Up the Creek without a Paddle
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- čas přidán 31. 05. 2024
- Happy Birthday (for yesterday!), Math Pesto! Here's where you can play their Red Lodge, Montana: sudokupad.app/3wy1bsu95d
Place the digits 0-8 once each in every row, column, and 3x3 box. A digit in a circle must appear at least once in the four surrounding cells. That digit also equals the number of surrounding cells that are shaded. A “?” represents the same number for both clues within a circle (i.e. that digit appears and that many cells are shaded). The value of “?” may differ between circles. Unshaded cells in the grid must be orthogonally connected, and unshaded cells that are orthogonally adjacent must have digits that differ by at least 5.
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▶ Contents ◀
0:00 Theme Music & News around the channel
2:28 Rules of today’s puzzle
3:50 Start Of Solve - Let's Get Cracking! - Zábava
![Pseudo Coup: Sudoku With a Difference! [Audio Fixed]](http://i.ytimg.com/vi/5speYaGTNcc/mqdefault.jpg)
Thanks so much Mark for solving this one! I commend you for persevering through! Delightful to watch as always. And much gratitude as well to my partner Mic for sending this one in! 😊
Happy Birthday!
A happy birthday to you.
A curse upon the puzzle too.
I've got Missoula on my mind,
and the girl I left behind.
Now go and crack a kettlehouse for two.
Hope you had a great birthday
Hi, this must have been tough to build. Great puzzle. Is there, by chance, a simpler way to disambiguate for parity the unshaded cells?
@@Paolo_De_Leva The intended step was still using the green cells in Row 8 to disambiguate the polarity, though a little more pencil-marking makes it easier to see.
This was a hard puzzle indeed. Everytime Mark has a 1h+ puzzle I get scared 😅
Thanks for the patience and for the solve Mark 😆
When a checkerboard rule applies, I like to place a light grey circle in every other cell. It works particularly well in colouring puzzles, when adding more colours can visually overwhelm the grid 46:28
Absolutely mesmerizing, Mark. I watched this video with great attention and loved it. I have always thought that Simon was the snake/pencil-puzzle guy, but you were fantastic in this solve. I admire so much about this, not least is the brilliance of figuring out how to use the question mark clues (which seemed to me to be very tricky). I will not be putting this one on my list, nope, not at all. Brilliant puzzle and brilliant solve, thank you, Mark.
Took me almost two hours to solve. The break-in I used was R8C8. If it was high, row 8 would need three low digits in columns 5, 7, and either 1 or 2. But it would also put an 1 into column 9, and 3 can't be in any of them, leaving only two digits to use.
After spending an age not looking at his 3 pencil marks he berates himself and then promptly proceeds to ignore the 0 pencil marks in box 8 putting 5678 in a cell with a 0 pencil mark and not using it.
Then in the final minutes "this looks unsolvable... let's just get rid of those [pencil marks that I'm ignoring that have the answer]"
By this point, I'm fairly sure both Mark and Simon put corner marks into grids as out of habit only. They're more for decoration than to assist with solving.
That was a really fun puzzle to solve. It wasn't easy but the logic wasn't buried too deep, it just needed a little bit of digging.
Never seen Mark struggle so much with a puzzle. I found it easier to determine where the 4's go more that the unshaded area. I think if he asked that question more often the solve would have been easier. Took me 90 minutes.
1:44:00 That was absolutely wonderful! Gorgeous construction with some really lovely logic. Thanks @MathPesto - Loved it!
That reassuring feeling where Mark gets stuck at the same place you did! This was very challenging, but very interesting. Happy birthday Math Pesto! I used a slightly different colouring system; I coloured the low/high for every cell that might not be shaded, and I think it helped me a bit more in the endgame to see when a cell had to be shaded because it's only possible values weren't valid for a shaded cell.
Blimey that was a tough puzzle, great to get a long Mark solve though. Happy Birthday MP !
Almost 2 hours from Mark? This puzzle must be brutal.
