Myxo here! This was an unexpected feature :D I Love Sleeping came up with the idea and we worked on the puzzle together. Simon, there was indeed a path through without (almost) bifurcation! It's possible to determine what digits (from A,B,C,D,5) have to go in all the cells without determining the total first, and then using parity to resolve it. But I am very impressed with your ability to see that all early in your head.
@@RyanAtOptimism I'm not sure if this is the intention: But what I saw at that point in the puzzle was - once we know there is a 5B combo in r34c1, we can ask if it can go in r4c1. That would put the 5 in the bottom row with the purple C. Since r4c2 is in A, that would make row 1 and row 4 the same digits. That puts the 5 in row 3 and the 34/B in row 4. The next thing I noticed was that r4c2 could then simply not be a 2. That makes A 46, which just solves everything at that point.
The same thing happened multiple times. "B can't be 2".... well, 5 + a V domino means that the purple C will always be the 7/8/9 in 17/18/19. So the moment 17 gets ruled out, C automatically becomes 8 or 9.
He has two modes, incredibly picking the most incredible and beautiful logic built into the puzzle, and missing the most obvious answers and getting there anyway by sheer power of brute mathing.
So according to Simon, if I write down 3+5=8 on a piece of paper it is a sum, but if I do the maths in my head then it isn't... maybe then it becomes a square root
@@sjm6280 the point is, as I said in my comment above, that acceptable bifurcation is a subjective thing. Most of us will say that if you prove something by contradiction it's ok, but if you randomly fill in number and explore aimlessly, it's not ok, even though if you encounter a contradiction, technically, it's a valid proof by contradiction. So, putting in an objective restriction such as "I'm allowed to bifurcate as much as my memory allows" is reasonable.
@@Ennar It is valid, but it gets funny when Simon says that "Bifurcation i would never ever do that it contradicts the fun in the puzzle and we might miss the setters magic" in every video he solves, and then ends up biforcation "in his head" in every puzzle. Simon just loved to make things alot more complicated than they are and its a beauty to watch. Just dont deny the biforcation. Embrace it.
Rules: 06:24 Let's Get Cracking: 07:25 Simon's time: 50m07s Puzzle Solved: 57:32 What about this video's Top Tier Simarkisms?! Three In the Corner: 2x (56:33, 56:36) Maverick: 2x (06:53, 06:55) Scooby-Doo: 1x (07:34) And how about this video's Simarkisms?! Sorry: 7x (14:18, 14:57, 15:33, 29:30, 30:53, 37:05, 39:40) Obviously: 6x (02:30, 08:24, 17:59, 32:20, 33:01, 33:07) Ah: 6x (01:39, 31:08, 37:53, 44:22, 48:02, 55:50) Hang On: 5x (17:17, 28:05, 28:07, 38:30, 49:56) Clever: 4x (39:20, 39:23, 55:30, 58:16) Shouting: 4x (03:07, 04:17, 05:31, 06:08) In Fact: 4x (35:31, 51:44, 52:34, 57:51) Unique: 3x (00:23, 06:19, 06:39) Goodness: 2x (48:42, 49:11) Axiomatically: 2x (08:43, 52:54) Beautiful: 2x (22:49, 33:49) Fascinating: 2x (57:05, 57:45) By Sudoku: 2x (09:56, 32:22) Wow: 2x (50:37, 50:40) Cake!: 2x (04:57, 06:11) What on Earth: 1x (48:20) Stuck: 1x (56:51) Lovely: 1x (05:31) Break the Puzzle: 1x (34:55) Incredible: 1x (04:52) Bizarre: 1x (00:19) Surely: 1x (37:25) What Does This Mean?: 1x (11:28) Have a Think: 1x (20:15) Pencil Mark/mark: 1x (11:52) Most popular number(>9), digit and colour this video: Ten (38 mentions) Two (60 mentions) Blue (4 mentions) Antithesis Battles: High (8) - Low (3) Even (8) - Odd (4) Row (49) - Column (38) FAQ: Q1: You missed something! A1: That could very well be the case! Human speech can be hard to understand for computers like me! Point out the ones that I missed and maybe I'll learn! Q2: Can you do this for another channel? A2: I've been thinking about that and wrote some code to make that possible. Let me know which channel you think would be a good fit!
I went down that rabbit hole to the extreme, sort of. I did switch to using a unique letter for each of the first two rows cells, which allowed me to get all the way to the end with the three unresolved dominos in columns 1, 2, and 4. Unfortunately I couldn't see a way to disambiguate it at this point and then I saw the logic Simon had used showing that a cell in row one and row two had to be both the same digit and yet it couldn't be the same from the logic, so I knew I'd eliminated a possibility somewhere. I was so close, though.
Loved this. A real feat to not only come up with a ruleset that can provide a 4x4 grid that takes an hour to solve, but then to actually execute the idea this well too. Brilliant work
Also interesting that for most videos it would be a cells/minute ratio, whereas for this it would be a minutes/cell ratio. 3.6+ at that. A 9x9 would need a 5 hour solve for that kind of ratio.
Does anybody remember the last time a 4x4 produced an hour-long video? It was April Fool's Day -- 2022 maybe? -- and it was Mark solving. Very, very funny.
Normally when you're explaining the puzzles, I can follow along and understand but this one I was trying to understand but I was not getting it. This is the puzzle that broke me 😂
19:08 Separate colours for each individual digit helped fill out most of the grid into sections, then slowly whittling down the options to find which digit had to missing broke open the finale. Very nice idea to make such a small grod contain so much
This channel has been in my recommended for a while, and I thoroughly enjoyed this video now watching one of these videos! It's a great break from a lot of the fast paced content on youtube, and you really get to think while you are watching the video which is always good and a lot of videos don't give you time to think nowadays. Definitely going to be watching more of these!
I'd have to call bifurcation on Simon, and I did essentially the same thing to disprove the 9 in r4c3, except that I can't visualize as much in my head. I just couldn't find the intended trick, so I settled for the solve using that bifurcation, and then watched hoping to see Simon show me the better way. I guess I'll upgrade my solve to successful since Simon's was pretty much the same. :)
solved in 18:24 - did a lot of coloring in the top two boxes until I narrowed the sum down to one of two numbers. I'm not sure what the best way is to proceed from there, as I ended up taking educated guesses for the lower half once I found the first digit. interesting puzzle edit: I figured out the way to sort out the bottom half afterwards. the color of r3c2 can't be the same as r2c1 because otherwise row 3 uses the same colors as row 2 due to the X in row 2. this was probably the actual way to clear it up. cool
I was surprised that once Simon got a 5 he didn't immediately tell that D had to be odd in order to make the top two rows add up to an even number (which is two times N).
I also found it interesting that he also did not notice that when he placed the 5 r1 had 3 numbers=10 and r2 had 2 numbers=10 so high A + low C = high C the combinations that work are 8+1=9 or 6+2=8
Took me about 2 hours, but I got there in the end. Basically had no digits other than the 5s placed until the very end. Worked in terms of high/low colour pairs adding to 10 (e.g. dark blue + light blue = 10, dark blue > 5, light blue
When I was studying linear algebra, I once set up and solved a 15x15 matrix to determine the magic square (up to a multiplicative constant!). I'm almost tempted to do the same thing here.
i've been binging a few of those video, usually trying myself then watching a few minutes when i'm stuck to get the point i've been missing, but here i completely gave up
This took me ages because I made it too complicated for myself. Once I asked the question "how many ways can you make 10 in 3 digits" I went from stumped to solved in about a minute. This is a real good puzzle. Kind of kicking myself for taking so long to realize the incredible structure behind it. More specifically, I had long since realized that X=18 or 19. R1C3 is a 5 (I forget how I came to that conclusion) so the top row is 10 + R1C4, making it 8 or 9. In addition R4C3 is also that same 8 or 9. I just didn't put together that as a result, this made rows 1 and 4 and columns 3 and 4 take up all the ways you can make 10 in three numbers, the two using a 5 were taken, and that R4C4 had to be the lone shared digit between the remaining ones. ie. a 1, which removed 9 as a possibility.
OK that's really nice. I was scrolling the comments trying to find a satisfying solution for this step and finally found yours. Thanks a bunch for explaining it so well.
Now Simon has awesome brain power. He causes those digits to flower. But I've ne'er seen before Someone solve four by four And take anywhere close to an hour. Seriously, though, Simon, that was a toughie. Great work!
