Three years ago, Simon introduced me to the art of Sudoku when I stumbled onto one of the CTC videos. Now, it is truly a great honor to have one of my own puzzles featured on this wonderful channel. Thank you, Simon, for the inspiration and for the great solve! -Yawnus
Where did you come up with this crazy idea? Great puzzle, and a a great solve by Simon. I do marvel at the ideas that these great setters come up with.
Immediately one of my favorite puzzles on the channel (thanks to you) and immediately one of my favorite videos on the channel (thanks to you and to Simon). Wonderfully brilliant. This is going into the "first few videos I send people to introduce them to CTC" list. Can't wait to see more of your puzzles featured!
53:00 "I get 9 in box 3 as a result of sudoku... a little known technique sometimes used in these puzzles" delivered with a straight face. Never gets old for me.😀
13:49 Leave it to simon to be on exactly the right track and all he had to do was keep going and include the 5 cages, but stops himself from finishing because he doesn't consider that it would do anything and then comes to the exact same conclusion from a sideways angle. knowing that there are 15 cages and the minimum is 6 in a 3 cell cage, going by 2s gets us to 34, and the maximum you can fit in a 5 cell cage is 35 leaving exactly 1 degree of freedom, meaning there can be one string of even digits and one string of odd digits, but no more than that. The minds of brilliant men function in mysterious ways :P
I actually followed Simon's method much more logically to me to actually get the 31,33,35 distributed and thought it flowed. I was worried seeing some of the comments it might be frustrating, but it did keep flowing and showed exactly where digits would come.
@@jonh6585That's true for the distribution of 31, 33 & 35, but finding that there must be a serie of even cages followed by a serie of odd cages could still have been made much faster by just counting the total number of cages. There could only be 15 even cages from 6 to 34, 15 odd cages from 7 to 35, or even cages staring at 6 and odd cages ending at 35. The secret tells us there are odd cages, so we are in the last case.
One of the many wonders of mathematics, you can define a set with 10 and 13 being consecutive inside that set. ...not the case for this Sudoku tho, silly Simon XD
If I were to guess he probably wanted to give an example different to the one in the rules like say 12 but said 10 by accident, didn't realise and kept going
I love these puzzles where 40 minutes in nothing is in the grid and time has flown by just listening to Simon's brain work. Beautiful puzzle and excellent solve!
One of these days CTC is going to do an AMA stream and someone is going to ask Mark what his hobbies outside of puzzle solving are and it'll go something like this: Mark: "Well I've got my pilots license and love to take my vintage plane out when I can." Simon: "🤔 That's interesting (don't talk to me at parties) there is a vintage pla... Hold on 😮 ... Mavrick? 😱" Mark: "I don't know what you're talking about. 😏"
thank you for the birthday wishes Simon :) 27 today! p.s. Peter subbed the chocolate cake out for tahini chocolate chip cookies... Although nothing truly beats a 1:1 ratio of cake to frosting, the cookies were still delicious. Such a great solve. Welcome to the channel Yawnus!
I have nowhere near the ability to play these versions of Sudoku. Yet I can’t get enough of Simon’s skill and logic. I was listening all along and watching Simon figure everything out. Yet as he was beginning to fill in the numbers my brain couldn’t figure out how he got there! Absolutely loved this puzzle.
Oh man, Simon just needed to keep on his first train of thought. Including the 5 cell cages would have made him realise that he only had one degree of freedom. Sorry if he realises quickly, I'm only at the 15 minute mark!
Yeah, I'm about 14m in now wondering when he will see the 1 degree of freedom given all 15 cages in a range from 6-35. I saw that in about a minute myself and can't figure out how to use it so watching to get a hint of what to do when he realizes it.
I am again flabbergasted & gob-smacked at Simon's problem-solving ability, ingenuity & tenacity. I'm a PhD mathematician, & I would love to see Simon spread his wings & fly from the 9x9 cage of sudoku to the wide skies of pure mathematics research (returning frequently to share his Sudoku songs with us, of course.) Maybe Full Deck & Missing a Few Cards & other mathematicians out there could give him some encouragement in this direction. I'm sure that within a few months he could be doing pure research in some area like combinatorics.
Minimum cage size is 6, maximum is 35. However, only 5 cages can be greater than 24 as boxes 3 and 9 can only have one cage each greater than 24. Which gives you all the even numbers from 6 to 24. So the 5 cell cages in boxes 3 and 9 have to be odd. Given that the 4 cell cage in box 6 is at least 26, this makes columns 8 and 9 very constrained. Which gives that the 5 cell cages in boxes 3 and 9 have to be 31 and 33 and the 5 cell cage in box 4 is 35. Also means that the 9 in the 4 cell cage in box 6 has to be in column 7 and the other cells in columns 8 and 9 are minimised. Flowed nicely from there.
I went a totally different direction on boxes 3 and 9. Once I realized that the 5 and 4 digit cages needed to be an odd and an even, 34 and 32 didn't work, because you can't fit 15 total cages while switching to odd somewhere down the line toward 6. So it has to be 31,33,35 for the 5 digit cages and between all the high numbers, it has to be 31 and 33 in box 3 and 9 - you need the flexibility that 35 doesn't give you to fit those digits in 2 columns. Vice versa with the 4 digit cages, as they need to be 12 and 14, and can't have any matching numbers beyond the required 1 and 2 (since 31+ can't have a 1 or 2). From there it was a fun exercise.... Great puzzle
Flowed nicely is putting it mildly. I quickly figured out the nature of the size 5 cages but I was 55 minutes in with nothing else before I realized the min/max solution to boxes 3,6,9. The puzzle then solved in 30 minutes from there. Quite possibly faster but I'm not very bright sometimes.
Never have I been screaming internally as much as at the beginning of this solve! The break in of this one I found super easy! 1 degree of freedom if you consider all the cages and two of the boxes need to be even/odd parity so the low cages have to be evens down to 6 and uppers have to be odds up to 35. I'm not that far into the video but Simons solution is really going round the houses so far! Which is not to say I could have done anything with that ‐ I never try to solve these myself, and Simons brain is phenomenal - but I'm now 28 mins into the video and the cage values haven't fully fallen out of it yet.
4:40 as Zven did a wonderful job, if you double clic an empty cell in a cage, it will select all cells of cages of the same size. Quick coloring mode : enabled ;D
I finished in 156 minutes. These rulesets are so simple, yet so effective in this puzzle. It's fantastic. It took me a long time to discover the break-in involving the two five-celled cages on the right, but it was amazing when I did. I think my favorite part was seeing that the three-celled cage in row 7 couldn't be a 20 or greater due to no 9s being available and a 78 pencil mark in r4c4 seeing two of the cells of the cage. Great Puzzle!
26:50 is my time. Probably one of my personal favorite solves of all time. The logic was just incredible. Searching for when the even cages turned to odd and realizing what that meant for the really big cages (sum-wise) is a little crazy. Loved it! Gonna watch the video now... just noticed it's an hour... WOW I'm guessing Simon took a while to figure out the big cages at the start because after that, it flowed really nicely. Gonna find out now...
