Thank you for the feature - it's an honour to be the balancing act to Jay Dyer!! The intention is to make a 9x9 version of this, on which the great Marty Sears himself intends to collaborate, so this was meant as a proof of concept puzzle just to check that something like this is in fact possible, but it ended up having lots of fascinating logic of its own so I'm glad I published it! And don't worry, I'm still fitting in some time for looking after my new family in between designing sudokus ;)

Simon noted at the start that "raw" is "war" reversed, more to the point the whole string "RAWFOGOFWAR" is a palindrome

"Ooh, a 24 minute Simon puzzle, I can do this one!" "No, I can't."

Surprising to me that Simon never deduced that the odd number in the palindrome had to be in the center.

Also, you got all of our names right (Stuart, Alannah and Orla) - we're all very flattered! :D

I approached this quite differently from Simon. I didn't immediately notice that even-length palindromes are impossible. Instead, I first noticed that there are at least 3 different digits on the palindrome, and this meant that at least one odd digit is there. And an odd digit can only be there (due to the self-counting rule) if it's in the middle - which means there could only be one odd digit there, together with two even ones. Trying to fit palindromes with 241s, it was just not possible without getting two 4s in a box or on a row/column. So I knew the 1 couldn't be on the palindrome, and could place it in column 3. After that I resorted to coloring.

Finding the 1 was approachable and quick. Then the middle digit of the palindrome has to be a 3. So plonk a 3 on the middle of the palindrome which is revealed after pacing the 1, and it solves quite quickly and nicely. Fun puzzle!

Wow! First time I’ve ever beaten Simon’s time in 6months of discovering this channel. 9:57 not spectacular but I’m normally double his times. Like many others went down route of odd number at centre with 2 and 4. 1 just not possible by trial and error. Rest falls into logic and sudoku. Lovey it!

Simon can explain and solve ultra difficult sudokus for which we can’t even figure out break in within an hour, and extremely easy ones in 15’50” minutes! The explanation and solve for this barely needs 5 minutes!

One interesting way to see the palindrome cannot be of length 7: Since there is a 4 on it, it must have a cell in every row and column. This is 8 requirements. The first cell of the palindrome takes 2 of these (it has a row and a column), and each cell along it can only take at most one of the remaining 6, since they are orthogonally connected. Thus there must be at least 7 cells along the palindrome. However, to keep getting a new row or column with each move requires going between two opposite corners (since doubling back either left-right or up-down would by necessity add a cell that doesn't get a new row or column in it), and such a path cannot visit all 4 quadrants (if it started at r1c1, then if it enters box 2, it can never get to box 3, and vice-versa; similar reasoning from any other corner), so it cannot be length 7, and is therefore length 9 (since it cannot be even)

Very interesting how different Simon approaches such a small puzzle compared to me. I was like: so I can put one 1 and/or three 3s on a line. Well for palindrome this only works with a middle digit, so there is either no 1 or no 3 on the line and therefore two 2s and four 4s x)

7 minutes to solve this one!!! So proud of myself! Thank you Simon for being my teacher for the last 3 years :D

Usually it takes me a while to figure out the core of the puzzle but this one just clicked! Got it in 1:54.

Usually I'm terrible at these, but I got it in 2:49! The only possible 1-2-4 palindrome is 4241424, but the three consecutive digits in column 3 prevent it. Therefore, 1 is off the palindrome, and the rest falls into place.

A simpler way if thinking of the palindrome is that it can only have one odd digit, because the odd digit appears necessarily at the middle of the palindrome (it's the only way of repeating it an odd number of times). And we know the palindrome has an odd digit because of column 3.

This was a very fun little puzzle! I stared at it for 19 minutes and then got a single digit and solved it within the minute. A sweet delight :)

5:45. The palindrome could either include 1 or 3 but not both, and it couldn't be 1 because the line would only be 7 characters long and there was no way to get four 4's on it, so the 1 had to be in R1C3, and everything else followed quite nicely.

