000,000,000,000,000,000,000 by my estimation, as the write-out of the word would need at least the number of letters as the lowest 8-chain
10^10^29, that's more than a googol! (And that's understating it by 10^27 orders of magnitude... I think... these numbers are so big it's hard to tell when you've made a mistake) Edit: changed "googolplex" to "googol," thank you David!
So I decided to try this out in irish and I got an 8 chain at 2474440 (Dó millúin ceathar Céad seachtar ceathar míle ceathar céad ceatharacha) and I didnt even check if it was the lowest 8 chain, lowest 7 chain is 41 (ceathracha haon) Lowest 6 chain is 14 (Ceathar déag) lowest 5 chain is 10 (deich) then it goes to 5(cuig then 4(Ceathar) then 6(sé) then 2(dó) where the chain ends
If we do have and's and count spaces (naming convention of num2words in python), then: Lowest 7 Chain: 113 Lowest 8 Chain: 477,777,777 And for 9 we already reach over the naming convention in the english language
funfact about number in Indonesian and Malay, they have pattern : 1 (Satu) & 9 (Sembilan) both begin with "S" 2 (Dua) & 8 (Delapan) both begin with "D" 3 (Tiga) & 7 (Tujuh) both begin with "T" 4 (Empat) & 6 (Enam) both begin with "E" and if you add each pair, you got 10 as total
You'll love this one: In Korean we have TWO number systems (Sino and Korean), and we can spell them out in TWO language scripts (Hangul [Korean] and English [Romaja])! That gives us FOUR potentials to find K! I'm also not counting our numbers using Chinese character (hanja) but that could work just by counting the strokes, and they are still pronounced the same way anyways. So.... * NAME = word (number it represents, number of letters) > next item ... * SINO-HANGUL (#1) - 일 (1, 3 letters) > 삼 (3, 3 letters) > loop * SINO-HANGUL (#2) - 사 (4, 2 letters) > 이 (2, 2 letters) > loop * SINO-HANGUL (#3) - 십일 (11, 6 letters) > 육 (6, 3 letters) > 삼 (3, 3 letters) > loop * SINO-HANGUL (#4) - 이십삼 (24, 8 letters) > 팔 (8, 3 letters) > 삼 (3, 3 letters) > loop * SINO-ROMAJA (#1) - o (5, 1 letter) > il (1, 2 letters) > i (2, 1 letter) > loop with 1&2 * SINO-ROMAJA (#2) - pal (8, 3 letters) > sam (3, 3 letters) > loop * KOREAN-HANGUL (#1) - 하나 (1, 4 letters) > 넷 (4, 3 letters) > 셋 (3, 3 letters) > loop * KOREAN-HANGUL (#2) - 여덟 (8, 6 letters) > 여섯 (6, 5 letters) > 다섯 (5, 5 letters) > loop * KOREAN-ROMAJA (#1) - hana (1, 4 letters) > net (4, 3 letters) > set (3, 3 letters) > loop * KOREAN-ROMAJA (#2) - yeoseot (6, 7 letters) > ilgop (7, 5 letters) > daseot (5, 6 letters) > loop with 6,7&5 All other words loop in these ways. So 7 is the largest number I could find using Latin characters (looped with 6 and 5), and 5 using Hangul. Hope this helps!
btw I don’t speak Japanese so please correct me. This is for all the numbers from 1 to 10. And I’m including the つ counter for the kunyomis. ONYOMI-KANA いち (1, 2 characters) -> に (2, 1 character) -> loop All the other numbers fall into that loop KUNYOMI-KANA All of them have 3 kanas, except ここの (9) which has 4. とお (10) has two but you can’t use it with a counter. ONYOMI-ROMAJI There are 2 end points: ni (2, 2 letters) and san (3, 3 letters) The longest chain is shichi (7, 6 letters) -> roku (6, 4 letters) -> shi (4, 3 letters) -> san (3, 3 letters) KUNYOMI-ROMAJI There are 3 end points: muttsu (6, 6 letters), nanatsu (7, 7 letters) and kokonotsu (9, 9 letters). All the numbers except 9 and 10 have either 6 or 7 letters. 9 has 9 letters and 10 has 2 letters
Isn't five "cinco" or something like that? EDIT: Okay, I've checked, it's written in six letters. In fact, it's pretty easy to see how in Italian, everything collapses to three pretty fast.
The absolute value of the positive part of the complex number [insert first number] plus the square root of negative [insert second number] all to the power of one...
Russian in Cyrillic I find has 2 stopping points and one loop: 3 (три) has 3 letters 11 (одиннадцать) has 11 letters and the loop 4 (четыре) has 6 letters 6 (шесть) has 5 letters 5 (пять) has 4 letters
I've also found that in Tagalog (Filipino) we have 2 stopping points: 4 (apat) and 12 (labing dalawa), and in numbers under 100, about 10% stop at 12 and the rest stop at 4
+Julián García of course they count. You can't just say "пять" and assume it has 3 letters, 'cause then it would be "пят" and that's a completely different word!
Hey there! In Lithuanian language, five (5) is *_penki_* so there are five letters in this word *:D* Oh and seven (7) is *_septyni_* with seven letters as well!! And two (2) is *_du_* with only two letters!!! Isn't that cool??? For those who have no idea how to pronounce these words, then here is how to do it: _penki_ pronounce as _panky, du_ - as _doo_ and _septyni_ read as _saptiny_ ( or use a Google translator if you don't trust me :D )
And all the other numbers up to ten reverts right back to those three, and no number up to 100 have more than ten letters. So the norwegian numbers are really booring. No loops and no lines longer than three steps. Im sad now
I'm sure loads of these can be found. "Ten open bracket eleven plus ten minus nine close bracket divided by two" is 60 letters and 10 (11 + 10 - 9) ÷ 2 = 60, and this only took me a few attempts to find.
Although a little vague a "few" might be considered to have as many letters as the word describes. And if you take "a baker's dozen" and include the apostrophe them you get 13.
@@TheHadMatters I deleted my comment because I thought I miscounted but I'm pretty sure now that there was no "a" when he first wrote the comment, note the fact that it's edited.
I think Russian gets the award for the largest number of loops. Eleven has eleven letters (одиннадцать) Three has three letters (три) And there is a three cycle of 6-5-4 Шесть (6) has five letters Пять (5) has four letters Четыре (4) has six letters. Three cycles,thats pretty good!
Same for Danish. Most numbers have quite short names. 1:En, 2:To, 3:Tre, 4:Fire, 5:Fem, 6:Seks, 7:Syv, 8:Otte, 9:Ni, 10:Ti, 11:Elleve, 12:Tolv, 13:Tretten, 14:Fjorten, 15:Femten etc....
brachypelmasmith are you thinking about 97? there's a fun comic with UK: 90 + 7, German: 7 +90, Japanese: 9'10+7 etc.. and then Danish: 7 + (-1/2 + 5) x 20,Looks very complicated, but spoken it is quite simple: """syv og halv fems""" (syv og halv fem sinde tyve) would be the "proper old danish" way to say it, but not many people does it that way, now a days.
I have now found a romance language with a 6-cycle (without spaces): Romansh (Vallader), spoken in Switzerland. The cycle is 3 -> 5 -> 8 -> 2 -> 4 -> 7 -> 3: trais -> tschinch -> ot -> duos -> quatter -> set -> trais
Of the 2,686 languages I have checked, this is the only one with a "true" 6-cycle among the numbers from 1 to 10 (where "true" means that no spaces or hyphens are counted).
You should check my post about the Henese-Flick: th-cam.com/video/LYKn0yUTIU4/w-d-xo.html&lc=z12dj3bgxvbyhzl3k23cgd5yzz2tcn1g104 A 12-chain containg a 7-loop. (Hoping I made no mistake)
Here are the cycles for 103 languages. Other information you asked about is at the bottom. Each set of brackets is a chain. Commas separate numbers in a chain. Afrikaans [4] Albanian [2][3][4,5][6,7][11,12][14][16] Amharic [3] Arabic [4,5] Armenian [5] Azerbaijani [2,3] Basque [2][3,4][9] Belarusian [3][4,6,5][11] Bengali [2] Bosnian [3] Bulgarian [3] Catalan [3,4,6] Cebuano [4][12] Cichewa [13,14][29] Chinese [1] Corsican [3] Croatian [3][4,6] Czech [3] Danish [2][3][4] Dutch [4] English [4] Esperanto [2][3][4] Estonian [4] Filipino [4][11] Finnish [5][8,9] French [4,6,3,5] Frisian [3,5,4,7] Galician [5] Georgian [4] German [4] Greek [4,7][5] Gujarati [1,2] Haitian Creole [2][3] Hausa [3][4][5][11] Hawaiian [5][11] Hebrew [5] Hindi [1,2] Hmong [2][3][4,5] Hungarian [4] Icelandic [3,4,6] Igbo [3] Indonesian [4,5] Irish [2,5,7,6][4,8][11,13] Italian [3] Japanese [1] Javanese [4,5] Kannada [2,3] Kazakh [2,3][4,7] Khmer [2] Korean [1,2] Kurdish(Kurmanji) [2] Kyrgyz [2,3][4] Lao [3] Latin [3,6][4,8] Latvian [5][7][11] Lithuanian [2][5][6][7] Luxembourgish [5,6] Macedonian [3][4,6] Malagasy [4,6,5][15][16] Malay [4,5] Malayalam [2] Maltese [5] Maori [4][5] Marathi [2] Mongolian [3,5] Myanmar(Burmese) [2] Nepali [2] Norwegian [2][3][4] Pashto [3][4] Persian [2][4] Polish [4,6,5] Portugese [5][4,6] Punjabi [1,2] Romanian [5] Russian [3][4,6,5][11] Samoan [2,3,4][12,13,14] Scots Gaelic [3] Serbian [3][4,6] Sesotho [9][10][16][23,27][25,28,26] Shona [4][5] Sindhi [2] Sinhala [2] Slovak [3] Slovenian [3] Somali [3,6][4][7][11][12] Spanish [4,6][5] Sundanese [4] Swahili [3,4] Swedish [3][4] Tajik [2,4,3] Tamil [3][4] Telugu [2,3] Thai [3] Turkish [2,3][4] Ukranian [3][4,6,5] Urdu [2][3] Uzbek [4,6][8] Vietnamese [3,4][5] Welsh [3][6] Xhosa [5,6,9][15,16,20][17,18][21][25][27] Yiddish [3,4] Yoruba [5] Zulu [10][12][14][16][19][22][23][27] The highest number with a higher letter count than the number is 29 in Zulu with 33 letters. French, Frisian, and Irish have the longest chain(4). Zulu has 8 chains, and all of them are single numbers.
So in Hungarian, every single number up to a million (where the names of the numbers become only constructed names) ends in a 2 -> 5 -> 2 chain EXCEPT 4, 20, 100 and 1000, which are the only numbers to contain exactly four letters, so they end differently. Therefore, we have two chains, the "general" one, and the ones constructed to contain exactly 20, 100, 1000 and so on letters (starting with 239). Great time-sink. :)
If you are counting Hungarian letters (ty, gy, ny, sz counts as one letter) instead of characters, the "Great Whirlpool of 5(öt)-2(kettő)" will swallow everything, but more interestingly, all the chains will reach it through 3(három). So begin a chain anywhere, it will end up 3-5-2-5-2-5... All but one: 1(egy) - 2(kettő) - 5(öt).
(Seven plus seven minus seven plus seven minus seven plus seven minus seven ) times ten 😎 (70 letters and adds up to 70, in memory of Trondor Does Stuff)
+inscribed Then we would get 4 possible interlooping numbers/cycles like the case of the binary system, where k=1 [一] , 2 [二], 3 [三], (4 [四] & 5 [五] loop with each other). I'm gonna randomly guess that most numbers falls into a 2 loop or a 4-5 loop, since 7 [七] ,8 [八] ,9 [九] ,10 [十] are all 2 strokes and 6 [六] is 4 stroke. Or we could go with the "phonetic system" where the k=2(er), 3 (san), which is so very /exciting/...
Count characters: 1 through 10 have 1 character 11 through 20 have 2 characters 21 through 100 have 3 characters (except multiples of 10, which have 2) 101 through 1000 have (call the number x): x mod 100 is between 1 and 9 -> 4 characters x mod 100 is between 11 and 19 -> 4 characters (except in formal situations, which have 5 characters) x is a multiple of 10 but not a multiple of 100 -> 4 characters x is a multiple of 100 -> 2 characters otherwise -> 5 characters etc.
Chinese characters can be transcribed to pinyin, a phonetic spelling using the same alphabet as English, and Chinese numbers under that transcription might be interesting to think about
Polish has a three loop. (4) cztery --> (6) sześć --> (5) pięć --> (4) cztery German has a fixed point at vier (4) Dutch has a fixed point at vier (4) Spanish has a pair (4) cuatro --> (6) seis --> (4) cuatro AND cinco (5) as a fixed point
In base harmonic series, most numbers explode violently. To convert a number to base harmonic series, write down a "1" if the number is positive, or a "0" if the number is negative, then either add (if negative) or remove (if positive) 1/n from the number where n is the number of digits you've written down so far. Repeat this until the number reaches 0. Here's an example of some common numbers in base harmonic series: -1 : 0 -0.5 : 01 0 : (empty string!) 0.5 : 10 1.5 : 11 2: 111101010101010101011001101001100101101001011001101001011010011001011001101001100101101001011001101001100101100110100110010110011010011001011001101001100101100110100110010110011010011001011001101001100101100110100110010110011010010110100110010110011010011001011010010110011010011001011001101001011010011001011001101001100101101001011001101001011010011001011010010110011010010110100110010110100101100110100110010110011010010110100110010110100101100110100101101001100101100110100110010110100101100110100110010110011010010110100110010110100101100110100101101001100101100110100110010110100101100110100101101001100101101001011001101001100101100110100101101001100101101001011001101001011010011001011001101001100101101001011001101001011010011001011010010110011010011001011001101001011010011001011001101001100101101001011001101001100101100110100101101001100101101001011001101001011010011001011001101001100101101001011001101001011010011001011010010110011010011001011001101001011010011001011010010110011010010110100110010110011010011001011010010110(........ it keeps going for a while) edit: Of course, if you only care about positive integers, stop as soon as your number is less than one, and you get the unique representation for that positive integer in base harmonic series. For example, 8 is exactly 1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111
Ok, "base harmonic" is pathological but works. For the original system, this means some values would have an infinite series as their representation. I somehow want to declare that cheating. The integers only representation is interesting! But I'm getting an off-by-one error when I try it. I have seven as a string of 616 1s and eight as 1674 1s. Either way, it means writing it down gets bigger faster than the value of the number. It's like a tally, but worse.
Could you write down what you're doing? It took me a little while to figure out Notch's system; I think he traded comprehensibility for conciseness. Take (number) and add or subtract 1/(number of digits so far generated) for a response. (number) is the last response generated. Determine addition or subtraction by flipping the sign of the number you're using for that round. All responses should look binary. For 7: 7 is positive, first digit 1 We have 1 digit so far and our number is 7 which is positive, so we subtract: 7-(1/1)=6. The second digit is a 1. We're 2 in and had a response of 6 last time, 6-(1/2)=11/2, 3rd digit 1 3 in and at 11/2, (11/2)-(1/3)=31/6, positive, 4th digit 1 4 and 31/6, (31/6)-(1/4)=118/24. positive, 5th digit 1 5 and 118/24, (118/24)-(1/5)=566/120, +, 6th digit 1 So the first 6 digits of this representation of 7 are actually 111111. I'm doing this by hand/brain though and so I'm getting a little sick of it and do not have at my disposal a way of quickly getting through a whole bunch of these. It's late so I don't feel like going on or thinking about it any further. Maybe tomorrow. Neat though What I THINK you're doing is always subtracting from the original number and using the 'responses' as the accompanying digit. That would mean it would be possible for one of the digits of the representations to be a fraction. If you'll notice, single digits do not tend to represent a whole fraction. (69/103) is not a numeral.