So amazing, Simon. An amazing puzzle, and I could not look away as you were solving. On the topic of bifurcation: I believe that the definition that you (and Mark) seem to have settled on is "if I can see it in my mind then it is not bifurcation." This is a bit hard on some of us who are not as good at visualizing things or running through maths and other matters in our minds. I am so very visual that I sometimes have to actually see something - not just see it in my mind - in order to determine its accuracy. Nonetheless, I think that it is a valid feature of pure logic to use if-then to eliminate options. That does not make things bifurcation, in my view, no matter if you write it down or do it in your mind. Otherwise I would have to admit that every single solve of a sudoku or other "logic" puzzle is bifurcation for me. If-then is legit, I believe, and requires no apology. (I am pleased that most of the comments I have read have not scolded you for bifurcation or even questioned it - well done, sudoku world.) Thanks, Simon, a fascinating hour spent in your company yet again.
When Simon pencils the D into box 1 (around 19:00), you can immediately rule out 20 as the total based on the letters in box 1. No matter how much Simon would like it to be 20. :)
I figured it couldn’t be 20 on the understanding that the rules implied the numbers 1-9 would appear, including 5. So once you know every box has an X that’s over But that interpretation of the rules turned out to be wrong
Isn't 20 immediately ruled out with the knowledge that the 2 V's must be distinct? The X in C2 implies that R1C1 and R2C1 must sum to 10 (as all the boxes, rows, and columns sum to the same number), and then R1C2 and R2C2 can't sum to 10 because of the X in R2 and the fact that the V's are distinct.
I think Simon overlooked that he had one X in every box, and that that meant that the two other digits would sum to the same total (either 8 or 9) in every box. It would have eliminated 5 from box 4 pretty early (since it would repeat column 3 to get to the right total), and I think you could deduce that the 8/9-total there had to be the same as the one in box 1.
Yes, that's how I did it. There were pairs of digits that added to the "outie" (n-10). I gave each digit its own colour. Knowing the pairs that added to 10 and the pairs that added to the outie, I was able to solve it pretty quickly once the 5 was found in row 1.
The step that D in r1c2 isn't 2 (or 4) has been on the board for about (edit:) 15 minutes, since r1c3 = 5 is deduced :) Simon even comes close with observing that the sum of the top two rows must be even, but then he phrases it in a very convoluted way. Instead, we can simply observe that the top two rows are AA+BB+CC+D+5, and as each AA, BB, CC is 10 which is even, D+5 must be even.
At 48:49 when you have the 345 in c1r4, you can rule out 5 because 24 is then canceled due to then 3 numbers in the 4th and first row being the same. Causing a repeat in rows since the rows have to be the same value the 4th number in both rows would also have to be the same. Which if the 5 was used in c1r4, then you get 5 plus 6 plus 8 or 9. This breaks the row from not being able to be 18 or 19. This gives you the 5 in c1r3. Always love watching you solve these.
You're a rockstar, I enjoy watching you solve these increasingly vexing puzzles.. you had eliminated the 1+9 from column 3 in the bottom right box, but didn't try the 2+8 that would have solved the remainder of the puzzle until you'd done some rather impressive brute forcing in boxes 1 and 3... but still. bravo! Keep making these, and we will keep watching them and throwing popcorn at our screens.
I solved it!!! In... 210 minutes lol. I must admit, listening to you solve puzzles while handling tasks around the house is very different from pausing at the start and solving it on my own. Completely fascinating and painstaking. Thanks for goading me into bizarre puzzles I have no background with, my brain is thoroughly exercised now hahaha ❤
Simon really has a knack for marking cells in the most convoluted way. Sp far (20:55) I think we have blue and pink b/c. and the orange and white A, B and D.
I didn't watch this solve, but I did color the cells to indicate they were specific digits AND I used different colored lines to indicate that the two cells the line joined summed to the same value as any other two cells joined by that line. And pencil marked, of course.
This is probably the first puzzle ive watched you solve where i was completely lost and after wwtching someone solve it, theres no way i could now do it
46:39 i dont know how to explain it well but this the moment i think i saw it. every row and column need to have 2 odd and 2 even digits. look at row 1. because there is a 5 and a V there is at least a minimum of 2 odd numbers in each row, column and box. now look at row 2. because simon proved there should be 6 in box 1 row 2 and the hidden V in box 2 row 2, there have to be a minimum of 2 even numbers in each row, column and box. so the possibilities were 18 and 19. 2 even digits and 2 odd digits can only sum to 18. therefore you know r2c2 must be a 7 because otherwise you would have 3 even digits in row 2 which cannot happen because of row 1 having a maximum of 2 even digits. i hope i am correct with this
My first try of a solve, worked out heaps of different logic, got stuck, watched a bit of the video up till you proved the 5 then finished it off!! Love this channel!!
18:38 but I'm not entirely sure how I did that. Running through the numbers needed to min and max the totals for each row, column and box helped a lot to see which totals could be achieved through a unique set of digits in the most ways. I've saved my replay, but now I need to figure out how to view it so I can see if I actually got through it without bifurcation (I'm highly suspect).
@@jacobcombs1106It's just not a very satisfying way to solve a puzzle. And, technically, if you bifurcate and guess correctly then you haven't proven uniqueness without going back and trying the other option and showing it fails, so if you think it is important to prove uniqueness then you have a bunch of extra work to do. I'm not sure how many people care about proving uniqueness, though.
@@thomasdalton1508 ah I see, what I've been doing and calling bifurcation isn't bifurcation then. Bifurcation in this context is exploring one possibility to the exclusion of others? I had been using the term to describe how if I realize there are only two possible combinations through logic using corner marks to show one and center marks to show the other until I find one breaks and then I can either control delete or shift delete erasing the disproven path. That would not meet the criteria for the taboo bifurcation since it is validly disproving the other path before committing to the correct one? For example this puzzle once I'd proven in my mind logically the sum of the rows columns and boxes must be 18 I logically worked my way to the realization I needed a five somewhere in the top half and there are only 2 valid cells for 5. So I corner marked 1 as 5 and center marked the other as 5 and began filling in corner marks with valid possibilities to make everything 18 following the Xs and Vs with the 5 in that spot and center marking the valid combinations for the 5 in the other spot. Quickly I worked out there was no valid combination for putting the 5 in box 2 because I would be forced into a repeated combination by the Xs and Vs and I deleted the invalid path. Eventually I had the top buttoned up and found myself in a similar bifurcating path on the bottom where it was either a 1/5 or 1/3 pair in r4c1 and r4c4 which meant an opposing pair of pairs in r3c1 and r3c4 to reach my sums and I used center marks to show one and corner marks to show the other and began stripping away the options that required repeated combinations which eventually disproved one of the pair of pairs and left the other pair of pairs resolved. I then filled in my cells certain it was the only possible combination.
@@jacobcombs1106 No, that's bifurcation. If you guess incorrectly and get a contradiction, it's completely solid logically. It is only if you guess correctly that you have arguably missed something out, since you don't actually know that the other option wouldn't have also worked.
@@thomasdalton1508 ok so it is indeed the guessing and arbitrarily excluding one of the paths without actually disproving it that makes it taboo bifurcation, so I feel like my method where I work both options in parallel which will always result in properly logically excluding one of the paths doesn't fit the taboo as I will still always show uniqueness and not just got lucky.
It took me a while (about 70 mins), but my strategy was to colour the top of the grid (after making Simon’s similar deduction that they all had to be unique) and then seeing how that fell into the lower half. Really tricky puzzle for sure!!
Simon saying he not gonna bifurcate, got me trying from 40:00. What I did is looking at the column 2&3 realising R1C2 must be odd. Then realising R2C2 can’t be 6, else it will create a sum more than 18
Solved in 23:46 and thus my first time commenting (after years of watching) because this is the most I've ever been faster than Simon by! Very fun puzzle!
He very clearly states that he only considers it bifurcation if he has to write it out to work it out. He doesn’t do that. Instead, he works it out in his head, then writes it out to show *us* why something doesn’t work.
I had n as 19 and worked out a possible grid. Got told it's wrong, so watch your vid till the 5 and then figured it out in 10 minutes after i knew n was 18. 37 min in total. What a brilliant puzzle
When trying the puzzle I came up with a crazy reasoning why the total cannot be 16. Of course there would have been an easier way, but I still like it. There are only 8 different ways that 4 different digits sum to 16, Howevrr, only one of them includes a 9. But Since 9 would appear in a row and a column, there need to be two different ways that four digits, including 9 sum up to 16.
