You Won't Believe How This Puzzle Ends
ฝัง
- เผยแพร่เมื่อ 8 ก.ย. 2024
- Today's puzzle showcases one of the great new constructors on the world sudoku scene - Joseph Nehme. This is another beauty - it's called "Encaged 9" and you can play it here:
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Rules:
In cages, digits must sum to the small clue given in the top-left of the cage. Digits cannot repeat within a cage. The digits in the circles are equal to the sum of the cells along their respective arrows.
logic-masters....
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'We have a puzzle today with a very rare quality...' Yes, it actually has a digit in it.
so true
I laughed so hard at it! lol
Title: You won't believe how this ends
Skips to end... all columns, rows and boxes contain digits 1-9 !!
@Anifco67 They do?
Hold on...
1... 2... 3... 4... 5... 6... 7... 8... 9........ WOW, YOU'RE RIGHT!
@Anifco67 There's a knowledge bomb.
What a SPOILER!1!! 11
Yes, but does the Phistomephel ring apply?
@@skarcade CTC, creating brand
new sentences
Simon leans back and exclaims "oh that's beautiful!" and then spends 15 seconds staring at the puzzle while we're all thinking "what? what's beautiful? tell us!"
Every time 😂
Sees other channel where title starts with "You Won't Believe" --> avoids, it's clickbait.
Sees the same on Craking The Cryptic twice a day --> Gets excited every single time.
Indeed 👍
Only a few channels can use such titles and not have their videos repulsively marked as clickbait by it.
Watched the whole video as soon as it came out and then noticed that the title doesn't make sense...
Lovely, isn't it
@@jpryan90
Agreed: clickbait-like titles are acceptable if (and only if) they actually deliver.
One time, I’d like to see Simon go full competition mode, don’t talk through the logic while solving one of these (maybe voice over afterwards?) and do it as quickly as he’s capable of.
Simon would be pretty fast but there are surely many faster people. Does Simon even enter the competitions anymore?
@@stephenbeck7222 sure, but those people don’t run CtC. Even if he doesn’t enter comps any more he’s surely faster than 99% of people who watch this channel?
Check out Bastien Vial-Jaime's channel, called 81Cells - he does exactly this, but instead of voice-over he just writes very succinct annotations of his solve path. th-cam.com/channels/uiNWH9CJdqRPt52CcQrBOA.html
I'd love to watch this as a side-by-side of Simon and Mark each solving the same puzzle in realtime. It'd be interesting to see how they tackle it differently (and, of course, who finishes it first!)
the problem is that "competition mode" usually means bifurcation, which Simon isn't a fan of. Mark has done it on the channel before and Simon has playfully made fun of him for it, but i don't think Simon would ever do a "competition solve" on his own.
So, I think there's actually a slight flaw in the logic in the break-in. The arrangement at 17:24 does actually work for adding up to 180, if my math is right. Because the entered squares are either arrow totals or intersection points, they all count double for the 180 total. Thus: (9+9+9+8+8+8+7+7)*2 = (27+24+14)*2 = (65)*2 = 130. Then you add in the 50 from the cages, and you get to 180, and everything is good to go.
I think what you're actually supposed to realize is that the 16 cage in row 2 must have at least one of 7 8 9, and that eats up all the degrees of freedom available for the intersections and arrows, since the best you can do is 986 in the relevant squares in row 2. Then you get ((9*4)+(8*2)+7+6)*2+50 = (36+16+7+6)*2+50 = (65)*2+50 = 130+50 = 180, and the only way to fit four of the same digit in the relevant squares is to put them in all the arrows. The solve proceeds in the same way from there.
Yeah, switching 8 and 9 does not decrease the degree of freedom because you lose 2 degree by putting 8 in the arrow, but you also gain 2 (not 1) degree by putting 9 in the middle.
I think your logic with considering 16 cage is indeed what we are supposed to think about
You can’t have 3 8s and 3 9s in the relevant positions. Only the outside circles can have more than two of the same digit. The centers, which are the only other position that 7, 8, or 9 could be in the relevant squares, can only have two of any digit. The only way to get to the required sum is with 4 9s, which must then be in the circles. You don’t need the 16 cage to conclude this.
@@renmaddox Not quite; again, the configuration at 17:24 is valid; 2 9s and 2 8s in the outside circles, and 9 8 7 7 in the centers.
@@codegorilla7120 Ah, yes. I missed the possibility of *not* putting a 9 in a center square, but still replacing a 9 with an 8 (and the 6 with a 7).
Oh god, I got stuck on this for so long thinking that 4+5+6 = 16 🤦♂️ and thinking there could be a 6 in that cage
*3:41** **_Rules_*
*4:10** **_Let's get cracking!_*
Thank you sir
This is what we all need
Thanks for that. I thought I was the only one that skipped the ever increasing list of stuff at the start.
@@malcolmjohnson4414 Agreed. I wish they would do their announcements at the end, but I understand that they want to make sure as many people know about the stuff as possible. I only hope that they don't mind that I've been posting these helpful time-savers for those of us who watch every post and already know all that info...