Things like this always make me wonder how the physical world is constructed, and if it relates to any of these systems (as in how atoms are constructed from similar parts, but have massive differences in their properties). Thanks for the info on harmonic series! :D
Zulu has two loops with relatively high k values: 19: nesishiyagalolunye 18: ishumi nesishiyagalombili 24: amashumi amabili nane Then back to 19: nesishiyagalolunye AND 8: eziyisishiyagalombili 21: amashumi amabili nanye 20: mabini 6: eziyisithupha 13: thirteen Then back to 8: eziyisishiyagalombili
On the other side of things, it can be fascinating sometimes to look at the etymologies of numbers across languages, and see what numbers are hiding inside them. For example, eleven and twelve go back to something that translates back as one-left and two-left. There's a cross linguistic tendency for words for five to be related to a word for hand, in English's case, five, fist, and finger seem to be connected. Seven is structurally really weird across the indo-european languages, and seems to go back to a proto-form akin to *septom, which looks suspiciously like the Akkadian word for seven (and other Afrasian languages, like Arabic, Ancient Egyptian, and others), which was something like sabhat. Four is a bit of a mystery, but has been made out by some linguists to be derived from a verb *kwet meaning "to join, pair", as a pairing of pairs. Eight is a dual stem of what was apparently an old measure of four fingers. Nine used to be thought to be related to new, and that our linguistic ancestors had a base 8 system, but the modern theory is that nine derives from a rare word *henu meaning "to lack", related to Greek aneu and German ohne, meaning "without". Ten and hundred are thought to derive from *kemt, the word for hand; *de-kmt evidently being something like "two hands", (de-)kmt-om being a collection of tens. One in the IE languages generally has one of two etymologies, as well: either from a word like *oinos, which was probably a pronoun originally, and a word for a collective unity, *sem, seen in words like "single", "same", "similar", etc. On the topic of other languages, maybe the neatest I can think of off the top of my head is Old Japanese's. Modern Japanese uses Chinese numbers in some contexts (like how we use Latin numbers in some contexts), and their native numbers have undergone some sound changes that obscure certain patterns, but there's a ton of good evidence to support Old Japanese's numbers looking like: 1: pito 2: puta 3: mi 4: yo 5: itu 6: mu 7: nana 8: ya 9: kokono 10: towo Note the relationships between 1 and 2, 3 and 6, 4 and 8, and maybe 5 and 10. 7 and 9 are then the odd ones out,
***** Nice! After this point, we get into hyphenated number words and things get contentious. But assume we stop at 20 - which is a pretty good run - now the question becomes, which of these 20 numbers is the most popular for all other number words to end on after reducing them like in the video?
How shall we represent the number 21 in this system? Twenty-one or Twentinapaloozopolis-one? If we're going with the latter, then 24 is another stopping point (unless we count the hyphen as a letter, in which case, the new stopping point becomes 25).
no 21 is Twentinapaloozopolis-a Twentinapaloozopolis-tw Twentinapaloozopolis-Thr Twentinapaloozopolis-Four Twentinapaloozopolis-Fiver Twentinapaloozopolis-Sixers Twentinapaloozopolis-Sevener Twentinapaloozopolis-Eighters Twentinapaloozopolis-Ninerrino thirty Making the train stop at 30 or 29 depending on how you look at it
1 - Yksi 2 - Kaksi 3 - Kolme 4 - Neljä 5 - Viisi 6 - Kuusi 7 - Seitsemän 8 - Kahdeksan 9 - Yhdeksän 10 - Kymmenen So I tested out with Finnish, and ended up with a constant, 5, which has five letters and leads up from most words, 1 to 4, and 2-6 leading to 5. There was also a loop, between 8, having 9 letters, and 9, having 8 letters. I hope you found this interesting :D
I tested out ordinal numbers: There is loop between 8. and 9. (kahdeksas and yhdeksäs). other ones are 6. (kuudes) 11. yhdestoista 12. kahdestoista Even though 21. (kahdeskymmenesensimmäinen) has 25 letters it goes to 25 --> 20 --> 14 --> 12
It has been nearly 20 years but I remember learning in school that "and" in a number denotes a decimal point. Four hundred ninety six would = 496. Four hundred AND ninety six would = 400.96. Likely things have changed in 20 years but this is how I was taught.
In the UK, it's standard practice to use "and" as a hundreds separator (e.g. 1,234,567 = one million, two hundred and thirty four thousand, five hundred and sixty seven) with "point" as the decimal separator, with decimals spelled out as individual numerals (e.g. 1,234,567.8901 = one million, two hundred and thirty four thousand, five hundred and sixty seven point eight nine zero one). However, I'd hazard a guess that UK / US aren't the only two countries separated by a common language (in this case, two countries sharing a common language, where standard practice results in different methods of enunciating numbers).
I was never taught this. And is used for anything below 100 and never else. Have always said 'point' for the decimal point and only say it other than a sequence of numbers for prices (and never with and). (UK/27)
Maybe thats what they teach in american schools, but to me that sounds wildly incorrect, like I think your teacher was wrong. I dunno though. I would have thought four hundred point ninety six would be universally used in english.
That doesn't really make sense. I mean maybe it was like that, but "and" isn't even shorter than saying "point", so it's just gonna bring you tons of confusion for no benefit.
I was taught that "and" can only indicate a decimal point if the reader of the words indicate a fraction. For example: One hundred and seventy three thousandths (One hundred) AND (seventy three thousandths) 100+ (73/1000) 100.073
Hebrew is crazy interesting about this. In Hebrew, you can count in 2 forms - male of female form. Most countable objects are male (building, wall, video, trophy, and many more, obviously) and less objects are referred to in the female form (finger, car, door and many more...). That Hebrew rule is the source of many grammar mistakes, even by native Israelis. If we count in the male form - you always end up circling around the number 5, which has 5 letters in the word "hamisha" - the male form of 5. If we count in the female form - you always end up circling around the number 4, which has 4 letters in the word "arba" - the female form of 5. Here is the chains from 1 to 10, the number of letters in Hebrew is stated in brackets. You'll see that Hebrew has less letters then English because most vowels are dropped when writing in Hebrew (and "sh" is one letter in Hebrew). Male form counting: 1. Ehad (3) > Shlosha (5) > Hamisha (5). 2. Shnaim (5) > Hamisha (5). 3. Shlosha (5) > Hamisha (5). 4. Arba'ah (5) > Hamisha (5). 5. Hamisha (5). 6. Shisha (4) > Arba'ah (5) > Hamisha (5. 7. Shiv'ah (4) > Arba'ah (5) > Hamisha (5). 8. Smona (5) > Hamisha (5). 9. Tish'ah (4) > Arba'ah (5) > Hamisha (5). 10. Asara (4) > Arba'ah (5) > Hamisha (5). All chains ending in 5! Female form counting: 1. Ahat (3) > Arba (4). 2. Shtaim (5) > Hamesh (3) > Shalosh (4) > Arba (4). 3. Shalosh (4) > Arba (4). 4. Arba (4). 5. Hamesh (3) > Shalosh (4) > Arba (4). 6. Shesh (2) > Shtaim (5) > Hamesh (3) > Shalosh (4) > Arba (4). 7. Sheva (3) > Shalosh (4) > Arba (4). 8. Shmoneh (5) > Hamesh (3) > Shalosh (4) > Arba (4). 9. Tesha (3) > Shalosh (4) > Arba (4). 10. Eser (3) > Shalosh (4) > Arba (4). All chains ending in 4!
I wonder how much different it would be if you didn't include vowel letters (e.g. the vav in shmoneh, making shmoneh 4 instead of 5) since they're technically optional. The first thing that comes to mind is that shalosh would become 3, making it another loop. I haven't thought through all the numbers yet, but it's interesting!
In Greek, 5 (πέντε) is the fixed point and we also have a 2 number loop that goes from 4 to 7. 4 (τέσσερα), 7 (εφτά) Most numbers end up in the 4-7 loop and not at 5! e.g. between the numbers 1-10 you can only go to 5 from 9 (εννιά). Every other number ends up at 4-7. If you go to 1-20, 16 of these end up in the 4-7 loop and 4 at 5.
So let's try Japanese Starting at 100 (hyaku) has five (go) has two (ni) which ends the cycle. However, there is another cycle in reverse starting at one (ichi) has four (yon) has three (san) which ends the cycle But wait there's more! In archaic counting one (hitotsu) has seven (nanatsu) which ends the cycle but in reverse 100 (momo) has four (yottsu) has six (muttsu) which ends the cycle Let's not go into the Japanese furigana characters shall we, I'm done
yeah but you're basing it off of romaji, if you instead base your counting based on how many syllables, all numbers result in an alternation between ichi and ni
That's just as arbitrary as going by the symbols used to transcribe (or shall I rather say "obfuscate" …) the sounds of English words, but you could just as well spin some arbitrarily silly scheme around the sums-of-digits of the ENIGMAed, LZ7-compressed, Morsed, ROT13ed words that you get from expressing the numbers in resistor colours. That said, for a bit of variety more you could go by stroke count of the Kanji numerals, for example.
Anvilshock That’s a good point, and I should have mentioned that yes it’s arbitrary, i’ve actually been doing this pattern from 1-10 in the numbers of lots of different languages recently and found some cool results. Such as German, Dutch, and Hebrew all go to 4 like english does. I like this pattern a lot because it combines my two favorite subjects ever: Linguistics and Math.
@@Anvilshock Everything is arbitrary. But as a matter of fact, we definitely should check kanji stroke order. And let's use the more archaic kanji for small numbers that have far far more strokes
Was thinking the same! They seem to know eachother already, as they have done videos together in the past. +standupmaths Matt: Get Tom Scott involved when he recovered from the Arctic!
The only problem with negatives is that you can only start with them no matter how you express it. So they don't end up being too interesting if you're looking at cycles and such. Library I'm using for this uses "minus" although I could change that. Either way this is a nice little thing.
Esperanto: Unu (3) Du (2) Tri (3) Kvar (4) Kvin (5) Ses (3) Sep (3) Ok (2) Naŭ (3) Dek (3) In Esperanto, two, three, and four all loop to each other, so every number ends in one of those three!
I guess most numbers in Esperanto have sequences that end in 3, seeing as 6 to 10 all have three letters, except for 8. And most numbers between 11 and 99 have 6 to 10 letters.
Zero (even w/even number of letters) One (odd/odd) Two (even/odd) Three (odd/odd) Four (even/even) Five (odd/even) Six (even/odd) Seven (odd/odd) Eight (even/odd) Nine (odd/even) Ten (even/odd) FWIW.
If we're doing number of strokes, Chinese has 3 1-loops (1,一;2,二;3,三; each has the same number of strokes as their number) and 1 2-loop (4,四 -> 5,五). On another note, zero has 13 strokes(零). The whole thing gets much more complicated if you use the "majuscule" numbers. (1-壹,12 strokes;2-贰,9;3-叁,8;&c.) EDIT: (copypasta from my next comment) Majuscule numbers have a k=28 (貳拾捌, 28 strokes total), which I think is the record so far. There are also five loops: 2 1-loops (27 貳拾柒, 27 strokes; the aforementioned 28), 2 2-loops (7 柒 9 strokes => 9 玖 7 strokes; 17 壹拾柒 30 strokes => 30 叁拾 17 strokes) and 1 4-loop (24 贰拾肆 31 strokes => 31叁拾壹 29 strokes => 29 贰拾玖 25 strokes => 25 贰拾伍 24 strokes). I'm using the accountant's system where numbers like 14 are written like one-ten four (壹拾肆) rather than ten four (拾肆). Also, I'm using simplified majuscule characters. Traditional majuscule characters have more strokes on 3 and 6, as well as the 10000 and 100000000 orders.
Majuscule numbers have a k=28 (貳拾捌, 28 strokes total), which I think is the record so far. There are also five loops: 2 1-loops (27 貳拾柒, 27 strokes; the aforementioned 28), 2 2-loops (7 柒 9 strokes => 9 玖 7 strokes; 17 壹拾柒 30 strokes => 30 叁拾 17 strokes) and 1 4-loop (24 贰拾肆 31 strokes => 31叁拾壹 29 strokes => 29 贰拾玖 25 strokes => 25 贰拾伍 24 strokes). I'm using the accountant's system where numbers like 14 are written like one-ten four (壹拾肆) rather than ten four (拾肆). Also, I'm using simplified majuscule characters. Traditional majuscule characters have more strokes on 3 and 6, as well as the 10000 and 100000000 orders.
+Litigious Society Chinese is different. In the PRC, for example, there is a State Language Commission that standardizes the number of strokes per character. Other countries using Chinese also have similar groups. (Really, the concept of strokes is not transferable to other scripts.)
You could also count characters: 1 through 10 have 1 character 11 through 20 have 2 characters 21 through 100 have 3 characters (except multiples of 10, which have 2) 101 through 1000 have (call the number x): x mod 100 is between 1 and 9 -> 4 characters x mod 100 is between 11 and 19 -> 4 characters (except in formal situations, which have 5 characters) x is a multiple of 10 but not a multiple of 100 -> 4 characters x is a multiple of 100 -> 2 characters otherwise -> 5 characters etc.
So I have taken years of Latin and never thought of this despite knowing the trick in English. Jumping into this i went of what are the terminal places. You end up with a 2-cycle of 4 (quattuor) and 8 (octo). well nothing interesting there but as I am checking I notice the number 5 (quinque) has a decent chain already with 5 (quinque) -> 7 (septem) -> 6 (sex) -> 3 (tres) -> 4 (quattuor) and the start of the cycle. A chain of 5 long already. building on this 9 (novem) is the first number that has 5 letters. so a 6 chain before we get to double digits. 15 (quindecim) is the first to have 9 letters; 24 (vīgintī quattuor) is the first to have 15 letters (not counting the space). finally 145 (centum quadrāgintā quīnque) gets you 24 letters. for a full chain of: 145 (centum quadrāgintā quīnque) -> 24 (vīgintī quattuor) -> 15 (quindecim) -> 9 (novem) -> 5 (quinque) -> 7 (septem) -> 6 (sex) -> 3 (tres) -> 4 (quattuor) So a chain of 9 numbers. I guess I kinda of combined two of the challenges. I found a chain longer than 7 in a language with a non single point end.
+Ducomors it occurs to me that you didn't count 8 in your loop for the final tally meaning a possible loop of 10 depending on if you count the loop or just the first number in the loop.
@@alauber86 I guess when looking for the longest chain, it has to be a chain where all numbers are different. So if there's a fixed point, you stop there, so in English the last number in a chain is FOUR. But when you hit the loop, you can continue without repeating numbers... That's until you list the whole loop, the next number would them be a repetition, so you stop.
