## Solving a hard one

Advanced methods and approaches for solving Sudoku puzzles

### Solving a hard one

(Thought i posted this yesterday - if it was removed, please tell me why)
As a compensation for the bad sample last time i want to show, how a pretty hard sudoko can be solved. Its nr 149 from the outlaws in the sudoko gallery.
Of course i know that there are many ways to solve it (rubylip's says, a 3 in r2c5 would lead to a contradiction), but i think, the steps might be interesting for others.

Code: Select all
`. . 1 | . 5 . | . . 2 6 . . | 4 . . | . . . . 9 . | . . . | 3 . . ----------------------. . 8 | . . 7 | . . 1 . . 5 | 6 . . | 4 . . 2 . . | . . . | 9 . . ----------------------. . 7 | . . . | . 8 . . . . | . . 3 | . . 5 4 . . | . 2 . | 6 . . `

The first time i got stuck here:

Code: Select all
`378   3478   1      3789   5       689     78   4679  2          6     23578  23     4      13789   1289    1578 1579  79         578   9      24     1278   1678    1268    3    14567 467        39    346    8      2359   349     7       25   2356  1          1379  137    5      6      1389    1289    4    237   378        2     13467  346    1358   1348    1458    9    3567  3678       135   12356  7      159    1469    14569   12   8     349        18    1268   269    1789   146789  3       127  12479 5          4     1358   39     15789  2       1589    6    1379  379     `

(The swordfish in 4 doesnt really help)

Elminate 3 from r9c3:
r9c3=3,r7c9=3
. r8c3=9,r6c3=6,r3c3=4,r1c8=4,r7c9=4
=> r9c3=9

Eliminate 2 from r4c7:
r4c7=2,r7c7=1,r8c7=7,r9c8={39},r9c9={39}
not possible, because r9c3={39}
=> (2 numbers for 3 cells) r4c7=5

Eliminate 2 from r8c3:
r2c3=3->r1c1<>3
r3c3=4,(r1c2<>3),r1c2={78},(r1c7={78}),r1c1<>{78}
=> (no number left for r1c1) r8c3=6

Eliminate 2 from r4c8:
r4c8=2,r5c6=2,r4c4=2,r3c3=4
. r4c2=6,r4c5=4,r6c3=4,r3c3<>4
Eliminate 3 from r4c1:
r4c1=3,r6c3=4
. r4c8=6,r4c2=4.r6c3<>4
=>r4c1=9, r4c5<>9

Elminate 3 from r6c3:
r6c3=3,r4c2=4,(r4c5<>9),r4c5=3,r1c4=3
r2c3=2, r3c3=4 (r1c2<>4), triple {378} in row 1 (c127)->r1c4=9
=>r6c3=4

With this i got here:
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`7   4   1     3   5    69     8  69  2          6   58  3     4   89   2      1  59  7          58  9   2     18  7    168    3  56  4          9   6   8     2   4    7      5  3   1          13  7   5     6   139  19     4  2   8          2   13  4     158 138  158    9  7   6          135 13  7     15  6    4      2  8   9          18  2   6     9   18   3      7  4   5          4   58  9     7   2    58     6  1   3          `

Matter of taste, eg
weak chain in 8: r2c2-r9c2~r9c6-r8c5~r2c5-r2c2 -> r2c2=8
or double forcing chain:
r8c1=8,r3c1=5,r2c2=8
r8c5=8,r2c5=9,r2c2=8
Last edited by Wolfgang on Mon Sep 05, 2005 6:12 am, edited 2 times in total.
Wolfgang

Posts: 208
Joined: 22 June 2005

Wolf

Is this grid really hard? I solved this one using Angus' program with nothing more than conjugate colours or a turbot fish. Have I missed out something? If what I have done was correct, then you have applied some very advanced techniques to a not too difficult puzzle.

By the way, you reasonings are very logical, though the first line should read 'Elminate 3 from r9c3'.
Jeff

Posts: 708
Joined: 01 August 2005

Jeff wrote:I solved this one using Angus' program with nothing more than conjugate colours or a turbot fish.

Maybe i have overlooked something (not the first time ). Please could you reproduce it?
By the way, you reasonings are very logical, though the first line should read 'Elminate 3 from r9c3'.

Yes, thanks.
Wolfgang

Posts: 208
Joined: 22 June 2005

BTW, a simple (noname?) pattern solves outlaw 8 (the rest is turbot fish and pairs)
Code: Select all
`Boxes 8 and 9:23479 234 2479    178 5   12         237   8   27      17  9   6          56    1   56      478 247 3  `

A=x->...->B=y, C={xy}, A~C~B
=> eliminate x from A (r7c456)
Wolfgang

Posts: 208
Joined: 22 June 2005

Wolfgang wrote:A=x->...->B=y, C={xy}, A~C~B
=> eliminate x from A (r7c456)

I'm cannot figure out what any of this means. Could you define

your shorthand? Are you using '->' to mean something different

from '=>'? In what sense are you using '~'? What are A, B and C?

Also, from the first post in the thread:

Wolfgang wrote:Elminate 3 from r9c3:
r9c3=3,r7c9=3
. r8c3=9,r6c3=6,r3c3=4,r1c8=4,r7c9=4
=> r9c3=9

-- What is the period at the beginning of the line?
-- What reasoning was there for choosing to suggest that r9c3=3?
-- Are these eliminations listed in a some sort of order?
-- I can't keep up -- it seems like your short hand is a little to short and variable.
tso

Posts: 798
Joined: 22 June 2005

tso wrote:Are you using '->' to mean something different
from '=>'?

no, both means "follows"
In what sense are you using '~'? What are A, B and C?

A, B, and C are cells, A~B means, A and B share a unit (are weakly linked)
9c3=3,r7c9=3
. r8c3=9,r6c3=6,r3c3=4,r1c8=4,r7c9=4
=> r9c3=9
-- What is the period at the beginning of the line?
-- What reasoning was there for choosing to suggest that r9c3=3?
-- Are these eliminations listed in a some sort of order?

The period: i wanted to put some blanks there to indicate thet the second line also starts with 9c3=3, but they were gone in the post, so i tried a period and some blanks
Why r9c3=3: because i wanted to eliminate the 3.
The order: You can do it in this order to solve the puzzle.
Wolfgang

Posts: 208
Joined: 22 June 2005

Grief - I thought this was getting sorted - but it's getting more confusing as people insist on using their own logic signs.

Can we PLEASE agree a common method?... it does nobody (not least the regulars) any favours to bring in more and more confusing nomenclature.

Rant over. Who will take responsibility for overseeing this?

stuartn
stuartn

Posts: 211
Joined: 18 June 2005

stuartn wrote:Grief - I thought this was getting sorted - but it's getting more confusing as people insist on using their own logic signs.

im sorry for that. It was not the first time i used it, for me it is just the shortest way to describe it, eg a xy chain would look similar:
A={x,y}, A=x->...->B=y, A~C~B
=> eliminate y from C.
Wolfgang

Posts: 208
Joined: 22 June 2005