This puzzle felt much more difficult as I watched you figure out the way the cages were going to work than the actual entering of the digits turned out to be. Thanks for this, Simon, always a pleasure.
1:49:06 - Wow! Congratulations Yawnus. What a fantastic debut puzzle It took me ages to work it through but I loved every minute of it. Brilliant puzzle.
Minimizing all cages non-consecutively, makes the lowest a 6 and the highest a 34, but there must be at least two odd cages. Since there is one degree of freedom, there can only be one gap where it switches from low even cages to high odd cages. If a 4 cell cage is odd, it is a max of 29 and all 5 cell cage have to be odd to make the 31, 33, and 35. So the 4 cell cages in boxes 3 and 9 cannot be odd, the 5 cell cages are odd and if either tries to be below 30 and odd it forces all 5 cell cages to be 31+ odd cages, so that breaks, so you can get the 5 cell cages in boxes 3 and 9 to both be 31-35 and odd. That makes the two 4 cell cages 10-14 and even.
That was a great feature -something Simon has not seen, that I did, and a multitude of things that Simon saw more quickly than I could think. And what a cool puzzle - to execute that ruleset so cleanly.
27:05 anyone else shouting about the cages? Or just me? ((15-1) * 2) + 6 = 34 ((15-1) * 2) + 7 = 35 So there is only 1 degree of freedom, so therefore the 5 cell cages must all be greater than 30. It doesn’t give you the even/odd split but if Simon had just continued his thought to include the 5 cell cages, he would have saved himself so much time. I had to 2x thru the video because I couldn’t unsee my deduction.
Wow! What a remarkable puzzle. I can't recall another puzzle that needed so much analysis to get to the break in. The key for me was realising that the 5-cell cage on the left had to be 35. Everything flowed quite smoothly from there, although I did have the benefit of a pad to cross off the sums which were accounted for. Well done keeping track of everything in your head.
Oh my goodness, I can't even really put my finger on exactly why, but I had so much fun doing this solve. Finding that logic in the right two columns that forced five cells to sum to a maximum of a single cell felt like such a brilliant discovery. And then all the rest of the way through the solve it was somehow shocking how everything only could work out one exact way, somehow feeling both unbelievable that it was the case, and yet not that hard to follow along with once you were past that first hurdle. 53:10 for me, I got the first digits after about 25 mins but my solving speed is nowhere near Simon's after, but I can say that I was grinning like a fool for the entire back half of the solve. Yawnus, thank you so much for gracing us with this one!
25:00 The way I used a completely different proof to get here is INSANE. Here is mine: 1. We know that box 3 comprises of 2 cages adding up to 45 (an odd number). So pne of the cages sums up to odd number and the other to an even number. 2. Let's try to assign the cage values across ALL cages starting from minimum all the way up to maximum one. Since we have 15 cages in total, we get: 6, 8,..., 32, 34. Since 5-cell cages max out at 35, we have one degree of freedom. Let's call that a "Switch". 3. When we use that Switch, we not only change the parity of the number we used in upon, but also the numbers greater than it. Here is the example with 14: 6, 8, 10, 12, 15, 17,..., 31, 33, 35. We can't Switch 14 to 16 as we run out of space. Keep that in mind as we will now prove 2 things: 1) 4-cell cage can't be odd; 2) 5-cell cage must add to 31, 33 or 35. 4. Let's assume 4-cell cage is odd. Now it maxes itself out at 29 (biggest odd number 4-cell cage can add up to). So by the time we get to 29, the Switch had already been activated. That also means that 31, 33 and 35 will be among cages' sums (see step 3). This trio of sums is 5-cell cage 'exclusive' and we have exactly 3 5-cell cages to distribute them in. But that will break the box 3 sum, since it will have 2 odd cages summarizing to even but 45 ain't even. So 5-cell cage is odd and 4-cell cage is even. 5. If a 5-cell cage sums to odd number below 31, then we would still have to distribute 31, 33 and 35 sums across 5-cell cages (because of a Switch being activated early, see step 3). But the are too few cages available for that. Thus a 5-cell cage must add up to an odd number that is at least 31 (31, 33 or 35). Proof is done. The way of Simon and me approaching this from 2 different sides and coming to the same conclusion is so beautiful.
Finished in 41:18. I needed to take a break in solving it as I hadn't had lunch yet, but as I was noodling over the puzzle while eating my lunch, I had a Eureka moment for the break-in with the 3 boxes on the right hand side. I made sure to finish my lunch though as otherwise, I don't think I would have had the fuel to continue from the break-in quickly. Excellent break-in and fun puzzle!
I love killer sudoku, but I confess I had difficulty completing this on my first attempt… but as a testament to having another go, I really enjoyed it! As others have said, with a total of 15 cages, there is a total of 15 possible cage totals, but they are in pairings of 6/7, 8/9, and so on up to 34/35. This leads to the observation that each time you select an odd number, you force all numbers higher than it to be odd and, conversely, each time you select an even number, you force all numbers less than it to be even. Considering boxes 3 and 9, it becomes clear that there must be both even and odd numbers, thus both 6 and 35 are cage totals. Working from the higher numbers first, and confirming 35 not 34… if you select 34, all numbers are even and boxes 3 and 9 break. Similarly, if you the select 32, you only have one odd number (35 already selected) and one or other of boxes 3 and 9 breaks. If you now select 30, you are effectively placing 33 and 35 (which must both be in 5 digit cages) in boxes 3 and 9, and you run into the problem that you don’t have enough value remaining to fill the remaining cells in columns 8 and 9. Therefore, not only does this mean selecting 31, but 31 must also be the total of one or other of the 5 cell cages in boxes 3 and 9, with the other being 33 or 35… this now also satisfies all three 5 cell cages. Noting that the 4cell cage corresponding with the 31 cage has a total of 14, all the numbers less than 14 are now even. Also, noting that the 5 cell cages are fully subscribed, 26/27 and 28/29 are both 4 cell cages, and the remaining 4 cell cage in box 3 or 9 must be 10 or 12… Thus, counting the 4cell cages, the totals of 10 and 12, together, represent a 3 cell and a 4 cell cage, meaning 16/17 to 24/25 must all be 3 cell cages - this also means selecting 24 not 25 (cannot put 25 in 3 cells), so all 3 cell cages are even. At this point the parity of 26/27 and 28/29 is yet to be determined, but the 4 cell cage in box 6 must be at least 26, so minimising the totals of the 5 cell cages in boxes 3 and 9, and also minimising the portion of the 4 cell cage in box 6 that also sits in columns 8 and 9, provides the break-in point to the puzzle. Great puzzle and very satisfying!
I don't normally attempt 4* difficulty puzzles, but the rules on this were too intriguing. Once I understood how everything fit together it was beautiful. Thanks for choosing this one, I would not have solved it otherwise!
I tried it, made a few realizations about the cage totals, and couldn't get one digit. I am impressed that you figured it out Simon. Plus I get to hear lines like "might as well do Sudoku".
I feel your pain. Congratulations for sticking with it. If I've gone nowhere after an hour, I usually give up and start getting hints from Simon. He has such incredible insight into the nature of any puzzle.