I think the parker conjecture is right. Any language that's worth its salt will have a system of numbers that reduces larger numbers into smaller words. If the way to say a number is comparable or longer to say than the value of the number, you're doing your job wrong. The only languages where the parker conjecture would be false are ones with bad counting systems (which shouldn't survive linguistically), or made up ones that have terrible counting systems. Japanese could be a fun one to look at. If you did it in terms of the local syllabary, you get a two loop between one and two (いち for one (two "letters") and に for two (one "letter"))but if you go into romaji (follows your examples more closely) you get the list: n name f(n) 1 ichi 4 2 ni 2 3 san 3 4 yon 3 5 go 2 6 roku 4 7 nana 4 8 hachi 5 9 kyuu 4 10 juu 3 Numbers do go above ten, but each iteration cuts the value down drastically. Below I have the largest possible strings for ten to ten thousand, as well as the format for large numbers in japanese: numbers above ten: n (above) juu (3) n (above) (largest family is hachi juu hachi (13) goes to juu san (6) goes to roku (4) to yon (3) back to san) numbers above a hundred: n hyaku (5) n juu(3) n (largest is hachihyakuhachijyuuhachi (23) nijuusan(8) hachi (5) go (2) ni(2) end loop) numbers above a thousand: n sen(3) n hyaku(3) n juu(3) n (largest is hachisenhachihyakuhachijuuhachi (31) sanjuuichi (10) juu (3) san(3) end loop) japanese also goes to ten thousand as a large number base, compared to english so: above ten thousand n man (3) n sen(3) n hyaku(3) n juu(3) n (largest is hachimanhachisenhachihyakuhachijuuhachi (39) sanjuukyuu(10) as above) this then repeats for numbers over ten thousand (ten ten thousands, a hundred ten thousands etc...) and is overall similar with small two or three letter prefixes marking each transition of ten thousands. Now, a bunch of numbers have odd quirks. For example, up there the number for eight thousand should be hassen and not hachisen, but things like this don't make enough of a difference to matter. The parker conjecture holds. You can see k = 3. everything should converge on that there. There are also a number of cases that converge on two. Pretty simple. Now, Japanese also has a local counting system for objects which is also used very frequently. Arguably as often as the numbers above. However, there are only a few of these, and they are used for small everyday numbers (two people, three elephants, five apples, that sort of thing.) interestingly, these counting numbers have three possible k values. Also interestingly, I do not know of a way to count above these possible values in this counting system, making the statistics easy. Here they are: n name f(n) 1 hitotsu 7 2 futatsu 7 3 mittsu 6 4 yottsu 6 5 itsutsu 7 6 muttsu 6 7 nanatsu 7 8 yattsu 6 9 kokonotsu 9 10 tou 3 as we can see, we've got k at 6,7, and 9. 4 numbers end directly at 7, 4 numbers end directly at 6. 10 goes to three then to six. only nine ends at nine. therefore we have a 50% chance of ending at 6, a 40% chance of ending at 7, and a 20% chance of ending at 9. of course, all this goes out the window if you decide to use hiragana (the local alphabet, of sorts, where a symbol represents what we might take two two right out ex: な for "na") to count "letters". This is really complicated, mostly because it's hard to tell what's a letter (is じゅう (jyuu, spelled "Ji" + small "yu"+ "u") three or two letters, since the yu is small and technically part of the ji? what about numbers like the above ハッセン(hassen, spelled "Ha" + small "tsu" + "se" + "n") is that three or four? since the small thing is technically not pronounced, and not really a letter like it is when it's normally used, but it still needs to be there to recognize the word.) Of course, these are all technicalities, and don't matter much even in the bigger picture. Even if you take the letter count by hiragana, all it does is shave the letter count (f(n)) for every number down compared to the english lettering or romaji (same thing). On average it should divide it by two, but I suspect the actual amount is closer to 2 and a half (since some hiragana have three letters in english, most have two, and one has one) You get the following first ten numbers: n name f(n) 1 いち 2 2 に 1 3 さん 2 4 よん 2 5 ご 1 6 ろく 2 7 なな 2 8 はち 2 9 きゅう 2 or 3 10 じゅう 2 or 3 You will get a loop between one and two ((1)いち↔に(2)). But the point is, regardless of what you do, for the Japanese language, the Parker conjecture holds. Just depends on how you test it.
Dang it, someone beat me to Japanese! I like was going to use the Romaji System as it would allow people to try and say the number. Not to mention, as you pointed out, writing it in hiragana vs romaji creates two separate systems for Parker's Conjecture here. In Hiragana, everything converges on 1 & 2. But in Romaji, everything will converge on either 2 or 3. As both of these numbers are self-feeding like English's "Four". Kanji would just be cheating. everything 10 and under would just be 1, and even very large numbers would drop to 1 in 2 or 3 steps.
In Vietnamese language, we have “hai” for 2 and “ba” for 3. Interestingly, these two numbers repeat itself no matter what number you start with, just like in English and other. What a fun fact!
Matt! I have semi proved your conjecture. Obviously, language systems are going to be extremely complicated and sometimes just unintuitive, but by putting some restraints on what you can do, I can prove that your conjecture is true. So, in any reasonable language, the number of letters needed to express a number should correspond linearly with how many digits it has. Formally, if we use your function f(x) as the length of a number when written out, and if d(x) is the number of digits that a number has, f(x)0 such that whenever x>r, we have C*log10(x)0, there exists an r>0such that whenever x>r, we have log10(x)/x
Note: I'm only 13, so this also might be wrong, but I've checked it with logical proof checkers and it isn't really super complicated so I am pretty sure it is devoid of holes.
Chris Daley XD Not at all. It took like 9 lines and I could condense it to 7 or even 6 if I made it a stand-alone LET expression instead of a proper function. Well, technically I could make it all 1 line but that wouldn't be too readable.
Chinese doesn't have a letter system, but there are strokes for each character. An analogue to the number letter mapping is then the number stroke mapping. Numbers one to three (一二三) are attractors, mapping to themselves, but four and five (四 and 五) map to each other in a loop. One is the lonely number. After that, all other numbers lead back to the numbers two through five, with most going to two, fee going to three, and the minority ending in the four/five loop.
Using "and" between the hundreds and the tens digits (i.e., 65536 is "sixty-five thousand, five hundred and thirty-six" but 1500 is "one thousand, five hundred") (and of course removing spaces and hyphens), I found the smallest number with a 7-chain: 124 (one hundred and twenty-four) has 23 letters, 23 (twenty-three) has 11 letters, 11 (eleven) has 6 letters, 6 (six) has 3 letters, 3 (three) has 5 letters, 5 (five) has 4 letters, and 4 (four) has 4 letters. The next smallest numbers that form 7-chains are 125, 129, 134, 135, 139, 143, 147, and 148, all with 23 letters. The smallest number with an 8-chain would therefore have at least 124 letters. Using long-scale (i.e., thousand, million, billion, trillion, etc.), I found that 113373373373 (one hundred and thirteen billion, three hundred and seventy-three million, three hundred and seventy-three thousand, three hundred and seventy-three) has 124 letters (which makes it an 8-chain), but I don't know whether this is minimal. I haven't checked whether there is a smaller number with (say) 125 or 129 letters. I am sure we will never make up enough names for powers of 10 to be able to get to a number with a 9-chain (which would by definition have at least 113373373373 letters). For context, the complete works of Shakespeare (including spaces, punctuation, and all) contain about 3.6 million characters. English Wikipedia currently contains about more than 2.9 billion words (approximately 15 billion characters, including spaces and punctuation). Wikipedia in all languages put together currently contain more than 27 billion words (approximately 130 billion characters (finally bigger than 113373373373, though including spaces and punctuation)). Omitting "and" (i.e., 65536 is "sixty-five thousand, five hundred thirty-six" and 1500 is still "one thousand, five hundred"), I found that the smallest number with a 7-chain is 323 ("three hundred twenty three" has 23 letters, then proceed as before to 11, 6, 3, 5, and 4). The next smallest are 327, 328, 333, 337, 338, 374, 375, and 379, all with 23 letters. Here, the smallest number with an 8-chain would therefore have at least 323 letters. Using long-scale again, I found that 1103323373373373373373373373373 (one nonillion, one hundred three octillion, three hundred twenty-three septillion, three hundred seventy-three sextillion, three hundred seventy-three quintillion, three hundred seventy-three quadrillion, three hundred seventy-three trillion, three hundred seventy-three billion, three hundred seventy-three million, three hundred seventy-three thousand, three hundred seventy-three) has 323 letters, but again I'm not sure whether this is minimal. There will never be a number with 1103323373373373373373373373373 letters. Typing the letters at a rate of one trillion letters per second, non-stop since the Big Bang, you would be be about 37% of the way done. One page in Microsoft Word at Times New Roman 1-point font has approximately 610000 characters (including spaces and punctuation). Typing a number with so many letters out in 1-point font, printing it out double-sided, and stacking the pages would produce a stack that goes from Earth to Alpha Centauri and back -- more than 1100 times!
Seems like a dynamic programming (DP) problem. You could build up the smaller solutions (for example: chains of 3) so that when you're investigating larger numbers, you can recognize that you've fallen into a chain you've seen before (and add it to an index)
German does too. but im pretty sure chains in german will be really long, considering 7 has 6 letters and 6 has 5 letters. zweihundertneunzehn -> zwanzig -> sieben -> sechs -> funf -> vier. is the longest i've found without going ridiculous. the chains actually turn out small, because 13-19 all have 8 letters, and 0-9, excluding six and 7 all have 4, so any 'teen' (zehn) goes to 8, which goes to 4 then loops.
Odd 'even' numbers: 7,11,17,27,37,47,57,67,71,73,75,77,79,87,97 Even 'even' numbers: 70,72,74,76,78 Then the pattern repeats. Ergo, 75% of 'even' numbers are odd, until you get to larger blocks of 'even' numbers (ie. 700, 7000, 70000, etc.) and then they progressively fall, tending towards 50% but never reaching it.
In italian, 6 has 3 letters ("sei"), 8 has 4 letters ("otto"), 10 has 5 letters ("dieci"), 12 has 6 letters ("dodici"). There is a very popular italian riddle based on this pattern
In italian you can do a chain of seven numbers starting from 15: 15: QUINDICI (8 letters) 8: OTTO (4 letters) 4: QUATTRO (7 letters) 7: SETTE (5 letters) 5: CINQUE (6 letters) 6: SEI (3 letters) 3: TRE (3 letters)
Now it is time to break the record :D In polish a 7 link chain starts with 14 14 (czternaście) -> 11 (jedenaście) -> 10 (dziesięć) -> 8 (osiem) -> 5 (pięć) -> 4 (cztery) -> 6 (sześć) and back to 5
I found out that 4(cuatro) and six(seis) go into a loop. I also found that 5(cinco) has five letters and 7(siete) and 9 (nueve) fall on it. Very cool idea in this video, congrats
For Russian: 1 - один (4 letters) 2 - два (3 let.) 3 - три (3 let.) 4 - четыре (6 let.) 5 - пять (4 let.) 6 - шесть (5 let.) 7 - семь (4 let.) 8 - восемь (6 let.) 9 - девять (6 let.) 10 - десять (6 let.) I can see a loop of numbers: 6 -> 5 -> 4 -> 6, and so on. Also, 2 gets us to 3, and that starts an infinite self-loop on number 3. But I dunno about the threshold exactly, so if any друг can help, that would be nice
11 - одиннадцать (11 let.) That is the third and the final loop for Russian, and also incidentally the k-threshold. I checked the numbers up to 1000, and 77% end up in the 6-5-4 loop, 22.7% in 11, and only 0.3% in 3. Also, out of 770 that end up stuck in 6-5-4, the vast majority (760) enters the loop at the number 6. Fun trivia.
Confirmed, output to this code is {18: 999904, 13: 96}, this is the most pythonic way I could think of quickly writing it (simple, to the point, no over abstracting): from collections import defaultdict def linguistic_bin_count(num): bin_num = bin(num) zeros = bin_num.count('0') - 1 # Python represents bin number as 0bxxxx # so we have to subtract 1 ones = bin_num.count('1') return 4*zeros + 3*ones def final_linguistic_bin_count(num): "I've assumed no loops, want to prove it?" next_num = linguistic_bin_count(num) if next_num == num: return num return final_linguistic_bin_count(next_num) tally = defaultdict(int) for n in range(1, 1000001): tally[final_linguistic_bin_count(n)] += 1 print(dict(tally))
Virgile Andreani Indeed, 97 if you include 0. Seems like a long winded proof would be to show that all numbers in a certain range go to 18, and then the smallest binary count is 11111...1111 and the longest binary count is 1000...0000 for a number of length n, and then show that because of this they end up eventually falling in to this range. You could more or less code this proof up. But I wonder if there's a proof from the book (a nice small elegant proof).
standupmaths Virgile Andreani Actually on 2nd thoughts I had an idea, the 96 numbers < 1000001 not including 0 are: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 31, 128, 271, 279, 283, 285, 286, 295, 299, 301, 302, 307, 309, 310, 313, 314, 316, 327, 331, 333, 334, 339, 341, 342, 345, 346, 348, 355, 357, 358, 361, 362, 364, 369, 370, 372, 376, 391, 395, 397, 398, 403, 405, 406, 409, 410, 412, 419, 421, 422, 425, 426, 428, 433, 434, 436, 440, 451, 453, 454, 457, 458, 460, 465, 466, 468, 472, 481, 482, 484, 488, 496, 767, 895, 959, 991, 1007, 1015, 1019, 1021 And these go up pretty big, so couldn't we artificially create a number that fell in to this? For example say we wanted to fall in to 271? Well that's simply 3 + 4*67 or : one zero zero ... zero (67 zeros). And indeed: > print(final_linguistic_bin_count(2**67)) 13 So it seems the numbers just become very sparse. So 2 questions, what is the next number after 1021 that resolves to 13? And can we use this infomation to show there are an infinite number that resolve to 13?
In grade school we had the little riddle "Four is the magic number" where your friends would try and figure out why exactly they could give you any number and you would always end up at the magic number four. Nice to see someone nerding out and expanding upon the concept to this degree.
In italian the fixed point is 3 (tre), and I found this quite long chain, starting from 44: 44 - quarantaquattro 15 - quindici 8 - otto 4 - quattro 7 - sette 5 - cinque 6 - sei 3 - tre
If you try Chinese, it will definitely be the most special one. In Chinese, if we count the syllable(s) of a number, it will always end up with one....... Coz every letter in Chinese only have one syllable, and numbers larger than ten would be a word combine with numbers and it's digit value. For example, 3795 in Chinese would be Three Thousand Seven Hundred Nine Ten Five, since each word starts block letter have only one syllable, it would then be 7 syllable in 3795, and 7 have only one syllable, then it ends at 1.
So the number would need to be at least a 21-digit number without any zeros to make a chain of 4. Eg, the number 111111111111111111111 will have 21 syllables and 21 have 3 and 3 have 1. A chain of 4 in Chinese would need a number with 111111111111111111111 syllables, let that number be x, then a chain of 5 would need a number with x syllables, let that be y; then a chain of 6 will need a number with y syllables, let that be z; chain of 7 will need a number with z syllables, let that be u; chain of 8 will need a number with u syllables, let that be w; think of how huge w would need to be.
