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keith
Joined: 19 Sep 2005 Posts: 3223 Location: near Detroit, Michigan, USA

Posted: Mon May 22, 2006 4:39 am Post subject: Two solutions? 


Some time ago, David Bryant (dcb) and I discussed puzzles that have a small number (two?) of solutions. We did not have any examples.
From another thread, Tarek tells me this one has exactly two solutions:
Code:  1 . .  . . .  5 8 .
8 . .  . . 1  . . 6
. 6 .  . 2 .  . . .
++
. . .  5 9 .  . . .
. . .  . 7 .  . . .
. 9 3  . . 2  . . .
++
. 1 .  . . .  8 . 4
. . .  9 . 7  . 5 .
6 . .  2 . .  7 1 .

(Not symmetric, but don't you think this looks like a tropical aquarium fish?)
This is a really tough puzzle!. My solver runs out of steam at this point:
Code: 
++++
 1 37 24  46 36 9  5 8 27 
 8 57 24  37 45 1  39 29 6 
 39 6 59  78 2 58  14 34 17 
++++
 47 24 16  5 9 68  23 37 18 
 25 8 16  14 7 3  29 46 59 
 57 9 3  18 46 2  14 67 58 
++++
 29 1 7  36 35 56  8 29 4 
 34 24 8  9 1 7  6 5 23 
 6 35 59  2 8 4  7 1 39 
++++

This is the first BUG pattern I have seen, rather than avoided. I was hoping to find some kind of lesson on uniqueness methods, and I am still looking.
Choosing <3> or <7> in R1C2 seems to give the two solutions.
Comments, anyone?
Keith 

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David Bryant
Joined: 29 Jul 2005 Posts: 559 Location: Denver, Colorado

Posted: Mon May 22, 2006 4:39 pm Post subject: Puzzles with multiple solutions 


Keith wrote:  Some time ago, David Bryant (dcb) and I discussed puzzles that have a small number (two?) of solutions. We did not have any examples.

Here's a link to that earlier discussion. I didn't have an example, but I did explain the simplest case of a puzzle with two solutions:
Code:  4 . 5 . . . . . . 5 . 4 . . . . . . (r1)
                 
                 
5 . 4 . . . . . . 4 . 5 . . . . . . (r9) 
This is just the "nonunique rectangle" pattern that has been discussed at great length already.
Keith wrote:  Choosing <3> or <7> in R1C2 seems to give the two solutions. 
Actually, setting one or the other permissible value in any one of the unresolved cells will yield two solutions  the "BUG" pattern is just like a "UR" in a certain sense  it can only exist in one of two possible states.
Code:  1 3/7 2/4 4/6 3/6 9 5 8 2/7
8 5/7 2/4 3/7 4/5 1 3/9 2/9 6
3/9 6 5/9 7/8 2 5/8 1/4 3/4 1/7
4/7 2/4 1/6 5 9 6/8 2/3 3/7 1/8
2/5 8 1/6 1/4 7 3 2/9 4/6 5/9
5/7 9 3 1/8 4/6 2 1/4 6/7 5/8
2/9 1 7 3/6 3/5 5/6 8 2/9 4
3/4 2/4 8 9 1 7 6 5 2/3
6 3/5 5/9 2 8 4 7 1 3/9 
We can understand why there are exactly two solutions by noticing that each "possible" value occurs twice in each column, and row, and 3x3 box. So setting one value in any unresolved cell will make a solution drop out.
Incidentally, your solver must be better than mine, Keith. After deriving the two possible solutions I set up a puzzle with 33 clues (the cells that are the same in both solutions).
Code:  **
1....958.
8....1..6
.6..2....
++
...59....
.8..73...
.93..2...
++
.17...8.4
..891765.
6..28471.
** 
After making all the obvious reductions I got the grid to this state.
Code:  1 3/7 2/4 3/4/6/7 3/4/6 9 5 8 2/7
8 3/5/7 2/4/5/9 3/4/7 3/4/5 1 2/3/4/9 2/3/4/7/9 6
3/9 6 4/5/9 3/4/7/8 2 5/8 1/3/4/9 3/4/7/9 1/7
2/4/7 2/4 1/6 5 9 6/8 1/2/3/4 3/4/6/7 1/2/7/8
2/4/5 8 1/6 1/4/6 7 3 1/2/4/9 2/4/6/9 1/2/5/9
4/5/7 9 3 1/4/6/8 4/6 2 1/4 4/6/7 1/5/7/8
2/9 1 7 3/6 3/5/6 5/6 8 2/9 4
2/3/4 2/4 8 9 1 7 6 5 2/3
6 3/5 5/9 2 8 4 7 1 3/9 
The only simple technique I can see for reducing this grid to the "BUG" pattern shown earlier is to assume the existence of the "BUG" pattern. and make eliminations on that basis. Can you explain how to get from this state to the "BUG" pattern _without_ just assuming it's a BUG? dcb 

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Angel
Joined: 26 Mar 2006 Posts: 31

Posted: Mon May 22, 2006 5:51 pm Post subject: 


This site regularly features puzzles that have exactly two or three solutions.
Look for the special puzzles. 