Usually (understatement) I am not able to solve the puzzles without peeking at the video and looking for a hint. This time, I managed to find the break in after a couple of minutes........... So proud of myself 🙂
Interesting concept for a puzzle, thanks. Reasonably happy with my solve, though I did have to make a few notes and as per Simon appreciated a few things I missed.
I tackled this from a slightly different angle ie there are a limited number of ways to fill 3, 4 and 5 cell cages within the constraint, boxes 3 and 9 have to have an odd even combo and that means totals with be three apart and there is only just enough different totals to be three apart once with just the highest totals being odd.
My 8 yo son just walked in, heard only one word, and said, “is that Simon, from Cracking the Cryptic? His first name should be Cracking, and his last name should be Cryptic, and his middle name should begin with T. It could be Timothy or something”
it is much more restrictive if you think about all the cages at the start :P its 15 sums that can't be consecutive between 6 and 35, and ignoring the degrees of freedom from even vs odd that is just 15 possibilities. so all the sums are in, it is consecutive in the evens odds, or first in the evens and then in the odds with an even and an odd in between.
13:40 "My intuition is so off" And I'm screaming at him (two weeks late, I might add), that if he just includes the 5-cell cages in the reasoning he started, he would get that there is a single degree of freedom on the cage values. It's not often I scream loudly enough to actually write a comment, but this time the thing I want to scream isn't "But look at this one thing that I noticed that you didn't" or "Remember that thing you forgot 5 seconds / 5 minutes ago", it's "Yes, keep doing the thing you're doing, don't stop". And I feel that that's a qualitatively different thing to scream.
The art of never using a pencil...wow Wish there was a notepad++ type editor on side so u can keep track of their mind working. For me, as pencilsolver , wrote down all the possible sums ...and then discovered how the Even and odd sums needed to jostle to suffice the rule (ie B6C7 digits) Got a bit stuck after that and came to watch the beginning only needed to watch the coloring of the 4 digit cages to realize what I forgot in my initial discoveries of the 4digitcagees then the puzzle went rather fast
51:14, but I had to look at the video to see how to use columns 8 and 9. I actually did the work to figure out what the sums of all the cages had to be, but didn't see how to start figuring out which sums went where. Still impressed how it the cages were resolved after the break in.
The 3 5-cages would have to have a total above 30. There are 15 cages, and to get the minimum sum of totals, you'd get 15 even totals from 6 to 34. So there's a gap of 1 here, because the 5-cages can go up to 35. This gap could go anywhere from 6 to 35, meaning either every total is odd if the gap is at 6, every total is even if the gap is at 35, or somewhere inbetween, it switches from even to odd totals because there'd be a gap of size 2. And it's clearly the third option, since boxes 3 and 9 both require even and odd totals. And from your logic that the totals are even from 6 to at least 24, and with the minimum odd number in boxes 3 and 9 being 31, it stands to reason that the remaining 2 totals above 31 have to be 33 and 35, and those can only be fulfilled by the 5-cages. So yeah, the 5-cages' totals have to be 31, 33, and 35.
from there, the 4 cages can be 30, 28,..,24 , 29,27,...,21 , 28,26,...,22 ... a lot of them necessarily have to land near the maximum consecutive sums. there is just one option for taking a 5 cage and mixing it in between the 4's and two options for mixing in three cage sums. interesting starting point :P
I feel like Simon got too focused in early and missed the big picture. The very first thing I noticed is that across all 15 cages, you have 1 degree of freedom, and that in turn will be eaten up by the flip from even to odd necessary by the cages in boxes 3 and 9. This means the values over 30 must be in the 5 cell cages in boxes 3 and 9 and both must be odd (so from 31 33 and 35). I’m not gonna pretend the rest of my solve was nearly as fluid as this but it feels like he tried to focus on the details before understanding the big picture.
It usually is a consolation to me that, whenever Simon may be stuck in a puzzle, the happy end is just guaranteed by the mere fact that we're watching the video! Indeed, there is balm in Gilead 🙂
there is 15 cages, 15 sums, non-consecutiveness forces a few things, first of all, if all the sums are even then there is no room at all to shift sums around all sums from 6 to 35( which is the maximum sum possible) must be included, there are 30 possible sums between 6 and 35 including both, 15 sums possible with no consecutive sum values. they can run from 6 to 34 if even, 7 to 35 if odd, and 6 to 35 if there is a transition from even to odd sums, in which case at the transition there is a gap of 3 between the sums instead of the 2 between the other sums. that characterizes all the possible arrangements of sums. it is not however possible for the lowest sum to be 7 and for there to be a transition to even sums, if the smallest sum is 7 then all sums have to be odd. it is also not possible to start with even and then switch to odd, and then switch back to even. so what we are looking at just from the rules and from the box numbers is one series of even sums and one series of odd sums, individually consecutive in their parity, with a transition between them, where the odd series starts 3 above where the even ends, or we are looking at just the consecutive even or odd sums. now the maximums become important for the ordering, the 5 cages have a maximum of 35, the 4 cages have a maximum of 30, so the 3 5 cages must have a value of either 34 or 35 for the largest sum, the second largest must have a value of 32 or 33 because of the possible sequences it can reside in, the third largest does not necessarily have to be as large, because a 4 cage could be 30, but if we live in a universe where this end section is odd, then the third 5 cage has to be 30 or 31, if we are in an even world up to and including the smallest 5 sum, then the smallest five sum can potentially have its minimum sum of 15 and any sum between 15 and 30 potentially. pretty cool puzzle :D
33m32s. I had a good time working out what the sums had to be and where they had to be located. I was pleasantly surprised with how the sums kind of fell into place once I got the 8 over in box 4
This software totally needs an aide memoir for puzzles like this, it hurt my brain a bit Awesome to have both zz top and R.E.M referenced by simon though
This is credible! A Sudoku grid with absolutely no digit to start: none in the grill, none outside the grill, none attached to the cages, none in the rules! OMG!
It’s a really clever set up but I’m not so sure it was the most obvious of solve paths (note that I’m referencing the thought process as opposed to difficulty level). I only realised the restriction on the pattern of odds/evens when looking at the comments and even then it took a moment to understand it correctly. An interesting debut but keeping track of the totals was a bit too troublesome. I enjoyed watching Simon’s solve though.
from there it is much easier to think about maxes and mins, two 5 cages must be above 30, two 3 cages must be 6,8 7,9 or 6,9, and two 4 cages must be between 24 and 29. so that is 6 cage sums already nailed down except from a choice of even odd boundary. much easier than to think about 8 3 cages independently :D. but it is always fun to watch you try to figure it out, sometimes you find stuff really easily and quickly and other times i think you miss super obvious stuff, so pretty much like my experience of my own reasoning lol.
There’s 15 cages. For they are 2 apart, that makes a range of 6 to 34., except 3 have to be odd. The 5 cages are the only ones that can exceed 30, so they can be 31, 33, 34, The remaining 12 cages have a range of 6 to 28, so they fit with the previous odd 30s. If you put any odds in the 3 or 4 cages, you lose room for a cage result, pushing it up, so you can’t use 30, but you still have to fit 3 above the lowest ones. Written at the 22 minute mark
An easier way to calculate the totals is to start with 6, since that’s the lowest number for the three cell cages. Since the minimum distance between the cages is two, you take the remaining seven and multiply them two. 14 + 6 =20
This is the kind of puzzle I do not like. It feels more like homework or a task that you have to brute force by calculating all possible alternatives to find your way around it.