Tang Chak Yan works in german and english as well, spanish is a 1 2 loop, and I bet most other languages have a similar pattern. Would be interesting to see if there are examples where it is not the case
In Chinese, surely it would make more sense to count the strokes rather than characters. 一， 二， 三， and 四 (1, 2, 3, 4) are all fixed points and 4 is the threshold. More interesting is if one considers the bankers' anti-fraud versions of the numbers. I'm a bit shaky on them, but I *think* the following is correct (could someone who actually knows Chinese check?): 拾伍, 拾陆, 贰拾柒, and 贰拾捌 (15, 16, 27 and 28) are all fixed points. 拾 and 玖 (7 and 9) form a 2-cycle, and 贰拾肆, 叁拾壹, 贰拾玖, and 贰拾伍 (24, 31, 29, 25) forms a 4-cycle. 31 seems to be the threshold. www.dropbox.com/s/xdxemlg0efqagzz/ChineseNumberStrokes.bmp?dl=0
Should we also look into languages that have the most unrelated chains? Albanian is pretty cool with that one. 2 is a single link chain that maps to itself (de); 3 has two links (nje [1] maps to tre [3]); 4 and 5 make a loop at the end of a chain (katër, pesë); 6 and 7 make a loop at the end of a chain (gjashtë, shtatë); 11 and 12 make a loop at the end of a chain (njëmbëdhjetë, dymbëdhjetë); 14 ends a chain (katërmbëdhjetë); 16 ends a chain (gjashtëmbëdhjetë). 7 individual chains. That's got to be the one to beat. And 16 is a contender for highest self-referencing number in a given spoken language. #albanianisamathsnerd
there are several in UK counting, Weardale has a 5 loop 3 Tethera 6 Yan-a-tic 7 Teyan-a-tic 9 Methera-tic 10 Bub back to 3, Teesdale counting has a 3 number loop 3 Tether 6 Lezar 5 Pip back to 3, Swaledale counting has a 2 number loop 4 Mether 6 Azer, Nidderdale counting has 2 stop points 5 Pitts 6 Tayter.
I've been playing about on MATLAB and I believe 113,373,373,373 (aka X) is the lowest number (including the word 'and', excluding any whitespace or punctuation) which will give you a 8 length chain to get to 4. As the lowest number to require 7 steps is 214, X must therefore be 214 letters long. After writing a function to tell me the letter length of any given number, I worked out that (777 and later 373) have the longest word length of any three digits at 27. It made sense to me to create a number which was an integer multiple of 3 long composed entirely of 7s. It turns out 12 7s give a letter length of 130 so all I had to do was change one of the groups of 3 7s to a group of three digits with a letter length 6 less than 777, ie 21. 113 turned out to be the lowest number with a letter length of 21 so I put that in the most significant position to make the number as small as possible. I then checked to make sure 777 was the smallest 3 digit group with a letter length of 27 and it turned out it was not, so I replaced all of those groups with 373. Just to make sure this wasn't a red herring I checked to makes sure that there weren't any lower value numbers than 113 with higher letter lengths than 21. I was lucky as there turned out not to be any. If there had been then the second most significant 3 digit group would have had to have had its letter value decreased accordingly. The number which will have a chain of 9 will need 113,373,373,373 letters which, assuming 5 letters per word, would take about 290 years for a single human being to finish saying outloud. You can view my code at github.com/osholt/NumberLengthMattParker/ . If you think I've done anything wrong or have suggestions please let me know.
I believe Matt is counting the "four" in the chain length, so in that case what you would have found is an 8-chain. "One hundred and thirteen billion three hundred and seventy three million three hundred and seventy three thousand three hundred and seventy three" has only 124 letters, but it does result in an 8-chain: X, 124, 23, 11, 6, 3, 5, 4. As he mentions, there are quite a few 7-chains less than 1,000,000 and that's without counting the "and".
Karl Muster I wrote 3 functions to deal with numbers less than 1000. In the ranges 1 to 15, 16 to 99 and 100 to 999. I then wrote a function to deal with numbers from 1,000 to 999,999 which basically just called the 3 functions I wrote to deal with numbers below 1000 but added on an extra number to account for 'thousand'. I repeated (ie copy pasted) this process up to the quintillions which is where 64 bit integers top out at. It basically just comes down to some logic around some lookup tables (or in this case switch-case statements).
In Polish: cztery (4) > sześć (6) > pięć (5) > cztery (4) by other numbers: dwa (2) > trzy (3) > cztery (4) jeden (1) >pięć (5) >cztery (4) siedem (7) > sześć (6) > pięć (5) > cztery (4) dziewięć (9) > osiem (8) >pięć (5) > cztery (4) The word "cztery" knows "four" is albo sinkhole in Polish Language.
“The least uninteresting positive integer” has 36 letters. I don’t know whether that qualifies as interesting, but the “function” can take paradoxical numbers as input.
In Polish there is no number below a hundred that has the same number of letters as its value For numbers from 1 to 10 there is: 5, 3, 4, 5, 4, 5, 6, 5, 8, 8 so there are some loops
"two cubed" has eight letters. It's the only fixed point in English prime factorizations.
There is a lot more fun to be had with cubed numbers!
First of all, you have a four link loop:
33³ = 35937 - thirty five thousand nine hundred and thirty seven - 43 letters (exluding whitespace)
43³ = 79547 - seventy nine thousand five hundred and seven - 38 letters
38³ = 54872 - fifty four thousand eight hundred and seventy two - 42 letters
42³ = 74088 - seventy four thousand and eighty eight - 33 letters!
On top of that there are also 4 stand alone end points!
41³ = 68921 - sixty eight thousand nine hundred and twenty one - 41 letters
51³ = 132651 - one hundred and thirty two thousand six hundred and fifty one - 51 letters
54³ = 157464 - one hundred and fifty seven thousand four hundred and sixty four - 54 letters
55³ = 166375 - one hundred and sixty six thousand three hundred and seventy five - 55 letters
And every number ends up at one of these fixed points or gets caught into this cycle with
f(x) = number of letters in x^3
I wonder what happens in my language, Slovene...
Eleventy-one also only has that many letters too ;p
@@Qril What do you mean? Eleventy-one (=111) has 11 letters
@Ian O'Keefe I wonder what the longest chain is...
Also, after "squared" and "cubed", do you use "TO THE n-th POWER"?
4096: TWO TO THE TWELFTH POWER so 20
What about prefixes?
Bi- 2
Tri- 3
Quad- 4
Penta- 5
thats cool
I love that. That is awesome
Broken Wave
HEXA- 6?
@@aaronconner8906 only if you spell it HEXXXA
A form of matter
Fun Fact: The word Octagons has 8 letters, and octagons have 8 sides.
"The absolute value of negative thirty seven" has |-37| letters
Edit about 2 years later, 6/7, 624 likes: why is this suddenly blowing up
i read that as -37 letters lol
Thats so good
sadly that's not how you'd read it, still good comment lol
now i know what these || brackets mean, it is to convert negatives to positives
@@thatonedynamitecuber in terms of pemdas, they're the same kind of operation as parentheses
Spanish has a loop of two at the numbers four and sick ("cuatro" and "seis"), but it also has a fixed point at five ("Cinco"). I find this interesting since it's never easy to tell if a number in Spanish will resolve to a fixed point at five or get trapped forever in the loop between six and four.
Four and sick🤨
portuguese also has that
phive has 5 letters.
Merritt That's brilliant
Merritt Congratulations! You can count!
phteven, is that you?
I think you mean Stephen
Phisix
In portuguese there's a loop inside a loop
4 (quatro) 6 (seis)
5 (cinco)
Yoki Popcorn sabia, sempre tem uns BR jogado pelo TH-cam KKKKKKKKK
6, 7, 8 and 9 all end up at 4. 10 breaks it ending up at 3 which goes to 4, and then 11 and 12 go to 4 again. And finally 13 going to the endless 5...
Same with spanish!
Works in spanish too! :3
Federico Marquez thats what i said! Sodium chloride!
1 = O
2 = TU
3 = TRI
4 = FOUR
5 = PHIVE
6 = SHINKS
7 = SEEVEEN
wait what am i doing
EEIGHEHT
Nneighnn
Tenneronie
Elleeveennh
Twelverhdofj
Thirdtheteeen
Fhourrttteeene
PphhaaaffftaAin
SsaxxxtaaaiiiInn
Savvvviiinnntainn
Using only numbers fifteen or smaller, Latin has an eight-length chain before the lasso:
Quindecim (XV) --> novem (IX) --> quinque (V) --> septem (VII) --> sex (VI) --> tres (III) --> quattuor (IV) --> octo (VIII), and then it loops between quattuor and octo.
The loop only consists of four and eight, that means that the main loop has the length of two
@@ttermit he said its a chain not a loop
The use of Roman Numerals is much appreciated 😂
Matt,
Having searched a couple of hundred languages, I finally found the first 5 cycle! Also, you were very lucky to have one of the few you checked have a four cycle (French), since of the ones I checked, I only found one other: Semelai, the cycle is 3 (hmpe) --> 4(hmpom) --> 5(mesong) --> 6(pru) --> 3(hmpe). This doesn't really count but also in Icelandic if the things you are counting are feminine, then 3(prjár) --> 5(fimm) --> 4(fjórar) --> 6(sex) --> 3(prjár).
The five cycle which I finally found is in Sardinian, where:
3(tres) --> 4(battoro) --> 7(sette) --> 5(chimbe) --> 6(ses) --> 3(tres)
Hope you are interested, and thank you for the video
wow.. this one is waaay below... but it's already 2 years ago, so i just disturb you with a notification XD
Tswana also has a 5-number loop:
3 (tharo) -> 5 (tlhano) -> 6 (thataro) -> 7 (supa) -> 4 (nne) -> 3...
Ligurian also has a 5 loop
3 (trei) - > 4 (quattro) - > 7 (sette) - > 5 (cinque) - > 6 (sei) - > 3 (trei)
I found another five loop in the Guanche language, which is the aboriginal language of the peoples from the Canary Islands.
5(simusetti) -> 9(alda-marava) -> 10(marava) -> 6(sesetti) -> 7(satti) -> 5 again
Quick note depends on how you spell them, because some conventions use k instead of ch to represent the word chimbe; plus, the word battoro, according to Sardinian orthographic rules, can also be spelt battor and this won't affect at all it's pronunciation
If you do this with Pokemon they all end at Blastoise.
For example:
Charmander (10)
10. Caterpie (8)
8. Wartortle (9)
9. Blastoise (9)
I think the longest chain possible is:
Fletchinder (11)
11. Metapod (7)
7. Squirtle (8)
8. Wartortle (9)
9. Blastoise (9)
Crabominable (12)
Butterfree (10)
Caterpie (9)
Blastoise (9)
This nerd thanks you.
Fletchinder (11)
11. Metapod (7)
7. Squirtle (8)
8. Wartortle (9)
9. Blastoise (9)
9. Wartortle (9)
9. Blastoise (9
9. Wartortle (9)
9. Blastoise (9)
wc
Never even thought of that, cool!
You don't go by number of letters, you go by number in national pokedex.
Here is a 7-cycle in Koasati (a native american language):
8 -> 12 -> 16 -> 19 -> 20 -> 9 -> 10 -> 8
ontotchiinan -> pokkol awah toklon -> pokkol awah hannaalin -> pokkol awah chakkaalin -> poltoklon -> chakkaalin -> pokkolin -> ontotchiinan
+renerpho That's amazing!
+
+
+
+
"Negative fifteen" has positive fifteen letters.
minus nine has 9
Positive fifteen... also has positive fifteen letters.
So does negative -17.
@@youtubeuserdan4017 so thats positive 17?
@@Sam-kj9ui Haha good one. It was a typo but I won't correct it so your joke will make sense.
In Russian you have 2 stopping points:
3 (три) -- 3 letters
11 (одиннадцать) -- 11 letters
And one circle:
4 (четыре) - 6 (шесть) - 5 (пять) - 4...
Polish has that same loop:
Cztery
Sześć
Pięć
Cztery
In Danish the numbers 2(to), 3(tre), 4(fire) are all stopping points. After this minefield every number has more letters than its value so there are no loops.
Swedish has quite a few 3-letter numerals in single digits and 2 stop points, tre (3) and fyra (4). But since the majority of single digits have 3 letters, I would imagine 3 is where most sequences would stop.
Ннайс
Portuguese too!
Cinco - 5 letters
Quatro - 6 letters -> seis - 4 letters
Assuming no "and"s and no counting spaces
Lowest 7 Chain: 323
Lowest 8 Chain: 1,103,323,373,373,373,373,373,373,373,373
The lowest number with a 9 chain would be super huge, somewhere around 10^100,000,000,000,000,000,000,000,000,000 by my estimation, as the write-out of the word would need at least the number of letters as the lowest 8-chain
u run this using code?
10^10^29, that's more than a googol! (And that's understating it by 10^27 orders of magnitude... I think... these numbers are so big it's hard to tell when you've made a mistake)
Edit: changed "googolplex" to "googol," thank you David!
Do we have a naming convention for such large numbers?
So I decided to try this out in irish and I got an 8 chain at 2474440 (Dó millúin ceathar Céad seachtar ceathar míle ceathar céad ceatharacha) and I didnt even check if it was the lowest 8 chain, lowest 7 chain is 41 (ceathracha haon) Lowest 6 chain is 14 (Ceathar déag) lowest 5 chain is 10 (deich) then it goes to 5(cuig then 4(Ceathar) then 6(sé) then 2(dó) where the chain ends
If we do have and's and count spaces (naming convention of num2words in python), then:
Lowest 7 Chain: 113
Lowest 8 Chain: 477,777,777
And for 9 we already reach over the naming convention in the english language
funfact about number in Indonesian and Malay, they have pattern :
1 (Satu) & 9 (Sembilan) both begin with "S"
2 (Dua) & 8 (Delapan) both begin with "D"
3 (Tiga) & 7 (Tujuh) both begin with "T"
4 (Empat) & 6 (Enam) both begin with "E"
and if you add each pair, you got 10 as total
What about 5?????
@@mikojel4736 5 is (Lima).. it starts with a unique letter
@@mikojel4736 Well five has the same letter as 5 and 5+5 = 10, the pattern is correct
Woooah that's useful for math, is it intensional or just a coincidence?
And what's the word for 10?
The was recommended to me 4 years later. I'm proud of the algorithm for it's humor.
"posted 4 days ago" here we go again
1 = N
2 = Tu
3 = Tre
4 = Four
5 = Phive
6 = Sihcks
7 = Sehvven
8 = Eaieghtt
9 = Neiaighnn
10 = Pteagheinn (p is silent)
11 = Ealleaveann
12 = Tchweiallvve
13 = Thieaghrteenn
14 = Foaiurghtteenn
15 = Pheiaghphtteenn
16 = Siehghckstteeinn
17 = Sceavveinntteeinn
18 = Eaieaightteaeinnde
Ugh this is tough. I’m done
Gj I think
20 = ptweiagghhnntteiaegh (the p is still silent)
19 = Naiiaeeeeiintteeinn
twenteeeeeeeeeeeeeee
@@janaki3829 twenteeeoooonnnnneeee
You'll love this one: In Korean we have TWO number systems (Sino and Korean), and we can spell them out in TWO language scripts (Hangul [Korean] and English [Romaja])! That gives us FOUR potentials to find K! I'm also not counting our numbers using Chinese character (hanja) but that could work just by counting the strokes, and they are still pronounced the same way anyways.
So....
* NAME = word (number it represents, number of letters) > next item ...
* SINO-HANGUL (#1) - 일 (1, 3 letters) > 삼 (3, 3 letters) > loop
* SINO-HANGUL (#2) - 사 (4, 2 letters) > 이 (2, 2 letters) > loop
* SINO-HANGUL (#3) - 십일 (11, 6 letters) > 육 (6, 3 letters) > 삼 (3, 3 letters) > loop
* SINO-HANGUL (#4) - 이십삼 (24, 8 letters) > 팔 (8, 3 letters) > 삼 (3, 3 letters) > loop
* SINO-ROMAJA (#1) - o (5, 1 letter) > il (1, 2 letters) > i (2, 1 letter) > loop with 1&2
* SINO-ROMAJA (#2) - pal (8, 3 letters) > sam (3, 3 letters) > loop
* KOREAN-HANGUL (#1) - 하나 (1, 4 letters) > 넷 (4, 3 letters) > 셋 (3, 3 letters) > loop
* KOREAN-HANGUL (#2) - 여덟 (8, 6 letters) > 여섯 (6, 5 letters) > 다섯 (5, 5 letters) > loop
* KOREAN-ROMAJA (#1) - hana (1, 4 letters) > net (4, 3 letters) > set (3, 3 letters) > loop
* KOREAN-ROMAJA (#2) - yeoseot (6, 7 letters) > ilgop (7, 5 letters) > daseot (5, 6 letters) > loop with 6,7&5
All other words loop in these ways. So 7 is the largest number I could find using Latin characters (looped with 6 and 5), and 5 using Hangul.
