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Rules of Snake:

  1. Shade some cells to form a non-intersecting path which does not touch itself, not even diagonally.
  2. Black circles must lie on one end of the path. White circles must lie somewhere along the path, but not at an end.
  3. A number outside the grid represents how many cells in the corresponding row or column are shaded.

The standard Snake rules apply. However, each number has been replaced by a letter. Different letters refer to different numbers.

enter image description here

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  • $\begingroup$ Does this mean that all circles are given? $\endgroup$
    – oAlt
    Jul 29 at 5:00
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    $\begingroup$ @oAlt I use the original rules of the Snake which may contain white circles in the puzzle. For this particular one, no white circles are given to the solvers. $\endgroup$
    – athin
    Jul 29 at 5:35
  • $\begingroup$ Can one letter correspond to number zero? $\endgroup$
    – daw
    Jul 29 at 8:36
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    $\begingroup$ Penpa link for those interested. $\endgroup$ Jul 29 at 12:46
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    $\begingroup$ @justhalf I think that's ITS; "it's a snake" read starting from bottom left $\endgroup$
    – oAlt
    Jul 29 at 16:00
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First of all,

Neither of S,N,A,K,E can be 0 because we couldn't connect the two ends in that case. Since there are only 5 cells for each column, each of S,N,A,K,E can be at max 5. And since the numbers need to be distinct, we know that each of 1,2,3,4,5 must appear exactly once.

Now,

One of S,N,A,K,E must be 5. It can't be the S because we need to get both to the leftmost column (remember, A cannot be 0) and to the other end on the right-hand side of the grid. The 5 cannot be any of N,A,K either, because a 5 there would force both adjacent numbers to be 1 and the numbers must be distinct. So we conclude that the only possibility is E=5. This gets us started:

enter image description here

From here we see that

K must be 1, because there are only two cells available and filling them both would prevent us from reaching the other end on the right. A must be at least 3 because the snake needs to turn around without touching itself in the leftmost column. If A=3, N must be 2 because 1,3 are taken by K,A and there are only 3 cells available without touching the snake in the A column. And if A=4, N must also be 2 because there are 2 free cells and the 1 is already taken by K. So we know that K=1, N=2 and E=5, and A,S are 3,4 in some order.

The leftmost A column

Must have a filled cell in the middle, regardless of whether A is 3 or 4.

enter image description here

If K is filled at the bottom,

We can't fill A=(3,4) and N=2 without touching the black circle on the S column, which means the snake goes straight from the S column to E without going to the leftmost A column first. We can't do that, so we conclude that K must be filled at the top. This gives us most of the N and A columns as well as the column to the right of the E.

enter image description here

Looking at the S column,

There is no legal way to fill in 4 cells without touching the snake in the adjacent N column. This means we know A=4 and S=3, which gives us the entire centre of the grid as well us the entire S row in the middle, which also conveniently blocks off all the cells in the top-right corner. And since we now know A=4, we can finish off the left-hand side of the grid as well.

enter image description here

Finishing it off,

The snake cannot turn up immediately because that would make I=T=7. It also cannot go into the corner since that would make T=E=5. So there's only one place where it can make that turn, giving T=6 and I=8. Finished!

enter image description here

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    $\begingroup$ Wish I could upvote twice. Excellent description of the logic used! $\endgroup$ Jul 29 at 15:12
  • $\begingroup$ Very well done! I ticked this answer as it really explains the logic and in a really nice way! Great job Jafe :D $\endgroup$
    – athin
    Jul 29 at 23:54
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completed snake puzzle Started by realizing E had to be 5 and K had to be 1 since the 5 must be bordered on 1s on both sides. A had to be either 3 or 4 such that it could duck back around. S had to be 3 since it crosses that many times and 4 would force lines to touch. That leaves 4 for A and 2 for N by elimination.

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    $\begingroup$ Didn't realise you had already answered when posting. Sorry! I'll leave my answer up since it has a few more intermediate steps in case anyone finds that useful. $\endgroup$
    – Jafe
    Jul 29 at 11:31

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