I wonder how much faster Simon could have broken into this if he'd gone the extra step of considering the 5 cell cages when doing the telephone pole calculus. 3 cell cages - 6-24, 8 cages, 4 degrees of freedom 3 AND 4 cell cages - 6-30, 12 cages, 2 degrees of freedom 3, 4, AND 5 cell cages, 6-35, 15 cages, 1 degree of freedom. So there is ONE gap of two where you change from the evens (low end) to odds (high end).
the much easier way to get counts on right 3 boxes is if boxes are 31 + 35 = 66 then remaining 4 cell cages in those boxes are 24 which puts 12 in col 8 and 123456 in col 7. That leaves 789 in box 6 col 7. Even if the 9 is used in 4 cell cage of box 6, you can't it make total 26 or more (9+6+5+4=24 ). This forces 7 or 8 into box 6 col 8 or 9 and makes 5 cell cages 31 (45679 or 35689) and 33 (xx789).
so it is then impossible for the sums of the 5 in the corner cages to be even at all. therefore there is either 3 maximized 5 cages, or an even 4 cage sum of 30, and the even 5 cage is smaller. that is possible. but then all other sums than the two 5 cages would be even with 100% certainty. but since we know these two cages now has to sum to 33, 35 or 31, then we know the even 4 cages has to sum to 14,12 or 10 in the corner boxes filled with cages. which means we found two 4 cages that interfere with the 3 cages, and the two others must be maximized, they must be 30, 28, 29 or 27. that nearly nails down the entire sequence :D. if the 5 cages are not all maximized then the remaining 5 cage has to be even, and all the other cages also have to be even. if not then the odd even transition can be anywhere, but all 5 cages are above 30 and two are odd. from there it is a little easier to proceed i think :P.
you end up with a series in the evens, which is equivalent to any series that is shifted to odd for the entire or part of it, that is in order first 8 3's, then 4 4's, then 3 5's, except from one shiftable 5 and two shiftable 4. the 4s can be shifted between 10 and 24 and to 30, but if the 5 cage sums are not maximized, the 4 cage sum must be 30, therefore no matter what you can only shift it there and so it is not really a degree of freedom as loose as the other ones. so a 4 cage 30 and a 5 and a 4 that can be shifted around to many positions, or two 4 cages that can be shifted around between 10 and 24 to move the 3 cage sums around. if the 30 sum is a 4 cage then ofc all the sums bellow are even. so many deductions to make just based on counting and the rules lol.
you have to have the sums of there are even sums and odd sums, to run from an even series from 6 to 2N, then from 2N+3 to 35. so you know the largest sum is 35, the next largest sum is and has to be 33. because then the two largest sums must be odd, because you can't have two odd 4's and two even 5's. at least one of the 5's in the corner cages must be odd, because two fives must be the largest sums, and the nature of the transition between even and odd is as such that it could not be between the two largest 5 sums, because then you would necessarily only have even 4 cages, and then both the 5 cages are forced to be odd anyway and must be 33 and 35. so if both were even then all sums bellow 32 must also be even. so two of the 5 cages then must sum to 35 and 33. then the last 5 cage can be 30, 31 or some lower number, if the largest 4 cage is 30, but if the largest 4 cage is 30, then all other cage sums are even, because of that there would be no other 4 cages that would be odd, and therefore... you would have to put even 4 cages with both the 5's in the completely filled boxes. if one 5 cage in these boxes is even, the 4 cage in there would be odd, then the 5 cage in that box would be smaller in sum than it, it would not be maximized, one 4 cage would have to be 30 exactly to displace it from its maximized position, which would make all the 4 cages even, and the odd 4 cage would be made impossible, which is jolly for sure.
It took me about an hour to get where you did at the 30min mark, but couldn't find the way to break in, deduced that the 3 big cages needed to be 31, 33, 35 respectively, but not which was which. EDIT: finished it in about 5 hours 😅.
Three years ago, Simon introduced me to the art of Sudoku when I stumbled onto one of the CTC videos. Now, it is truly a great honor to have one of my own puzzles featured on this wonderful channel. Thank you, Simon, for the inspiration and for the great solve! -Yawnus
Impressive puzzle.
That was a very enjoyable puzzle.
Where did you come up with this crazy idea? Great puzzle, and a a great solve by Simon. I do marvel at the ideas that these great setters come up with.
Immediately one of my favorite puzzles on the channel (thanks to you) and immediately one of my favorite videos on the channel (thanks to you and to Simon). Wonderfully brilliant. This is going into the "first few videos I send people to introduce them to CTC" list. Can't wait to see more of your puzzles featured!
One of the most enjoyable break-ins in a while :). Thank you Yawnus. Great puzzle!
53:00 "I get 9 in box 3 as a result of sudoku... a little known technique sometimes used in these puzzles" delivered with a straight face. Never gets old for me.😀
I don't even play Sudoku, I'm just addicted to these videos. It's like getting to watch someone crack the Enigma Machine every day.
Listening on headphones, I can occasionally hear a bird singing outside Simon's open window. It's rather charming.
I love to hear the birds singing.
I heard the birds singing too, but when I paused the video I realized that these are the ones singing outside of my window)
13:49 Leave it to simon to be on exactly the right track and all he had to do was keep going and include the 5 cages, but stops himself from finishing because he doesn't consider that it would do anything and then comes to the exact same conclusion from a sideways angle. knowing that there are 15 cages and the minimum is 6 in a 3 cell cage, going by 2s gets us to 34, and the maximum you can fit in a 5 cell cage is 35 leaving exactly 1 degree of freedom, meaning there can be one string of even digits and one string of odd digits, but no more than that. The minds of brilliant men function in mysterious ways :P
I know? He came up with some brilliant thoughts I would never find, and I was sitting here thinking "But you could have just kept counting.'
Came here to mention the same thing 😀
I actually followed Simon's method much more logically to me to actually get the 31,33,35 distributed and thought it flowed.
I was worried seeing some of the comments it might be frustrating, but it did keep flowing and showed exactly where digits would come.
@@jonh6585That's true for the distribution of 31, 33 & 35, but finding that there must be a serie of even cages followed by a serie of odd cages could still have been made much faster by just counting the total number of cages.
There could only be 15 even cages from 6 to 34, 15 odd cages from 7 to 35, or even cages staring at 6 and odd cages ending at 35.
The secret tells us there are odd cages, so we are in the last case.
Crying and throwing up, begging Simon to just count all the cages.
For the record, I feel honored at being told the secret every single time.
10’s consecutive with 13 now - The wonders of mathematics
Cracking the Cryptic knowledge bomb there :D
One of the many wonders of mathematics, you can define a set with 10 and 13 being consecutive inside that set.