Hope this helps!
ill try this for japanese
btw I don’t speak Japanese so please correct me. This is for all the numbers from 1 to 10. And I’m including the つ counter for the kunyomis.
ONYOMI-KANA
いち (1, 2 characters) -> に (2, 1 character) -> loop
All the other numbers fall into that loop
KUNYOMI-KANA
All of them have 3 kanas, except ここの (9) which has 4. とお (10) has two but you can’t use it with a counter.
ONYOMI-ROMAJI
There are 2 end points: ni (2, 2 letters) and san (3, 3 letters)
The longest chain is shichi (7, 6 letters) -> roku (6, 4 letters) -> shi (4, 3 letters) -> san (3, 3 letters)
KUNYOMI-ROMAJI
There are 3 end points: muttsu (6, 6 letters), nanatsu (7, 7 letters) and kokonotsu (9, 9 letters). All the numbers except 9 and 10 have either 6 or 7 letters. 9 has 9 letters and 10 has 2 letters
In Norwegian:
2 = To
3 = Tre
4 = Fire
(and 0 = null but that has nothing to do in this video)
XDvard help there's a forest 4
Cameron Crego ok-
3 is "tre" also in Italian, and I think it is the only number to have the same amount of letters of its value
Isn't five "cinco" or something like that?
EDIT: Okay, I've checked, it's written in six letters. In fact, it's pretty easy to see how in Italian, everything collapses to three pretty fast.
In Danish:
2 = To
3 = Tre
4 = Fire
"Positive fifteen" has 15 letters, and "seven plus seven" has 14 letters.
Yeah, I was also thinking of positive fifteen
Also just realised I'm a year lates :|
The absolute value of the positive part of the complex number [insert first number] plus the square root of negative [insert second number] all to the power of one...
Russian in Cyrillic I find has 2 stopping points and one loop:
3 (три) has 3 letters
11 (одиннадцать) has 11 letters
and the loop
4 (четыре) has 6 letters
6 (шесть) has 5 letters
5 (пять) has 4 letters
I've also found that in Tagalog (Filipino) we have 2 stopping points:
4 (apat) and 12 (labing dalawa), and in numbers under 100, about 10% stop at 12 and the rest stop at 4
I found it earlier (in Russian) but good work :-)
And try to get the next number after 100 that comes down to three in Russian. Good luck!
Nice. I was shuffling around Russian in my head and suspected that it had a more complex structure.
Do the strong sign and soft sign count as letters?
+Julián García of course they count. You can't just say "пять" and assume it has 3 letters, 'cause then it would be "пят" and that's a completely different word!
Hey there! In Lithuanian language, five (5) is *_penki_* so there are five letters in this word *:D* Oh and seven (7) is *_septyni_* with seven letters as well!! And two (2) is *_du_* with only two letters!!! Isn't that cool??? For those who have no idea how to pronounce these words, then here is how to do it: _penki_ pronounce as _panky, du_ - as _doo_ and _septyni_ read as _saptiny_ ( or use a Google translator if you don't trust me :D )
In Norwegian, we have three numbers like the English four:
2 = To
3 = Tre
4 = Fire
If 4 is fire, is 5 water? :)
Same in Danish but although the languages are close its not the same in swedish where "to" becomes "två".
And all the other numbers up to ten reverts right back to those three, and no number up to 100 have more than ten letters. So the norwegian numbers are really booring. No loops and no lines longer than three steps.
Im sad now
Nobody:
Norwegians: 1 2 3 🔥 5
@@tegneren in danish 74 (fireoghalvfjerds) has 16 letters so we can manage a 4 number line despite having the same 3 early loops.
This was about 50% more interesting than I suspected.
1-hydrogen
8-oxygen
6-carbon
Crap
2-Helium
6-Carbon
Crap
3-Lithium
7-Nitrogen
8-Oxygen
6-Carbon
OH CRAP
104-RUTHERFORDIUM
13-MAGNESIUM
9-FLUORINE
8-OXYGEN
6-CARBON
FUYU
Ceca Siahaan I found an element that doesn't end in carbon: Boron has 5 letters
Chunky Ragu i know,it's either carbon or boron
18-ARGON 54-XENON
5-BORON 5-BORON
SUCCESS SUCCESS
...they're the only ones.
@@limecyanizer4394 Xenon is feeling left out
@@MotoCat91 yes i edited it
I just realised boron is #5 on the periodic table and it has five letters
It works for carbon too
BRILLLIANT!!! How did I not catch that?? Absolutely amazing!
It also works for Oganesson! :D I'm so smart.
Dalynnrocks123 are you a fan of Studio C?
Heck Yeah!
"absolute value of open parenthesis negative thirty two times two minus twenty one close parenthesis" is 85 letters and |(-32*2-21)| = 85
I'm sure loads of these can be found. "Ten open bracket eleven plus ten minus nine close bracket divided by two" is 60 letters and 10 (11 + 10 - 9) ÷ 2 = 60, and this only took me a few attempts to find.
Although a little vague a "few" might be considered to have as many letters as the word describes.
And if you take "a baker's dozen" and include the apostrophe them you get 13.
@@Sednas You must have missed the A or the S?
@@TheHadMatters I deleted my comment because I thought I miscounted but I'm pretty sure now that there was no "a" when he first wrote the comment, note the fact that it's edited.
@@Sednas Ah.
I think Russian gets the award for the largest number of loops.
Eleven has eleven letters (одиннадцать)
Three has three letters (три)
And there is a three cycle of 6-5-4
Шесть (6) has five letters
Пять (5) has four letters
Четыре (4) has six letters.
Three cycles,thats pretty good!
toki pona has 6 cycles
"2: to", "3: tre", "4: fire" in Norwegian
fire.
Tre is Italian as well, cinco (5) works in Spanish, vier (4) in German
Same for Danish. Most numbers have quite short names.
1:En, 2:To, 3:Tre, 4:Fire, 5:Fem, 6:Seks, 7:Syv, 8:Otte, 9:Ni, 10:Ti, 11:Elleve, 12:Tolv, 13:Tretten, 14:Fjorten, 15:Femten etc....
what was that weird hard number in danish? 50 something?
brachypelmasmith are you thinking about 97?
there's a fun comic with UK: 90 + 7, German: 7 +90, Japanese: 9'10+7 etc.. and then Danish: 7 + (-1/2 + 5) x 20,Looks very complicated, but spoken it is quite simple: """syv og halv fems"""
(syv og halv fem sinde tyve) would be the "proper old danish" way to say it, but not many people does it that way, now a days.
I have now found a romance language with a 6-cycle (without spaces): Romansh (Vallader), spoken in Switzerland.
The cycle is 3 -> 5 -> 8 -> 2 -> 4 -> 7 -> 3:
trais -> tschinch -> ot -> duos -> quatter -> set -> trais
Of the 2,686 languages I have checked, this is the only one with a "true" 6-cycle among the numbers from 1 to 10 (where "true" means that no spaces or hyphens are counted).
+renerpho True dedication at its best.
Well done!
This is good.
You should check my post about the Henese-Flick:
th-cam.com/video/LYKn0yUTIU4/w-d-xo.html&lc=z12dj3bgxvbyhzl3k23cgd5yzz2tcn1g104
A 12-chain containg a 7-loop. (Hoping I made no mistake)
Here are the cycles for 103 languages. Other information you asked about is at the bottom.
Each set of brackets is a chain. Commas separate numbers in a chain.
Afrikaans [4]
Albanian [2][3][4,5][6,7][11,12][14][16]
Amharic [3]
Arabic [4,5]
Armenian [5]
Azerbaijani [2,3]
Basque [2][3,4][9]
Belarusian [3][4,6,5][11]
Bengali [2]
Bosnian [3]
Bulgarian [3]
Catalan [3,4,6]
Cebuano [4][12]
Cichewa [13,14][29]
Chinese [1]
Corsican [3]
Croatian [3][4,6]
Czech [3]
Danish [2][3][4]
Dutch [4]
English [4]
Esperanto [2][3][4]
Estonian [4]
Filipino [4][11]
Finnish [5][8,9]
French [4,6,3,5]
Frisian [3,5,4,7]
Galician [5]
Georgian [4]
German [4]
Greek [4,7][5]
Gujarati [1,2]
Haitian Creole [2][3]
Hausa [3][4][5][11]
Hawaiian [5][11]
Hebrew [5]
Hindi [1,2]
Hmong [2][3][4,5]
Hungarian [4]
Icelandic [3,4,6]
Igbo [3]
Indonesian [4,5]
Irish [2,5,7,6][4,8][11,13]
Italian [3]
Japanese [1]
Javanese [4,5]
Kannada [2,3]
Kazakh [2,3][4,7]
Khmer [2]
Korean [1,2]
Kurdish(Kurmanji) [2]
Kyrgyz [2,3][4]
Lao [3]
Latin [3,6][4,8]
Latvian [5][7][11]
Lithuanian [2][5][6][7]
Luxembourgish [5,6]
Macedonian [3][4,6]
Malagasy [4,6,5][15][16]
Malay [4,5]
Malayalam [2]
Maltese [5]
Maori [4][5]
Marathi [2]
Mongolian [3,5]
Myanmar(Burmese) [2]
Nepali [2]
Norwegian [2][3][4]
Pashto [3][4]
Persian [2][4]
Polish [4,6,5]
Portugese [5][4,6]
Punjabi [1,2]
Romanian [5]
Russian [3][4,6,5][11]
Samoan [2,3,4][12,13,14]
Scots Gaelic [3]
Serbian [3][4,6]
Sesotho [9][10][16][23,27][25,28,26]
Shona [4][5]
Sindhi [2]
Sinhala [2]
Slovak [3]
Slovenian [3]
Somali [3,6][4][7][11][12]
Spanish [4,6][5]
Sundanese [4]
Swahili [3,4]
Swedish [3][4]
Tajik [2,4,3]
Tamil [3][4]
Telugu [2,3]
Thai [3]
Turkish [2,3][4]
Ukranian [3][4,6,5]
Urdu [2][3]
Uzbek [4,6][8]
Vietnamese [3,4][5]
Welsh [3][6]
Xhosa [5,6,9][15,16,20][17,18][21][25][27]
Yiddish [3,4]
Yoruba [5]
Zulu [10][12][14][16][19][22][23][27]
The highest number with a higher letter count than the number is 29 in Zulu with 33 letters.
French, Frisian, and Irish have the longest chain(4). Zulu has 8 chains, and all of them are single numbers.
Wow that’s awesome
One thing to note is the Irish chains seem to use the counting system we use for people, which uses longer words than the regular counting system
I'd argue, that japanese is also [1,2] not only [1]. I guess you used the kanji 1(ー). But the hiragana 1(いち) 2(に) also works
@@freakymidgetI was puzzling over the numbers given for Irish for a while before coming to the same conclusion.
You forgot Catalan fixed point 1 = "u"
So in Hungarian, every single number up to a million (where the names of the numbers become only constructed names) ends in a 2 -> 5 -> 2 chain EXCEPT 4, 20, 100 and 1000, which are the only numbers to contain exactly four letters, so they end differently. Therefore, we have two chains, the "general" one, and the ones constructed to contain exactly 20, 100, 1000 and so on letters (starting with 239). Great time-sink. :)
Lol the only numbers that don't work that arent powers of ten are 4 and 20
Weed confirmed
Bullshit. Négy is 3 letters (and four characters). Gy is a letter.
If you are counting Hungarian letters (ty, gy, ny, sz counts as one letter) instead of characters, the "Great Whirlpool of 5(öt)-2(kettő)" will swallow everything, but more interestingly, all the chains will reach it through 3(három). So begin a chain anywhere, it will end up 3-5-2-5-2-5...
All but one: 1(egy) - 2(kettő) - 5(öt).
On the "medium pleasing" level: TWOCUBED
Wolf Elkan two cubed is 2 * 2 * 2 or 8... two cubed has 8 letters...
OH.
MY.
MATH.
Seven plus seven
Nine Plus Two (11 letters)
Both “eleven plus two” and “twelve plus one” work for 13. They are also anagrams of each other.
(Seven plus seven minus seven plus seven minus seven plus seven minus seven ) times ten 😎 (70 letters and adds up to 70, in memory of Trondor Does Stuff)
Chinese has no letters. done.
Stroke number
So, the word for zero maps to zero (because every number does).
Cool.
+inscribed Then we would get 4 possible interlooping numbers/cycles like the case of the binary system, where k=1 [一] , 2 [二], 3 [三], (4 [四] & 5 [五] loop with each other). I'm gonna randomly guess that most numbers falls into a 2 loop or a 4-5 loop, since 7 [七] ,8 [八] ,9 [九] ,10 [十] are all 2 strokes and 6 [六] is 4 stroke. Or we could go with the "phonetic system" where the k=2(er), 3 (san), which is so very /exciting/...
Count characters:
1 through 10 have 1 character
11 through 20 have 2 characters
21 through 100 have 3 characters (except multiples of 10, which have 2)
101 through 1000 have (call the number x):
x mod 100 is between 1 and 9 -> 4 characters
x mod 100 is between 11 and 19 -> 4 characters (except in formal situations, which have 5 characters)
x is a multiple of 10 but not a multiple of 100 -> 4 characters
x is a multiple of 100 -> 2 characters
otherwise -> 5 characters
etc.
Chinese characters can be transcribed to pinyin, a phonetic spelling using the same alphabet as English, and Chinese numbers under that transcription might be interesting to think about
Polish has a three loop.
(4) cztery --> (6) sześć --> (5) pięć --> (4) cztery
German has a fixed point at vier (4)
Dutch has a fixed point at vier (4)
Spanish has a pair
(4) cuatro --> (6) seis --> (4) cuatro
AND cinco (5) as a fixed point
portuguese has the same perks as spanish
In base harmonic series, most numbers explode violently.
To convert a number to base harmonic series, write down a "1" if the number is positive, or a "0" if the number is negative, then either add (if negative) or remove (if positive) 1/n from the number where n is the number of digits you've written down so far. Repeat this until the number reaches 0.
Here's an example of some common numbers in base harmonic series:
-1 : 0
-0.5 : 01
0 : (empty string!)
0.5 : 10
1.5 : 11
2: 111101010101010101011001101001100101101001011001101001011010011001011001101001100101101001011001101001100101100110100110010110011010011001011001101001100101100110100110010110011010011001011001101001100101100110100110010110011010010110100110010110011010011001011010010110011010011001011001101001011010011001011001101001100101101001011001101001011010011001011010010110011010010110100110010110100101100110100110010110011010010110100110010110100101100110100101101001100101100110100110010110100101100110100110010110011010010110100110010110100101100110100101101001100101100110100110010110100101100110100101101001100101101001011001101001100101100110100101101001100101101001011001101001011010011001011001101001100101101001011001101001011010011001011010010110011010011001011001101001011010011001011001101001100101101001011001101001100101100110100101101001100101101001011001101001011010011001011001101001100101101001011001101001011010011001011010010110011010011001011001101001011010011001011010010110011010010110100110010110011010011001011010010110(........ it keeps going for a while)
edit:
Of course, if you only care about positive integers, stop as soon as your number is less than one, and you get the unique representation for that positive integer in base harmonic series. For example, 8 is exactly 1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111
Ok, "base harmonic" is pathological but works. For the original system, this means some values would have an infinite series as their representation. I somehow want to declare that cheating.