7:07 For me, now the palindrome time fits with the puzzle

As I tried to place 1,2,3,4, The break in stumped me to the core. But I solved this great riddle By finding the middle And ended with quite a good score! (Leaves a partridge in a pear tree)

I watched 5 minutes of Simon's solve and wasn't following so I skimmed the comments to try and see how on earth to get started and then decided to give it a shot with barely a hint from the comments. Solved it in 13:01!! I'm very proud of that considering I'm still new to variant sudoku

Love the long videos! Watching Simon struggle for two plus hours is time well spent!

The moment I realized that the palindrome was extremely restricted, I got all the 1's and solved the puzzle in 7:38. Very lovely puzzle.

7 minute solve. It took me a bit to realize 1 couldn't be the central digit of the palindrome because your line would never be long enough to get into all 4 boxes... That was a pleasure to think through.

Went through the whole Crux video without being bored once! I'd love a longer format now and then as a way of introducing a bit of variety, and as a shoutout to the brilliant minds behind the harder puzzles. Great job on the Jay Dyer solve!

Solved in 31:31. I was stupid at the beginning, needed forever to understand that the palindrome has to have an odd length (so 7 or 9). Then I somehow found a path and pencilmarked the whole thing so I had a solution but I didn't know if it was the only one. So I kind of semi-guessed there! 😍😍😍😍

I decided to have a bit more fun with this puzzle by not placing a single digit until I solved the whole thing, completely bypassing the help from removing fog of war. It made it really fun and was very satisfying placing all the digits at the end revealing it correctly underneath the fog. Really nice. I finished in 69 minutes.

Finished in 12:40. Interesting constraints which led to a fun break-in. Obviously filling out the numbers is trivial once you figure out where the palindrome line can be. I dunno where Simon started, but I realized that the 4 had to be on the palindrome line which forces each row, column and box to have part of the line. And either the 1 or the 3 but not both can be on the palindrome line. So, I focused on the top row and tried to see which palindrome lines could enter the top row and found that the top right corner was the only possible square the palindrome line could enter. This made it trivial to know where the 1 and 3 are in column 3 since the center cannot be the only square that's at first visible since the opposite sides of it have to be different numbers and therefore we know that the 3 had to be at the very bottom since it's the only center possible and the line is of length 9. Fun puzzle!

Release as much long video's as you can! I love them! For example: I've rewachted Gliperal's puzzle from December 2022 two times. I love those so much!

2:16 for me. Something about it just clicked and all the sudden I was done! Great little puzzle

Dyer's puzzle broke my brain. This is what I needed to repair it. Thanks!

Solved in only 10:28, i figured there was not much vombinations of digits that could be the palindrome so figured that out on paper first, then started overlaying on the grid and when i found a path for the palindrome that suited i could start filling in. Lovely puzzle! Reading about a 9x9 posiible puzzle coming up i guess i will need more logic for that one, so now to watch Simon solve this one.

Loved it! My time: 19:12. I was able to figure out the entire line without placing a single digit, then I just filled it in.

This was fun. I worked out the length and shape of the palindrome and penciled in all the numbers before clearing any of the fog.

Very proud of my 3:47 time! What a cute little puzzle!

1:54* Definitely need the asterisk, haha! Looked at the puzzle for a couple of seconds before my walk, then thought about it the whole time. I figured out either 1 or 3 couldn't be on the line (only one odd digit allowed), and then did a bit of thinking of how many cells we'd need to hit to calculate that the 1 had to be off the line. Put a 1 in column 3, saw I was correct, and continued on my walk. The rest of my walk, I thought about how to get four 4s on the line, that the line on the left side couldn't hit Row 1, and kinda of figured the rest of it out. Took like 30 seconds to type it all in when I was done walking. My walk was 50 minutes long, haha.

13:49 for me. I didn't catch things like the length of the palindrome, but I still worked it out thinking about where 1 could go if possible. Quite pleasant!

2:37 for me. Lovely little GAS Fog of war puzzle!

Thank you for realeasing the very long solve of 2h42mins. I must say myself that I love long and especially very long solves a lot. For a while I was only watching solves that are over 40-50mins but love seeing 2h+ solves.

Congratulations on the feature and the new baby, Pedaling Pianist. It is a beautiful puzzle.

4:52 here. Fun and approachable. Definitely met the proof of concept it set out to do.