@@buzzly108 they should divide the video in chapters. I'm sure it's easy to accomplish.
I punched the air when Simon finally filled in the remaining cell in box 7 lol
Thank you Simon for the great solve :) I'm so glad you liked it. And for everyone who's asking what's special about the ending, it's the fact that all 9s are encaged either in cages or in arrow heads or in the red cross shape.
That's a very nice puzzle. Thanks for making it.
Beautiful puzzle!
Maybe the most beautiful puzzle I've seen. The break-in was just so clever...couldn't get it myself until Simon said "let's look at THESE columns and THESE rows together...then everything fell in place. Just so cool to see all the cages falling one after another...not difficult once you got started, but very cool.
@@MichaelGreen831 Thank you. I'm glad you liked it :)
@@dwarfigplays9806 Thank you :)
Simon delaying while his brain downshifts from complex Sudoku to counting from 1 to 9: "This... square... looks... like... it... should... be... a... 2"
Of course I know how it ends. With me breaking the puzzle and watching the video.
The fact that we get 2 videos a day of this... amazingness! Idk how you do what you do but never stop. You make me want to learn how to do these crazy puzzles myself.
It was so obvious from the beginning that the geometry is the key. But there are so many possible geometric patterns that give nice, but useless formulae (I played with the phistophemel ring for a long time - and only after a long time found exactly Simons path). What is so stunning in Simons solve is the fact that he instantly focusses exactly on the crucial spot. Well, he does not spot the critical pattern in every of his solves - but still in incredibly often. Cudos!
Simon did his math wrong in this one. The configuration shown at 16:55 gives a total of exactly 180 if you remember to double count the central cells (30 from box one + 34 from box three + 34 from box seven + 32 from box nine + 50 from cages). He needs to do more logic to rule it out fairly.
17:29 I think you got lucky there, because the bulb AND the green cell are both counted twice, so exchanging them doesn't result in any loss of freedom.
The key is in the 16-cage on row 2 which must have at least one of 7,8,9, meaning that the initial assumed freedom does in fact not exist because you can't put 7&8 in the green cells on row 2 combined with a 9 in the bulb.
This puzzle ends with a box with 8 cells filled already, not having its last digit placed for what seems like five minutes.
Which is completely believable on this channel 🤣.
so funny
"All the bulbs of the arrows must be 9" - I don't think you can reach that conclusion at this point in the solve. Note that _both_ the bulbs _and_ the green cells are counted double towards the total. So swapping a bulb with a green cell does NOT loose a degree of freedom.
For example, the situation that you have at 16:57, with two of the bulbs an 8, works correctly for the total sum.
(that particular solution is not possible, because you cannot place a 7, 8, or 9 in the 16 cage anymore, but that is a different reason)
Won't believe how it ends???🙈
Can't believe how it starts 😂
next time on CTC: "Sodoku Secrets The World's Best Puzzlers Don't Want YOU 2 Know!"
We want you to know them, we just want to be the ones to show you =D
Mathematicians hate him!
9:40 we all know that moment when the dvd doesn‘t work proberly.
At 15:05, you say that by putting a 9 in the center instead of an 8, you "gain one here." Don't you gain two, since that square is counted twice? So, at 17:02, when you have two 8's and two 9's in the circles, with 9/8/7/7 in the center of the boxes, the total for the rows and columns (counting those centers twice) is 180.
I haven't watched the rest of the video yet, but it seems that using this logic to conclude the circles must all be 9's is flawed. Did I miss something there?
True but it doesn't really matter, there is still a problem with the 9s being in the same row or column...
@@francoisduez601 Not at 17:02 cause in box 1 is no 9. And still the 2 columns and 2 rows would add to 180.
It seemed so to me. Both the arrow total and the middle square are double counted so switching them doesn't actually change the total. It does however affect the other square in the row od collumn, but the example Simon tried I think still works. It might be broken a step or two ahead, but not with just what was shown.
I think you are correct and the logic at that point is wrong, digits in the arrow circles are counted twice, but intersections between the columns and the rows are also counted twice
@@Draddar yes it’s broken because 7, 8, 9 are outside of the 16 cage (if somewhere in a circle is a 8) And 6+5+4 only adds up to 15.
"BOB- BOB- BOBBINS! ...oh, this is very clever, this is lovely." Classic Simon.
I just want to say, this is incredibly relaxing to watch as you wind down for the night after a massive amount of homework. Thanks for the brain switch Simon!
It's the same for me, except I teach and have to write all the tests and assign the homework (which all of it really is homework for the teacher), and I teach math, lol, how bizarre is it to relax by seeing numbers manipulated
At the start, the sum of the the 2nd and 8th rows and columns is 2x the circles + 2x the intersections + the 4 groups of 3 for a total of 180. Thus, the sum of twice the circles and intersections is 130. The circles and intersections add up to 65, of which there are 8. The average is greater than 8. The maximum configuration is 9x4 + 8x2 + 7x2 = 66. By substituting a 7 with 6, you get 65.