I suspect he probably could have solved it faster if he had pencilmarked more like we'd expect, but for some reason he decided to start off with almost no pencilmarking. On the other hand, he spent the longest time on things like "what if r3c5 is a 3", ignoring the pencilmarks on the right showing it couldn't be... perhaps he had a really bad day.
It's not that bad. Or at least, it doesn't require any huge conceptual leaps. More so just a whole lot of smaller insights. It's a very dense puzzle. But do-able.
It took me ages .... but I really enjoyed it. Great and very exciting puzzle.
Nice to see a longer solve. 😊
129:04 Tough but doable. Thank you and Happy Birthday @mathpesto!
Oh wow, battled through it, but it was achievable! As always breathtaking Math Pesto! Happy birthday!
Solving this puzzle did kind of feel like strolling through a forest with many creeks to cross. Sometimes you find a nice path through the woods where you can walk some small distance, but the next treacherous creek is right around the corner. Eventually it took just over an hour to complete the hike.
I've been working on this for a few hours and I'm glad to see Mark struggling as much as I have been. Super hard puzzle, but it's been fun.
1:12:40 finish. I decided to just go with three solid colors instead of flashing (Shaded/Unshaded High/Unshaded Low). This made it easier to differentiate high vs low (in row 8, the color I had three of couldn't have 1 or 3, so therefore had to be high). This was an excellent puzzle, very fun!
Mark needs to not look to go into beast mode so quickly on these long solves. He has a tendency to delete pencil marks before he uses them. 😁😁😁
One clever bit of logic which would have made the tail of Mark's solve much easier is to do with the 4's early on (available at least from 1:11:00). The simplest way to see it is that if you put a 4 in r1c5 or r2c5, that forces a 4 in c7 in box 3 - and this breaks the 45 pair down below! This ends up giving you the 2 in box 3 much earlier, making all that logic around where the 8 leads a lot easier.
If you prefer working in cases, you can also see it as: whether the circled 4 is placed in c6 or c7, you still get a 4 in r7c5, either by making r7c5 the only place for a 4 in the column, or by making r7c7 a 5.
I guess the three clue in the top right should be flashing in neon colors, depending on how often he overlooks it 😅
I really like 0-8 sudokus and this was one of the best i've ever seen 👍
Great solve as always 😁
As soon as Mark had filled in the third, fourth, fifth, sixth, eighth, and ninth cells of column 5 as green, there was a surprisingly easy way to prove r7c7 was not a 4: If it were, then the "4" clue between boxes 2 and 3 would then have to be in box 2. But this would leave no place for the 4 in column 5!
Very good puzzle. Always rather intimidating when you see a long video by Mark but then they are the only ones where I usually have a chance of beating his time.
Solved this much more quickly than Mark for one simple reason: I was much less reluctant to start pencil-marking like crazy!
40:54 for me. Wonderful puzzle!!
It wasn't an easy puzzle, and I couldn't do it without resorting to the video occasionally. The issue for me was, I was constantly overlooking consequences of the whisper-like rule. On the other hand I found a few parts of the solve much quicker than Mark, who didn't seem to have his best day at making use of his corner marks. Usually that is something I consider more of a Simon weakness (if you can call it that). As with several puzzles that don't have an extremely well telegraphed solve path it's always a tradeoff between going deep on a particular idea or solving tactic, or change for some alternative attempt. I think at some points trying an alternative attempt after a few solve steps would have sped up Mark's solve, on the other hand it's his very persistence and perseverance that made him succeed after all.
44:16 for me. this is the one i guessed the polarity nice
34:46 🙂
Happy belated birthday, Math Pesto!!!
Dear Sven: Can't like the puzzle if I can't finish the puzzle, but I DO like the puzzle and wanna like it.
For now, I would suggest to like the videos and especially the livestreams 😉
At 48:57 cell 8 of box 2 is said to have to be 3. But row 3 already has its 3 in box 3 ! Disproving the hypothesis...
Oooh a real tough one
Definitely one to watch
Awesome puzzle. 98:45 for me
Mark would have found it easier if he had done the sudoku that he eventually did in columns 6 and 7 *before* trying to determine the green path through box 1. In particular, it would give him a 2 in r3c7, which answers the question he spent so long agonising over, with regard to whether there was a 2 in r3c3 or not.
