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Brick wall

 
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Marty R.



Joined: 12 Feb 2006
Posts: 5162
Location: Rochester, NY, USA

PostPosted: Wed May 31, 2006 5:18 pm    Post subject: Brick wall Reply with quote

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:
-------------
|6--|-3-|8--|
|---|--4|--2|
|-9-|---|-1-|
-------------
|--5|2--|---|
|7--|5-1|--6|
|---|--9|5--|
-------------
|-2-|---|-4-|
|1--|6--|---|
|--3|-8-|--9|
-------------
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Steve R



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

PostPosted: Wed May 31, 2006 7:21 pm    Post subject: Reply with quote

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
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Marty R.



Joined: 12 Feb 2006
Posts: 5162
Location: Rochester, NY, USA

PostPosted: Thu Jun 01, 2006 5:02 pm    Post subject: Reply with quote

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.
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Steve R



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

PostPosted: Thu Jun 01, 2006 9:02 pm    Post subject: Reply with quote

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
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keith



Joined: 19 Sep 2005
Posts: 3175
Location: near Detroit, Michigan, USA

PostPosted: Thu Jun 01, 2006 9:58 pm    Post subject: Upside Down? Reply with quote

Marty,

I am told these puzzles are easier if you are upside down. (Or, rotate it the other way.) Laughing

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 6-link 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 7-link 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 (non-intersecting) 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 6-link 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 8-link 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 X-Wing 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 8-link 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 4-link 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 6-link 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 (non-intersecting) 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 4-link 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 4-link 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 (non-intersecting) 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 4-link 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 4-link 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 X-Wing
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
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Marty R.



Joined: 12 Feb 2006
Posts: 5162
Location: Rochester, NY, USA

PostPosted: Fri Jun 02, 2006 12:10 am    Post subject: Reply with quote

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