...not the case for this Sudoku tho, silly Simon XD
Naughty brain
Ps: I don't like when Simon says it on himself, he has a wonderful brain
If I were to guess he probably wanted to give an example different to the one in the rules like say 12 but said 10 by accident, didn't realise and kept going
I love these puzzles where 40 minutes in nothing is in the grid and time has flown by just listening to Simon's brain work. Beautiful puzzle and excellent solve!
Every time you fly out of London, take a moment to think that your plane could be disrupting Simon's thoughts 😮
I'm still convinced that MAveRicK is well aware of Simon's schedule...
Made the same joke in my head. How often do you hear him fly by multiple times..
Does anyone else picture Mark, doing his best Bond villain impersonation, “No Simon… I expect you to be mildly annoyed”?
Just me?
One of these days CTC is going to do an AMA stream and someone is going to ask Mark what his hobbies outside of puzzle solving are and it'll go something like this:
Mark: "Well I've got my pilots license and love to take my vintage plane out when I can."
Simon: "🤔 That's interesting (don't talk to me at parties) there is a vintage pla... Hold on 😮 ... Mavrick? 😱"
Mark: "I don't know what you're talking about. 😏"
There was a video in which Mark picked up the phone and said:"hi Maverick, he just started recording, you can take off". It was hilarious 😂
@@msx80 For real? I need to know which one, if it really happened! 🤣
thank you for the birthday wishes Simon :) 27 today!
p.s. Peter subbed the chocolate cake out for tahini chocolate chip cookies... Although nothing truly beats a 1:1 ratio of cake to frosting, the cookies were still delicious.
Such a great solve. Welcome to the channel Yawnus!
I have nowhere near the ability to play these versions of Sudoku. Yet I can’t get enough of Simon’s skill and logic. I was listening all along and watching Simon figure everything out. Yet as he was beginning to fill in the numbers my brain couldn’t figure out how he got there! Absolutely loved this puzzle.
Oh man, Simon just needed to keep on his first train of thought. Including the 5 cell cages would have made him realise that he only had one degree of freedom.
Sorry if he realises quickly, I'm only at the 15 minute mark!
One degree of freedom that he also solved by analysing that at least two of the cages have an odd total
Yeah, I'm about 14m in now wondering when he will see the 1 degree of freedom given all 15 cages in a range from 6-35. I saw that in about a minute myself and can't figure out how to use it so watching to get a hint of what to do when he realizes it.
44 minutes in, 2 digits placed. ...
I am again flabbergasted & gob-smacked at Simon's problem-solving ability, ingenuity & tenacity. I'm a PhD mathematician, & I would love to see Simon spread his wings & fly from the 9x9 cage of sudoku to the wide skies of pure mathematics research (returning frequently to share his Sudoku songs with us, of course.) Maybe Full Deck & Missing a Few Cards & other mathematicians out there could give him some encouragement in this direction. I'm sure that within a few months he could be doing pure research in some area like combinatorics.
The sudoku community will never let him go :)
"Solve counter 69. Like the puzzle: yes, it's very... nice."
Nice.
Minimum cage size is 6, maximum is 35. However, only 5 cages can be greater than 24 as boxes 3 and 9 can only have one cage each greater than 24. Which gives you all the even numbers from 6 to 24. So the 5 cell cages in boxes 3 and 9 have to be odd. Given that the 4 cell cage in box 6 is at least 26, this makes columns 8 and 9 very constrained. Which gives that the 5 cell cages in boxes 3 and 9 have to be 31 and 33 and the 5 cell cage in box 4 is 35. Also means that the 9 in the 4 cell cage in box 6 has to be in column 7 and the other cells in columns 8 and 9 are minimised. Flowed nicely from there.
That's the way, 45 minutes for me, now I'll watch the video
I went a totally different direction on boxes 3 and 9. Once I realized that the 5 and 4 digit cages needed to be an odd and an even, 34 and 32 didn't work, because you can't fit 15 total cages while switching to odd somewhere down the line toward 6. So it has to be 31,33,35 for the 5 digit cages and between all the high numbers, it has to be 31 and 33 in box 3 and 9 - you need the flexibility that 35 doesn't give you to fit those digits in 2 columns. Vice versa with the 4 digit cages, as they need to be 12 and 14, and can't have any matching numbers beyond the required 1 and 2 (since 31+ can't have a 1 or 2). From there it was a fun exercise.... Great puzzle
This was pretty much my break-in as well.
Flowed nicely is putting it mildly. I quickly figured out the nature of the size 5 cages but I was 55 minutes in with nothing else before I realized the min/max solution to boxes 3,6,9. The puzzle then solved in 30 minutes from there. Quite possibly faster but I'm not very bright sometimes.
Never have I been screaming internally as much as at the beginning of this solve! The break in of this one I found super easy!
1 degree of freedom if you consider all the cages and two of the boxes need to be even/odd parity so the low cages have to be evens down to 6 and uppers have to be odds up to 35. I'm not that far into the video but Simons solution is really going round the houses so far!
Which is not to say I could have done anything with that ‐ I never try to solve these myself, and Simons brain is phenomenal - but I'm now 28 mins into the video and the cage values haven't fully fallen out of it yet.
And here I am, struggling to solve a "hard" sudoku in under 30 minutes … the mental gymnastics in these videos baffles me.
This is a *quality* puzzle. Thanks Yawnus!!!
I would never have figure that one out. Simon is to Sudoku what Magnus Carlsen is to chess. Truly amazing mental abilities.
4:40 as Zven did a wonderful job, if you double clic an empty cell in a cage, it will select all cells of cages of the same size.
Quick coloring mode : enabled ;D
I believe it's just selecting like-colored cells while on color mode
I finished in 156 minutes. These rulesets are so simple, yet so effective in this puzzle. It's fantastic. It took me a long time to discover the break-in involving the two five-celled cages on the right, but it was amazing when I did. I think my favorite part was seeing that the three-celled cage in row 7 couldn't be a 20 or greater due to no 9s being available and a 78 pencil mark in r4c4 seeing two of the cells of the cage. Great Puzzle!
26:50 is my time. Probably one of my personal favorite solves of all time. The logic was just incredible. Searching for when the even cages turned to odd and realizing what that meant for the really big cages (sum-wise) is a little crazy. Loved it! Gonna watch the video now... just noticed it's an hour... WOW I'm guessing Simon took a while to figure out the big cages at the start because after that, it flowed really nicely. Gonna find out now...
Nice, that would have been about my time too but I got stuck for almost an hour on part 2 of the break-in. Great flow, brilliant construction.
This puzzle felt much more difficult as I watched you figure out the way the cages were going to work than the actual entering of the digits turned out to be. Thanks for this, Simon, always a pleasure.
1:49:06 - Wow! Congratulations Yawnus. What a fantastic debut puzzle It took me ages to work it through but I loved every minute of it. Brilliant puzzle.
1:23 - "So do have a go at it - If you have *_a few minutes_* spare"
Simon, you crack me up!