So, I think there's a slight flaw in the logic around 16 minutes in. There *are* ways to arrange the 4 circles and 4 central squares in the corner boxes that satisfy the math WITHOUT putting 9s in each circle, since both the circles and the central squares are double counted, albeit in different ways. (There are ways to fit two circle 9s, two circle 8s, a central box 9, central box 8, and two central box 7s.) A better path would be to notice that in row 2, you can't put 7/8/9 in the 3 candidate cells, because when combined with the 16 cage, you then need to put 5 in the 3 remaining cells. This reduces your max to 23, plus the 24 in the three cells in row 8, plus 9+9 in the two last circles for a 65 total. Double count, add the 50 cage totals to reach 180.
I think you Made a small mistake at 15:40! With the 9 in R2C8 you are still able to get to 180! The reason ist doesn‘t work is the 16 cage, which can‘t be a 16 Cage without 7,8 and 9. But 4*7+6*8*6*9+50=180
The restriction is that the intersections and circles in the rows and columns 2 and 8 are 7-8-9 and the circles outside are 8 and 9. Exactly as Simon shows.
Or one column and row 7-8-9 and the other 6-8-9 and all the circles are 9. The 16 requires a 7 which solves the correct option.
@@rmahu But he didn't actually carry through the logic correctly - you can exactly put one '9' in a center cell. The four circle cells and the four middle cells are all doubled to make up 130 - so the eight cells must add up to 65. 9+9+9+9+8+8+7+6 works... but so does 9+9+9+8+8+8+7+7. He made an error in his logic at this stage, but it was luckily saved by the 16 cage requiring a 7+, which he hadn't considered yet.
his mistake was taking one degree of freedom from box 3 because 9 went to middle and 8 to arrow. but actually there is no lose of degree because both arrow and middle of box is counted 2 times
@@CircleTheSkies Doubly lucky, he chose the correct configuration by mistake and did not deviate from the correct solution path. The 16 tells you that there is a 6 and in which row it is, Simon only discovered where the 6 was.
However once you realize the 16 cage needs a 7. It stops the 3*9 3*8 2*7 arrangement and forces 9*4 8*2 7*1 6*1
So the logic in the beginning should be something like this:
The arrow circles and the four cells r2c2, r2c8, r8c2, r8c8 (the "8 cells") each counted twice plus the 10, 10, 14 and 16 cages need to equal 180. Therefore the 8 cells need to sum up to 65. Theoretically, four 9s, two 8s and two 7s are possible for a sum of 66, but this leaves the 16 cage without a valid solution and we need one 6 (or lower, but that would break the 65 total).
Therefore, the 65 total needs to be four 9s, two 8s, one 7 and one 6. To be able to place four 9s, they need to be in the arrow circles.
From here, Simon's logic should pick up again.
I start pathetically, watch for help, then finish the puzzle myself, but I like to watch to the end just to hear Simon's superlatives!
I really thought Phistomefel's trick was going to be used with the red cages effectively almost entirely covering the ring around the central box. Great puzzle and solve nonetheless!
I used a variant of Phistomefel’s theorem to conclude that the circles and centers of corner boxes sum to 65. I also wondered that, but didn’t see how to use it even in the endgame. See the middle diagram here. scontent-sjc3-1.xx.fbcdn.net/v/t1.0-0/p526x296/133504493_1980350375438118_8896927471456975670_o.jpg?_nc_cat=109&ccb=2&_nc_sid=730e14&_nc_ohc=W5umyev9ATIAX_HqdH7&_nc_ht=scontent-sjc3-1.xx&tp=6&oh=041db465bc1c6025637e12b954ed1d88&oe=60278959
Hey, I know this is a long shot, but I keep seeing people commenting mentioning this Phistomefel. I don't suppose you can point me to the right videos to understand the context? I can see one which is a live battle, but I would assume that's, like... the season finale of their rivalry. Can you - or anyone else who sees this - help?
Your videos makes me happy, Simon. Keep it going mate. Greetings from Germany.
From 32:38-35:24, almost 3 whole minutes, anybody else shouting BOX 7?!? On a real note great work Simon, I always love your videos :)
Uff... the beginning of this one was quite difficult, I couldn't do it without Simon's help! But this was a lovely sudoku, I especially liked the placement of the 9's on the red cross! ♥
I’m always baffled by how you can find the logic so quickly
Friday, Friday, gotta get down on Friday.
The most frustrating thing is I can never break into puzzles like these, but after the break-in it's usually pretty accessible
14:44 You don't directly lose any degrees of freedom from the switch directly, due to the double-counting. The cells with stuff in them need to add to 65, and that's it.
the 5 in box 7 was empty forever!!! such a pretty puzzle tho
Possibly slightly safer break in (at least I'm more comfortable with it):
1 - argument that the 8 cells that make the arrow totals and centres of boxes 1,3,7 & 9 must sum to 65
(as previous, consider R2+R8+C2+C8 totals, subtract cages, pull arrows into their totals counted twice, half everything)
2 - then consider those same 8 cells but now think of them row by row:
R1: single cell, max 9
R2: three cells, max 24
R8: similarly max 24
R9: similarly max 9
Total: max 66, so one degree of freedom
3 - then focus on row 2:
there's the three cells which sum to 23 or 24
there's the 16 cage
remaining 3 cells: at least 6
hence can't be the 24, must be 23!
that's the degree of freedom used up!