Unrelated to this video, but I just found 8 Harry Potter themed Cryptic Crosswords on Wizarding World's website. I think that would make an excellent bonus video for all us CtC Potterheads! ⚡⚡⚡
6d Under centre of passage, you and I join aliens to find Chocolate Frogs, say (6) 🐸
when i realized that cells no matter the way taken need even or odd spaces (i think MArk called that the diagonal something), i could delve into the solving.
1:26:00 and thereabouts: Reasons the ? in box 1 cannot be a 4:
Consider the missing digits in box 5 (3/4/5/7). Of these, only 4 and 5 can go in R5C6 forming with R8C6 a 4/5 pair in the column. This forces the 4 from the circle between boxes 2 and 3 to be in box 3. 4 cannot be in the bottom row of box 2, so box 2 must also have a 4 in rows 1 and 2, forming an x-wing with the 4 in box 3. So the 4 in box 1 must be in row 3, not around the circled ?.
A completely different proof involves marking the 4 gray cells around the circled 4, then R1C1 to complete the 3 (graying R2C1 instead would leave an isolated green cell). R2C1 is green and needs to connect so R3C1 is also green and must be a low digit. R3C3 would also be green (and a 2) to connect all the green. The only low digit available in the row for R3C1 is a 3 but that has to be in the cells around the circled 3 (and it also leads to two 8's seeing each other trying to whisper through a 3).
68:32 for me. Felt like the main thing was disambiguating the low/high which I got pretty quick but Mark took awhile to see. Other than that, I do t think anything about this puzzle is too tricky.
im hung up on the difference between the question mark clues and the given clues, they mean the same thing right? the number always works for both clues? but the extra line of clarification/explanation for the question marks got me worried
Took me 2 and a half hour..insane puzzle
Finished in 174:38 after MANY, MANY resets. I don't know where I kept going wrong, but I kept getting into a situation where I had an error.
Frustrating, but fun puzzle!
Only took me 4 hours total, but I did it!
For future thought, since you know there's the bishop's polarity in the puzzle, using two different greens for the non-shaded cells might have helped. (I gave up after 15 minutes. What do I know?)
Very early on when getting stuck Mark mentions the long straight of unshaded cells in C5 if R7C5 was unshaded and questioned whether that was a problem.
Unfortunately he didn't delve deeper into it due to the digits being 0-8 making it hard to reason about but if I'm not mistaken that would result in 3 distinct digits of one polarity and 4 distinct digits of the other polarity. The 4 distinct digits would have to include the monogamous digit (3/5) but all of the cells have at least 2 neighboring cells that see each other (same column or box) so there wouldn't be 2 valid digits to pair up with.
This would force the continuation of the path. R7C5 needs to be shaded, the path needs to get out so R8C4 needs to be unshaded which defines the 3 shaded cells of the circle clue. Taking it further the unshaded cells still need to get out so R7C4 is unshaded as well.
I haven't looked into it further but this might have given more information or even a significantly different path for finding the solution and get unstuck more quickly.
I don't think r3c5 has two neighbouring green cells that see each other?
It couldn't be the monogamous 3, because of 3 being in row 3 in box 3 for the quadruple clue. But it could have been a monogamous 5 with neighbouring zeros in r3c6 and r5c4.
@@RichSmith77 oh yeah, you are absolutely right, I missed that :o
72:24 and the last bit was a slog. Did enjoy it though. Flicking through the video, looks ironically like Mark would have been better served with some more pencil marking.
I also had huge problems seeing the 3s from box 3 pointing at R3C5. Since the solution path is narrow there one cannot advance before one realizes R3C5 must be a 0. This did cost me quite some minutes, also. Great puzzle!
wow what a doozy
Nice design logic/dd is simple but gutwrenching. Interestingly you dont need the given 8 to solve
Rarely see mark above 1.5hr
My mind thought 'Up POOP creek without a paddle.'