The integers only representation is interesting! But I'm getting an off-by-one error when I try it. I have seven as a string of 616 1s and eight as 1674 1s. Either way, it means writing it down gets bigger faster than the value of the number. It's like a tally, but worse.
NOTCH I AM A BIG FAN!!!!
Could you write down what you're doing? It took me a little while to figure out Notch's system; I think he traded comprehensibility for conciseness. Take (number) and add or subtract 1/(number of digits so far generated) for a response. (number) is the last response generated. Determine addition or subtraction by flipping the sign of the number you're using for that round. All responses should look binary. For 7:
7 is positive, first digit 1
We have 1 digit so far and our number is 7 which is positive, so we subtract: 7-(1/1)=6. The second digit is a 1.
We're 2 in and had a response of 6 last time, 6-(1/2)=11/2, 3rd digit 1
3 in and at 11/2, (11/2)-(1/3)=31/6, positive, 4th digit 1
4 and 31/6, (31/6)-(1/4)=118/24. positive, 5th digit 1
5 and 118/24, (118/24)-(1/5)=566/120, +, 6th digit 1
So the first 6 digits of this representation of 7 are actually 111111. I'm doing this by hand/brain though and so I'm getting a little sick of it and do not have at my disposal a way of quickly getting through a whole bunch of these. It's late so I don't feel like going on or thinking about it any further. Maybe tomorrow. Neat though
What I THINK you're doing is always subtracting from the original number and using the 'responses' as the accompanying digit. That would mean it would be possible for one of the digits of the representations to be a fraction. If you'll notice, single digits do not tend to represent a whole fraction. (69/103) is not a numeral.
Minecraft hits Math! I love it :)
I'm another big fan of the Markus! :)
Things like this always make me wonder how the physical world is constructed, and if it relates to any of these systems (as in how atoms are constructed from similar parts, but have massive differences in their properties). Thanks for the info on harmonic series! :D
Zulu has two loops with relatively high k values:
19: nesishiyagalolunye
18: ishumi nesishiyagalombili
24: amashumi amabili nane
Then back to 19: nesishiyagalolunye
AND
8: eziyisishiyagalombili
21: amashumi amabili nanye
20: mabini
6: eziyisithupha
13: thirteen
Then back to 8: eziyisishiyagalombili
Higher k values and longer chains.
I like this one for the high k values. In the second sequence though, '13' is translated as 'thirteen' (English)
High marks for knowing Zulu, it's not everyday you see that.
Teacher [in Zulu]: "OK children, let's count to 100 together and we'll break for lunch at '34.'"
konkani
2 goes to 3
3 goes to 4
4 goes to 5
5 goes to 6
6 goes to 2
numbers one through six: ek, don, teen, chaar, paaMch, sa
+
Even more than you said first, it goes 2-3-4-5-6, so you've got one more than French!
+
+
On the other side of things, it can be fascinating sometimes to look at the etymologies of numbers across languages, and see what numbers are hiding inside them.
For example, eleven and twelve go back to something that translates back as one-left and two-left. There's a cross linguistic tendency for words for five to be related to a word for hand, in English's case, five, fist, and finger seem to be connected. Seven is structurally really weird across the indo-european languages, and seems to go back to a proto-form akin to *septom, which looks suspiciously like the Akkadian word for seven (and other Afrasian languages, like Arabic, Ancient Egyptian, and others), which was something like sabhat. Four is a bit of a mystery, but has been made out by some linguists to be derived from a verb *kwet meaning "to join, pair", as a pairing of pairs. Eight is a dual stem of what was apparently an old measure of four fingers. Nine used to be thought to be related to new, and that our linguistic ancestors had a base 8 system, but the modern theory is that nine derives from a rare word *henu meaning "to lack", related to Greek aneu and German ohne, meaning "without". Ten and hundred are thought to derive from *kemt, the word for hand; *de-kmt evidently being something like "two hands", (de-)kmt-om being a collection of tens. One in the IE languages generally has one of two etymologies, as well: either from a word like *oinos, which was probably a pronoun originally, and a word for a collective unity, *sem, seen in words like "single", "same", "similar", etc.
On the topic of other languages, maybe the neatest I can think of off the top of my head is Old Japanese's. Modern Japanese uses Chinese numbers in some contexts (like how we use Latin numbers in some contexts), and their native numbers have undergone some sound changes that obscure certain patterns, but there's a ton of good evidence to support Old Japanese's numbers looking like:
1: pito
2: puta
3: mi
4: yo
5: itu
6: mu
7: nana
8: ya
9: kokono
10: towo
Note the relationships between 1 and 2, 3 and 6, 4 and 8, and maybe 5 and 10. 7 and 9 are then the odd ones out,
That's a Parker Square of a conjecture, Matt
don't bring that up
Cordialmanx233 Ignoring it???... that's how we got here in the beginning.
Parkerly intelligent remark
Four, Fiver, Sixers, Sevener, Eighters, Ninerrino, Tenerrinos, Elevenerrino, Twelverrinos...hey this works pretty well!
*****
I was thinking "O", "Tw", "Thr" because that sounds sillier
Thirteenerino, Fourteenerrino, Fifteenapalooza, Sixteenapaloozas, Seventeenapalooza, Eighteenerinopolis, Nineteenerrinopolis, Twentinapaloozopolis
***** Nice! After this point, we get into hyphenated number words and things get contentious. But assume we stop at 20 - which is a pretty good run - now the question becomes, which of these 20 numbers is the most popular for all other number words to end on after reducing them like in the video?
How shall we represent the number 21 in this system? Twenty-one or Twentinapaloozopolis-one? If we're going with the latter, then 24 is another stopping point (unless we count the hyphen as a letter, in which case, the new stopping point becomes 25).
no 21 is
Twentinapaloozopolis-a
Twentinapaloozopolis-tw
Twentinapaloozopolis-Thr
Twentinapaloozopolis-Four
Twentinapaloozopolis-Fiver
Twentinapaloozopolis-Sixers
Twentinapaloozopolis-Sevener
Twentinapaloozopolis-Eighters
Twentinapaloozopolis-Ninerrino
thirty
Making the train stop at 30 or 29 depending on how you look at it
(WARNING: BAD PROGRAMMING JOKE)
In English you get a 'four' loop.
lul
your warning doesn't prevent cringes
lol
That's a TERRIBLE pun and I LOVE it!
No, why 😂
I appreciate this video being put on my feed 4 years after it was released.
same
same
1 - Yksi
2 - Kaksi
3 - Kolme
4 - Neljä
5 - Viisi
6 - Kuusi
7 - Seitsemän
8 - Kahdeksan
9 - Yhdeksän
10 - Kymmenen
So I tested out with Finnish, and ended up with a constant, 5, which has five letters and leads up from most words, 1 to 4, and 2-6 leading to 5. There was also a loop, between 8, having 9 letters, and 9, having 8 letters. I hope you found this interesting :D
I tested out ordinal numbers:
There is loop between 8. and 9. (kahdeksas and yhdeksäs).
other ones are
6. (kuudes)
11. yhdestoista
12. kahdestoista
Even though 21. (kahdeskymmenesensimmäinen) has 25 letters it goes to
25 --> 20 --> 14 --> 12
Doesn't this disprove his conjecture? As 5 is the threshold and then 6 has fewer letters but then 7 more again?
then seven is the threshold not 5
It has been nearly 20 years but I remember learning in school that "and" in a number denotes a decimal point. Four hundred ninety six would = 496. Four hundred AND ninety six would = 400.96. Likely things have changed in 20 years but this is how I was taught.
In the UK, it's standard practice to use "and" as a hundreds separator (e.g. 1,234,567 = one million, two hundred and thirty four thousand, five hundred and sixty seven) with "point" as the decimal separator, with decimals spelled out as individual numerals (e.g. 1,234,567.8901 = one million, two hundred and thirty four thousand, five hundred and sixty seven point eight nine zero one).
However, I'd hazard a guess that UK / US aren't the only two countries separated by a common language (in this case, two countries sharing a common language, where standard practice results in different methods of enunciating numbers).
I was never taught this. And is used for anything below 100 and never else. Have always said 'point' for the decimal point and only say it other than a sequence of numbers for prices (and never with and). (UK/27)
Maybe thats what they teach in american schools, but to me that sounds wildly incorrect, like I think your teacher was wrong. I dunno though. I would have thought four hundred point ninety six would be universally used in english.
That doesn't really make sense. I mean maybe it was like that, but "and" isn't even shorter than saying "point", so it's just gonna bring you tons of confusion for no benefit.
I was taught that "and" can only indicate a decimal point if the reader of the words indicate a fraction.
For example:
One hundred and seventy three thousandths
(One hundred) AND (seventy three thousandths)
100+ (73/1000)
100.073
Hebrew is crazy interesting about this.
In Hebrew, you can count in 2 forms - male of female form.
Most countable objects are male (building, wall, video, trophy, and many more, obviously) and less objects are referred to in the female form (finger, car, door and many more...).
That Hebrew rule is the source of many grammar mistakes, even by native Israelis.
If we count in the male form - you always end up circling around the number 5, which has 5 letters in the word "hamisha" - the male form of 5.
If we count in the female form - you always end up circling around the number 4, which has 4 letters in the word "arba" - the female form of 5.
Here is the chains from 1 to 10, the number of letters in Hebrew is stated in brackets.
You'll see that Hebrew has less letters then English because most vowels are dropped when writing in Hebrew (and "sh" is one letter in Hebrew).
Male form counting:
1. Ehad (3) > Shlosha (5) > Hamisha (5).
2. Shnaim (5) > Hamisha (5).
3. Shlosha (5) > Hamisha (5).
4. Arba'ah (5) > Hamisha (5).
5. Hamisha (5).
6. Shisha (4) > Arba'ah (5) > Hamisha (5.
7. Shiv'ah (4) > Arba'ah (5) > Hamisha (5).
8. Smona (5) > Hamisha (5).
9. Tish'ah (4) > Arba'ah (5) > Hamisha (5).
10. Asara (4) > Arba'ah (5) > Hamisha (5).
All chains ending in 5!
Female form counting:
1. Ahat (3) > Arba (4).
2. Shtaim (5) > Hamesh (3) > Shalosh (4) > Arba (4).
3. Shalosh (4) > Arba (4).
4. Arba (4).
5. Hamesh (3) > Shalosh (4) > Arba (4).
6. Shesh (2) > Shtaim (5) > Hamesh (3) > Shalosh (4) > Arba (4).
7. Sheva (3) > Shalosh (4) > Arba (4).
8. Shmoneh (5) > Hamesh (3) > Shalosh (4) > Arba (4).
9. Tesha (3) > Shalosh (4) > Arba (4).
10. Eser (3) > Shalosh (4) > Arba (4).
All chains ending in 4!
That's really cool. Obviously they are used in different situations, but/so which one is taught first usually?
@@TheRenegade... well for counting it's always female, but for other use it's taught at the same time because you use it all the time
I wonder how much different it would be if you didn't include vowel letters (e.g. the vav in shmoneh, making shmoneh 4 instead of 5) since they're technically optional. The first thing that comes to mind is that shalosh would become 3, making it another loop. I haven't thought through all the numbers yet, but it's interesting!
In Greek, 5 (πέντε) is the fixed point and we also have a 2 number loop that goes from 4 to 7.
4 (τέσσερα), 7 (εφτά)
Most numbers end up in the 4-7 loop and not at 5!
e.g. between the numbers 1-10 you can only go to 5 from 9 (εννιά). Every other number ends up at 4-7. If you go to 1-20, 16 of these end up in the 4-7 loop and 4 at 5.
So let's try Japanese
Starting at 100 (hyaku)
has five (go)
has two (ni)
which ends the cycle.
However, there is another cycle in reverse
starting at one (ichi)
has four (yon)
has three (san)
which ends the cycle
But wait there's more!
In archaic counting
one (hitotsu)
has seven (nanatsu)
which ends the cycle
but in reverse
100 (momo)
has four (yottsu)
has six (muttsu)
which ends the cycle
Let's not go into the Japanese furigana characters shall we, I'm done
All according to keikaku.
yeah but you're basing it off of romaji, if you instead base your counting based on how many syllables, all numbers result in an alternation between ichi and ni
That's just as arbitrary as going by the symbols used to transcribe (or shall I rather say "obfuscate" …) the sounds of English words, but you could just as well spin some arbitrarily silly scheme around the sums-of-digits of the ENIGMAed, LZ7-compressed, Morsed, ROT13ed words that you get from expressing the numbers in resistor colours. That said, for a bit of variety more you could go by stroke count of the Kanji numerals, for example.
Anvilshock That’s a good point, and I should have mentioned that yes it’s arbitrary, i’ve actually been doing this pattern from 1-10 in the numbers of lots of different languages recently and found some cool results. Such as German, Dutch, and Hebrew all go to 4 like english does. I like this pattern a lot because it combines my two favorite subjects ever: Linguistics and Math.
@@Anvilshock Everything is arbitrary. But as a matter of fact, we definitely should check kanji stroke order. And let's use the more archaic kanji for small numbers that have far far more strokes
You should contact Tom Scott about this. I seem to remember he's done videos in the past about odd counting systems and he's also a trained linguist.
Ah, you're right! That's why this feels so familiar.
Was thinking the same! They seem to know eachother already, as they have done videos together in the past.
+standupmaths Matt: Get Tom Scott involved when he recovered from the Arctic!
Eli Reid
Yes, that :D Thanks
What if we use absolute value and go into negative numbers? "Negative seventeen" has 17 letters in it!
I believe all number find their way to -17 if you take the negative of the amount of letters in the word as your next number to plug in.
I like this
If you use another notation, minus nine is also an example.
The only problem with negatives is that you can only start with them no matter how you express it. So they don't end up being too interesting if you're looking at cycles and such. Library I'm using for this uses "minus" although I could change that. Either way this is a nice little thing.
But it doesn't have minus 17 letters
Esperanto:
Unu (3)
Du (2)
Tri (3)
Kvar (4)
Kvin (5)
Ses (3)
Sep (3)
Ok (2)
Naŭ (3)
Dek (3)
In Esperanto, two, three, and four all loop to each other, so every number ends in one of those three!
Same in Danish
1 (en)
2 (to)
3 (tre)
4 (fire)
In Esperanto 8 is ok? Lol
I guess most numbers in Esperanto have sequences that end in 3, seeing as 6 to 10 all have three letters, except for 8. And most numbers between 11 and 99 have 6 to 10 letters.
Here's a 9-cycle in the Native American language Urgubonsiti!!
12 --> 7 --> 20 --> 23 --> 6 --> 18 --> 24 -->21 --> 22 --> 12
gakawopu -->wokka bontu cokawallah --> wurtha pokkaloo chakkalob --> guntbu --> ottawona chitschnia --> ottawona cokawallah buhnti --> wurtha pokkaloo cheehab --> wuntha gobaluk cheehatsu --> wokka noomibu
woah
gakawopu (12) has 8 letters...
Opps lol!!!
Oops *
It starts at 7
Turkish has two structures:
Four has four letters: dört
Three has two: üç
Two has three: iki
So you end up in either 2>3>2>3...
Or 4>4>4>4...
Zulu's "AMASHUMI AMABILI NA MBILI" (22) is the longest occurrence I've found,
Zulu also has "ISHUMI NESIHLANU" (15).