Yes!!! I LOVE your longer solve videos. You must’ve understand that they allow me to watch at my leisure while enjoying super intense logic. It also allows me to bank future videos which I’ll be able to essentially binge watch, also at my lose.

Took me 31:29, I'm not convinced it was an airtight solve, but after spending 30 minutes drawing different paths and seeing a recurring pattern for where some of the digits would go, I was pretty convinced and felt vindicated when I placed the correct digit in the only open space and revealed the map was exactly how I had drawn it.

3:02 for me - I quickly deduced that it couldn't be 7 long, so must me 9 long with 234, and then could fill in the 1, expose fog, determine the middle, and run with it.

Solved in 9:44 and very happy with that! Went a little something like: 1) There must be an odd digit on the palindrome bc of column 3. 2) To appear on the palindrome an odd number of times (i.e. 'itself' number of times) the digit must be the middle cell on the palindrome. 3) Therefore, only one of the odd digits can appear on the palindrome and r1c3 cannot be palindrome. 4) Satisfied myself that a 7-cell palindrome was never big enough to pick up all four 4s. 5) Palindrome contains 2, 3, 4 (to make 9 cells, not 7) so r1c3 must be a 1. 6) 5th cell along palindrome must be a 3 to place the odd digit in the middle of the string. 7) Some simple '1 cannot be on the palindrome' stuff to place the remaining 1s and 3s. 8) Wasn't sure how to resolve the 24 pairs until I highlighted one 'set' and realised it placed the digit on the palindrome 4 times and that finished the puzzle. I probably missed something at the start where my logic was possibly not actually thorough enough to completely rule out all alternatives, but it felt pretty good.

I'm absurdly pleased that your solve time was, itself, a palindrome

"Raw fog of war" backwards reads "Raw fog of war" Noice👌

I love the longer videos because I can sit and crochet while watching even though sometimes I get distracted by yelling that there's something obvious you can do

4:01 for me ^^ i seem to have landed on all the right questions pretty quickly =)

I thought this one would be a good video when I completed it yesterday- I also knew it would be a fast solve but so glad you guys did it. As it was both very easy and complex at the same time. Loved it when I solved it!

3 had to be in the middle of the palindrome. It makes the solve even faster.

YES. MORE LONG SOLVES. I’d happily sit and watch a 5 hour epic

I watched the whole of the long video. It was awesome.

You and I solved this puzzle almost exactly backwards. The second digit I put in was r4c3=3, which was your final entry.

This is packing a lot of wonderful logic into a tiny little package. 😊

solved in 14:55. this puzzle's logic was some pair of clever and humorous, thinking about how harshly the circles rule constrains the size of palindrome lines and makes only two possible sizes for the visible palindrome line, which only one works.

Gee these setters are clever! Such a fun, clever and approachable puzzle. Loved it. 8:31 for me

Please do more of these very long solves - maybe once every two weeks?

I really liked the longer video! Heck, even if it was 8 hours, I would still watch it. You are very entertaining and quick with your reasoning. Why wouldn't we want longer videos?

I loved this puzzle. It shows they don't have to be hard to be interesting, although often the harder puzzles are also the interesting ones. I also liked that Simon's logic was different to mine, which again suggests it is interesting. For the record mine was: There are at least 3 different digits. There can be only one odd digit since it must occupy the central cell of the palindrome. Therefore it has 2 and 4. Since it has all four 4s it visits each edge which requires three horizontal and three vertical moves. Together with the starting cell this gives seven cells as a minimum. But it must also visit every box we must double back either horizontally or vertically. For exampe if we visit boxes in the order 1,2,4,3 we must double back horizontally. Therefore the minimum is nine cells and the digits are 2, 3 and 4. Since this logic is based on visiting all the 4s the start and end of the palindrome is 4. The central cell is 3. R1c3 is 1... And then it resolves beautifully. Now to watch Crux which stumped me.

00:06:14 Took me a bit to figure out the trick and realize you could have both a 1 and 3 on. Then a bit of trying to figure out "ok, how can I manipulate this line" and realize the final bit to then go for a relatively quick solve.

I really like the long 2.5+ hour solves but only have time to watch them every now and again. Maybe once a month or scattered upload long solves would be incredible!