Minimizing all cages non-consecutively, makes the lowest a 6 and the highest a 34, but there must be at least two odd cages. Since there is one degree of freedom, there can only be one gap where it switches from low even cages to high odd cages. If a 4 cell cage is odd, it is a max of 29 and all 5 cell cage have to be odd to make the 31, 33, and 35. So the 4 cell cages in boxes 3 and 9 cannot be odd, the 5 cell cages are odd and if either tries to be below 30 and odd it forces all 5 cell cages to be 31+ odd cages, so that breaks, so you can get the 5 cell cages in boxes 3 and 9 to both be 31-35 and odd. That makes the two 4 cell cages 10-14 and even.
That was a great feature -something Simon has not seen, that I did, and a multitude of things that Simon saw more quickly than I could think. And what a cool puzzle - to execute that ruleset so cleanly.
Took me 184 minutes, but i solved it without needing help.
My time also. Haven't watched the video yet but I was able to solve without set theory.
Extremely fun puzzle! All the deductions seemed very logical and flowed very nicely. Took me almost exactly 60 mins.
What a spectacular solve of a great puzzle! You continue to amaze me, Simon.
27:05 anyone else shouting about the cages? Or just me?
((15-1) * 2) + 6 = 34
((15-1) * 2) + 7 = 35
So there is only 1 degree of freedom, so therefore the 5 cell cages must all be greater than 30.
It doesn’t give you the even/odd split but if Simon had just continued his thought to include the 5 cell cages, he would have saved himself so much time.
I had to 2x thru the video because I couldn’t unsee my deduction.
Wow! What a remarkable puzzle. I can't recall another puzzle that needed so much analysis to get to the break in. The key for me was realising that the 5-cell cage on the left had to be 35. Everything flowed quite smoothly from there, although I did have the benefit of a pad to cross off the sums which were accounted for. Well done keeping track of everything in your head.
Oh my goodness, I can't even really put my finger on exactly why, but I had so much fun doing this solve. Finding that logic in the right two columns that forced five cells to sum to a maximum of a single cell felt like such a brilliant discovery. And then all the rest of the way through the solve it was somehow shocking how everything only could work out one exact way, somehow feeling both unbelievable that it was the case, and yet not that hard to follow along with once you were past that first hurdle.
53:10 for me, I got the first digits after about 25 mins but my solving speed is nowhere near Simon's after, but I can say that I was grinning like a fool for the entire back half of the solve. Yawnus, thank you so much for gracing us with this one!
Wow, that was interesting how you had to deduce so much before you could fill in.
25:00 The way I used a completely different proof to get here is INSANE. Here is mine:
1. We know that box 3 comprises of 2 cages adding up to 45 (an odd number). So pne of the cages sums up to odd number and the other to an even number.
2. Let's try to assign the cage values across ALL cages starting from minimum all the way up to maximum one. Since we have 15 cages in total, we get:
6, 8,..., 32, 34.
Since 5-cell cages max out at 35, we have one degree of freedom. Let's call that a "Switch".
3. When we use that Switch, we not only change the parity of the number we used in upon, but also the numbers greater than it. Here is the example with 14:
6, 8, 10, 12, 15, 17,..., 31, 33, 35.
We can't Switch 14 to 16 as we run out of space. Keep that in mind as we will now prove 2 things: 1) 4-cell cage can't be odd; 2) 5-cell cage must add to 31, 33 or 35.
4. Let's assume 4-cell cage is odd. Now it maxes itself out at 29 (biggest odd number 4-cell cage can add up to). So by the time we get to 29, the Switch had already been activated. That also means that 31, 33 and 35 will be among cages' sums (see step 3). This trio of sums is 5-cell cage 'exclusive' and we have exactly 3 5-cell cages to distribute them in. But that will break the box 3 sum, since it will have 2 odd cages summarizing to even but 45 ain't even. So 5-cell cage is odd and 4-cell cage is even.
5. If a 5-cell cage sums to odd number below 31, then we would still have to distribute 31, 33 and 35 sums across 5-cell cages (because of a Switch being activated early, see step 3). But the are too few cages available for that. Thus a 5-cell cage must add up to an odd number that is at least 31 (31, 33 or 35). Proof is done.
The way of Simon and me approaching this from 2 different sides and coming to the same conclusion is so beautiful.
Finished in 41:18. I needed to take a break in solving it as I hadn't had lunch yet, but as I was noodling over the puzzle while eating my lunch, I had a Eureka moment for the break-in with the 3 boxes on the right hand side. I made sure to finish my lunch though as otherwise, I don't think I would have had the fuel to continue from the break-in quickly.
Excellent break-in and fun puzzle!
I love killer sudoku, but I confess I had difficulty completing this on my first attempt… but as a testament to having another go, I really enjoyed it!
As others have said, with a total of 15 cages, there is a total of 15 possible cage totals, but they are in pairings of 6/7, 8/9, and so on up to 34/35. This leads to the observation that each time you select an odd number, you force all numbers higher than it to be odd and, conversely, each time you select an even number, you force all numbers less than it to be even. Considering boxes 3 and 9, it becomes clear that there must be both even and odd numbers, thus both 6 and 35 are cage totals.
Working from the higher numbers first, and confirming 35 not 34… if you select 34, all numbers are even and boxes 3 and 9 break. Similarly, if you the select 32, you only have one odd number (35 already selected) and one or other of boxes 3 and 9 breaks.
If you now select 30, you are effectively placing 33 and 35 (which must both be in 5 digit cages) in boxes 3 and 9, and you run into the problem that you don’t have enough value remaining to fill the remaining cells in columns 8 and 9. Therefore, not only does this mean selecting 31, but 31 must also be the total of one or other of the 5 cell cages in boxes 3 and 9, with the other being 33 or 35… this now also satisfies all three 5 cell cages.
Noting that the 4cell cage corresponding with the 31 cage has a total of 14, all the numbers less than 14 are now even. Also, noting that the 5 cell cages are fully subscribed, 26/27 and 28/29 are both 4 cell cages, and the remaining 4 cell cage in box 3 or 9 must be 10 or 12… Thus, counting the 4cell cages, the totals of 10 and 12, together, represent a 3 cell and a 4 cell cage, meaning 16/17 to 24/25 must all be 3 cell cages - this also means selecting 24 not 25 (cannot put 25 in 3 cells), so all 3 cell cages are even.
At this point the parity of 26/27 and 28/29 is yet to be determined, but the 4 cell cage in box 6 must be at least 26, so minimising the totals of the 5 cell cages in boxes 3 and 9, and also minimising the portion of the 4 cell cage in box 6 that also sits in columns 8 and 9, provides the break-in point to the puzzle.
Great puzzle and very satisfying!
I don't normally attempt 4* difficulty puzzles, but the rules on this were too intriguing. Once I understood how everything fit together it was beautiful. Thanks for choosing this one, I would not have solved it otherwise!
Love hearing the birds outside Simon’s window! Enjoyed the puzzle!
I tried it, made a few realizations about the cage totals, and couldn't get one digit. I am impressed that you figured it out Simon. Plus I get to hear lines like "might as well do Sudoku".
Beautiful puzzle. Took me almost 3 hours to crack it. I loved that the ruleset is so simple and that it gives rise to a unique logic.
I feel your pain. Congratulations for sticking with it. If I've gone nowhere after an hour, I usually give up and start getting hints from Simon. He has such incredible insight into the nature of any puzzle.
