The New Sudoku Players' Forum

[m_b_metcalf]

27 digits appears to be the max for a symmetrical pi puzzle; you can get 32 - all the digits before the first 0 - without symmetry, though I haven't looked very far to see if there's actually an interesting puzzle there.

mith, do you have an example of a valid puzzle with all the first 32 non-zero clues? My first attempt was

Code: Select all

 . . . 3 1 . 4 . .
 1 . . . . . . 5 .
 9 2 6 . . 5 3 . .
 5 . 8 . . . 9 . .
 . . 7 9 . 3 . 2 .
 . . 3 . 8 4 6 . .
 . . . 2 6 . . 4 3
 . . . . 3 8 . . .
 3 . 2 . 7 9 5 . .  No. of givens =  32

found 9 solution(s)

Adding 6r1c6 yields a valid puzzle.

Thanks,

Mike

[mith]

I misspoke; there was a bug in my script causing it to return as soon as it found a valid puzzle, even if it didn't have all the digits. I'll have to fix it at some point and run it again.

[m_b_metcalf]

I misspoke; there was a bug in my script causing it to return as soon as it found a valid puzzle, even if it didn't have all the digits. I'll have to fix it at some point and run it again.

Thanks. I'm getting closer:

Code: Select all

 . . 3 . 1 . . 4 .
 1 . . . . . . 5 9
 . 2 6 . 5 . . 3 .
 . 5 . 8 . 9 . . 7
 . . 9 3 . 2 . . .
 3 8 . . 4 . . 6 2
 . . . 6 . . 4 . 3
 . . . . . 3 . . 8
 . 3 2 . . 7 . 9 5
No. of givens =  32, found 3 solution(s)

[mith]

I quite like that one. The decimal point between the 3 and the 1 gives the three unique solutions if assigned.

[m_b_metcalf]

Bingo!

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 . . 3 . 1 . 4 . .
 . . 1 5 . . 9 2 .
 . 6 5 . . . . 3 .
 . 5 . . 8 . . . 9
 . . 7 9 . 3 . . 2
 3 . 8 . . 4 . 6 .
 . 2 6 . 4 . . . 3
 . . . 3 . . . 8 .
 . 3 . . 2 7 . 9 5

No. of givens =  32

singles found 1 solution

[mith]

Very nice!

[Mathimagics]

I like π, as you can imagine.

You might be able to extend the number of clues beyond 32 by simply treating that pesky 0 in position 33 as a normal non-given, but displayed as 0, not a dot, in the puzzle string.

So, can you do 34 digits (33 clues)? 35?

36 could be a real problem, with the repeating digit pair 35 = 36 = "8", and you are probably in the last row by that stage ...

[mith]

You actually can't go any farther (other than adding the 0 at the end, if there is room).

Digit 2 (1) forces Digit 4 (1) to row 2.
Digit 5 (5) forces Digit 9 (5) to row 3, which forces Digit 11 (5) to row 4.
Digit 13 (9) forces Digit 15 (9) to row 5.
Digit 16 (3) forces Digit 18 (3) to row 6.
Digit 21 (6) forces Digit 23 (6) to row 7.
Digit 25 (3) forces Digit 26 (3) to row 8, which forces Digit 28 (3) to row 9.
Digit 29 (2) rules out placing Digit 34 (2).

[Mathimagics]

Nicely put! 32 it is ...

[m_b_metcalf]

I like π, as you can imagine.

And e as well, I assume:

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 2 . 7 . 1 . . . 8
 . . . 2 8 . . . 1
 . . 8 . . . . 2 .
 . 8 . 4 . . . . 5
 . . . . . . . . 9
 0 . 4 . 5 2 . 3 .
 . . 5 . . . 3 . 6
 0 . 2 . . 8 7 . 4
 . 7 1 . 3 . 5 . 2  31 digits of e (including 2 zeros)

I'm now going to fall into a rabbit hole and work on i.

Regards,

Mike

[Mathimagics]

i before e, except after π ... so they say

That's nice, Mike!

Meanwhile, I have searched through the first billion digits of π and can find no 81-digit sequence that forms a valid puzzle.

I looked also at e, ϕ, and even Euler's constant (thought I just might get lucky there!) ... no cigar!

Probability theory is a hard task master ... monkeys, typewriters, Shakespeare, etc ...

Would anyone care to look at the 22 trillion digits of π that are available?

[m_b_metcalf]

Sad news: I have searched through the first billion digits of π and can find no 81-digit sequence that forms a valid puzzle.

Interesting. Was it even close? What was the maximum number of complete rows that you achieved?

[Mathimagics]

That information is not available, sorry!

I just looped over the data, advancing the grid pointer by one at each step, and simply called my fsss2 solver. Got zero (invalid) every time ...

[JPF]

Bingo!

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 . . 3 . 1 . 4 . .
 . . 1 5 . . 9 2 .
 . 6 5 . . . . 3 .
 . 5 . . 8 . . . 9
 . . 7 9 . 3 . . 2
 3 . 8 . . 4 . 6 .
 . 2 6 . 4 . . . 3
 . . . 3 . . . 8 .
 . 3 . . 2 7 . 9 5

No. of givens =  32

singles found 1 solution

Excellent!
This reminds me of a good old RW riddle here

JPF

[m_b_metcalf]

You actually can't go any farther (other than adding the 0 at the end, if there is room).

Still working on that, but the best so far is

Code: Select all

 . . 3 1 . . . 4 .
 1 5 . . . . 9 . .
 2 . 6 5 . . . . 3
 5 . 8 . . . . 9 7
 . . 9 3 . . 2 . .
 3 . . . 8 4 . 6 .
 . . . 2 6 . 4 3 .
 . . . . 3 8 . . .
 . 3 2 7 9 . . 5 0

No. of givens =  32, plus a zero
found 4 solution(s)