Me: first 65 minutes - staring at the monitor... then got a crazy idea and solved it in 16 minutes! Beautiful puzzle this is!
That's some perseverance. I tend to give up after some 15 minutes if I can't figure out the break-in.
And that's assuming I even bother trying it myself.
I have seen hundreds of solves on this channel. This is by far my favourite. The geometry of this puzzle is astounding.
Simon: "Please leave a comment, especially a nice one."
Me screaming about how nice this comment is
Was anyone else SUPER excited when simon unclicked a highlight!
We're an easy crowd to please.
Definitely felt some relief
What a wonderful setting! And great solve too!
I mentioned in a puzzle last week that the red background in cells made it impossible for some people to do that puzzle. At least in that puzzle the red had a reason for being there. Here it is just for show.
The ink is blue, the background is red, can't make out a thing on there!
If anyone else is having a similar problem, click the button on the left that opens the puzzle in the old format, in that format you can change the background colour. It does then allow you to try the puzzle.
First time in one of these that it got to a point where I was solving it before Simon, quite interesting how fun it can be.
Had the sudoku open for 2 days, got nowhere. Watched the video and realised I wasted 2 days for nothing. That logic at the beginning, no chance I'd have come up with it. But now I've added it to my "known techniques" list, so maybe next time. Great puzzle
Great solve of a great puzzle!
Another brilliant solve of a beautiful puzzle! While I threw in the towel after 42 minutes and started the video, I got at least some satisfaction from the fact that Simon was initially following the exact same train of thought as I had until then.
Simon i really apreciate you so much!
I have tried this one and gave up not long after...
Watching the way you solve these puzzles while explaining every little thought that comes to your mind is just wonderful!
Keep going :D
And there's another "encaged 9" Simon didn't mention... there are exactly nine cells inside the "cage cross".
This is a nice comment for Simon :)
39:37 ... I tried using the same strategy Simon used, and even got the same conclusion, but I couldn't 'prove' (to myself, anyway) why it had to be what it was; even after watching Simon's reasoning, it's hard for me to convince myself of such. So while I did solve this one, I can't give myself 'full credit'. Still, that I solved it at all gives me some satisfaction.
Tough puzzle!
where in the hierarchy is the Bob-Bob-Bobbins? so far it's: Bobbins, Archbobbins and Uberbobbins.
One of these days he'll accidentally throw out a Bobble and it'll be the greatest day of my life.
@@bobblebardsley
Or “bobber!”, a portmanteau of “bobbins!” and “bother!”
I'm familiar with archbobbins but never seen the Uberbobbins video, where is it?
Didn't we have a "triple bobbins" the other day?
Just once, I want to hear Simon exclaim, "oh, fuck me..."
I think I have a new favourite puzzle. Absolutely loved this one, thank you so much.
Oh, and I didn't notice anything special about how it ends. Maybe I should watch the video...
This sudoku breaks the Geneva convention
The reasoning you use to place the nines in the arrow circles isn't quite complete - you can still have three nines, three eights, and two sevens in the circles and centers of the corner squares, which adds up to 65. Since all of those cells get doubled, that gives you exactly 180 as the sum of the columns and rows you looked at. To eliminate that possibility, you need to look at the top 16-cage, which can only work if it contains a 9, 8, or 7
Another knowledge bomb from CtC.. "That's got to be a something"
terrifically elegant puzzle!
i think this is a more logically sound way to look at the break-in: highlight all 12 of the cells on the four three-cell arrows (i made them green). let's consider the minimum total sum of all the green cells. three are in column 2 which also contains a three-cell 10 cage, so the minimum sum of the three green cells in that column is 11, because the minimum sum of any six cells in a column is 21. likewise, the three green cells in column 8 must also add up to at least 11. the remaining green cells are three in row 2 (minimum sum 6) and three in row 8 (minimum sum 7, since there's a 14 cage in the row). so the absolute minimum sum of all arrow cells is 11+11+6+7 = 35. that means the four circles must contain at least 9/9/9/8, and there's only one degree of freedom to increase one of the previously mentioned sums: the green cells in c2 or c8 could add up to at most 12, or r2 could add up to at most 7, or r8 could add up to at most 8.
but if any three of the arrow circles are 9 (and at least three must be 9), then the fourth must also be 9, because there'd be nowhere else to place the 9 in the row or column of the fourth circle. for example, if you put 9s into r1c8, r8c9, and r9c2, where can a 9 go in r2? r2c2 and r2c8 are ruled out by sudoku. you certainly can't put a 9 into a green cell (on an arrow). and you can't put 9 into the 16 cage, because it would have to go with two cells adding to 7, which would make the three green cells in r2 necessarily add up to at least 8; otherwise you'd have five cells in the row (three green cells and the two non-9s in the 16 cage) adding to less than 15. but the green cells can't add up to 8, because there is only one degree of freedom allowed from their minimum of 6. so the only place for 9 is in the circle in r2c1.
the cases of c2 and c8 are more straightforward because it's obvious that 9 can't go in a 10 cage or on an arrow. in r8, 9 can't go in the 14 cage because it would need to go with two cells adding to only 5, which would make the three green cells in r8 add up to at least 10; but their maximum is 8.
thus all four circles must contain 9, and the green cells add up to 36. from there you can get the x-wing on 8s in c2/c8/r2/r8 and proceed much as simon did. note that putting 1/2/7 into one of the 10 cages (which you must do once, because there's nowhere else 7 can go in c2 and c8 except r8 and only one 7 can go in r8) necessarily uses your one degree of freedom, since the green cells in that column now must be at least 3+4+5 = 12 instead of 11.