Wow this is an incredible puzzle- hope to see more from Yawnus!
Usually (understatement) I am not able to solve the puzzles without peeking at the video and looking for a hint. This time, I managed to find the break in after a couple of minutes........... So proud of myself 🙂
Interesting concept for a puzzle, thanks. Reasonably happy with my solve, though I did have to make a few notes and as per Simon appreciated a few things I missed.
Oh, this was a tough, but incredible puzzle. Glad to see it featured.
32:25
Some very beautiful logic which flowed smoothly from one step to the next. Incredible debut.
I love the puzzles with simple rules
"Solve counter, 69. Do I like the puzzle? Yes it's very... Nice"
Well played, Simon. Well played.
Hey Simon, love watching your puzzles and solves. You can't turn off Maverick, but maybe change the notification settings on your phone? Buzz buzz!
I tackled this from a slightly different angle ie there are a limited number of ways to fill 3, 4 and 5 cell cages within the constraint, boxes 3 and 9 have to have an odd even combo and that means totals with be three apart and there is only just enough different totals to be three apart once with just the highest totals being odd.
My 8 yo son just walked in, heard only one word, and said, “is that Simon, from Cracking the Cryptic? His first name should be Cracking, and his last name should be Cryptic, and his middle name should begin with T. It could be Timothy or something”
it is much more restrictive if you think about all the cages at the start :P its 15 sums that can't be consecutive between 6 and 35, and ignoring the degrees of freedom from even vs odd that is just 15 possibilities. so all the sums are in, it is consecutive in the evens odds, or first in the evens and then in the odds with an even and an odd in between.
13:40 "My intuition is so off" And I'm screaming at him (two weeks late, I might add), that if he just includes the 5-cell cages in the reasoning he started, he would get that there is a single degree of freedom on the cage values. It's not often I scream loudly enough to actually write a comment, but this time the thing I want to scream isn't "But look at this one thing that I noticed that you didn't" or "Remember that thing you forgot 5 seconds / 5 minutes ago", it's "Yes, keep doing the thing you're doing, don't stop". And I feel that that's a qualitatively different thing to scream.
Whoa, Last Call BBS and Dungeons and Diagrams? Sign me up! I've been having a blast with it a while ago.
50:48 for me. One of the most satisfying puzzles in a while!
20:58 for me. Wow, loved this one so much. Awesome puzzle!!
Amazing sudoku and deductions!
The art of never using a pencil...wow
Wish there was a notepad++ type editor on side so u can keep track of their mind working.
For me, as pencilsolver , wrote down all the possible sums ...and then discovered how the Even and odd sums needed to jostle to suffice the rule (ie B6C7 digits) Got a bit stuck after that and came to watch the beginning only needed to watch the coloring of the 4 digit cages to realize what I forgot in my initial discoveries of the 4digitcagees then the puzzle went rather fast
I took a real pencil and a piece of paper instead:)
51:14, but I had to look at the video to see how to use columns 8 and 9. I actually did the work to figure out what the sums of all the cages had to be, but didn't see how to start figuring out which sums went where.
Still impressed how it the cages were resolved after the break in.
Anyone else under the impression that Maverick has seen the channel and just flies circles around Simon's house to spite him?
The 3 5-cages would have to have a total above 30. There are 15 cages, and to get the minimum sum of totals, you'd get 15 even totals from 6 to 34. So there's a gap of 1 here, because the 5-cages can go up to 35.
This gap could go anywhere from 6 to 35, meaning either every total is odd if the gap is at 6, every total is even if the gap is at 35, or somewhere inbetween, it switches from even to odd totals because there'd be a gap of size 2. And it's clearly the third option, since boxes 3 and 9 both require even and odd totals.
And from your logic that the totals are even from 6 to at least 24, and with the minimum odd number in boxes 3 and 9 being 31, it stands to reason that the remaining 2 totals above 31 have to be 33 and 35, and those can only be fulfilled by the 5-cages.
So yeah, the 5-cages' totals have to be 31, 33, and 35.
Video title: "Is It Possible To Solve This?"
Me seeing a 1:02:06 video of it being solved: "Hmm is it?"
from there, the 4 cages can be 30, 28,..,24 , 29,27,...,21 , 28,26,...,22 ... a lot of them necessarily have to land near the maximum consecutive sums. there is just one option for taking a 5 cage and mixing it in between the 4's and two options for mixing in three cage sums. interesting starting point :P
I feel like Simon got too focused in early and missed the big picture.
The very first thing I noticed is that across all 15 cages, you have 1 degree of freedom, and that in turn will be eaten up by the flip from even to odd necessary by the cages in boxes 3 and 9.
This means the values over 30 must be in the 5 cell cages in boxes 3 and 9 and both must be odd (so from 31 33 and 35).
I’m not gonna pretend the rest of my solve was nearly as fluid as this but it feels like he tried to focus on the details before understanding the big picture.
Yes it is! I got it in 46:46. Nice logic to deduce the cage sums.
What a lovely idea. Quite amazing how it all starts falling into place! 56:47 for me.
Is it possible to solve this?
Well, I think yes. Otherwise we would not be able to see this
It usually is a consolation to me that, whenever Simon may be stuck in a puzzle, the happy end is just guaranteed by the mere fact that we're watching the video! Indeed, there is balm in Gilead 🙂
😂
Another ludicrous hour-long puzzle where 40 minutes of brainstorming are required to place the first digit. :D
"Have a go if you have a few minutes to spare"... several hours later, and I have absolutely nothing. Not even a pencil mark.
there is 15 cages, 15 sums, non-consecutiveness forces a few things, first of all, if all the sums are even then there is no room at all to shift sums around all sums from 6 to 35( which is the maximum sum possible) must be included, there are 30 possible sums between 6 and 35 including both, 15 sums possible with no consecutive sum values. they can run from 6 to 34 if even, 7 to 35 if odd, and 6 to 35 if there is a transition from even to odd sums, in which case at the transition there is a gap of 3 between the sums instead of the 2 between the other sums. that characterizes all the possible arrangements of sums. it is not however possible for the lowest sum to be 7 and for there to be a transition to even sums, if the smallest sum is 7 then all sums have to be odd. it is also not possible to start with even and then switch to odd, and then switch back to even. so what we are looking at just from the rules and from the box numbers is one series of even sums and one series of odd sums, individually consecutive in their parity, with a transition between them, where the odd series starts 3 above where the even ends, or we are looking at just the consecutive even or odd sums. now the maximums become important for the ordering, the 5 cages have a maximum of 35, the 4 cages have a maximum of 30, so the 3 5 cages must have a value of either 34 or 35 for the largest sum, the second largest must have a value of 32 or 33 because of the possible sequences it can reside in, the third largest does not necessarily have to be as large, because a 4 cage could be 30, but if we live in a universe where this end section is odd, then the third 5 cage has to be 30 or 31, if we are in an even world up to and including the smallest 5 sum, then the smallest five sum can potentially have its minimum sum of 15 and any sum between 15 and 30 potentially. pretty cool puzzle :D
18:30 When Simon goes two tone and makes the board look like the TISWAS logo.
