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 Post subject: Sujiko
PostPosted: Thu Aug 01, 2019 6:45 am 
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Grand Master
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Sujiko 1

I've been doing these in the Sunday Telegraph - with difficulty. Hence I thought to create them so as to understand the solution methodology better.

The clues are the sum of the adjacent four squares.
The numbers 1-9 cannot repeat in the blue nonets.
Numbers can repeat in rows and columns.
Numbers can repeat in clue sums.

Given the central clue not that hard hence I have left out a few sums. I am never sure whether to do this as leaving out clues makes the solution path clearer, if more difficult.


Image

Apologies to Andrew and anyone else who tried this: I transposed the first two clues, corrected now.


Last edited by HATMAN on Sat Aug 10, 2019 5:17 pm, edited 2 times in total.

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 Post subject: Re: Sujiko
PostPosted: Tue Aug 06, 2019 6:34 pm 
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Thanks! Will give it a go this weekend!


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 Post subject: Re: Sujiko
PostPosted: Sun Aug 11, 2019 7:53 pm 
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Location: Lethbridge, Alberta, Canada
Thanks HATMAN for this fun puzzle.

Here is how I solved it:
Clues are the totals of the 4 adjacent cells.

Nonets are normal, containing 1-9.

Numbers may repeat in clue totals, but only when some of the numbers are in different nonets.

Numbers may repeat in rows and columns.

1a. R34C34 = 4 = [1111]
1b. R23C23 = 25 contains 1 in R3C3 -> R2C23 + R3C2 = 24 = {789}, locked for upper-left nonet
1c. R45C23 = 11 contains 1 in R4C3 -> R4C2 + R5C23 = 10 = {235}, locked for lower-left nonet
1d. R45C45 = 10 contains 1 in R4C4 -> R4C5 + R5C45 = 9 = {234}, locked for lower-right nonet
1e. R45C34 = 10, R4C34 = [11] = 2 -> R5C34 = 8 = [53]
1f. R45C12 = 18, R45C2 = {23} = 5 -> R45C1 = 13 = {49/67}, no 8
1g. R45C56 = 21, R45C5 = {24} = 6 -> R45C6 = 15 = {69/78}, no 5
1h. R56C12 = 26, R5C2 = {23} -> R5C1 + R6C12 = 23,24 = {689/789}, no 4, 8,9 locked for lower-left nonet
1i. R23C34 = 14, R3C34 = [11] = 2 -> R2C34 = 12 = [75/84/93]
1j. R12C34 = 20, R2C34 = 12 -> R1C34 = 8 = {26/35}/[44], no 7,8,9

2. The four clues in upper-left nonet total 96 which includes R13C13 once, R13C2 + R2C13 twice and R2C2 four times
2a. Subtract 45 for R123C123 once; this leaves R13C2 + R2C13 + three times R2C2 = 51
2b. Subtract 24 for R2C23 + R3C2 (R3C3 has already been subtracted by step 2a); this leaves R1C2 + R2C1 + twice R2C2 = 27
2c. R1C2 + R2C1 + twice R2C2 = 27, R12C12 = 24 -> R2C2 = R1C1 + 3 -> R1C1 + R2C2 = [69] (cannot be [47] = 11 because R1C2 + R2C1 cannot total 13, cannot be [58] = 13 which clashes with R1C2 + R2C1 = 11 = {56}) -> R1C1 = 6, R2C2 = 9
2d. R12C12 = 24, R1C1 + R2C2 = [69] = 15 -> R1C2 + R2C1 = 9 = {45}, locked for upper-left nonet
2e. R12C23 = 22 with R1C2 from {45}, R1C3 from {23}, R2C2 = 9, R2C3 from {78} can only be [4297] -> R2C1 = 5, R3C12 = [38]

[I originally used ‘algebra’ for the lower-left nonet, but no need for that now with the corrected puzzle.]
3a. R34C12 = 18, R3C12 = [38] = 11 -> R4C12 = 7 = [43]
3b. R45C12 = 18, R4C12 = 7 -> R5C12 = 11 = [92]
3c. R56C12 = 26, R5C12 = 11 -> R6C12 = 15 = {78}, locked for lower-left nonet
3d. R56C23 = 20, R5C23 = [25] = 7 -> R6C23 = 13 = [76], R6C1 = 8

