View previous topic :: View next topic 
Author 
Message 
Marty R.
Joined: 12 Feb 2006 Posts: 5160 Location: Rochester, NY, USA

Posted: Wed May 31, 2006 5:18 pm Post subject: Brick wall 


My present position:
Code:  
6 1457 1247 179 3 257 8 579 457 
38 13578 178 1789 159 4 3679 35679 2 
2348 9 2478 78 25 6 347 1 3457 

349 1346 5 2 67 8 3479 379 1347 
7 38 289 5 4 1 239 2389 6 
248 1468 12468 3 67 9 5 278 1478 

89 2 6789 179 159 357 367 4 3578 
1 478 4789 6 29 2357 237 23578 3578 
5 67 3 4 8 27 1 267 9 
 
As published:
Code:  
638
42
91

52
7516
95

24
16
389
 


Back to top 


Steve R
Joined: 24 Oct 2005 Posts: 289 Location: Birmingham, England

Posted: Wed May 31, 2006 7:21 pm Post subject: 


Marty
The computer suggested that the next step was to eliminate 2 from r56c3 via
r5c3 = 2 or r6c3 = 2 => r1c3 ≠ 2 => r1c6 = 2 => r9c6 = 7
r9c6 = 7 => r9c8 = 2 => r5c8 & r6c8 ≠ 2 => r5c7 = 2 => r5c3 ≠ 2
r9c6 = 7 => r9c2 = 6 => r7c3 ≠ 6 => r6c3 = 6
So r5c1 = 2. There are probably beeter ways but I hope this helps.
Steve 

Back to top 


Marty R.
Joined: 12 Feb 2006 Posts: 5160 Location: Rochester, NY, USA

Posted: Thu Jun 01, 2006 5:02 pm Post subject: 


Steve R wrote:  Marty
The computer suggested that the next step was to eliminate 2 from r56c3 via
r5c3 = 2 or r6c3 = 2 => r1c3 ≠ 2 => r1c6 = 2 => r9c6 = 7
r9c6 = 7 => r9c8 = 2 => r5c8 & r6c8 ≠ 2 => r5c7 = 2 => r5c3 ≠ 2
r9c6 = 7 => r9c2 = 6 => r7c3 ≠ 6 => r6c3 = 6
So r5c1 = 2. There are probably beeter ways but I hope this helps.
Steve 
Thank you Steve but, unfortunately, I was unable to do anything beyond eliminating the other "2s" in c1 and r6, but I'll study it further. (I think you meant r6c1, not r5c1).
But that logic was interesting. It looked like the start of a DIC, as in what happens if the "2" in box 7 is in c3 or in c1. Except we found out that if the "2" was in c3, then it couldn't be in c3.
That is something that looks pretty difficult for a human solver to spot. 

Back to top 


Steve R
Joined: 24 Oct 2005 Posts: 289 Location: Birmingham, England

Posted: Thu Jun 01, 2006 9:02 pm Post subject: 


You’re right, Marty. Apologies for the typo.
The computer used a second chain to complete the solution I am well out of my depth here but it seams to be a challenging puzzle.
You are most welcome to more info; otherwise perhaps the grid should be left to the kangaroos,
Steve 

Back to top 


keith
Joined: 19 Sep 2005 Posts: 3173 Location: near Detroit, Michigan, USA

Posted: Thu Jun 01, 2006 9:58 pm Post subject: Upside Down? 


Marty,
I am told these puzzles are easier if you are upside down. (Or, rotate it the other way.)
Anyway, here is what Sudoku Susser says:
Quote: 
Code: 
++++
 6 1457 1247  179 3 257  8 579 457 
 358 13578 178  1789 15679 4  3679 35679 2 
 23458 9 2478  78 2567 25678  3467 1 3457 
++++
 3489 13468 5  2 467 3678  13479 3789 13478 
 7 348 2489  5 4 1  2349 2389 6 
 2348 13468 12468  3478 467 9  5 2378 13478 
++++
 589 2 6789  1379 1579 357  1367 4 13578 
 1 4578 4789  6 24579 2357  237 23578 3578 
 45 4567 3  147 8 257  1267 2567 9 
++++