That's similar to my approach. I enumerated all the ways to make sums from 17 to 23 (excluding 20, for obvious reasons, and others because there weren't enough to be unique) and found the only sum that permitted enough combinations with elements satisfying all the (explicit and hidden) X/V sum rules. Once I got the sum, the rest fell out pretty quickly.
some insights on the general structure I gained after solving: (1) if two vertically adjacent 2x2 boxes have a digit in common, they have a vertical domino in common, and both boxes are identical to the columns (2) in our case, boxes 2 and 3 have a digit in common, so boxes 1 and 2 have no digits in common, otherwise row 1 or 2 is identical to column 3 or 4. Boxes 3 and 4 are disjoint for the same reason (3) rows 1+2 and rows 3+4 are 8 disjoint digits out of 9 with the same sum, hence the same digits (4) the vertical dominos in rows 1+2 also show up in rows 3+4
Solved it! Hooray! Usually I feel too intimidated to try these prior to watching the video. Attempted solving with x at a fairly high value, and felt it was just too hard very quickly. Dropped to the right x value by luck and pushed through quite fast with trial and error. Thanks for that :)
what saved me from the guessing and checking was i noticed the top right digit in the top left box had to be odd due to the fact that there were 3 other 10s and an odd in the top 2 boxes but the sum of all the digits in the top 2 boxes had to be even as it was 2x the amount each row/column is set to
I spent an hour proving various thing about what various cells could or couldn’t be but I never found the trick that you did to not row 1 column 3 was either a 5 or a complement. But I had already narrowed down the top right cell to not be 9 and had proven N was either 18, 19, 21, or 22 so it couldn’t be 20. Most of my deductions were from the relationship between r1 c2 and r2 c2. After ages of no more deductions I had to give up and watch your video, but when you made your deduction I immediately knew it was a 5 because I had already disproven N being 20 and thanks to all my other deductions the puzzle had no more resistance and everything fell into place. The pivotal thing for me would be knowing N was at least 18 and the top right cell couldn’t be 9 so once the 5 got placed the top row gave me N=18 and the top right cell was 8 which permeated across the grid. Great solve and I had a wonderful time solving with help from your 1 deduction.
After getting the 5, it was all quite easy. There was no need to consider parity. The minimum value of N was 18 (not 17 as you suggested). This is because you'd assumed in C2 that 6 could go with 1, but the 6 would make blue 14, so orange would be 23, so the minimum is 10 + 6 + 2. Similarly, the maximum was 22. However, once you have the 5, in R1, the maximum comes down to 19. Therefore hi C is 89 and lo C is 12. This makes lo B 34, and hi B 67. Working this through, you end up with hi A also 67, so N=18, and the rest is just a matter of ensuring you don't repeat a set of digits. This was quite a fiendish little beasty. I can't recall the specific deduction I used to get N, but it was quite simple, and related to the unique sets. Unfortunately, replaying my solve didn't include any pencil-marking which would reveal what I was thinking at the time. I did have 34 and 12 in the first two cells, rather than 24 and 13, and I got that by eliminating 3 from R1C2. There was some problem with R1C2=3 and R2C3=6. It may have been effectively the same restriction you spotted, although it certainly wasn't as convoluted.
After N = 18 it also immediately follows C = 8 from the top row, R1C2 = 1 (from the sum of top two rows, other fields sum to 35), and the whole upper left solution from the relationship R1C2 = 5 + R2C1 (the column and the square containing R2C2 differ by 5, so must the complement set of fields) resolving B and then the rest is trivial.
I havent seen someone else solve it my way. Once the beginning is done. So you determined all x,v and same numbers. You can determine that r1 amd r2 consists of 8 different digits. And the missing one is 5,6,7,8,9 which means the sume of r1+r2=[36,40] so the only options are 18,19,20. 20 is easily sort out because of b and c. And 19 follows quickly once the numbers are reduced to their posibility. How to say that alle digets are different: r1c2 cant be r1c3 because then r2c2 must be the same as r1c4. Which makes the the same sum but box 1 is now the same as r1 :)
Writing this as I am watching the video: One thing I noticed around 20:00 is that the possible options for N is 18, 19, 20 and 21. My logic is: Box1 + Box2 = 2N Subtracting the two vertical X-sums (r1c1 r4c1 r1c4 r4c4) and the horizontal X-sum (r2c2 r2c3) leaves you with the squares r1c2 r1c3 and they sum to (2N - 3X) = (2N - 30). Now I just tried all the different sum totals of the r1c2 r1c3 - pair, and using the logic that (2N - 30) = (r1c2 + r1c3): 15 pair (summing to 6) gives you N=18. 16 or 25 pair (summing to 7) gives you N=18.5 which isn't possible. 17 or 26 or 35 pair (summing to 8) gives you N = 19. 18 or 27 or 36 or 45 pair (summing to 9) gives you N=19.5 which isn't possible. 19 or 28 37 or 46 pair (summing to 10) gives you N=20. 29 or 38 or 47 pair (summing to 11) gives you N=20.5 which isn't possible. 39 or 48 pair (summing to 12) gives you N=21. 49 pair (summing to 13) gives you N=21.5 which isn't possible. So the possible options for N is 18, 19, 20 and 21. And we also can know that these two square sum to an even total.
Great solve - I ended up giving up and enjoyed following along. Crazy that a 4x4 can be this hard. To disambiguate things at 51:15 in a more straightforward fashion, (I think): - r3c1 must be 5 or low B (as by column 3, n = 15 + low B) - row 2 has the digits {high A, low B, high B, low C} - if r3c1 is low B, then r3c2 must be high A (low A leaves r3 too low, breaking the puzzle by forcing box 4 to be way too high), so then row 3 has at least the digits {high A, high B, low C} - so 3 digits overlap between r2 and r3 meaning the 4th must as well, breaking the puzzle So then r3c1 cannot be low B and must be 5, also putting low B in r4c1, hopefully allowing some subsequent deductions.
Here is a slightly simpler way to proceed at 50:30 Let m be the 1s digit of the mystery total. We see from row 2 that the high A digit plus the low C digit equals m. Suppose for a contradiction r3c2 is the high A digit. This is next to the low C digit so the other two cells in the row add to 10. This places 5 in r4c1. But now row 4 contains 5, the low A digit, and the high C digit, making it the same as row 1.
At 17:30 he had just gotten the top left domino sums to 10. There is a very useful algebra that can be done then: Call R1C3 as A and call R2C1 as B. The top two rows total 10 + 10 + 5 + A + B. Thus, 2N = 25 + A + B. Another way of looking at the top two rows is as 10 + 10 +10 + 5 + A - B (we counted the B twice but not the A at all). Thus, 2N = 35 + A - B. Those equations give B = 5 and 2T - 30 = A which gives T < 20 and gives T = 19 with A = 8 OR T = 18 with A =6. (I think I had already eliminated 17 as T at that point).
Took me about 55 minutes. I used the fact that the remaining 3 cells of the bottom row added up to 10 a little earlier. This basically left 3 different permutations that solved themselves when you wrote all the numbers in
In the end, instead of eliminating 4 from R3C1 and R4C1, it is easier to put 5 in R3C1 by eliminating it from R4C1. A 5 in R4C1 would force Row 1 and 4 to have the same set of digits and is therefore impossible. From there it gets easier again once you see R3C2 can't be 68 as it would force the placed 5 in R3C1 to be part of a 10 sum (because of the X and the digits in R2). The rest just falls in place from there.
Let's use upper and lower case to denote big and small. At 25:11, now that we have the 5, the key insight that solves the puzzle is algebra. From column 3 (or row 1) we have n=C+10 (because of the v and the 5), then row 2 gives A+c=C. Likewise, column 2 gives d+B=C. This rules out C=7 because there has to be two distinct big small sums that add to C. The rest is easy.
48:30 after deducing that 5 in c1, you could not put 5 in the corner. Because then you can't put A-high in r4 (because total will be bigger than 19), and if put A-low, r4 will repeat r1. So r3c1 is 5, r4c1 is B-low.
00:18:07. Holy shit I can't believe I just did that... so i treated it as a polarity puzzle when I saw the Vs and Xs like oh gotta have two highs and 2 lows in order to balance it out. Surely that means it's all 20 right? Realized nope to do that would automatically force every row and column practically to be the same. 19? No same problem. 18? Very quickly resolved the only apparent way to make the top two boxes work for 18, then worked out the only two possible combinations to keep the bottom boxes also 18 with every row and column 18 and realized one of those combinations had repeat rows but the other didn't and bing bango bongo bobs your uncle the puzzle was done.
This was one of the first puzzles featured on this channel that I actually tried solving myself (or, more precisely, the first after the disaster of trying some when I first found Cracking the Cryptic and discovered that they were, in fact, much harder than Simon made them seem). Watching this video upon giving the puzzle up is really funny to me, because I placed three digits in the first couple minutes (using logic that I can’t actually recall), and then proceeded to make very little progress for another couple minutes, subsequently staring at it without learning anything for about half an hour. I go to watch Simon’s solve, and he reveals about three dynamics I had not noticed in the first five minutes, but without placing a digit quickly at all. I haven’t finished watching this, so I’m not even convinced the logic I used to place the digits was at all sound, but if it was then the comparison between my attempted solve and Simon’s successful one would be quite funny. Edit: yep. Idk what kind of oversight happened at the beginnings of my solve, but my placed digits were in fact wrong. I think it was so early into the puzzle that I hadn’t completely understood the rules and their implications yet.