Fist thing I noticed was the Phistomophel Ring being fully known, thanks to the 1 cell arrows.
Title: "You Won't Believe How This Puzzle Ends!"
Me: "I Can't Fathom How This Puzzle Starts!"
This is one of the most beautiful puzzles I have seen on this channel. Just could not make the decisive opening move that Simon made without which there were way too many options and one went in circles in the grid. Jospeh Nehme congrats and thanks.
Thank you. :)
I understood your calculations for 180, but if you had 2 9s and 2 8s inside the circles, with 9 7 7 8 in the middle of boxes 13 7 and 9 wouldn't this also make 180?
18 + 18 + 16 + 16 + 50 + (9+7+7+8)*2
There must have been another reason why this wasn't possible
Correct. It's the 16-cage in row 2 that soaks up a degree of freedom, which then forces there to be four 9s in the four circles.
The 16 cage in box 2 needs a 7 8 or 9.
I have no clue how to solve this puzzle, but I feel like Simon chose the hardest way possible.
21:41 I love when Simon starts explaining something, and then admits he has no clue what it means or if it helps in the least.
Here's my relatively simple break-in:
Each of rows 2 & 8 and columns 2 & 8 contains a circle, 3 cells that lie on one arrow or another, and a 3-cell cage. In each such row or column, the 3 caged cells & the 3 arrow cells contain 6 different digits, which must sum to at least 21. In each of the columns, the caged cells sum to 10, so the arrow cells must sum to at least 11. r8 has a 14 cage, so the arrow cells must sum to at least 7. In r2, the arrow cells can sum to as little as 6, since together with the 16 cage they will sum to at least 22.
11 + 11 + 7 + 6 = 35, which is the least that the 12 arrow cells can total, and thus the least that the 4 circles can total. So the circles must contain either 4 9’s, or 3 9’s & an 8. As Simon discovered, the latter won’t work, so 4 9’s it is, totaling 36. Since this is only 1 more than the minimum, the center cells of the corner boxes will have to have their maximum values; decreasing any one of them will affect both a row & a column, pushing the arrow/circle total over 36.
So the center cells in boxes 1 & 3 must = 6 & 8, the arrow cells in r2 = 1-2-3, and the 16 cage = 4-5-7. The center cells in boxes 7 & 9 must = 7 & 8, the arrow cells in r8 = 1-2-4, and the 14 cage = 3-5-6. Columns 2 & 8 still have to be worked out. In one of them, the top & bottom center cells will = 6 & 8, and the arrow cells will total 12; in the other, the center cells will = 7 & 8, and the arrow cells will total 11.
So here's how I use these videos to try to get better myself at sudoku:
- First I open the puzzle and try to solve as much as I can for myself. I give myself a good time. Most of the time, I need help lmao
Interesting, most of the time I find things and start them in a different way. Sometimes very ineffective perspectives haha but sometimes just other alternatives I suppose.
- I watch a little bit of them. like 10% to 25%. I rescue some methodology, and although I see some digits...
- I pick it up later, maybe the next day, and then apply what I remember from the methodology. I recall the place of the digit (or some digits) most of the time, yet, unless I the logic by myself, I'll treat it as a mistake from Simon's part (such a bold move, I know)
- Normally that's enough to keep me going and once you're cracking the answers just come. But if not, I rewatch some of WHAT I ALREADY WATCHED (Yes that initial10-25%) and then ask: "What is Simon missing here?".
- If the pattern is still fuzzy, I'll just watch more and repeat the process.
Thank you Simon!
You're hillarious and enlightening
Another incredible solve, Simon. I don't know how you come up with how to start some of these puzzles. Question though (for anybody) About 17:00, I thought it was possible to have one of the outside circles be an 8. Why would that have not been possible? Ahh, got it .There would need to be TWO 8's outside and that's not possible. Leaves 9/9/9/8/8/8/7/7 as first eight digits (must add to 65) and now 16 cage won't work. Brilliant puzzle!
It has nothing to due with the 16 cage. Each of the circle cell see two of the centers in boxes 1,3,7 & 9 which of course each see yet anothers center. So the short answer is by sudoku all the circles have to be the same. The next part is that for the maths by decreasing the circle you lose two in the reach for 180, but this is offset by that you gain back two for each increased center square digit. It seems like a wash! So if you move the circles all to 8 then you need to the centers to add to (180-(8*8+50))/2=33 instead of 29. But alas due to sudoku only two out of the four the centers can be increased from 8 to 9. So you lost two degrees of freedom when we could only afford 1. i.e. you can't make 33 with 9+9+7+7.