So cool! Got it in 12 mins - frustrated for a while, and then it clicked. Now to see if I had the same solve path as Simon...

The longer videos are definitely my favorites, haven’t watched the new one yet but will do tonight. But please keep posting long videos ^^

my first deduction was that 1 and 3 could only go in the middle of a palindrome and then realized that we already had 3 given cells in one row. meaning the last one could only be 1 or 3 and the center of the line had to be one of the given cells. you can't put the center in the middle cells since the 2 adjacent ones see each other breaking the palindrome/sudoku rule so it's either top or bottom. if you try to make the top or bottom cell center containing a 1 (length 7 palindrome) you quickly realize that there is no possible working constellation since only the row/column/box with the center can have 3 cells you either run out of room to fill in the line or have double numbers in a row/column. knowing that 3 is the center gives you a 1 at the top, showing you how the line runs to the corner. just count to the 5th cell (bottom given one) to enter a 3 and solve from there is easy because 1 is never on the line, 2 only twice (which has to be in box 2 and 3) and rest is simple sudoku

I love these longer videos. Thank you for publishing it

Fun Puzzle I forgot about the nature of the line and needed to remember about the number 1.

I got 18:32! After I realized why 4241424 was impossible, the first digit came quickly, and the rest of the puzzle unraveled. Very fun

Very happy with this puzzle, only needed to take o e hint from simon about the length needing to be 9 cells long

When I saw that Simon's video was a 4x4 fog puzzle, I immediately tried it myself, and got it in under 10 minutes. It was fun. I saw quickly that 1 couldn't be on the 3 palindrome cells we already knew about (the visible one and its neighbors), because it would have to be at the center of the line (I forget how I ruled out putting it in r4c3), and it wasn't too hard from there. I didn't use Simon's counting argument, though -- what I did was rather less elegant.

Break in logic summary: I think the break in of “The orthogonal palindrome must have an odd length” is key. To get that with 4 digits, you need 1 odd digit & 0-2 even digits. This one clearly has 3 digits, so both evens, one odd. Figuring out line length is the second necessary realization for break in.

I do want to see longer videos. Please! Here is the thing. I don't have enough time to solve every puzzle I see here. So my filter for them is: I pick only puzzles from 1 hour+ videos and try to solve them faster than you. Sometimes I even succeed. One out of twenty maybe :)

Oh, that was interesting. 2:13, plus a minute or two of thinking about it while Simon was explaining the rules and observing that his 32123 sample palindrome didn't work, plus a clue from skimming comments about how the center needed to be an odd digit (which got me on the right thinking path). My logical path was figuring out that the palindrome could only contain _one_ odd digit, and that the visible one needed to contain three digits, and then convincing myself that a palindrome with 4, 2, and 1 wasn't possible because the 4s couldn't be far enough away. After that it was quite straightforward; the remaining notable logical trick was noticing that a palindrome with 4, 3, and 2 that has 3s at the ends is equivalent to a palindrome 4, 2, and 1 plus a couple of extra cells, so the same logic that rules out the 4, 2, and 1 palindrome also rules that out. Likewise if the 2s are at the ends; that's even more constrained. So it needs 4s at the ends.

20:49 for me. Nice puzzle. Amazing what can be done with only 4x4

I started by asking where the odd numbers go. I knew the middle of the palindrome had to be an odd number, and the other odd had to be excluded. That led mew to find the middle of the palindrome and I was able to work from there. The break in was heck until I figured that out, but after that, it was a lot of fun.

At 20:10 a bit of logic I kept thinking you'd jump on any moment is that the middle of the odd length palindrome can only be an odd number, so it has to be 3. And now that we've seen 5 steps of the 9 length palindrome from one of the ends, we know r4c3 is the center space and thus a 3.

Very nice puzzle, one of the very few I managed to solve faster the simon. My time is 12:06. And my way to solve it was thinking about the possible combinations of the line. I was thinking that there couldn't be 1 and 3 cause I would have to have a middle digit and they would collide and so I figured out it was 3 and from there it was an easy solve.

