A Lesson From The Sudoku Professor

Today's puzzle is a gorgeous snake-based sudoku by The Sudoku Professor himself, Richard Stolk! You can play it here:

Rules:
Normal sudoku rules apply. Draw a snake in the grid that starts and ends in the cells with the circles. The snake wriggles horizontally and vertically and never touches itself, not even diagonally. Clues outside the grid indicate the sum of the digits that are part of the snake in the respective row or column.

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Hi! We're Simon Anthony and Mark Goodliffe, two of the UK's most enthusiastic puzzle solvers. We have both represented the UK at the World Sudoku Championships and the World Puzzle Championships.

Thank you for watching!
Simon and Mark

[richardstolk]

You really took the long road before placing the first digits since the 2/3 pair was sitting silent in row 8 for ages...
But I enjoyed your more elegant route a lot! 👍
Placing the snake is a matter of wiggling as you did. Well done again, it was fun to watch!

[FirenWithLime]

"Wiggle your snake, pack your snake into a certain area and then allow the snake to expand"
How would the naked singles react to that, I wonder.

[kevinlund]

Alright, I'll have a go at explaining how I logically broke into this puzzle.

As Simon mentioned, c6 and c7 can have a maximum of 3 and 2 non-snake cells respectively or 5 between them. This means there must be at least 13 snake cells in total in c6 and c7.

So, imagine an invisible line between c6 and c7. We know that the head and tail are on separate sides of this line, meaning that the snake must cross this line an odd number of times. In rows where the snake crosses the line, there must be a snake cell on either side of the invisible line, meaning these rows contribute 2 snake cells to c6 and c7. In rows where the snake does not cross the line, there can be a maximum of 1 snake cell in c6 and c7.

Imagine the snake crosses the invisible line exactly 3 times. This means that 3 of the rows will contribute 2 snake cells to c6 and c7, while the other 6 rows will contribute at most 1 snake cell. if you add these up, that's 12 total snake cells in c6 and c7, which is lower than the required 13. Thus, the snake will need to cross the line more than three times in order to collect the required number of snake cells. We also know that the snake cannot cross the line in two consecutive rows, because that would cause the snake to touch itself (which it mustn't do). This means that the snake can cross the line a maximum of 5 times. Since the snake can't cross an even number of times, we know that the snake must cross the line exactly 5 times.

The only way to cross 5 times would be for the snake to cross in every other row, so we can fill in snake cells in r1, r3, r5, r7, and r9 in c6 and c7.

Next, look at r7c8. This cell cannot be snake because that would force r6c7 and r8c7 to both be non-snake, which would force r4c7 and r2c7 to both be snake, which would cause our snake to branch. This consumes one of the possible non-snake cells in r7, the other of which - as Simon pointed out - must be in c2, c3 or c4. Thus, we can fill in r7c1, r7c5, and r7c9 as snake.

Now, r7c9 must escape, so we fill in r6c9 and r8c9 as snake. Because the snake can't touch diagonally, r8c8 can't be snake. This gives us r9c8 and r9c9 as snake.

Fill in a few more obvious non-snake cells r8c6, r8c7 and we're now left with trying to determine how the snake along r9 connects to the rest of the snake. It must either attach to r7c1 or r7c5. Based on row totals, we can have at most 2 snake cells in r8 and at most 7 snake cells in r9. With these constraints, it's easy to deduce that there is no way for the bottom snake to reach r7c1, so it must reach r7c5. With some minor deductions, we can see that it must get there by claiming r9c5 and r8c5.

Now the only way to satisfy r7 is for r7c2 and r7c3 to both be snake. Finally, in order to satisfy c6 and c7, the snake must alternate back and forth up the two columns, which eventually leaves us where Simon left off and made his request for someone to explain logically.

Its a bit long winded, but I believe this is a totally logical way to break into this puzzle! Let me know if I missed anything!

[nadines.]

‘Pack your snake in certain areas and allow it to expand?’ Valentine’s Day still in the air?

[GreatUSTreasureHunt]

My anaconda don't want none if you can't do sums, hon!

[AlRoderick]

The snake jokes are absolutely charming 🐍 🎺

[littleschnitzel]

In case you wonder, Simon was not birfucating throughout the video, he merely cosplayed Mark Godliffe.

[khayman]

I just love those knowledge bombs. They're always the same but the way they're told just puts a smile on my face every single time. Keep 'em coming, please 🤣🤣

[Joeevr]

Pretty early on, when you pointed out the 1-2-3-4 quadruple in row 8, you should've noticed the 3's were only in snake cells. That would've allowed you to immediately deduce the 5 sum of that row had to be a 2-3 pair, and you could've placed the 1-4 in the other 2 cells right away.