We can easily use Excel to reduce a lot of repetitive calculations. 1234 represents rows, ABCD represents columns It can be simply known that C3 can only be 1-4 We assume that the sum of the rows and columns is S, and we can calculate all the numbers in rows 12 and column C. S is a moderate number around 17-21 By using quick calculations in Excel, it can be found that 21 is too large, while 20 has many duplicate numbers. I initially thought it was 19, but I couldn't get the correct result. After switching to 18, the remaining numbers were quickly calculated
Spencer here and me and aisha listened to the birthday message today and both had a good laugh its awesome to get a little ahout out her and my daughter made brownies for me instead of chocolate cake and my favorite dinner an awesome birthday for sure
You can also exclude r2c1 to be 7 by summing rows 1 and 2. Since the sum of both rows' sums should be even (2*N) and the sum is X+X+5+V+r2c1, i.e. 30+r2c1, it can not be 7 (since 7 is odd in most of the times).
Okay, I'm commenting before watching with a solve-time of 28min. My method of solving: Brute-Force R1C2 & R2C2 and play through every scenario provided by those two, proving 2, 3 & 4 can't go into R1C2 and from there on out it was more or less just a chain-reaction to solve^^
Another similar but shorter way to get through the end - you can determine that r4c2 can’t be the low A. If it is, then r4c1+4 adds to the same as r1c2+3 but can’t be the same digits. If low A is 4, that sum is 6 but can’t be 1,5 or 2,4. If low A is 2, then that sum is 8 but can’t be 3,5 or 2,6. That leaves 1,7 but one of those must pair with 15 (the rest of column 1) to get to either 18 or 19.
I am so happy I did this! Really amazing puzzle.. My approach was to use letters for all digits and then solve it algebraically. I got stuck coz I assumed all 9 digits must be used and I couldnt see a way to do that while 8 digits gave an elegant solution. So I came back to watch the solve and realized that was an artificial constraint in my head..
I mean, depending on which definition you have of "Normal sudoku rules", in this puzzle normal sudoku rules may apply. (they apply in the version fill every cell with digits 1-9 so that no number repeats in any box, row or column)
When we created this puzzle, we used the words 'non-standart sudoku rules apply'. But as Simon said, the amount of sudoku-logic used in this grid is about none, so I can understand why they chose to word it like that.
@@I_Love_Sleeping-Leeor Yeah, I was referring to the rules on screen, it's understandable Simon reformulated the rules that way for entertaining, editorial and clarity reasons, but still what he was saying about not following standard sudoku techniques even if it can have the same formulation of (some) rules is amusing to me
Solved in 30:04, but I was only able to deduce the row/column/box sum to 2 options, started guessing and a few digits later I was done. At least I found one correct digit without guessing 🙂
AT 40:52 another way to see why R2C1 can't be 7 is the following: Because the sum of the two upper rows is 2N it must be even, but this sum consists of 3 X-pairs (totaling 30) plus the 5 in R1C3 and the number in R1C2. To make this total even R1C2 must be odd, so cannot be 2. Thus R1C1 cannot be 3 and R2C1 cannot be 7 (by following the X's and V's). The A-pair is both even digits and the D-pair is both odd digits. It doesn't tell us the parity of B and C pairs though, which would crack the puzzle.
I'm actually proud of the fact that I had a solid element of logic Simon never used. Each of the top boxes has a 10 domino and a 5 domino. If you label the other two dominos, it becomes a LOT easier to eliminate options. The problem is I still got stuck on the bottom two boxes outside dominos.
@@super_7710 I finally got it last night! There was some minor bifurcation followed by reverse engineering the logic XD But in the end I had an actual logic path through the whole puzzle. But yeah, those summed pairs I mentioned are almost always a low/high pair where 5 and one other digit are the weird ones.
I just solved this in 35 minutes but feel very much like I lucked into it. I was trying to determine the min/max of the rows and initially had 17-23 which quickly changed to 17-21. The 21 felt not that useful so looked at the 17 and determined it had to be at least 18. When determining it must be at least 18 I had almost half the board filled in my head as a hypothetical and was like wth might as well see if I can finish it from there. So I proved that that could be the answer but did nothing to prove the other ways were wrong. Definitely the strangest "solve" I've ever done but the genius of the actual logic is well over my head.
72:14, I spun my wheels for like 20 minutes, looked at the video for the logic that r1c3 had to be a 5 (just wasn't seeing it), then looked again for a little help with box 3 (guess I just had to try and see what my options were). I still felt a bit stuck at that point, got up to do something else, and realized that I was only thinking about the columns, not how they interacted with Rows 3 and 4. Sat back down and finished it in like 1 minute at that point.
The way Simon solves a puzzle is by spending a minute actually solving it and then spending two minutes explaining his reasoning so everyone else can follow. The way I solve a puzzle is by staring at the screen for ten minutes, then giving up and watching his two minute explanation
I tried to do this with algebra and spent 3 hours on it ^^; Managed to get the answer after by trying different values of B and F (cells named alphabetically, starting from A from the top left, left to right top to bottom) and checking to see if the solutions follow the rules
Great puzzle. I think an easier way to get into the lower half is to think about the low digits in the dominos r34c1 and r34c4. In this way you can see (like simon did), that r34c1 has to be a 5B domino. Because 5 cannot go with the high C it has to be in r3c1. Then it is not too complicated to remove the possibilities for low AB and C in the right domino, so it has to be lowD and highA.
a 4x4 in 1 hour. This will be a must watch
144 ^??°
If "digits per unit of time," how fast he figures out digits was the same for 9x9 grid, it would be five hours long video.
Myxo here! This was an unexpected feature :D I Love Sleeping came up with the idea and we worked on the puzzle together. Simon, there was indeed a path through without (almost) bifurcation! It's possible to determine what digits (from A,B,C,D,5) have to go in all the cells without determining the total first, and then using parity to resolve it. But I am very impressed with your ability to see that all early in your head.
Fascinating puzzle! Well done to you and your collaborator.
as a lover of Australian wildlife, I thank you for your service.😊
I'd really love to see this path! I hope Simon demonstrates it on the channel. Or maybe Myxo, you could post the path somewhere?
I loved the parity bit at the end. That was a beautiful way of clearing the puzzle. great work and I loved it.
@@RyanAtOptimism I'm not sure if this is the intention: But what I saw at that point in the puzzle was - once we know there is a 5B combo in r34c1, we can ask if it can go in r4c1. That would put the 5 in the bottom row with the purple C. Since r4c2 is in A, that would make row 1 and row 4 the same digits. That puts the 5 in row 3 and the 34/B in row 4. The next thing I noticed was that r4c2 could then simply not be a 2. That makes A 46, which just solves everything at that point.
Did anyone else start to get nervous when he didn’t remove his 1&9 for the example half way through
I may have started yelling at my screen.
When he asked if he had the 1 9 domino I got a bit nervous 😅
I was sure he was going to go with it then discover it broke the puzzle before realizing.
OOOH yeah. And it nearly bit him, too -- but fortunately he remembered just in time.
I know I was shouting at the screen
Simon if you think I am going to watch you solve a 4x4 sudoku in one hour instead of starting my homework, you are absolutely right.
Simon finding out that R1C4 can't be 6 using the most intricate logic when he could have considered that 16 < 17 is classic.
The same thing happened multiple times. "B can't be 2".... well, 5 + a V domino means that the purple C will always be the 7/8/9 in 17/18/19. So the moment 17 gets ruled out, C automatically becomes 8 or 9.
Shortly followed by backing in to ruling out 7 from r4c3 when he could have considered 5+5+7 < 18 or 19.
He has two modes, incredibly picking the most incredible and beautiful logic built into the puzzle, and missing the most obvious answers and getting there anyway by sheer power of brute mathing.
That's Simon for ya. If there's a more convoluted, difficult way... he's going to find another even more difficult one...
Simon: Its not biforcation if I did it in my head
Also Simon: fills 90% of grid in head
It is a small grid 😂
Meanwhile, every single proof by contradiction is bifurcation. The whole thing is completely subjective.
So according to Simon, if I write down 3+5=8 on a piece of paper it is a sum, but if I do the maths in my head then it isn't... maybe then it becomes a square root
@@sjm6280 the point is, as I said in my comment above, that acceptable bifurcation is a subjective thing. Most of us will say that if you prove something by contradiction it's ok, but if you randomly fill in number and explore aimlessly, it's not ok, even though if you encounter a contradiction, technically, it's a valid proof by contradiction. So, putting in an objective restriction such as "I'm allowed to bifurcate as much as my memory allows" is reasonable.
