 # SudokuSolver Forum

A forum for Sudoku enthusiasts to share puzzles, techniques and software
 It is currently Sun Dec 08, 2019 9:07 pm

 All times are UTC

 Print view Previous topic | Next topic
Author Message
 Post subject: Sujiko Posted: Thu Aug 01, 2019 6:45 am Grand Master Joined: Wed Apr 30, 2008 9:45 pm
Posts: 596
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. 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.

Top Post subject: Re: Sujiko Posted: Tue Aug 06, 2019 6:34 pm Addict Joined: Mon Apr 21, 2008 11:35 pm
Posts: 47
Thanks! Will give it a go this weekend!

Top Post subject: Re: Sujiko Posted: Sun Aug 11, 2019 7:53 pm Grand Master Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1685
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 = 
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 =  = 2 -> R5C34 = 8 = 
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 =  = 2 -> R2C34 = 12 = [75/84/93]
1j. R12C34 = 20, R2C34 = 12 -> R1C34 = 8 = {26/35}/, 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 =  (cannot be  = 11 because R1C2 + R2C1 cannot total 13, cannot be  = 13 which clashes with R1C2 + R2C1 = 11 = {56}) -> R1C1 = 6, R2C2 = 9
2d. R12C12 = 24, R1C1 + R2C2 =  = 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  -> R2C1 = 5, R3C12 = 

[I originally used ‘algebra’ for the lower-left nonet, but no need for that now with the corrected puzzle.]
3a. R34C12 = 18, R3C12 =  = 11 -> R4C12 = 7 = 
3b. R45C12 = 18, R4C12 = 7 -> R5C12 = 11 = 
3c. R56C12 = 26, R5C12 = 11 -> R6C12 = 15 = {78}, locked for lower-left nonet
3d. R56C23 = 20, R5C23 =  = 7 -> R6C23 = 13 = , 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 =  (cannot be  = 12 because R5C6 + R6C5 cannot total 8)
4d. R5C6 + R6C6 =  = 8 -> R5C6 + R6C5 = 12 = 
4e. R45C56 = 21, R45C5 =  = 6, R5C6 = 7 -> R4C6 = 8, R6C4 = 9

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

Top Post subject: Re: Sujiko Posted: Mon Aug 12, 2019 6:31 pm Grand Master Joined: Wed Apr 30, 2008 9:45 pm
Posts: 596
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.

Top Post subject: Re: Sujiko Posted: Tue Aug 13, 2019 9:00 pm Grand Master Joined: Wed Apr 30, 2008 9:45 pm
Posts: 596
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. Top Post subject: Re: Sujiko Posted: Tue Aug 13, 2019 11:34 pm Addict Joined: Mon Apr 21, 2008 11:35 pm
Posts: 47
Thanks, HATMAN! That was a fun change - especially after your correction! Top Post subject: Re: Sujiko Posted: Wed Aug 14, 2019 7:22 pm Grand Master Joined: Wed Apr 30, 2008 9:45 pm
Posts: 596
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.

Top Display posts from previous: All posts1 day7 days2 weeks1 month3 months6 months1 year Sort by AuthorPost timeSubject AscendingDescending

 All times are UTC

#### Who is online

Users browsing this forum: No registered users and 5 guests

 You cannot post new topics in this forumYou cannot reply to topics in this forumYou cannot edit your posts in this forumYou cannot delete your posts in this forumYou cannot post attachments in this forum

Search for:
 Jump to:  Select a forum ------------------ Puzzles and Solutions    Standard Sudoku    Killer Puzzles    Samurai    Clueless Specials    Clueless Explosions    Gattai Puzzles    Other Variants Solvers & Techniques    Standard Techniques    Killer Techniques    Jigsaw Techniques    Gattai Techniques    Software General Discussion Recreation Area 