Even has an even amount of letters and odd has an odd amount of letters.
That’s odd
And prime has a prime amount of letters
That's how I know which one is which
French isn't so kind for that
It's an even(/odd/prime) _number_ of letters, not an even(/odd/prime) _amount_ because they're countable.
Zero (even w/even number of letters)
One (odd/odd)
Two (even/odd)
Three (odd/odd)
Four (even/even)
Five (odd/even)
Six (even/odd)
Seven (odd/odd)
Eight (even/odd)
Nine (odd/even)
Ten (even/odd)
FWIW.
I'm going to start saying "fivearino."
And I'm getting recommended this video 4 years later.
Neat.
If we're doing number of strokes, Chinese has 3 1-loops (1,一;2,二;3,三; each has the same number of strokes as their number) and 1 2-loop (4,四 -> 5,五). On another note, zero has 13 strokes(零). The whole thing gets much more complicated if you use the "majuscule" numbers. (1-壹,12 strokes;2-贰,9;3-叁,8;&c.)
EDIT: (copypasta from my next comment)
Majuscule numbers have a k=28 (貳拾捌, 28 strokes total), which I think is the record so far.
There are also five loops: 2 1-loops (27 貳拾柒, 27 strokes; the aforementioned 28), 2 2-loops (7 柒 9 strokes => 9 玖 7 strokes; 17 壹拾柒 30 strokes => 30 叁拾 17 strokes) and 1 4-loop (24 贰拾肆 31 strokes => 31叁拾壹 29 strokes => 29 贰拾玖 25 strokes => 25 贰拾伍 24 strokes).
I'm using the accountant's system where numbers like 14 are written like one-ten four (壹拾肆) rather than ten four (拾肆). Also, I'm using simplified majuscule characters. Traditional majuscule characters have more strokes on 3 and 6, as well as the 10000 and 100000000 orders.
Majuscule numbers have a k=28 (貳拾捌, 28 strokes total), which I think is the record so far.
There are also five loops: 2 1-loops (27 貳拾柒, 27 strokes; the aforementioned 28), 2 2-loops (7 柒 9 strokes => 9 玖 7 strokes; 17 壹拾柒 30 strokes => 30 叁拾 17 strokes) and 1 4-loop (24 贰拾肆 31 strokes => 31叁拾壹 29 strokes => 29 贰拾玖 25 strokes => 25 贰拾伍 24 strokes).
I'm using the accountant's system where numbers like 14 are written like one-ten four (壹拾肆) rather than ten four (拾肆). Also, I'm using simplified majuscule characters. Traditional majuscule characters have more strokes on 3 and 6, as well as the 10000 and 100000000 orders.
+Litigious Society Chinese is different. In the PRC, for example, there is a State Language Commission that standardizes the number of strokes per character. Other countries using Chinese also have similar groups. (Really, the concept of strokes is not transferable to other scripts.)
You could also count characters:
1 through 10 have 1 character
11 through 20 have 2 characters
21 through 100 have 3 characters (except multiples of 10, which have 2)
101 through 1000 have (call the number x):
x mod 100 is between 1 and 9 -> 4 characters
x mod 100 is between 11 and 19 -> 4 characters (except in formal situations, which have 5 characters)
x is a multiple of 10 but not a multiple of 100 -> 4 characters
x is a multiple of 100 -> 2 characters
otherwise -> 5 characters
etc.
you can relate the number of strokes to the number of character, so in the end its the same
***** True.
The algorithm recommended this to me after 4 years of being uploaded.
So I have taken years of Latin and never thought of this despite knowing the trick in English. Jumping into this i went of what are the terminal places. You end up with a 2-cycle of 4 (quattuor) and 8 (octo). well nothing interesting there but as I am checking I notice the number 5 (quinque) has a decent chain already with 5 (quinque) -> 7 (septem) -> 6 (sex) -> 3 (tres) -> 4 (quattuor) and the start of the cycle. A chain of 5 long already. building on this 9 (novem) is the first number that has 5 letters. so a 6 chain before we get to double digits. 15 (quindecim) is the first to have 9 letters; 24 (vīgintī quattuor) is the first to have 15 letters (not counting the space). finally 145 (centum quadrāgintā quīnque) gets you 24 letters.
for a full chain of:
145 (centum quadrāgintā quīnque) -> 24 (vīgintī quattuor) -> 15 (quindecim) -> 9 (novem) -> 5 (quinque) -> 7 (septem) -> 6 (sex) -> 3 (tres) -> 4 (quattuor) So a chain of 9 numbers.
I guess I kinda of combined two of the challenges. I found a chain longer than 7 in a language with a non single point end.
This is cool. and yeah it looks right.
Thanks.
+Ducomors it occurs to me that you didn't count 8 in your loop for the final tally meaning a possible loop of 10 depending on if you count the loop or just the first number in the loop.
+Ducomors That is great! I figured someone would check Latin.
@@alauber86 I guess when looking for the longest chain, it has to be a chain where all numbers are different. So if there's a fixed point, you stop there, so in English the last number in a chain is FOUR.
But when you hit the loop, you can continue without repeating numbers... That's until you list the whole loop, the next number would them be a repetition, so you stop.
I think the parker conjecture is right. Any language that's worth its salt will have a system of numbers that reduces larger numbers into smaller words. If the way to say a number is comparable or longer to say than the value of the number, you're doing your job wrong. The only languages where the parker conjecture would be false are ones with bad counting systems (which shouldn't survive linguistically), or made up ones that have terrible counting systems.
Japanese could be a fun one to look at. If you did it in terms of the local syllabary, you get a two loop between one and two (いち for one (two "letters") and に for two (one "letter"))but if you go into romaji (follows your examples more closely) you get the list:
n name f(n)
1 ichi 4
2 ni 2
3 san 3
4 yon 3
5 go 2
6 roku 4
7 nana 4
8 hachi 5
9 kyuu 4
10 juu 3
Numbers do go above ten, but each iteration cuts the value down drastically. Below I have the largest possible strings for ten to ten thousand, as well as the format for large numbers in japanese:
numbers above ten: n (above) juu (3) n (above) (largest family is hachi juu hachi (13) goes to juu san (6) goes to roku (4) to yon (3) back to san)
numbers above a hundred: n hyaku (5) n juu(3) n (largest is hachihyakuhachijyuuhachi (23) nijuusan(8) hachi (5) go (2) ni(2) end loop)
numbers above a thousand: n sen(3) n hyaku(3) n juu(3) n (largest is hachisenhachihyakuhachijuuhachi (31) sanjuuichi (10) juu (3) san(3) end loop)
japanese also goes to ten thousand as a large number base, compared to english so:
above ten thousand n man (3) n sen(3) n hyaku(3) n juu(3) n (largest is hachimanhachisenhachihyakuhachijuuhachi (39) sanjuukyuu(10) as above)
this then repeats for numbers over ten thousand (ten ten thousands, a hundred ten thousands etc...) and is overall similar with small two or three letter prefixes marking each transition of ten thousands.
Now, a bunch of numbers have odd quirks. For example, up there the number for eight thousand should be hassen and not hachisen, but things like this don't make enough of a difference to matter. The parker conjecture holds.
You can see k = 3. everything should converge on that there. There are also a number of cases that converge on two. Pretty simple. Now, Japanese also has a local counting system for objects which is also used very frequently. Arguably as often as the numbers above. However, there are only a few of these, and they are used for small everyday numbers (two people, three elephants, five apples, that sort of thing.)
interestingly, these counting numbers have three possible k values. Also interestingly, I do not know of a way to count above these possible values in this counting system, making the statistics easy. Here they are:
n name f(n)
1 hitotsu 7
2 futatsu 7
3 mittsu 6
4 yottsu 6
5 itsutsu 7
6 muttsu 6
7 nanatsu 7
8 yattsu 6
9 kokonotsu 9
10 tou 3
as we can see, we've got k at 6,7, and 9.
4 numbers end directly at 7, 4 numbers end directly at 6. 10 goes to three then to six. only nine ends at nine.
therefore we have a 50% chance of ending at 6, a 40% chance of ending at 7, and a 20% chance of ending at 9.
of course, all this goes out the window if you decide to use hiragana (the local alphabet, of sorts, where a symbol represents what we might take two two right out ex: な for "na") to count "letters".
This is really complicated, mostly because it's hard to tell what's a letter (is じゅう (jyuu, spelled "Ji" + small "yu"+ "u") three or two letters, since the yu is small and technically part of the ji? what about numbers like the above ハッセン(hassen, spelled "Ha" + small "tsu" + "se" + "n") is that three or four? since the small thing is technically not pronounced, and not really a letter like it is when it's normally used, but it still needs to be there to recognize the word.) Of course, these are all technicalities, and don't matter much even in the bigger picture.
Even if you take the letter count by hiragana, all it does is shave the letter count (f(n)) for every number down compared to the english lettering or romaji (same thing). On average it should divide it by two, but I suspect the actual amount is closer to 2 and a half (since some hiragana have three letters in english, most have two, and one has one)
You get the following first ten numbers:
n name f(n)
1 いち 2
2 に 1
3 さん 2
4 よん 2
5 ご 1
6 ろく 2
7 なな 2
8 はち 2
9 きゅう 2 or 3
10 じゅう 2 or 3
You will get a loop between one and two ((1)いち↔に(2)).
But the point is, regardless of what you do, for the Japanese language, the Parker conjecture holds. Just depends on how you test it.
I'm so sorry for how long this is.
+Andriy Predmyrskyy That is a good analysis!
If you need a major, this could be a linguistics thesis.
Dang it, someone beat me to Japanese! I like was going to use the Romaji System as it would allow people to try and say the number. Not to mention, as you pointed out, writing it in hiragana vs romaji creates two separate systems for Parker's Conjecture here. In Hiragana, everything converges on 1 & 2. But in Romaji, everything will converge on either 2 or 3. As both of these numbers are self-feeding like English's "Four".
Kanji would just be cheating. everything 10 and under would just be 1, and even very large numbers would drop to 1 in 2 or 3 steps.
that is without contest the greatest, and most meaningful TH-cam comment I have ever seen. Props dude
In Frisian there is a chain from 7 > 3 > 5 > 4 > 7
EDIT: I don't speak frisian, I just found the chain on google translate
There are only 4 links in that chain.
It's not looking for the longest chain, but the longest base loop.
+Raguna163
Terminology aside, it's still not better than French's 3 > 5 > 4 > 6 > 3. It's not worse either of course.
Yeah, but the french loop is 5 > 4 > 6 > 4 > 5, which is the same length. So it's not really an improvement to the one in the video. It's still neat.
Fryslan boppe, frankryk yn e groppe
I never knew how bad my procrastination is for my finals until I realized I watched a whole video about the number four
In high school, we played this as a game called "four is magic".
Me too, except it was called four is cosmic
Thank you! I was looking for this comment.
We played this as a riddle and called it “Four is the law”.
In Vietnamese language, we have “hai” for 2 and “ba” for 3. Interestingly, these two numbers repeat itself no matter what number you start with, just like in English and other. What a fun fact!
Matt! I have semi proved your conjecture. Obviously, language systems are going to be extremely complicated and sometimes just unintuitive, but by putting some restraints on what you can do, I can prove that your conjecture is true.
So, in any reasonable language, the number of letters needed to express a number should correspond linearly with how many digits it has. Formally, if we use your function f(x) as the length of a number when written out, and if d(x) is the number of digits that a number has, f(x)0 such that whenever x>r, we have C*log10(x)0, there exists an r>0such that whenever x>r, we have log10(x)/x
Note: I'm only 13, so this also might be wrong, but I've checked it with logical proof checkers and it isn't really super complicated so I am pretty sure it is devoid of holes.
That's impressive for a 13-year old. And I thought my longest-chain-finding Lisp function was cool...
You use Lisp? What are you? A masochist?
Chris Daley XD
Not at all. It took like 9 lines and I could condense it to 7 or even 6 if I made it a stand-alone LET expression instead of a proper function.
Well, technically I could make it all 1 line but that wouldn't be too readable.
Człowiek Wiking
The language is inherently unreadable. XD
7:18 I guess semi-consistent is the same as "Parker consistent"
Chinese doesn't have a letter system, but there are strokes for each character. An analogue to the number letter mapping is then the number stroke mapping. Numbers one to three (一二三) are attractors, mapping to themselves, but four and five (四 and 五) map to each other in a loop. One is the lonely number. After that, all other numbers lead back to the numbers two through five, with most going to two, fee going to three, and the minority ending in the four/five loop.
*fewer, not fee
*fewer, not fee
*fewer, not fee
Well there is pinyin, where er (2), and san (3) would map to theirself
+vlanoik *themselves, by the way
Using "and" between the hundreds and the tens digits (i.e., 65536 is "sixty-five thousand, five hundred and thirty-six" but 1500 is "one thousand, five hundred") (and of course removing spaces and hyphens), I found the smallest number with a 7-chain:
124 (one hundred and twenty-four) has 23 letters,
23 (twenty-three) has 11 letters,
11 (eleven) has 6 letters,
6 (six) has 3 letters,
3 (three) has 5 letters,
5 (five) has 4 letters, and
4 (four) has 4 letters.
The next smallest numbers that form 7-chains are 125, 129, 134, 135, 139, 143, 147, and 148, all with 23 letters.
The smallest number with an 8-chain would therefore have at least 124 letters. Using long-scale (i.e., thousand, million, billion, trillion, etc.), I found that 113373373373 (one hundred and thirteen billion, three hundred and seventy-three million, three hundred and seventy-three thousand, three hundred and seventy-three) has 124 letters (which makes it an 8-chain), but I don't know whether this is minimal. I haven't checked whether there is a smaller number with (say) 125 or 129 letters.
I am sure we will never make up enough names for powers of 10 to be able to get to a number with a 9-chain (which would by definition have at least 113373373373 letters). For context, the complete works of Shakespeare (including spaces, punctuation, and all) contain about 3.6 million characters. English Wikipedia currently contains about more than 2.9 billion words (approximately 15 billion characters, including spaces and punctuation). Wikipedia in all languages put together currently contain more than 27 billion words (approximately 130 billion characters (finally bigger than 113373373373, though including spaces and punctuation)).
Omitting "and" (i.e., 65536 is "sixty-five thousand, five hundred thirty-six" and 1500 is still "one thousand, five hundred"), I found that the smallest number with a 7-chain is 323 ("three hundred twenty three" has 23 letters, then proceed as before to 11, 6, 3, 5, and 4).
The next smallest are 327, 328, 333, 337, 338, 374, 375, and 379, all with 23 letters.
Here, the smallest number with an 8-chain would therefore have at least 323 letters. Using long-scale again, I found that 1103323373373373373373373373373 (one nonillion, one hundred three octillion, three hundred twenty-three septillion, three hundred seventy-three sextillion, three hundred seventy-three quintillion, three hundred seventy-three quadrillion, three hundred seventy-three trillion, three hundred seventy-three billion, three hundred seventy-three million, three hundred seventy-three thousand, three hundred seventy-three) has 323 letters, but again I'm not sure whether this is minimal.
There will never be a number with 1103323373373373373373373373373 letters. Typing the letters at a rate of one trillion letters per second, non-stop since the Big Bang, you would be be about 37% of the way done. One page in Microsoft Word at Times New Roman 1-point font has approximately 610000 characters (including spaces and punctuation). Typing a number with so many letters out in 1-point font, printing it out double-sided, and stacking the pages would produce a stack that goes from Earth to Alpha Centauri and back -- more than 1100 times!
In Italian: otto (8) -> quattro (4) -> sette (7) -> cinque (5) -> sei (6) -> tre (3).