@@Ennar It is valid, but it gets funny when Simon says that "Bifurcation i would never ever do that it contradicts the fun in the puzzle and we might miss the setters magic" in every video he solves, and then ends up biforcation "in his head" in every puzzle. Simon just loved to make things alot more complicated than they are and its a beauty to watch. Just dont deny the biforcation. Embrace it.
Rules: 06:24
Let's Get Cracking: 07:25
Simon's time: 50m07s
Puzzle Solved: 57:32
What about this video's Top Tier Simarkisms?!
Three In the Corner: 2x (56:33, 56:36)
Maverick: 2x (06:53, 06:55)
Scooby-Doo: 1x (07:34)
And how about this video's Simarkisms?!
Sorry: 7x (14:18, 14:57, 15:33, 29:30, 30:53, 37:05, 39:40)
Obviously: 6x (02:30, 08:24, 17:59, 32:20, 33:01, 33:07)
Ah: 6x (01:39, 31:08, 37:53, 44:22, 48:02, 55:50)
Hang On: 5x (17:17, 28:05, 28:07, 38:30, 49:56)
Clever: 4x (39:20, 39:23, 55:30, 58:16)
Shouting: 4x (03:07, 04:17, 05:31, 06:08)
In Fact: 4x (35:31, 51:44, 52:34, 57:51)
Unique: 3x (00:23, 06:19, 06:39)
Goodness: 2x (48:42, 49:11)
Axiomatically: 2x (08:43, 52:54)
Beautiful: 2x (22:49, 33:49)
Fascinating: 2x (57:05, 57:45)
By Sudoku: 2x (09:56, 32:22)
Wow: 2x (50:37, 50:40)
Cake!: 2x (04:57, 06:11)
What on Earth: 1x (48:20)
Stuck: 1x (56:51)
Lovely: 1x (05:31)
Break the Puzzle: 1x (34:55)
Incredible: 1x (04:52)
Bizarre: 1x (00:19)
Surely: 1x (37:25)
What Does This Mean?: 1x (11:28)
Have a Think: 1x (20:15)
Pencil Mark/mark: 1x (11:52)
Most popular number(>9), digit and colour this video:
Ten (38 mentions)
Two (60 mentions)
Blue (4 mentions)
Antithesis Battles:
High (8) - Low (3)
Even (8) - Odd (4)
Row (49) - Column (38)
FAQ:
Q1: You missed something!
A1: That could very well be the case! Human speech can be hard to understand for computers like me! Point out the ones that I missed and maybe I'll learn!
Q2: Can you do this for another channel?
A2: I've been thinking about that and wrote some code to make that possible. Let me know which channel you think would be a good fit!
Wow, this is so cool! How long did it take you?
Missing “ah” at 23:34
@@kiwi5675 computer code (probably a long time to write the code though)
@@illv3 isn't that just part of the following "actually?"
Wow this is such a cool bot
Simon: "one thing I will not do is bifurcate!"
Simon 5 minutes in the future: "in the world where N is 18, we know that this is a 1-7 pair..."
"I think were gonna find n is 20, because.. it's such a beautiful number."
pure logic
I went down that rabbit hole to the extreme, sort of. I did switch to using a unique letter for each of the first two rows cells, which allowed me to get all the way to the end with the three unresolved dominos in columns 1, 2, and 4. Unfortunately I couldn't see a way to disambiguate it at this point and then I saw the logic Simon had used showing that a cell in row one and row two had to be both the same digit and yet it couldn't be the same from the logic, so I knew I'd eliminated a possibility somewhere. I was so close, though.
Loved this. A real feat to not only come up with a ruleset that can provide a 4x4 grid that takes an hour to solve, but then to actually execute the idea this well too. Brilliant work
16 cells? One hour? This must surely be the highest cell-to-minute ratio in a long time
Also interesting that for most videos it would be a cells/minute ratio, whereas for this it would be a minutes/cell ratio. 3.6+ at that. A 9x9 would need a 5 hour solve for that kind of ratio.
@@rgoyal107 you know that both cells/minute and minute/cells are valid ratios?
Does anybody remember the last time a 4x4 produced an hour-long video? It was April Fool's Day -- 2022 maybe? -- and it was Mark solving. Very, very funny.
@@EnnarYeah, my bad. I meant in most puzzles cells/minute would be a greater than 1 value. And in this puzzle minutes/cells would be greater than 1.
@@rgoyal107 that is true, yes :)
Normally when you're explaining the puzzles, I can follow along and understand but this one I was trying to understand but I was not getting it. This is the puzzle that broke me 😂
"It's not bifurcation if you fill in the entire grid in your head to realize it doesn't work out!" 🤣
Did it in 16 minutes. I have never been that much faster than Simon, now I feel proud of myself.
That's insane, this would take me many hours to solve (then I would give up)
19:08
Separate colours for each individual digit helped fill out most of the grid into sections, then slowly whittling down the options to find which digit had to missing broke open the finale.
Very nice idea to make such a small grod contain so much
This channel has been in my recommended for a while, and I thoroughly enjoyed this video now watching one of these videos! It's a great break from a lot of the fast paced content on youtube, and you really get to think while you are watching the video which is always good and a lot of videos don't give you time to think nowadays. Definitely going to be watching more of these!
Welcome to the club!
I'd have to call bifurcation on Simon, and I did essentially the same thing to disprove the 9 in r4c3, except that I can't visualize as much in my head. I just couldn't find the intended trick, so I settled for the solve using that bifurcation, and then watched hoping to see Simon show me the better way. I guess I'll upgrade my solve to successful since Simon's was pretty much the same. :)
solved in 18:24 - did a lot of coloring in the top two boxes until I narrowed the sum down to one of two numbers. I'm not sure what the best way is to proceed from there, as I ended up taking educated guesses for the lower half once I found the first digit. interesting puzzle
edit: I figured out the way to sort out the bottom half afterwards. the color of r3c2 can't be the same as r2c1 because otherwise row 3 uses the same colors as row 2 due to the X in row 2. this was probably the actual way to clear it up. cool
I was surprised that once Simon got a 5 he didn't immediately tell that D had to be odd in order to make the top two rows add up to an even number (which is two times N).
I was shouting that at my screen
Simon is the sort of guy who develops a new branch of algebra with irrational numbers to figure out that 2+2 = 4.
years of complex sudoku has taken its toll
I also found it interesting that he also did not notice that when he placed the 5 r1 had 3 numbers=10 and r2 had 2 numbers=10 so high A + low C = high C the combinations that work are 8+1=9 or 6+2=8
The 7 in r2c2 was also much easier with oddness/evenness logic. Could have saved him a lot of time.
Took me about 2 hours, but I got there in the end. Basically had no digits other than the 5s placed until the very end. Worked in terms of high/low colour pairs adding to 10 (e.g. dark blue + light blue = 10, dark blue > 5, light blue
When I was studying linear algebra, I once set up and solved a 15x15 matrix to determine the magic square (up to a multiplicative constant!). I'm almost tempted to do the same thing here.
i've been binging a few of those video, usually trying myself then watching a few minutes when i'm stuck to get the point i've been missing, but here i completely gave up
This took me ages because I made it too complicated for myself. Once I asked the question "how many ways can you make 10 in 3 digits" I went from stumped to solved in about a minute. This is a real good puzzle. Kind of kicking myself for taking so long to realize the incredible structure behind it.
More specifically, I had long since realized that X=18 or 19. R1C3 is a 5 (I forget how I came to that conclusion) so the top row is 10 + R1C4, making it 8 or 9. In addition R4C3 is also that same 8 or 9. I just didn't put together that as a result, this made rows 1 and 4 and columns 3 and 4 take up all the ways you can make 10 in three numbers, the two using a 5 were taken, and that R4C4 had to be the lone shared digit between the remaining ones. ie. a 1, which removed 9 as a possibility.
That is nice. I wish Simon had seen it. I was stuck at that point and didn't really find an elegant solution.
exactly, very elegant.
pink +10 forces 1 in the bottom corner, and solves the puzzle. Great puzzle, loved it.
OK that's really nice. I was scrolling the comments trying to find a satisfying solution for this step and finally found yours. Thanks a bunch for explaining it so well.
Now Simon has awesome brain power.
He causes those digits to flower.
But I've ne'er seen before
Someone solve four by four
And take anywhere close to an hour.
Seriously, though, Simon, that was a toughie. Great work!