He never actually came to the conclusion, he didn't really look at the 16 cage until he'd placed the 9's. The configuration he had at 17:00 works if you don't look at the 16 cage, so he had a lucky guess there.
@@RKelleyCook No, all 4 circles (by Sudoku) do not need to be the same number. Look at Simon's solve at the 17:00 mark. He has 8's in boxes 1&3, 9's in boxes 7&9.
@@Waggles1123 I'll take his lucky guess over my logical ability any day.
Yes you all are right, it does add exactly to 180
Adjust glasses, think again! that's a good idea for a tshirt
It sounded like an anime move
The adventures of Bobbins the seemingly underpowered protagonist who gains incredible powers by adjusting his/her glasses.
I'd watch that😁
I let the puzzle title and the relatively short video length fool me into thinking I could at least _start_ this one, however hard the end might be. I was woefully mistaken; I got nowhere.
Solved it with much help from the video.
For a moment, I believed Normal Sudoku Rules DIDNT apply because it was suspiciously missing despite having been included in the past 100+ videos or so.
Can someone correct me because I think there are arrangements possible without all circle cells being 9, which seems to be the logical conclusion that Simon uses to place the first digits in.
If the circle cell in box 1 and 3 are 9, and the circle cells in box 7 and 9 are 9, you can place a 7 in box 1, an 8 in box 3, a 9 in box 7 and a 7 in box 9, *all adjacent to the respective circle cells*
So it would look something like CIRCLE CELLS : (9 + 9 + 8 + 8) x 2 = 68 *coutned twice because of the arrow sums*
CELLS ADJACENT TO CIRCLES : (7 + 8 + 9 + 7) x 2 = 62 *counted twice because the cells are considered twice when taking rows 2 and 8 and colums 2 and 8 *
KILLER CAGE SUM = 50
Total = 68 + 62 + 50 = 180 = sum of rows and columns 2 & 8, with 4 intersecting cells taken twice.
So you *can* have a 9 in the adjacent cell in box 7, without breaking the puzzle, to my knowledge. Which seems to be hte reason Simon concluded all circles are 9s.
I'm going crazy lol help
I'd like to know why this arrangement isn't possible
I'm sure I'm making an obvious mistake somewhere.
You can't use all of 7, 8 and 9 in r2 in the corner boxes - then the 16 cage doesn't work
I took a different path, but also looking at rows two and eight and columns two & eight. There are three arrow cells in each of these columns and rows, all different. The minimums are therefore 123, but the ten cages restrict the two columns. The minimum for those columns add up to eleven. 11 + 11 + 6 +6 = 34, 4 X 9 = 36, and we've only got two degrees of freedom. But the fourteen cage in row eight also doesn't work with 123 in the row because 456 adds to 15, so 124 is the lowest for that row. Now we're down to one degree of freedom. Finally, realizing that you have to put a 7 in the sixteen cage and then trying to figure out how to get 7's and 8's in all the rows and columns given their limited opportunities eliminates the final degree of freedom when you realize one ten cage must be a 127 cage, and the rest solves exactly like Simon or some minor variation thereof.
My guess before watching: 9 rows, 9 columns, and 9 boxes all containing the digits 1-9
YOU WERE RIGHT?!?!?!?!?!?!?!???!?!?!,!,!,!,,!?!?!?!?!?!?!?!?!??!?!?!?!
How did you know?!!!??????!!?!?!?!?
This needs to be the top comment
40:07 but this was one of my favorite's on the channel because of how beautifully everything flowed once you got the 9's at the beginning.
36:37 I like how this implies that you like the mean comments as well as the nice ones.
always enjoy your videos and enthusiasm, keep up the great content Simon
took me about 15 minutes to realize the arrows werent thermos and i kept wondering how simon was missing you couldnt have put a 7, 8, or 9 in the bulb.
Enjoyed this one, though I fiddled around with the circles way longer than I needed.
This puzzle made me happy
big mistake in 17:26. if you count numbers you have ist actually perfect 180. your conclusion that 9 has to be in arrows is there for incorrect. you can actually have one 9 in middle because when you swap 8 and 9 in box 3 you dont lose the degree of freedom because both arrow and middle cell are counter 2 times ( so it doent matter if you have 9 in arrow and 8 in middle or 8 in arrow and 9 in middle you still end up with 34 from that box). the 9 in middle only cause arrow from box 3 and 1 to be cut from 9 but that still works with one 9
it might be possible to make 9 in middle of box impossible but not in way you did
The way I approached this, is by considering the minimum values of all the arrow cells combined, by looking at them from row and column perspective.
In c2 and c8 we know that the six cells that contain the 10-cage and three arrow cells have to add to a minimum of 21. Therefore the three arrow cells in c2 sum to at least 11. Equivalently the three arrow cells in c8 sum to 11.
In r8 and r2 we again have six cells comprised of a 3-cell killer cage and three arrow cells, so we use the same trick. In r8 this results in the arrow cells summing to at least 7 (21-14). For r2 it would result in a minimum of 5 (21-16) for the arrow cells, but three cells in the same row would have to sum to at least 6 anyway.