I'm firmly in the "more than happy to watch long videos" camp, but I got to the channel late yesterday and so I watched Crux rather than yesterday's video. Thanks for the short video today, buys me time to go back and catch up on yesterday's. XD

I do love your long videos!! Even though I cannot solve those, watching you logic your way through them is amazing to watch!

Finished at 02:34, pretty straight forward but interesting logic!

Cool that I think this puzzle may be solvable without the fog ever clearing, super cool puzzle!!

I stand by my comment haha! Stu is probably the loveliest evil genius you will ever meet. Loved this puzzle, really happy to see it featured, and yes we are looking forward to hopefully making a full size version of this idea soon :D

I admit, the longer a video is the harder it is for me to squeeze it into my day. I have built up a backlog of several months' worth of videos, which is...daunting. The 40 minute solves hit the sweet spot for me, but as I say that I have rarely if ever achieved such a solve time myself.

I loved the long video havnt quite finished it yet but wouldnt complain about getting more :p

Love the long videos, even 2h plus

This took me almost an hour and I'm still not sure if all my deductions were rock solid. I got myself to a point where I could only see one way forward and had to hope I hadn't missed something important. Luckily (or perhaps skillfully) I was corrected and finished the puzzle.

Solved it.. but after determining the palindrome had 234 on it, but no 1, I didn't think to place the 1. lol Instead I found a solution before entering a single digit, and then replaced the pencil marks with filled in digits all at once. Beautiful design to be able to solve it from knowing only one cell.

11m41s. Very fun! Reasoning about where this palindromic line could go was delightful

I started this this morning, thought I figured out a logical deduction to the first digit, entered it, and it was wrong🙃 Haven’t had a chance to try it again today but I’ll give it another bath now before I watch the solve.

I didn’t even know you’d get the additional palindrome clues. I solved the palindrome route by going through all options. Instead of putting in a 1 at the top. 28 min.

Thank you for the shout out, it means a lot to Karamella and me :)

very fun! Quick solve after you place the 1 in R1C3.

got it in 3:10. Short and sweet but surprisingly nice logic and steps

Very nice puzzle!! ❤

Ruling out a length 7 line was the key. This was easy. It was clear that the line was of odd length, and contained three different digits. Therefore it had to be length 7 or 9, with either 124 or 234 on it. This meant it had have a 4 on it, and therefore had to visit every row, column, and box. There couldn't be three cells on the line in C2, or R1C2 would have to be the same as R1C3, so there had to be three cells on the line in box 3 in order to make it into box 1. This means there are two sets of the three different digits that must be on the line, and therefore there would be at least two of the odd digit, which must therefore be 3. That meant R1C3=1, with the fog fixing R4C3 as the centre, which had to be 3. This revealed that R4C1=1, R2C2=1, and R3C4=1. Now all the fog's clear. It only takes a minute to fill in the other digits.

5:04 yay that was a neat puzzle!

Three digits are needed, so one odd digit is needed. That odd digit is the single center, and all other digits (besides repeating the center digit) must be even. The cleared cell can't be the center, and a center above it broke the puzzle. A center of 1 placed two many 4s in too few blocks. The center had to be 3, but I may have made an unwarranted assumption that it was immediately below the cleared cell. Alternatives didn't occur to me, but it worked out correct. 17:10 Turning right at the bottom gets you nowhere. You must turn right at the top. If you continue to R1C4, you need three digits in the block, none of which is the center. That means that the odd digit must be 3, and the line is nine cells long. If you stop at R2C2, you need to go left at the bottom, and reach column 1 and row 1. That needs at least nine cells total.

Wow, this looked really hard, but I managed it in 33:06 with no help, although I did need pen and paper to visualise some things. Now to watch Simon's no doubt paperless solve...

That's was a very smart 4x4 fog sudoku! Thank you for the solve!

ive played chants of sennar is amazing. ill try to catch the stream. ive watched crux solve this morning, what an adventure.

How I knew it had length > 7: you have to traverse every row and column, which is 3 horizontal movements + 3 vertical movements, + the 1 cell you started in. So the palindrome has to visit 7 cells as long as it doesn't backtrack. But to go into each box, you have to backtrack in at least one direction to make the "U" shape, which means the palindrome must have length > 7.