33m32s. I had a good time working out what the sums had to be and where they had to be located. I was pleasantly surprised with how the sums kind of fell into place once I got the 8 over in box 4
What a devilishly holistic puzzle
This software totally needs an aide memoir for puzzles like this, it hurt my brain a bit
Awesome to have both zz top and R.E.M referenced by simon though
This is credible! A Sudoku grid with absolutely no digit to start: none in the grill, none outside the grill, none attached to the cages, none in the rules! OMG!
It’s a really clever set up but I’m not so sure it was the most obvious of solve paths (note that I’m referencing the thought process as opposed to difficulty level). I only realised the restriction on the pattern of odds/evens when looking at the comments and even then it took a moment to understand it correctly. An interesting debut but keeping track of the totals was a bit too troublesome. I enjoyed watching Simon’s solve though.
from there it is much easier to think about maxes and mins, two 5 cages must be above 30, two 3 cages must be 6,8 7,9 or 6,9, and two 4 cages must be between 24 and 29. so that is 6 cage sums already nailed down except from a choice of even odd boundary. much easier than to think about 8 3 cages independently :D. but it is always fun to watch you try to figure it out, sometimes you find stuff really easily and quickly and other times i think you miss super obvious stuff, so pretty much like my experience of my own reasoning lol.
I liked this puzzle. Thank you !
There’s 15 cages. For they are 2 apart, that makes a range of 6 to 34., except 3 have to be odd. The 5 cages are the only ones that can exceed 30, so they can be 31, 33, 34,
The remaining 12 cages have a range of 6 to 28, so they fit with the previous odd 30s.
If you put any odds in the 3 or 4 cages, you lose room for a cage result, pushing it up, so you can’t use 30, but you still have to fit 3 above the lowest ones.
Written at the 22 minute mark
An easier way to calculate the totals is to start with 6, since that’s the lowest number for the three cell cages. Since the minimum distance between the cages is two, you take the remaining seven and multiply them two. 14 + 6 =20
This is the kind of puzzle I do not like. It feels more like homework or a task that you have to brute force by calculating all possible alternatives to find your way around it.
Simon to two girls at a party: Let'see, you're both odd and both over thirty...
I wonder how much faster Simon could have broken into this if he'd gone the extra step of considering the 5 cell cages when doing the telephone pole calculus.
3 cell cages - 6-24, 8 cages, 4 degrees of freedom
3 AND 4 cell cages - 6-30, 12 cages, 2 degrees of freedom
3, 4, AND 5 cell cages, 6-35, 15 cages, 1 degree of freedom. So there is ONE gap of two where you change from the evens (low end) to odds (high end).
That you have published the video I suspect that you can solve it
I feel your pain, Simon.
Maverick distracted me too, and I just can't get it back.
the much easier way to get counts on right 3 boxes is if boxes are 31 + 35 = 66 then remaining 4 cell cages in those boxes are 24 which puts 12 in col 8 and 123456 in col 7. That leaves 789 in box 6 col 7. Even if the 9 is used in 4 cell cage of box 6, you can't it make total 26 or more (9+6+5+4=24 ). This forces 7 or 8 into box 6 col 8 or 9 and makes 5 cell cages 31 (45679 or 35689) and 33 (xx789).
Yup, that's the key. Wish it hadn't taken me 55 minutes to see it.
so it is then impossible for the sums of the 5 in the corner cages to be even at all. therefore there is either 3 maximized 5 cages, or an even 4 cage sum of 30, and the even 5 cage is smaller. that is possible. but then all other sums than the two 5 cages would be even with 100% certainty. but since we know these two cages now has to sum to 33, 35 or 31, then we know the even 4 cages has to sum to 14,12 or 10 in the corner boxes filled with cages. which means we found two 4 cages that interfere with the 3 cages, and the two others must be maximized, they must be 30, 28, 29 or 27. that nearly nails down the entire sequence :D. if the 5 cages are not all maximized then the remaining 5 cage has to be even, and all the other cages also have to be even. if not then the odd even transition can be anywhere, but all 5 cages are above 30 and two are odd. from there it is a little easier to proceed i think :P.
All that work repaid with two songs!
you end up with a series in the evens, which is equivalent to any series that is shifted to odd for the entire or part of it, that is in order first 8 3's, then 4 4's, then 3 5's, except from one shiftable 5 and two shiftable 4. the 4s can be shifted between 10 and 24 and to 30, but if the 5 cage sums are not maximized, the 4 cage sum must be 30, therefore no matter what you can only shift it there and so it is not really a degree of freedom as loose as the other ones. so a 4 cage 30 and a 5 and a 4 that can be shifted around to many positions, or two 4 cages that can be shifted around between 10 and 24 to move the 3 cage sums around. if the 30 sum is a 4 cage then ofc all the sums bellow are even. so many deductions to make just based on counting and the rules lol.
great solve
Maverick always knows. I really think Simon should check his office for bugs 🛫😈
Simon, Simon, Simon, follow the rules. If you use everything from 6 to 24 on three cages, you can't have ANY MORE CAGES AT ALL!
Since you've been okay with Islands of Insight in terms of motion sickness, you'd be okay with The Talos Principle. TPP has even less movement.
48:30 I think Simon is making an error here - I don't see a 6 cage at this moment yet. Did I miss anything?
Actually, never mind. It can't be the 6 cage, because the upper cage can't be either 2 or 3
you have to have the sums of there are even sums and odd sums, to run from an even series from 6 to 2N, then from 2N+3 to 35. so you know the largest sum is 35, the next largest sum is and has to be 33. because then the two largest sums must be odd, because you can't have two odd 4's and two even 5's. at least one of the 5's in the corner cages must be odd, because two fives must be the largest sums, and the nature of the transition between even and odd is as such that it could not be between the two largest 5 sums, because then you would necessarily only have even 4 cages, and then both the 5 cages are forced to be odd anyway and must be 33 and 35. so if both were even then all sums bellow 32 must also be even. so two of the 5 cages then must sum to 35 and 33. then the last 5 cage can be 30, 31 or some lower number, if the largest 4 cage is 30, but if the largest 4 cage is 30, then all other cage sums are even, because of that there would be no other 4 cages that would be odd, and therefore... you would have to put even 4 cages with both the 5's in the completely filled boxes. if one 5 cage in these boxes is even, the 4 cage in there would be odd, then the 5 cage in that box would be smaller in sum than it, it would not be maximized, one 4 cage would have to be 30 exactly to displace it from its maximized position, which would make all the 4 cages even, and the odd 4 cage would be made impossible, which is jolly for sure.
It took me about an hour to get where you did at the 30min mark, but couldn't find the way to break in, deduced that the 3 big cages needed to be 31, 33, 35 respectively, but not which was which.
EDIT: finished it in about 5 hours 😅.
beautiful puzzle
I started by assuming that every cage is as minimum as possible gives 6,8,10,12…..32, 34
Which also means that you can only switch polarity once. Which i think is a good point that Simon missed.
@@awebmate I got interrupted in the middle of writing this comment, but that was exactly what I wanted to say