Combining these results we get a minimum of 11+11+7+6 = 35, which is only one below the maximum sum of the arrow totals (4x9=36), so we have one degree of freedom.
This means all cells in r2, r8, c2, c8 that contain an arrow cell or killer cage will contain the number 1-5, 6 or 1-5, 7. This leaves the blank cells to be 6, 7, 8 or 9 and we need the four arrow totals to sum to at least 35, so either four 9s or three 9s and one 8. However, like mentioned in the video, every blank cell in these columns/rows sees two arrow totals, so neither of the blank cells can be a 9, which leaves the only option for a 9 in these rows/columns to be in the four arrow total cells.
Still took me 90 minutes to solve the puzzle though xD
Okay, what am I missing? In his hypothetical where one circle doesn't have a 9, box one has 32 in the hatch (2x9, doubel count the seven), box two has 34, box seven has 34, box 9 has 32. Added up thats 132. Plus the 50 in the cages thats 182. Meaning you still have a degree of freedom. Am I wrong here, or is Simon's logic flawed and he just got lucky?
When the 9 moves out of the circle it rules out 9 in one of the other circles by sudoku.
You are correct. Simon's logic is flawed. You can't rule out that configuration without considering the 16 cage.
17:29 Broken? I calculate that this *just* works... you forgot that you're double counting the digits in the middle of each box, so you lost 2 degrees of freedom by taking 9 off the arrow in box 3, but you also gained *2* degrees of freedom by changing the 8 to 9 in the middle *since it's double counted.* So there's no net loss for that box. You do lose 2 degrees of freedom in box 1 by the 9 affecting that arrow, but you still have 7 in the middle, and that brings you to exactly 180 for the total, which is exactly what it should be.
You've got 8 in two arrows and 9 in two, and the middle digits are 7, 7, 8, & 9. That's two 7s, three 8s, and three 9s. *All* of them get double counted, so the total is 4*7+6*8+6*9 = 130, add the four cages which sum to 50 and that's 180.
I know everybody has already commented on the logic flaw around 19:00, but I didn't see it explained in detail, probably because I'm a noob and don't just understand the whole logic by a single detail, so I'll still go though it. Idk, maybe some other noob may find my explanation clearer on one correct way to solve it without just trying numbers in.
What I did is:
-take rows 2 and 8 and columns 2 and 8, adding up to 180;
-remove the cages, so the total in the "+" patterns (counting the center cell of each twice) is 130;
-knowing that in each "+" one of the arms is the sum of the other three, the count is the same a taking the circle cell twice, so diving the total by two you get 65, which is exactly the sum of the four circle cells plus the four middle cells of each corner box;
-we already know that the 8 and 9 in columns 2 and 8 are in four of the counted cells so you can remove them and obtain 31, which is therefore the sum of the circle cells in box 1 and 9 plus the "third" cell in columns 2 and 8.
-because the "third" cells can at most be two 7's, the two circle boxes must at least add up to 17, so they must at least be an 8 and a 9.
-because the two circle cells can at most be two 9's, the sum of the "third" cell in columns 2 and 8 must at least be 13 to get to the total, so these cells must at least be a 6 and a 7. (because again 8s and 9s are unavailable)
So what you get is that in the highlighted cells in columns 2 and 8 must be 6 7 8 9's (with no more than one 6), and in the circles in columns 1 and 9 there must be two 9's or an 8 and a 9.
Any other restriction comes from other clues. (so Simon was mistaking in thinking there MUST be four 9's in the circles based on geometry alone)
25:40 for me. Liked the logic at the start to get the initial break. From there, it becomes quite nice
Regarding the logic at 17:24 - why there must be four 9s in the four circles:
You have four dominoes that must add up to 65 (r2c12, r12c8, r8c89, r89c2).
The largest possible domino is 17 (8/9). Therefore the smallest possible domino is 14, because 14, 17, 17, and 17 makes 65. But you can't have two 17 dominoes in adjacent corners, because both cells of one domino "see" one of the cells of the next domino going round clockwise, and you can't put a total of three 8s or 9s in a single row or single column.
Therefore, the actual smallest domino is 15. And the two remaining dominoes must be 17, 16, and 17. The two 17s have to be on opposite corners, and the 15 and 16 have to be on opposite corners.
In addition, we know that we must have a 6 on one of the dominoes, because row 2 must have a 6, and the 16 cage must have a 7, 8, or a 9. Therefore, our 15 domino must be a 6/9 domino (in either of the two upper boxes), the other dominoes are 8/9, 7/9, and 8/9.
You can pencil in the dominoes using normal sudoku and the four 9s must go in the four cells that are not "seen" by their neighboring dominos, i.e. the four circled squares. Then you have a 6, 8, 7, and 8 in the remaining four squares, with the 6 in one of the two at the top.
The stuff of nightmares.
I’m impressed with myself how quickly I realized the 9s break the puzzle if in any other position
Love this channel and it's always thrilling to watch you solve the puzzles. It's also good for my psyche, to get reminded just how stupid I am from time to time. Helps to put my feet on the ground, I guess.