So amazing, Simon. An amazing puzzle, and I could not look away as you were solving. On the topic of bifurcation: I believe that the definition that you (and Mark) seem to have settled on is "if I can see it in my mind then it is not bifurcation." This is a bit hard on some of us who are not as good at visualizing things or running through maths and other matters in our minds. I am so very visual that I sometimes have to actually see something - not just see it in my mind - in order to determine its accuracy. Nonetheless, I think that it is a valid feature of pure logic to use if-then to eliminate options. That does not make things bifurcation, in my view, no matter if you write it down or do it in your mind. Otherwise I would have to admit that every single solve of a sudoku or other "logic" puzzle is bifurcation for me. If-then is legit, I believe, and requires no apology. (I am pleased that most of the comments I have read have not scolded you for bifurcation or even questioned it - well done, sudoku world.) Thanks, Simon, a fascinating hour spent in your company yet again.
When Simon pencils the D into box 1 (around 19:00), you can immediately rule out 20 as the total based on the letters in box 1. No matter how much Simon would like it to be 20. :)
I figured it couldn’t be 20 on the understanding that the rules implied the numbers 1-9 would appear, including 5. So once you know every box has an X that’s over
But that interpretation of the rules turned out to be wrong
Isn't 20 immediately ruled out with the knowledge that the 2 V's must be distinct? The X in C2 implies that R1C1 and R2C1 must sum to 10 (as all the boxes, rows, and columns sum to the same number), and then R1C2 and R2C2 can't sum to 10 because of the X in R2 and the fact that the V's are distinct.
I think Simon overlooked that he had one X in every box, and that that meant that the two other digits would sum to the same total (either 8 or 9) in every box. It would have eliminated 5 from box 4 pretty early (since it would repeat column 3 to get to the right total), and I think you could deduce that the 8/9-total there had to be the same as the one in box 1.
Yes, that's how I did it. There were pairs of digits that added to the "outie" (n-10). I gave each digit its own colour. Knowing the pairs that added to 10 and the pairs that added to the outie, I was able to solve it pretty quickly once the 5 was found in row 1.
The step that D in r1c2 isn't 2 (or 4) has been on the board for about (edit:) 15 minutes, since r1c3 = 5 is deduced :)
Simon even comes close with observing that the sum of the top two rows must be even, but then he phrases it in a very convoluted way. Instead, we can simply observe that the top two rows are AA+BB+CC+D+5, and as each AA, BB, CC is 10 which is even, D+5 must be even.
yessss this!
At 48:49 when you have the 345 in c1r4, you can rule out 5 because 24 is then canceled due to then 3 numbers in the 4th and first row being the same. Causing a repeat in rows since the rows have to be the same value the 4th number in both rows would also have to be the same. Which if the 5 was used in c1r4, then you get 5 plus 6 plus 8 or 9. This breaks the row from not being able to be 18 or 19. This gives you the 5 in c1r3. Always love watching you solve these.
You're a rockstar, I enjoy watching you solve these increasingly vexing puzzles.. you had eliminated the 1+9 from column 3 in the bottom right box, but didn't try the 2+8 that would have solved the remainder of the puzzle until you'd done some rather impressive brute forcing in boxes 1 and 3... but still. bravo! Keep making these, and we will keep watching them and throwing popcorn at our screens.
I solved it!!! In... 210 minutes lol. I must admit, listening to you solve puzzles while handling tasks around the house is very different from pausing at the start and solving it on my own.
Completely fascinating and painstaking. Thanks for goading me into bizarre puzzles I have no background with, my brain is thoroughly exercised now hahaha ❤
Simon really has a knack for marking cells in the most convoluted way. Sp far (20:55) I think we have blue and pink b/c. and the orange and white A, B and D.
I didn't watch this solve, but I did color the cells to indicate they were specific digits AND I used different colored lines to indicate that the two cells the line joined summed to the same value as any other two cells joined by that line. And pencil marked, of course.
Usually in this scenario I make the V digits A, B, C and D, and their counterparts on the Xs are E, F, G and H.
This is probably the first puzzle ive watched you solve where i was completely lost and after wwtching someone solve it, theres no way i could now do it
46:39 i dont know how to explain it well but this the moment i think i saw it. every row and column need to have 2 odd and 2 even digits. look at row 1. because there is a 5 and a V there is at least a minimum of 2 odd numbers in each row, column and box. now look at row 2. because simon proved there should be 6 in box 1 row 2 and the hidden V in box 2 row 2, there have to be a minimum of 2 even numbers in each row, column and box. so the possibilities were 18 and 19. 2 even digits and 2 odd digits can only sum to 18. therefore you know r2c2 must be a 7 because otherwise you would have 3 even digits in row 2 which cannot happen because of row 1 having a maximum of 2 even digits. i hope i am correct with this
My first try of a solve, worked out heaps of different logic, got stuck, watched a bit of the video up till you proved the 5 then finished it off!! Love this channel!!
And my first 3 in the corner was so exiting!🎉
18:38 but I'm not entirely sure how I did that. Running through the numbers needed to min and max the totals for each row, column and box helped a lot to see which totals could be achieved through a unique set of digits in the most ways. I've saved my replay, but now I need to figure out how to view it so I can see if I actually got through it without bifurcation (I'm highly suspect).
What is the taboo around bifurcation?
@@jacobcombs1106It's just not a very satisfying way to solve a puzzle. And, technically, if you bifurcate and guess correctly then you haven't proven uniqueness without going back and trying the other option and showing it fails, so if you think it is important to prove uniqueness then you have a bunch of extra work to do. I'm not sure how many people care about proving uniqueness, though.
@@thomasdalton1508 ah I see, what I've been doing and calling bifurcation isn't bifurcation then. Bifurcation in this context is exploring one possibility to the exclusion of others? I had been using the term to describe how if I realize there are only two possible combinations through logic using corner marks to show one and center marks to show the other until I find one breaks and then I can either control delete or shift delete erasing the disproven path. That would not meet the criteria for the taboo bifurcation since it is validly disproving the other path before committing to the correct one?
For example this puzzle once I'd proven in my mind logically the sum of the rows columns and boxes must be 18 I logically worked my way to the realization I needed a five somewhere in the top half and there are only 2 valid cells for 5. So I corner marked 1 as 5 and center marked the other as 5 and began filling in corner marks with valid possibilities to make everything 18 following the Xs and Vs with the 5 in that spot and center marking the valid combinations for the 5 in the other spot. Quickly I worked out there was no valid combination for putting the 5 in box 2 because I would be forced into a repeated combination by the Xs and Vs and I deleted the invalid path. Eventually I had the top buttoned up and found myself in a similar bifurcating path on the bottom where it was either a 1/5 or 1/3 pair in r4c1 and r4c4 which meant an opposing pair of pairs in r3c1 and r3c4 to reach my sums and I used center marks to show one and corner marks to show the other and began stripping away the options that required repeated combinations which eventually disproved one of the pair of pairs and left the other pair of pairs resolved. I then filled in my cells certain it was the only possible combination.
@@jacobcombs1106 No, that's bifurcation. If you guess incorrectly and get a contradiction, it's completely solid logically. It is only if you guess correctly that you have arguably missed something out, since you don't actually know that the other option wouldn't have also worked.
@@thomasdalton1508 ok so it is indeed the guessing and arbitrarily excluding one of the paths without actually disproving it that makes it taboo bifurcation, so I feel like my method where I work both options in parallel which will always result in properly logically excluding one of the paths doesn't fit the taboo as I will still always show uniqueness and not just got lucky.
It took me a while (about 70 mins), but my strategy was to colour the top of the grid (after making Simon’s similar deduction that they all had to be unique) and then seeing how that fell into the lower half. Really tricky puzzle for sure!!
I think I'm gonna give this a try first, but I fully plan on needing to watch the video to solve it!
Good luck!
As you can tell by the video length, it is not an easy puzzle.
Oh, this is such a clever puzzle! I'm impressed that it solves uniquely.
Thank you! :)
Simon saying he not gonna bifurcate, got me trying from 40:00. What I did is looking at the column 2&3 realising R1C2 must be odd. Then realising R2C2 can’t be 6, else it will create a sum more than 18
Solved in 23:46 and thus my first time commenting (after years of watching) because this is the most I've ever been faster than Simon by! Very fun puzzle!
Simon...says "I refuse to bifurcate" and then bifurcates his way through the rest of the puzzle.
He very clearly states that he only considers it bifurcation if he has to write it out to work it out. He doesn’t do that. Instead, he works it out in his head, then writes it out to show *us* why something doesn’t work.
Ah i definitely need to get back into this channel and this is the perfect way to start. will watch when im home
41:13 Another way to see this is if you add the x dominoes with the 5 and the V you get 30. And you know the sum has to be even.
I could never solve this but sometimes I see something a minute or so before simon and it makes me feel clever
Only discovered today - after a year. Great tiny puzzle.
38:49 I actually really love that you refuse to bifurcate and solve these awesome puzzles solely through common sense and maths.
I had n as 19 and worked out a possible grid. Got told it's wrong, so watch your vid till the 5 and then figured it out in 10 minutes after i knew n was 18. 37 min in total.