* The following immediate observations can be made:
R5C5 must be <4>.
Deduction pass 1; 23 squares solved; 58 remaining.
* R9C4 is the only square in column 4 that can be <4>. It is thus pinned to that value.
From this deduction, the following moves are immediately forced:
R9C1 must be <5>.
Deduction pass 2; 25 squares solved; 56 remaining.
* R9C7 is the only square in row 9 that can be <1>. It is thus pinned to that value.
Deduction pass 3; 26 squares solved; 55 remaining.
* Squares R4C5 and R6C5 in column 5 form a simple naked pair. These 2 squares both contain the 2 possibilities <67>. Since each of the squares must contain one of the possibilities, they can be eliminated from the other squares in the column.
R2C5  removing <67> from <15679> leaving <159>.
R3C5  removing <67> from <2567> leaving <25>.
R7C5  removing <7> from <1579> leaving <159>.
R8C5  removing <7> from <2579> leaving <259>.
Deduction pass 4; 26 squares solved; 55 remaining.
* R3C6 is the only square in block 2 that can be <6>. It is thus pinned to that value.
Deduction pass 5; 27 squares solved; 54 remaining.
* R4C6 is the only square in column 6 that can be <8>. It is thus pinned to that value.
Deduction pass 6; 28 squares solved; 53 remaining.
* R6C4 is the only square in block 5 that can be <3>. It is thus pinned to that value.
Deduction pass 7; 29 squares solved; 52 remaining.
* Found a 6link Comprehensive Chain. If we assume that square R6C1 is <8> then we can make the following chain of conclusions:
R3C1 must be <2> (C1 pin), which means that
R3C5 must be <5> (force), which means that
R8C5 must be <2> (C5 pin), which means that
R8C3 must be <9> (R8 pin), which means that
R7C1 must be <8> (force), which means that
R6C1 can't be <8> (buddy contradiction).
Since this is logically inconsistent, R6C1 cannot be <8>.
(5 links were considered before finding this chain)
Deduction pass 8; 29 squares solved; 52 remaining.
* Found a 7link Comprehensive Chain. If we assume that square R6C8 is <2> then we can make the following chain of conclusions:
R6C1 must be <4> (force), which means that
R3C1 must be <2> (C1 pin), which means that
R3C5 must be <5> (force), which means that
R8C5 must be <2> (C5 pin), which means that
R9C6 must be <7> (force), which means that
R9C8 must be <2> (R9 pin), which means that
R6C8 can't be <2> (buddy contradiction).
Since this is logically inconsistent, R6C8 cannot be <2>.
(5 links were considered before finding this chain)
Deduction pass 9; 29 squares solved; 52 remaining.
* Intersection of row 6 with block 4. The value <2> only appears in one or more of squares R6C1, R6C2 and R6C3 of row 6. These squares are the ones that intersect with block 4. Thus, the other (nonintersecting) squares of block 4 cannot contain this value.
R5C3  removing <2> from <289> leaving <89>.
Deduction pass 10; 29 squares solved; 52 remaining.
* Found a Nishio contradiction. After 2 cycles, it became clear that R8C5 could not be a <5>.
R8C5  removing <5> from <259> leaving <29>.
Deduction pass 11; 29 squares solved; 52 remaining.
* Found a 6link Comprehensive Chain. If we assume that square R2C8 is <5> then we can make the following chain of conclusions:
R9C8 must be <6> (C8 pin), which means that
R9C6 must be <2> (R9 pin), which means that
R8C5 must be <9> (force), which means that
R3C5 must be <2> (C5 pin), which means that
R3C9 must be <5> (R3 pin), which means that
R2C8 can't be <5> (buddy contradiction).
Since this is logically inconsistent, R2C8 cannot be <5>.
(9 links were considered before finding this chain)
Deduction pass 12; 29 squares solved; 52 remaining.
* Found a 8link Comprehensive Chain. If we assume that square R3C1 is <2> then we can make the following chain of conclusions:
R6C1 must be <4> (force), which means that
R6C3 must be <2> (R6 pin), which means that
R7C3 must be <6> (C3 pin), which means that
R9C2 must be <7> (force), which means that
R9C6 must be <2> (force), which means that
R8C5 must be <9> (force), which means that
R3C5 must be <2> (C5 pin), which means that