Finished in 26m38s, and I don't understand the reference to the ending... after finding the value for r7c3 it unwinds...
Huh. So, the way I got the puzzle started (you know, after staring at it for a while and wondering why I couldn't get it going) was basically the same but I think a bit more elegant: in columns 2 and 8, the ten cage plus the three digits on the arrows have to sum to at least 21 (1+2+3+4+5+6), so the arrow numbers sum to at least 11. In row 8, the three arrow cells sum to at least 7 for the same reason (21-14 = 7.) In row 2, the three arrow cells have to sum to at least 6 (1+2+3). All together, the cells on the arrows have to sum to at least 35. But the *most* they can sum to is 36, if all four arrows sum to 9. That gives us at least three arrows that are 9.
Then, because there still can't be an 8 or a 9 in either 10 cage or on an arrow, two of the intersections of rows 2 and 8 and columns 2 and 8 have to be 9 or 8, either upper left/lower right or upper right/lower left. But each of them sees two different arrow circles. So if one was a 9, then it wouldn't be possible for three arrow circles to be 9. So they must both be 8, and all four arrow circles are 9.
No adding up to a three digit number this way.
What a lovely puzzle... loved the break in!
Just realized I'd worked for like 2 hours thinking the corner arrows were thermos not spirals! I had posted: I got a break in with a couple ones and nines doing set through rows 2 & 8 and then columns 2 & 8, motivated by the 16 & 14 in the two rows and the 10 & 10 in the two columns. This made the vertical remainders in the corner spirals a max compared to the horizontal and placed some ones and nines. Then complete standstill. There's a phistomofel ring screaming at me with cages adding to 72 but I'm having trouble relating that to the outside corner 2/2 boxes. And there's a giant plus sign clamoring even louder geometrically. But I have no idea how to connect these things. The title is you won't believe how it ends. I'm thinking I'd be impressed with how it starts - or a little more of how it starts.
Simon, your conclusion at 17:00 wasn't complete. You lucked out a bit because in the end, the 9 does go in the outside circles, but your way of arriving at that conclusion was incomplete.
The sum centers of boxes boxes 1, 3, 7, and 9, plus the outside circles is exactly equal to 65, with no degrees of freedom due to the cages and the arrows (180-10-10-14-16 = 130, then divide by 2 since each are double counted). The configuration of your puzzle at 17:00 would be correct except for the 16 cage in box 2. The 16 cage needs a 7, 8, or a 9. This means that one of the digits in the sum equaling 65 is a 6. At that point, you'd find that if you didn't have four 9's, you couldn't get to 65 (6+9+9+9+8+8+8+7 = 64), so from there you have to put 9 in the circles. Further, you'd find that you need two 8s because 9+9+9+9+8+7+7+6 = 64, but you got to that.
Yes, this. In counting the freedom of degrees, he seemed to forget that both the circled cell and the center cell are double counted. So the 9+9+9+9+8+8+7+6 can be in any order (obeying the sudoku rules of course) as long as the 6 is on row 2. If the arrow circles were less than 9, that might later cause problems in placing the small digits on the arrows, but he didn´t know that at the time.
@@mikkimakki4436 To have four 9s, they have to be in the circle cells. Once you put a 9 in a center cell in box 1,3,7 or 9, you can only now have 3 9s contributing to your 65, which means you can't get to 65 including a 6.
I feel like there's a false assumption in the break in...
He only tried making an arrow an 8 where there was already an 8 in the centre, in box 3 in his example. Couldn't there be an 8 in an arrow where there's a 7 in the centre of the box? Say box 1. 8 on the arrow, 7 in the centre.
Oh no, then I guess the centre of box 3 couldn't be 7,8, or 9. My version only works with 3 9 arrows and 1 8 arrow for 70, and then box 1,3,7,9 centres need to add up to 30 and that 6 would spoil it because you can't have 3 8s.
@@snarfmason There are ways to put two 8s in two of the circles and get the overall math to work right. But then it breaks row 2 of the grid via the 16-cage.
@@KAS132000 Thank you, this is the piece I was missing :)
The arrangement of the 15:40 is valid, when considering the cross-sum information by itself (which is how he ruled it out).
This one was not solvable by me without the help of Simon. I needed to look up the start, but after that, it went quite smoothly and took 50 minutes.
1:12.27. Very nice puzzle, I like how symmetrical the start was
Solved it with a friend, the breakthrough near the start seems to be what does it. I had to watch the video after to see whether we'd missed something amazing at the end because we were just doing normal sudoku to finish it. Turns out the clickbait title is a bit misleading, it's the start that's interesting : )
I have seen you struggle with this in several videos, and realised that I never have the problem and figured it may be helpful to say: so you would turn the big digits options off, Simon, you could always read the cage totals ^^ :))
Simon also got it flawed at 35:42, you can't tell r6c5 is a 3 -- I think Simon got confused by the unshadowed 3 at r5c7 and thought that 3 is in the same sudoku box, but it isn't. However, it is trivial to solve it at this point.