What a brilliant puzzle
When trying the puzzle I came up with a crazy reasoning why the total cannot be 16. Of course there would have been an easier way, but I still like it. There are only 8 different ways that 4 different digits sum to 16, Howevrr, only one of them includes a 9. But Since 9 would appear in a row and a column, there need to be two different ways that four digits, including 9 sum up to 16.
That's similar to my approach. I enumerated all the ways to make sums from 17 to 23 (excluding 20, for obvious reasons, and others because there weren't enough to be unique) and found the only sum that permitted enough combinations with elements satisfying all the (explicit and hidden) X/V sum rules. Once I got the sum, the rest fell out pretty quickly.
some insights on the general structure I gained after solving:
(1) if two vertically adjacent 2x2 boxes have a digit in common, they have a vertical domino in common, and both boxes are identical to the columns
(2) in our case, boxes 2 and 3 have a digit in common, so boxes 1 and 2 have no digits in common, otherwise row 1 or 2 is identical to column 3 or 4. Boxes 3 and 4 are disjoint for the same reason
(3) rows 1+2 and rows 3+4 are 8 disjoint digits out of 9 with the same sum, hence the same digits
(4) the vertical dominos in rows 1+2 also show up in rows 3+4
Solved it! Hooray! Usually I feel too intimidated to try these prior to watching the video. Attempted solving with x at a fairly high value, and felt it was just too hard very quickly. Dropped to the right x value by luck and pushed through quite fast with trial and error. Thanks for that :)
what saved me from the guessing and checking was i noticed the top right digit in the top left box had to be odd due to the fact that there were 3 other 10s and an odd in the top 2 boxes but the sum of all the digits in the top 2 boxes had to be even as it was 2x the amount each row/column is set to
This puzzle took me about 3 hours to solve, its crazy to watch the video and see how they solve it contrast to the way you solved it.
31:36 I spent a lot of time searching for sums that can be the same X + low, V + high, and X + high + something
I spent an hour proving various thing about what various cells could or couldn’t be but I never found the trick that you did to not row 1 column 3 was either a 5 or a complement. But I had already narrowed down the top right cell to not be 9 and had proven N was either 18, 19, 21, or 22 so it couldn’t be 20. Most of my deductions were from the relationship between r1 c2 and r2 c2. After ages of no more deductions I had to give up and watch your video, but when you made your deduction I immediately knew it was a 5 because I had already disproven N being 20 and thanks to all my other deductions the puzzle had no more resistance and everything fell into place. The pivotal thing for me would be knowing N was at least 18 and the top right cell couldn’t be 9 so once the 5 got placed the top row gave me N=18 and the top right cell was 8 which permeated across the grid.
Great solve and I had a wonderful time solving with help from your 1 deduction.
Loved the puzzle :D 10:49 for me; just a little confusion when I took colouring too far.
I'm glad you liked it! Well done for solving!
After getting the 5, it was all quite easy. There was no need to consider parity. The minimum value of N was 18 (not 17 as you suggested). This is because you'd assumed in C2 that 6 could go with 1, but the 6 would make blue 14, so orange would be 23, so the minimum is 10 + 6 + 2. Similarly, the maximum was 22. However, once you have the 5, in R1, the maximum comes down to 19. Therefore hi C is 89 and lo C is 12. This makes lo B 34, and hi B 67. Working this through, you end up with hi A also 67, so N=18, and the rest is just a matter of ensuring you don't repeat a set of digits.
This was quite a fiendish little beasty. I can't recall the specific deduction I used to get N, but it was quite simple, and related to the unique sets. Unfortunately, replaying my solve didn't include any pencil-marking which would reveal what I was thinking at the time. I did have 34 and 12 in the first two cells, rather than 24 and 13, and I got that by eliminating 3 from R1C2. There was some problem with R1C2=3 and R2C3=6. It may have been effectively the same restriction you spotted, although it certainly wasn't as convoluted.
After N = 18 it also immediately follows C = 8 from the top row, R1C2 = 1 (from the sum of top two rows, other fields sum to 35), and the whole upper left solution from the relationship R1C2 = 5 + R2C1 (the column and the square containing R2C2 differ by 5, so must the complement set of fields) resolving B and then the rest is trivial.
Simon: “I haven’t got a Scooby Doo”
Me: “Ruh roh”
Every. Time.
At 25:32 it would've been possible to deduce that D is not 2 or 4, since its complement must be the odd number not used in the first two rows.
4x4? I'll give it a try!
Saw the video length...
well, or maybe not 😅
"That's three in the corner..."
I literally said that as you did...So happy to see someone have the exact same thought as me sometime!
I havent seen someone else solve it my way. Once the beginning is done. So you determined all x,v and same numbers. You can determine that r1 amd r2 consists of 8 different digits. And the missing one is 5,6,7,8,9 which means the sume of r1+r2=[36,40] so the only options are 18,19,20. 20 is easily sort out because of b and c. And 19 follows quickly once the numbers are reduced to their posibility.
How to say that alle digets are different: r1c2 cant be r1c3 because then r2c2 must be the same as r1c4. Which makes the the same sum but box 1 is now the same as r1 :)
Simon's thought process is so amazing. I wish I could conclude logic as fast as he does.
65:45 - clever, had a lot of trouble!
I don't know where that "3 in the corner" joke came from but I it got me laughing so hard
very entertaining video as a whole. insane puzzle too
Writing this as I am watching the video:
One thing I noticed around 20:00 is that the possible options for N is 18, 19, 20 and 21.
My logic is:
Box1 + Box2 = 2N
Subtracting the two vertical X-sums (r1c1 r4c1 r1c4 r4c4) and the horizontal X-sum (r2c2 r2c3) leaves you with the squares r1c2 r1c3 and they sum to (2N - 3X) = (2N - 30).
Now I just tried all the different sum totals of the r1c2 r1c3 - pair, and using the logic that (2N - 30) = (r1c2 + r1c3):
15 pair (summing to 6) gives you N=18.
16 or 25 pair (summing to 7) gives you N=18.5 which isn't possible.
17 or 26 or 35 pair (summing to 8) gives you N = 19.
18 or 27 or 36 or 45 pair (summing to 9) gives you N=19.5 which isn't possible.
19 or 28 37 or 46 pair (summing to 10) gives you N=20.
29 or 38 or 47 pair (summing to 11) gives you N=20.5 which isn't possible.
39 or 48 pair (summing to 12) gives you N=21.
49 pair (summing to 13) gives you N=21.5 which isn't possible.
So the possible options for N is 18, 19, 20 and 21. And we also can know that these two square sum to an even total.
So at 24:46 when you got the 5, you can eliminate the options of r1c2 being 2 or 4.
Great solve - I ended up giving up and enjoyed following along. Crazy that a 4x4 can be this hard. To disambiguate things at 51:15 in a more straightforward fashion, (I think):
- r3c1 must be 5 or low B (as by column 3, n = 15 + low B)
- row 2 has the digits {high A, low B, high B, low C}
- if r3c1 is low B, then r3c2 must be high A (low A leaves r3 too low, breaking the puzzle by forcing box 4 to be way too high), so then row 3 has at least the digits {high A, high B, low C} - so 3 digits overlap between r2 and r3 meaning the 4th must as well, breaking the puzzle
So then r3c1 cannot be low B and must be 5, also putting low B in r4c1, hopefully allowing some subsequent deductions.
Here is a slightly simpler way to proceed at 50:30
Let m be the 1s digit of the mystery total. We see from row 2 that the high A digit plus the low C digit equals m.
Suppose for a contradiction r3c2 is the high A digit. This is next to the low C digit so the other two cells in the row add to 10. This places 5 in r4c1. But now row 4 contains 5, the low A digit, and the high C digit, making it the same as row 1.
At 17:30 he had just gotten the top left domino sums to 10. There is a very useful algebra that can be done then:
Call R1C3 as A and call R2C1 as B. The top two rows total 10 + 10 + 5 + A + B. Thus, 2N = 25 + A + B.
Another way of looking at the top two rows is as 10 + 10 +10 + 5 + A - B (we counted the B twice but not the A at all). Thus, 2N = 35 + A - B.
Those equations give B = 5 and 2T - 30 = A which gives T < 20 and gives T = 19 with A = 8 OR T = 18 with A =6. (I think I had already eliminated 17 as T at that point).
Took me about 55 minutes. I used the fact that the remaining 3 cells of the bottom row added up to 10 a little earlier. This basically left 3 different permutations that solved themselves when you wrote all the numbers in
Wow that was an impressive solve of a an impressive puzzle.
It is a Semimagic Square. The positive diagonal adds up to 18 too. But the negative diagonal only sums to 14.
Amazing solve of an amazing puzzle.