R3C1 can't be <2> (buddy contradiction).
Since this is logically inconsistent, R3C1 cannot be <2>.
(9 links were considered before finding this chain)
Deduction pass 13; 29 squares solved; 52 remaining.
* R6C1 is the only square in column 1 that can be <2>. It is thus pinned to that value.
Deduction pass 14; 30 squares solved; 51 remaining.
* Squares R3C1 and R4C1 in column 1 and R3C7 and R4C7 in column 7 form a Simple XWing pattern on possibility <4>. All other instances of this possibility in rows 3 and 4 can be removed.
R4C2  removing <4> from <1346> leaving <136>.
R3C3  removing <4> from <2478> leaving <278>.
R3C9  removing <4> from <3457> leaving <357>.
R4C9  removing <4> from <1347> leaving <137>.
Deduction pass 15; 30 squares solved; 51 remaining.
* Found a 8link Comprehensive Chain. If we assume that square R1C3 is <4> then we can make the following chain of conclusions:
R1C6 must be <2> (R1 pin), which means that
R3C5 must be <5> (force), which means that
R8C5 must be <2> (C5 pin), which means that
R8C3 must be <9> (R8 pin), which means that
R5C3 must be <8> (force), which means that
R4C1 must be <9> (B4 pin), which means that
R3C1 must be <4> (C1 pin), which means that
R1C3 can't be <4> (buddy contradiction).
Since this is logically inconsistent, R1C3 cannot be <4>.
(9 links were considered before finding this chain)
Deduction pass 16; 30 squares solved; 51 remaining.
* Found a 4link Comprehensive Chain. If we assume that square R8C3 is <7> then we can make the following chain of conclusions:
R6C3 must be <4> (C3 pin), which means that
R7C3 must be <6> (C3 pin), which means that
R9C2 must be <7> (force), which means that
R8C3 can't be <7> (buddy contradiction).
Since this is logically inconsistent, R8C3 cannot be <7>.
(7 links were considered before finding this chain)
Deduction pass 17; 30 squares solved; 51 remaining.
* Found a 6link Comprehensive Chain. If we assume that square R8C3 is <9> then we can make the following chain of conclusions:
R6C3 must be <4> (C3 pin), which means that
R7C3 must be <6> (C3 pin), which means that
R9C2 must be <7> (force), which means that
R9C6 must be <2> (force), which means that
R8C5 must be <9> (force), which means that
R8C3 can't be <9> (buddy contradiction).
Since this is logically inconsistent, R8C3 cannot be <9>.
(9 links were considered before finding this chain)
Deduction pass 18; 30 squares solved; 51 remaining.
* R8C5 is the only square in row 8 that can be <9>. It is thus pinned to that value.
Deduction pass 19; 31 squares solved; 50 remaining.
* R3C5 is the only square in column 5 that can be <2>. It is thus pinned to that value.
Deduction pass 20; 32 squares solved; 49 remaining.
* R1C3 is the only square in row 1 that can be <2>. It is thus pinned to that value.
Deduction pass 21; 33 squares solved; 48 remaining.
* R3C9 is the only square in row 3 that can be <5>. It is thus pinned to that value.
Deduction pass 22; 34 squares solved; 47 remaining.
* R8C8 is the only square in column 8 that can be <5>. It is thus pinned to that value.
Deduction pass 23; 35 squares solved; 46 remaining.
* Squares R3C3 and R3C4 in row 3 form a simple naked pair. These 2 squares both contain the 2 possibilities <78>. Since each of the squares must contain one of the possibilities, they can be eliminated from the other squares in the row.
R3C1  removing <8> from <348> leaving <34>.
R3C7  removing <7> from <347> leaving <34>.
Deduction pass 24; 35 squares solved; 46 remaining.
* Intersection of column 8 with block 6. The value <8> only appears in one or more of squares R4C8, R5C8 and R6C8 of column 8. These squares are the ones that intersect with block 6. Thus, the other (nonintersecting) squares of block 6 cannot contain this value.
R6C9  removing <8> from <1478> leaving <147>.
Deduction pass 25; 35 squares solved; 46 remaining.
* Found a 4link Comprehensive Chain. If we assume that square R1C2 is <7> then we can make the following chain of conclusions:
R1C4 must be <1> (R1 pin), which means that