A chain of length six with all the numbers between 3 and 8.
Seems like a dynamic programming (DP) problem. You could build up the smaller solutions (for example: chains of 3) so that when you're investigating larger numbers, you can recognize that you've fallen into a chain you've seen before (and add it to an index)
fun fact: Dutch has the same loop as English
German does too. but im pretty sure chains in german will be really long, considering 7 has 6 letters and 6 has 5 letters. zweihundertneunzehn -> zwanzig -> sieben -> sechs -> funf -> vier. is the longest i've found without going ridiculous. the chains actually turn out small, because 13-19 all have 8 letters, and 0-9, excluding six and 7 all have 4, so any 'teen' (zehn) goes to 8, which goes to 4 then loops.
jack rycroft, yeah, i used to know that, but that was a long 2 years ago
MegaMGstudios and also Filipino
Swedish has two of these loops, with 3(Tre) and 4(Fyra)
hebrew as well, four is ארבע
You should make Reddit forums for these challenges instead of using the comments system.
A TH-cam commenter saying reddit commenters are trolls? u wot m8?
there is a Matt Parker subreddit. www.reddit.com/r/MattParker/comments/523e65/four_has_four_letters/
+jothamvw
could be worse they could say 4chan is good
***** ew
What about the people that haven't got a Reddit account and do not want to start one?
People like me
The first two numbers which contain the word 'EVEN' are both odd - SEVEN and ELEVEN.
He said "contain," not "end." Seventy counts for his purposes.
SEVENTEEN is also odd, and it contains 'even', so i guess its the first 3 numbers which contain 'even'
as is twentyseven, thirtyseven, fortyseven, fifty seven and sixty seven
Odd 'even' numbers: 7,11,17,27,37,47,57,67,71,73,75,77,79,87,97
Even 'even' numbers: 70,72,74,76,78
Then the pattern repeats.
Ergo, 75% of 'even' numbers are odd, until you get to larger blocks of 'even' numbers (ie. 700, 7000, 70000, etc.) and then they progressively fall, tending towards 50% but never reaching it.
The first prime number is even.
In italian,
6 has 3 letters ("sei"),
8 has 4 letters ("otto"),
10 has 5 letters ("dieci"),
12 has 6 letters ("dodici").
There is a very popular italian riddle based on this pattern
Am italian
Didn't know
I've investigated that in Polish, and below 1,000,000 numbers I've found lowest and longest chain :)
269 = dwieście sześćdziesiąt dziewięć [29] => dwadzieścia dziewięć [19] => dziewiętnaście [14] => czternaście [11] => jedenaście [10] => dziesięć [8] => osiem [5] => pięć [4] => cztery [6] => sześć [5] => pięć [4] => loop :)
0:59 What the Hell, i just literally thinking about that 1 SECOND BEFORE THAT!
In Finnish, anything smaller than seven leads to five (viisi) while seven and above leads to looping eight and nine (kahdeksan, yhdeksän).
Thanks for the programming challenge. Even though I suck at programming, it'll pass the time before I have to go back to school.
In italian you can do a chain of seven numbers starting from 15:
15: QUINDICI (8 letters)
8: OTTO (4 letters)
4: QUATTRO (7 letters)
7: SETTE (5 letters)
5: CINQUE (6 letters)
6: SEI (3 letters)
3: TRE (3 letters)
+Alessandro Gatti I think Italian now holds the record for lowest starting number seven chain.
Now it is time to break the record :D
In polish a 7 link chain starts with 14
14 (czternaście) -> 11 (jedenaście) -> 10 (dziesięć) -> 8 (osiem) -> 5 (pięć) -> 4 (cztery) -> 6 (sześć) and back to 5
And a smallest 8 link chains starts with 19 (14 letters) and smallest 9 link chain starts with 29 (19 letters)
I found out that 4(cuatro) and six(seis) go into a loop. I also found that 5(cinco) has five letters and 7(siete) and 9 (nueve) fall on it. Very cool idea in this video, congrats
In Spanish*
I was initially presented this as a word game that you bug people with around campfires and such. It was called "4 is cosmic"
For Russian:
1 - один (4 letters)
2 - два (3 let.)
3 - три (3 let.)
4 - четыре (6 let.)
5 - пять (4 let.)
6 - шесть (5 let.)
7 - семь (4 let.)
8 - восемь (6 let.)
9 - девять (6 let.)
10 - десять (6 let.)
I can see a loop of numbers: 6 -> 5 -> 4 -> 6, and so on. Also, 2 gets us to 3, and that starts an infinite self-loop on number 3. But I dunno about the threshold exactly, so if any друг can help, that would be nice
11 - одиннадцать (11 let.)
That is the third and the final loop for Russian, and also incidentally the k-threshold.
I checked the numbers up to 1000, and 77% end up in the 6-5-4 loop, 22.7% in 11, and only 0.3% in 3. Also, out of 770 that end up stuck in 6-5-4, the vast majority (760) enters the loop at the number 6. Fun trivia.
@@Yulenka- Holy gosh, you've done a great job, tovarisch!
Confirmed, output to this code is {18: 999904, 13: 96}, this is the most pythonic way I could think of quickly writing it (simple, to the point, no over abstracting):
from collections import defaultdict
def linguistic_bin_count(num):
bin_num = bin(num)
zeros = bin_num.count('0') - 1 # Python represents bin number as 0bxxxx
# so we have to subtract 1
ones = bin_num.count('1')
return 4*zeros + 3*ones
def final_linguistic_bin_count(num):
"I've assumed no loops, want to prove it?"
next_num = linguistic_bin_count(num)
if next_num == num:
return num
return final_linguistic_bin_count(next_num)
tally = defaultdict(int)
for n in range(1, 1000001):
tally[final_linguistic_bin_count(n)] += 1
print(dict(tally))
+Damian Shaw Nice code! Glad it's been verified. It is indeed very pythony to use a "for n in range".
It seems that actually only 96 numbers tumble down to 13. Hypothesis verified up to 10^9 with a similar code.
Virgile Andreani
Indeed, 97 if you include 0. Seems like a long winded proof would be to show that all numbers in a certain range go to 18, and then the smallest binary count is 11111...1111 and the longest binary count is 1000...0000 for a number of length n, and then show that because of this they end up eventually falling in to this range. You could more or less code this proof up.
But I wonder if there's a proof from the book (a nice small elegant proof).
standupmaths
Virgile Andreani
Actually on 2nd thoughts I had an idea, the 96 numbers < 1000001 not including 0 are: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 31, 128, 271, 279, 283, 285, 286, 295, 299, 301, 302, 307, 309, 310, 313, 314, 316, 327, 331, 333, 334, 339, 341, 342, 345, 346, 348, 355, 357, 358, 361, 362, 364, 369, 370, 372, 376, 391, 395, 397, 398, 403, 405, 406, 409, 410, 412, 419, 421, 422, 425, 426, 428, 433, 434, 436, 440, 451, 453, 454, 457, 458, 460, 465, 466, 468, 472, 481, 482, 484, 488, 496, 767, 895, 959, 991, 1007, 1015, 1019, 1021
And these go up pretty big, so couldn't we artificially create a number that fell in to this? For example say we wanted to fall in to 271? Well that's simply 3 + 4*67 or : one zero zero ... zero (67 zeros).
And indeed:
> print(final_linguistic_bin_count(2**67))
13
So it seems the numbers just become very sparse. So 2 questions, what is the next number after 1021 that resolves to 13? And can we use this infomation to show there are an infinite number that resolve to 13?
Good point. Indeed, the next number which falls down to 13 is 4294967303 (= 2^32 + 7), with 29 zeroes and 4 ones (takes under a minute in C).
In Hebrew:
אחד = 1
שתיים = 2
שלוש = 3
ארבע = 4
חמש = 5
שש = 6
שבע = 7
שמונה = 8
תשע = 9
עשר = 10
In grade school we had the little riddle "Four is the magic number" where your friends would try and figure out why exactly they could give you any number and you would always end up at the magic number four. Nice to see someone nerding out and expanding upon the concept to this degree.
In italian the fixed point is 3 (tre), and I found this quite long chain, starting from 44:
44 - quarantaquattro
15 - quindici
8 - otto
4 - quattro
7 - sette
5 - cinque
6 - sei
3 - tre
If you try Chinese, it will definitely be the most special one. In Chinese, if we count the syllable(s) of a number, it will always end up with one....... Coz every letter in Chinese only have one syllable, and numbers larger than ten would be a word combine with numbers and it's digit value. For example, 3795 in Chinese would be Three Thousand Seven Hundred Nine Ten Five, since each word starts block letter have only one syllable, it would then be 7 syllable in 3795, and 7 have only one syllable, then it ends at 1.
So the number would need to be at least a 21-digit number without any zeros to make a chain of 4. Eg, the number 111111111111111111111 will have 21 syllables and 21 have 3 and 3 have 1. A chain of 4 in Chinese would need a number with 111111111111111111111 syllables, let that number be x, then a chain of 5 would need a number with x syllables, let that be y; then a chain of 6 will need a number with y syllables, let that be z; chain of 7 will need a number with z syllables, let that be u; chain of 8 will need a number with u syllables, let that be w; think of how huge w would need to be.
yeah..
yi, er, san, si, wu, liu, qi, ba, jiu, shi..
my chinese is not the best but...that's clearly not a syllabus
Tang Chak Yan but a 21 digit number with no zeros has 41 syllables then right? Cuz the million, thousand, hundred, ten etc count right?
Tang Chak Yan works in german and english as well, spanish is a 1 2 loop, and I bet most other languages have a similar pattern. Would be interesting to see if there are examples where it is not the case
In Chinese, surely it would make more sense to count the strokes rather than characters. 一， 二， 三， and 四 (1, 2, 3, 4) are all fixed points and 4 is the threshold. More interesting is if one considers the bankers' anti-fraud versions of the numbers. I'm a bit shaky on them, but I *think* the following is correct (could someone who actually knows Chinese check?): 拾伍, 拾陆, 贰拾柒, and 贰拾捌 (15, 16, 27 and 28) are all fixed points. 拾 and 玖 (7 and 9) form a 2-cycle, and 贰拾肆, 叁拾壹, 贰拾玖, and 贰拾伍 (24, 31, 29, 25) forms a 4-cycle. 31 seems to be the threshold. www.dropbox.com/s/xdxemlg0efqagzz/ChineseNumberStrokes.bmp?dl=0
You should do this with Tom Scott... maybe he'll have some ideas about languages and numbers
Should we also look into languages that have the most unrelated chains? Albanian is pretty cool with that one.
2 is a single link chain that maps to itself (de);
3 has two links (nje [1] maps to tre [3]);
4 and 5 make a loop at the end of a chain (katër, pesë);
6 and 7 make a loop at the end of a chain (gjashtë, shtatë);
11 and 12 make a loop at the end of a chain (njëmbëdhjetë, dymbëdhjetë);
14 ends a chain (katërmbëdhjetë);
16 ends a chain (gjashtëmbëdhjetë).
7 individual chains. That's got to be the one to beat. And 16 is a contender for highest self-referencing number in a given spoken language.
#albanianisamathsnerd
This discovery was the best thing I've ever done. And Matt didn't even notice.
there are several in UK counting, Weardale has a 5 loop 3 Tethera 6 Yan-a-tic 7 Teyan-a-tic 9 Methera-tic 10 Bub back to 3, Teesdale counting has a 3 number loop 3 Tether 6 Lezar 5 Pip back to 3, Swaledale counting has a 2 number loop 4 Mether 6 Azer, Nidderdale counting has 2 stop points 5 Pitts 6 Tayter.
Me: Let's see what's in my recommended today
TH-cam: Four has four letters
I've been playing about on MATLAB and I believe 113,373,373,373 (aka X) is the lowest number (including the word 'and', excluding any whitespace or punctuation) which will give you a 8 length chain to get to 4. As the lowest number to require 7 steps is 214, X must therefore be 214 letters long.
After writing a function to tell me the letter length of any given number, I worked out that (777 and later 373) have the longest word length of any three digits at 27. It made sense to me to create a number which was an integer multiple of 3 long composed entirely of 7s. It turns out 12 7s give a letter length of 130 so all I had to do was change one of the groups of 3 7s to a group of three digits with a letter length 6 less than 777, ie 21. 113 turned out to be the lowest number with a letter length of 21 so I put that in the most significant position to make the number as small as possible. I then checked to make sure 777 was the smallest 3 digit group with a letter length of 27 and it turned out it was not, so I replaced all of those groups with 373.
Just to make sure this wasn't a red herring I checked to makes sure that there weren't any lower value numbers than 113 with higher letter lengths than 21. I was lucky as there turned out not to be any. If there had been then the second most significant 3 digit group would have had to have had its letter value decreased accordingly.
The number which will have a chain of 9 will need 113,373,373,373 letters which, assuming 5 letters per word, would take about 290 years for a single human being to finish saying outloud.
You can view my code at github.com/osholt/NumberLengthMattParker/ .
If you think I've done anything wrong or have suggestions please let me know.
I believe Matt is counting the "four" in the chain length, so in that case what you would have found is an 8-chain. "One hundred and thirteen billion three hundred and seventy three million three hundred and seventy three thousand three hundred and seventy three" has only 124 letters, but it does result in an 8-chain: X, 124, 23, 11, 6, 3, 5, 4. As he mentions, there are quite a few 7-chains less than 1,000,000 and that's without counting the "and".
Oh, I missed that. Thanks for pointing it out. I'll change my post. 😊
This probably required some clever programming.
This probably required some clever programming.
Karl Muster I wrote 3 functions to deal with numbers less than 1000. In the ranges 1 to 15, 16 to 99 and 100 to 999. I then wrote a function to deal with numbers from 1,000 to 999,999 which basically just called the 3 functions I wrote to deal with numbers below 1000 but added on an extra number to account for 'thousand'. I repeated (ie copy pasted) this process up to the quintillions which is where 64 bit integers top out at. It basically just comes down to some logic around some lookup tables (or in this case switch-case statements).
1 has one syllable. Its the only fixed number parallel to its diaphragmatic shaping
hey what about ta-woo :(
In Polish:
cztery (4) > sześć (6) > pięć (5) > cztery (4)
by other numbers:
dwa (2) > trzy (3) > cztery (4)
jeden (1) >pięć (5) >cztery (4)
siedem (7) > sześć (6) > pięć (5) > cztery (4)
dziewięć (9) > osiem (8) >pięć (5) > cztery (4)
The word "cztery" knows "four" is albo sinkhole in Polish Language.
“The least uninteresting positive integer” has 36 letters. I don’t know whether that qualifies as interesting, but the “function” can take paradoxical numbers as input.
That makes my head hurt.
In Polish there is no number below a hundred that has the same number of letters as its value
For numbers from 1 to 10 there is: 5, 3, 4, 5, 4, 5, 6, 5, 8, 8 so there are some loops
Let's see, in Latvian we have 5, 7, 11 holes all on their own. In Russian it's 3, 4-5-6 cycle, 11. Don't know any more
Searched Danish from 0-100, the longest sequence I could find was 33-11-6-4. We have multiple loops so sequences are cut short pretty quick.
Sequences by length:
1-2
18-5-3
24-10-2
33-11-6-4
Length of numbers:
2: to (2)
3: tre (3)
4: fire (4)
5: atten (18)
6: elleve (11)
7: tretten (13)
8: enogtyve (21)
9: treogtyve (22)
10: fireogtyve (24)
11: treogtredve (33)
12: fireogtredve (34)
13: tooghalvtreds (52)
14: treoghalvtreds (53)
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