[markp]

30:35 finish. Here is the logic I came up with for determining the snake. The fact that the 38 requires at least 6 snake cells and 42 requires at least 7 snake cells, logic dictates that they must weave back and forth to avoid the snake touching itself, otherwise one of the columns would not contain enough snake cells. No matter the direction, this requires r1c6-7, r3c6-7, r5c6-7, r7c6-7, and r9c6-7 all to be snake cells. Since the 5 in r8 requires exactly two cells (one to get to r9, another to exit r9), therefore all of the snake cells in r9 are adjacent. A 33 clue requires at least 5 cells, therefore r9c5 is on the snake. Any attempt to loop back to c9 after leaving it creates too many non-snake cells in c7, therefore the snake cells in c9 are all adjacent. If the 7 clue in c8 was two or three adjacent snake cells, it would either make c7 or c6 impossible without the snake touching. Therefore, the 7 clue is a single cell. If the entry for c7 is anywhere other than r1c7 or r9c7, the c6-c7 weave breaks (c9 needs at least 3 adjacent snake cells, so must enter c7 at one of the 5 previously marked weave connectors, but entering at r5 would require a return to c9, which is impossible, and entering at either r3 or r7 would force the weave back to c8 at some point, which is impossible). Identifying that r7 must have at least 7 snake cells, and that having all of r7c2-4 would split the snake, either r7c2 or r7c4 must be non-snake (r7c3 must always be a snake to fulfill r7 while avoiding a snake touching at a corner). This leaves a maximum of 1 more potential non-snake in r7. If the snake goes up from r9c5, r7 would have at least 3 non-snake cells. Therefore the snake goes down from r9c5 to r9c9. The only two places for the snake to exit r9 and not touch the other head/tail of the snake are c1 and c5, and therefore the snake turns up at r9c5 to r7c5. The snake then weaves between c6 and c7 (requiring r4c6 to achieve the minimum number of snake cells in c6). The rest of the logic is as determined by Simon.

[frostplatypus]

Love the classic proof by "if you don't believe me, try it yourself" ;)

[QuantumUniverseUvA]

My break-in for the snake: there are at most 5 cells missing from columns 6 and 7. Therefore, there must be at least four places where the snake is in both columns - that is, it must move 'from left to right' at least four times. But in fact, it must do so an odd number of times to start on one side and end on the other, so it must move from left to right five times. This is only possible if the five horizontal pairs are in row 1, 3, 5, 7 and 9. Now you can proceed as Simon did, looking at row 7 first and figuring out which two cells must be missing there, and from there the exact wiggles of the snake around columns 7 and 8 are easily found.

(Edit: oh, I see this is quite similar to what Mark P wrote below.)

[Coyotek]

27:11, which included one guess early regarding the placement of the snake; after solving, I spent about 1 1/2 minutes convincing myself why my other possibility could never work. All told, under 30 minutes for a 'pure-logic' solve, which very much pleases me.

Nice puzzle!

[Coyotek]

I started differently. I saw the '5' clue in r8, which told me the only way for the snake to get to r9 was 'one way in, one way out' through r8. That, combined with the clue for r7 and the location of one end of the snake (as well as the two large-numbered column clues) allowed me to locate the most of the snake.

[curtissumner]

51:35. The snake was very tricky to get in the beginning, but once you get the first solid step the rest falls in place nicely. I found that a lot of progress kind of slid into place without me noticing, which I very much enjoyed. Excellent puzzle!

[okubi]

14:31 i'm not sure if Simon was trying to dig out or dig deeper into the innuendo

[Mark]

Not only do we get Suduko, we get words of the day:
Otiose - serving no practical purpose or result

[loganvedders]

I just want someone to talk to me the way he talks to his puzzles😔

[bobblebardsley]

I don't have a really elegant explanation but I think looking at the bottom three rows, you have a very small clue sandwiched between two larger clues, so you can easily deduce that the snake must cross row 8 twice (partly to avoid stranding the head of the snake in the bottom row). It also needs to reach the top half of the grid so it can't do it with only one green cell in row 7. Row 7 cannot have more than two green cells because of the 41 sum. So you can deduce that the snake must cross row 7 twice, leaving only two green cells (ie it must cross it at the edges of the grid) and from there it very quickly becomes clear how the head and tail join to it, and the c6 and c7 clues restrict the way it wiggles up those two columns, but I'm not sure how specifically they restrict it without just spotting it the way you did in your solve. Like I said, I think I can see how the logic works but I haven't fully figured it out in my own head!

[richardglover]

Really, really lovely puzzle which was so enjoyable to watch solve and whilst I didn't have go at, I now want to very much to try one with this same rule set.