In the end, instead of eliminating 4 from R3C1 and R4C1, it is easier to put 5 in R3C1 by eliminating it from R4C1. A 5 in R4C1 would force Row 1 and 4 to have the same set of digits and is therefore impossible. From there it gets easier again once you see R3C2 can't be 68 as it would force the placed 5 in R3C1 to be part of a 10 sum (because of the X and the digits in R2). The rest just falls in place from there.
Let's use upper and lower case to denote big and small.
At 25:11, now that we have the 5, the key insight that solves the puzzle is algebra. From column 3 (or row 1) we have n=C+10 (because of the v and the 5), then row 2 gives A+c=C. Likewise, column 2 gives d+B=C. This rules out C=7 because there has to be two distinct big small sums that add to C. The rest is easy.
first puzzle we have tried and managed to complete in under 24 minutes!
I am thankful 9 made a solid appearance in the video even though it did absolutely nothing for the solve.
48:30 after deducing that 5 in c1, you could not put 5 in the corner. Because then you can't put A-high in r4 (because total will be bigger than 19), and if put A-low, r4 will repeat r1.
So r3c1 is 5, r4c1 is B-low.
00:18:07. Holy shit I can't believe I just did that... so i treated it as a polarity puzzle when I saw the Vs and Xs like oh gotta have two highs and 2 lows in order to balance it out. Surely that means it's all 20 right? Realized nope to do that would automatically force every row and column practically to be the same. 19? No same problem. 18? Very quickly resolved the only apparent way to make the top two boxes work for 18, then worked out the only two possible combinations to keep the bottom boxes also 18 with every row and column 18 and realized one of those combinations had repeat rows but the other didn't and bing bango bongo bobs your uncle the puzzle was done.
228% faster than Simon this is a unicorn moment for me.
This was one of the first puzzles featured on this channel that I actually tried solving myself (or, more precisely, the first after the disaster of trying some when I first found Cracking the Cryptic and discovered that they were, in fact, much harder than Simon made them seem).
Watching this video upon giving the puzzle up is really funny to me, because I placed three digits in the first couple minutes (using logic that I can’t actually recall), and then proceeded to make very little progress for another couple minutes, subsequently staring at it without learning anything for about half an hour. I go to watch Simon’s solve, and he reveals about three dynamics I had not noticed in the first five minutes, but without placing a digit quickly at all.
I haven’t finished watching this, so I’m not even convinced the logic I used to place the digits was at all sound, but if it was then the comparison between my attempted solve and Simon’s successful one would be quite funny.
Edit: yep. Idk what kind of oversight happened at the beginnings of my solve, but my placed digits were in fact wrong. I think it was so early into the puzzle that I hadn’t completely understood the rules and their implications yet.
We can easily use Excel to reduce a lot of repetitive calculations.
1234 represents rows, ABCD represents columns
It can be simply known that C3 can only be 1-4
We assume that the sum of the rows and columns is S, and we can calculate all the numbers in rows 12 and column C.
S is a moderate number around 17-21
By using quick calculations in Excel, it can be found that 21 is too large, while 20 has many duplicate numbers.
I initially thought it was 19, but I couldn't get the correct result. After switching to 18, the remaining numbers were quickly calculated
It took half an hour to calculate S=19 by myself. and 3mins to calculate S=18 by Excel XD
Spencer here and me and aisha listened to the birthday message today and both had a good laugh its awesome to get a little ahout out her and my daughter made brownies for me instead of chocolate cake and my favorite dinner an awesome birthday for sure
You can also exclude r2c1 to be 7 by summing rows 1 and 2. Since the sum of both rows' sums should be even (2*N) and the sum is X+X+5+V+r2c1, i.e. 30+r2c1, it can not be 7 (since 7 is odd in most of the times).
Okay, I'm commenting before watching with a solve-time of 28min. My method of solving: Brute-Force R1C2 & R2C2 and play through every scenario provided by those two, proving 2, 3 & 4 can't go into R1C2 and from there on out it was more or less just a chain-reaction to solve^^
When we do “WHAT-IF” analysis, isn’t it the bifurcation anyway? It seems that it’s an essential part of sudoku
Another similar but shorter way to get through the end - you can determine that r4c2 can’t be the low A. If it is, then r4c1+4 adds to the same as r1c2+3 but can’t be the same digits. If low A is 4, that sum is 6 but can’t be 1,5 or 2,4. If low A is 2, then that sum is 8 but can’t be 3,5 or 2,6. That leaves 1,7 but one of those must pair with 15 (the rest of column 1) to get to either 18 or 19.
This stumped me so hard, I had to watch the video.
I am so happy I did this! Really amazing puzzle.. My approach was to use letters for all digits and then solve it algebraically.
I got stuck coz I assumed all 9 digits must be used and I couldnt see a way to do that while 8 digits gave an elegant solution. So I came back to watch the solve and realized that was an artificial constraint in my head..
Good job for solving! I'm glad you enjoyed :)
"How hard can a 4x4 be?"
*one hour and a flood of sweat and tears later*
I mean, depending on which definition you have of "Normal sudoku rules", in this puzzle normal sudoku rules may apply. (they apply in the version fill every cell with digits 1-9 so that no number repeats in any box, row or column)
When we created this puzzle, we used the words 'non-standart sudoku rules apply'. But as Simon said, the amount of sudoku-logic used in this grid is about none, so I can understand why they chose to word it like that.
@@I_Love_Sleeping-Leeor Yeah, I was referring to the rules on screen, it's understandable Simon reformulated the rules that way for entertaining, editorial and clarity reasons, but still what he was saying about not following standard sudoku techniques even if it can have the same formulation of (some) rules is amusing to me
25:11 for me, very cute puzzle.
"How hard could a 4x4 sudoko be?"
Me: Checks time stamp and sees its an hour for Simon. "Good lord. Well, definitely gotta watch this."
Solved in 30:04, but I was only able to deduce the row/column/box sum to 2 options, started guessing and a few digits later I was done. At least I found one correct digit without guessing 🙂
AT 40:52 another way to see why R2C1 can't be 7 is the following:
Because the sum of the two upper rows is 2N it must be even, but this sum consists of 3 X-pairs (totaling 30) plus the 5 in R1C3 and the number in R1C2. To make this total even R1C2 must be odd, so cannot be 2. Thus R1C1 cannot be 3 and R2C1 cannot be 7 (by following the X's and V's). The A-pair is both even digits and the D-pair is both odd digits. It doesn't tell us the parity of B and C pairs though, which would crack the puzzle.
I'm actually proud of the fact that I had a solid element of logic Simon never used. Each of the top boxes has a 10 domino and a 5 domino. If you label the other two dominos, it becomes a LOT easier to eliminate options. The problem is I still got stuck on the bottom two boxes outside dominos.
I think you gotta do some magic square stuff where lows are paired with highs. I couldn't solve it myself.
@@super_7710 I finally got it last night! There was some minor bifurcation followed by reverse engineering the logic XD But in the end I had an actual logic path through the whole puzzle. But yeah, those summed pairs I mentioned are almost always a low/high pair where 5 and one other digit are the weird ones.
I just solved this in 35 minutes but feel very much like I lucked into it. I was trying to determine the min/max of the rows and initially had 17-23 which quickly changed to 17-21. The 21 felt not that useful so looked at the 17 and determined it had to be at least 18. When determining it must be at least 18 I had almost half the board filled in my head as a hypothetical and was like wth might as well see if I can finish it from there. So I proved that that could be the answer but did nothing to prove the other ways were wrong. Definitely the strangest "solve" I've ever done but the genius of the actual logic is well over my head.
72:14, I spun my wheels for like 20 minutes, looked at the video for the logic that r1c3 had to be a 5 (just wasn't seeing it), then looked again for a little help with box 3 (guess I just had to try and see what my options were). I still felt a bit stuck at that point, got up to do something else, and realized that I was only thinking about the columns, not how they interacted with Rows 3 and 4. Sat back down and finished it in like 1 minute at that point.
The way Simon solves a puzzle is by spending a minute actually solving it and then spending two minutes explaining his reasoning so everyone else can follow.
The way I solve a puzzle is by staring at the screen for ten minutes, then giving up and watching his two minute explanation
I tried to do this with algebra and spent 3 hours on it ^^; Managed to get the answer after by trying different values of B and F (cells named alphabetically, starting from A from the top left, left to right top to bottom) and checking to see if the solutions follow the rules
Great puzzle. I think an easier way to get into the lower half is to think about the low digits in the dominos r34c1 and r34c4. In this way you can see (like simon did), that r34c1 has to be a 5B domino. Because 5 cannot go with the high C it has to be in r3c1. Then it is not too complicated to remove the possibilities for low AB and C in the right domino, so it has to be lowD and highA.