[There is no total for R56C34, so I’ll use my original ‘algebra’ for this step.]
4. The four clues in lower-right nonet total 70 which includes R46C46 once, R46C5 + R5C46 twice and R5C5 four times
4a. Subtract 45 for R456C456 once; this leaves R46C5 + R5C46 + three times R5C5 = 25
4b. Subtract 9 for R4C5 + R5C45 (R4C4 has already been subtracted by step 4a); this leaves R5C6 + R6C5 + twice R5C5 = 16
4c. R5C6 + R6C5 + twice R5C5 = 16, R56C56 = 20 -> R6C6 = R5C5 + 4 -> R5C5 + R6C6 = [26] (cannot be [48] = 12 because R5C6 + R6C5 cannot total 8)
4d. R5C6 + R6C6 = [26] = 8 -> R5C6 + R6C5 = 12 = [75]
4e. R45C56 = 21, R45C5 = [42] = 6, R5C6 = 7 -> R4C6 = 8, R6C4 = 9

5a. R34C56 = 28, R4C56 = [48] = 12 -> R3C56 = 16 = {79}, locked for upper-right nonet
5b. R23C34 = 14, R2C3 = 7, R3C34 = [11] = 2 -> R2C4 = 5
5c. R12C34 = 20, R1C3 = 2, R2C34 = [75] = 12 -> R1C4 = 6
5d. R23C56 = 26, R3C56 = {79} = 16 -> R2C56 = 10 = {28}, locked for upper-right nonet
5e. R12C45 = 17, R12C4 = [65] = 11 -> R12C5 = 6 = [42] -> R12C6 = [38]
5f. R23C45 = 15, R23C4 = [51] = 6, R2C5 = 2 -> R3C5 = 7, R3C6 = 9

Solution:
6 4 2 6 4 3
5 9 7 5 2 8
3 8 1 1 7 9
4 3 1 1 4 8
9 2 5 3 2 7
8 7 6 9 5 6

Note:
that my 'algebra' method only worked so efficiently because R34C34 were fixed. That would, of course, also apply for a Sujiko with central total 36.


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 Post subject: Re: Sujiko
PostPosted: Mon Aug 12, 2019 6:31 pm 
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Puzzle Question

I've just published this question on the players forum (where they are into theoretical analysis) however, I'm not sure how many of you visit there, so:

If the 25 sums give a feasible solution is it automatically unique in all cases? I.e. necessary is the same as sufficient.

Whether it is solvable without heavy number crunching is a separate matter - although even then the analysis is only of spreadsheet level.

I am attempting currently to create a counter example.


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 Post subject: Re: Sujiko
PostPosted: Tue Aug 13, 2019 9:00 pm 
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As a counter example this has multiple solutions, however what I have found so far are just symmetry variations based on one nonet solution repeated.

Image


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 Post subject: Re: Sujiko
PostPosted: Tue Aug 13, 2019 11:34 pm 
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Thanks, HATMAN! That was a fun change - especially after your correction! :dance:


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 Post subject: Re: Sujiko
PostPosted: Wed Aug 14, 2019 7:22 pm 
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I've been spending too much time on counter examples but it is definitely not unique when feasible in all cases, but I am sure that in the vast majority when clues are not symmetrical it is.
Where the 25 sums are all 20 then then the four 20 in a nonet give seven solutions plus symmetries:
C1 C2 C3 C4 C5 C6 C7 C8 C9
1 8 3 9 2 7 4 5 6
1 6 7 9 4 3 2 5 8
3 5 2 8 4 9 7 1 6
1 6 7 8 5 2 3 4 9
2 5 8 7 6 1 3 4 9
3 2 7 9 6 5 4 1 8
4 3 7 5 8 2 6 1 9

So centre values of 1 3 7 9 have no solutions 2 5 8 have a single solution and 4 and 6 have two slightly linked solutions.

Using the 5 centre set in the 4 nonets we can form a solution with the four centre cells 1991 and another with four centre cells 3773, plus of course their symmetries.

There may be other solutions using mix and match but as I am doing this semi-manually, life is too short.


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