R2C5 must be <5> (force), which means that
R1C6 must be <7> (force), which means that
R1C2 can't be <7> (buddy contradiction).
Since this is logically inconsistent, R1C2 cannot be <7>.
(10 links were considered before finding this chain)
Deduction pass 26; 35 squares solved; 46 remaining.
* Found a 4link Comprehensive Chain. If we assume that square R7C3 is <7> then we can make the following chain of conclusions:
R7C1 must be <9> (R7 pin), which means that
R2C1 must be <8> (C1 pin), which means that
R3C3 must be <7> (force), which means that
R7C3 can't be <7> (buddy contradiction).
Since this is logically inconsistent, R7C3 cannot be <7>.
(10 links were considered before finding this chain)
Deduction pass 27; 35 squares solved; 46 remaining.
* Intersection of column 3 with block 1. The value <7> only appears in one or more of squares R1C3, R2C3 and R3C3 of column 3. These squares are the ones that intersect with block 1. Thus, the other (nonintersecting) squares of block 1 cannot contain this value.
R2C2  removing <7> from <13578> leaving <1358>.
Deduction pass 28; 35 squares solved; 46 remaining.
* Found a 4link Comprehensive Chain. If we assume that square R8C7 is <7> then we can make the following chain of conclusions:
R9C8 must be <2> (B9 pin), which means that
R9C2 must be <6> (R9 pin), which means that
R8C2 must be <7> (C2 pin), which means that
R8C7 can't be <7> (buddy contradiction).
Since this is logically inconsistent, R8C7 cannot be <7>.
(11 links were considered before finding this chain)
Deduction pass 29; 35 squares solved; 46 remaining.
* Found a 4link Comprehensive Chain. If we assume that square R1C2 is <4> then we can make the following chain of conclusions:
R1C4 must be <1> (R1 pin), which means that
R2C5 must be <5> (force), which means that
R1C6 must be <7> (force), which means that
R1C2 must be <5> (R1 pin (which contradicts our original choice)).
Since this is logically inconsistent, R1C2 cannot be <4>.
(12 links were considered before finding this chain)
Deduction pass 30; 35 squares solved; 46 remaining.
* R1C9 is the only square in row 1 that can be <4>. It is thus pinned to that value.
From this deduction, the following moves are immediately forced:
R3C7 must be <3>.
R3C1 must be <4>.
R8C7 must be <2>.
R5C7 must be <9>.
R5C3 must be <8>.
R5C2 must be <3>.
R3C3 must be <7>.
R8C3 must be <4>.
R3C4 must be <8>.
R2C3 must be <1>.
R5C8 must be <2>.
R4C1 must be <9>.
R2C5 must be <5>.
R6C3 must be <6>.
R1C2 must be <5>.
R2C2 must be <8>.
R7C5 must be <1>.
R1C6 must be <7>.
R7C1 must be <8>.
R6C5 must be <7>.
R7C3 must be <9>.
R4C2 must be <1>.
R6C8 must be <8>.
R6C9 must be <1>.
R4C5 must be <6>.
R6C2 must be <4>.
R2C1 must be <3>.
R8C2 must be <7>.
R7C4 must be <7>.
R8C6 must be <3>.
R9C2 must be <6>.
R8C9 must be <8>.
R9C8 must be <7>.
R9C6 must be <2>.
R1C8 must be <9>.
R4C8 must be <3>.
R7C7 must be <6>.
R7C9 must be <3>.
R7C6 must be <5>.
R2C4 must be <9>.
R1C4 must be <1>.
R2C8 must be <6>.
R2C7 must be <7>.
R4C9 must be <7>.
R4C7 must be <4>.
Deduction pass 31; 81 squares solved; 0 remaining.
Solution found!
Heuristics used:
1 x Nishio
11 x Comprehensive Forcing Chains
1 x Simple XWing
3 x Intersection Removal
2 x Simple Naked Sets
12 x Pinned Squares
Deduction completed...
Code: 
++++
 6 5 2  1 3 7  8 9 4 
 3 8 1  9 5 4  7 6 2 
 4 9 7  8 2 6  3 1 5 
++++
 9 1 5  2 6 8  4 3 7 
 7 3 8  5 4 1  9 2 6 
 2 4 6  3 7 9  5 8 1 
++++
 8 2 9  7 1 5  6 4 3 
 1 7 4  6 9 3  2 5 8 
 5 6 3  4 8 2  1 7 9 
++++

Solution found!

Best wishes,
Keith 

Back to top 


Marty R.
Joined: 12 Feb 2006 Posts: 5160 Location: Rochester, NY, USA

Posted: Fri Jun 02, 2006 12:10 am Post subject: 


My head is swimming! I'm gonna kick the habit and stick to crosswords and Jumbles! 

Back to top 




You cannot post new topics in this forum You cannot reply to topics in this forum You cannot edit your posts in this forum You cannot delete your posts in this forum You cannot vote in polls in this forum

Powered by phpBB © 2001, 2005 phpBB Group