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

Posted: Mon May 22, 2006 7:02 pm Post subject: Two solutions. 


The most familiar puzzle with two solutions is Professor Royle’s:
Code: 
5 . 2 . . . 4 . .
. . . 7 1 . . . 3
. . . . . . . . .
. . . . . 4 6 . .
. 7 . 2 . . . . .
. 1 . . . . . . .
6 . . . . 2 . . .
. . . . 3 . . 1 .
4 . . . . . . . .

At the time I came across it, no twosolution puzzle with fewer than 16 clues was known. It would be interesting to know if things have changed.
The puzzle is worth trying. A few tricks are required and the end of the road is blindingly obvious.
Steve 

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ravel
Joined: 21 Apr 2006 Posts: 536

Posted: Mon May 22, 2006 8:02 pm Post subject: Re: Two solutions. 


Steve R wrote: 
At the time I came across it, no twosolution puzzle with fewer than 16 clues was known. It would be interesting to know if things have changed.

Nothing changed, gfroyles 16clue is still the only one that has only 2 solutions. It seems, that no one anymore believes that there could be a 16clue with one solution, but they are far away from being able to prove that.
On the other side, Ocean and coloin then have recently found minimal sudokus (that have no redundant clues) with 35 clues. 

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keith
Joined: 19 Sep 2005 Posts: 3223 Location: near Detroit, Michigan, USA

Posted: Mon May 22, 2006 10:16 pm Post subject: 


David, starting from here:
Code: 
++++
 1 37 24  3467 346 9  5 8 27 
 8 357 2459  347 345 1  2349 23479 6 
 39 6 459  3478 2 58  1349 3479 17 
++++
 247 24 16  5 9 68  1234 3467 1278 
 245 8 16  146 7 3  1249 2469 1259 
 457 9 3  1468 46 2  14 467 1578 
++++
 29 1 7  36 356 56  8 29 4 
 234 24 8  9 1 7  6 5 23 
 6 35 59  2 8 4  7 1 39 
++++

Credits: The solver is Sudoku Susser 2.5.3 by Robert Woodhead (MadOverlord), with the uniqueness tests (Unique Loops and BUG) turned off.
Here are the details. This is not for the squeamish!
Code: 
The current puzzle is invalid; it has multiple solutions. Deductive methods may make some pointless progress, but some of them may take a very long time to run (in particular the more complex ones). Remember that you can hold down cmd. to cancel deducing at any time.
Deduction pass 1; 33 squares solved; 48 remaining.
* Found a 6link Simple Forcing Loop. If we assume that square R1C2 is <7> then we can make the following chain of conclusions:
R1C9 must be <2>, which means that
R8C9 must be <3>, which means that
R9C9 must be <9>, which means that
R9C3 must be <5>, which means that
R9C2 must be <3>, which means that
R1C2 must be <7>.
This forms a inherently bidirectional loop through the puzzle that permits reductions. On each edge of the loop, one of the two squares must have a particular value, so their common buddies cannot contain that value, as follows:
One of R1C2 and R1C9 must be <7>.
One of R1C9 and R8C9 must be <2>.
One of R8C9 and R9C9 must be <3>.
One of R9C9 and R9C3 must be <9>.
One of R9C3 and R9C2 must be <5>.
One of R9C2 and R1C2 must be <3>.
Thus we can deduce that:
R1C4  cannot contain <7> because of R1C2 and R1C9.
R4C9  cannot contain <2> because of R1C9 and R8C9.
R5C9  cannot contain <2> because of R1C9 and R8C9.
R2C2  cannot contain <3> because of R9C2 and R1C2.
(3 links were considered before finding this loop)
Deduction pass 2; 33 squares solved; 48 remaining.
* Found a 4link Comprehensive Chain. If we assume that square R4C9 is <7> then we can make the following chain of conclusions:
R6C9 must be <8> (C9 pin), which means that
R6C1 must be <5> (R6 pin), which means that
R4C1 must be <7> (C1 pin), which means that
R4C9 can't be <7> (buddy contradiction).
Since this is logically inconsistent, R4C9 cannot be <7>.
(14 links were considered before finding this chain)
Deduction pass 3; 33 squares solved; 48 remaining.
* Squares R4C3<16>, R4C6<68> and R4C9<18> in row 4 form a comprehensive naked triplet. These 3 squares can only contain the 3 possibilities <168>. Since each of the squares must contain one of the possibilities, they can be eliminated from the other squares in the row.
R4C7  removing <1> from <1234> leaving <234>.
R4C8  removing <6> from <3467> leaving <347>.
Deduction pass 4; 33 squares solved; 48 remaining.
* Found a 4link Simple Forcing Loop. If we assume that square R4C9 is <1> then we can make the following chain of conclusions:
R6C7 must be <4>, which means that
R6C5 must be <6>, which means that
R4C6 must be <8>, which means that
R4C9 must be <1>.
This forms a inherently bidirectional loop through the puzzle that permits reductions. On each edge of the loop, one of the two squares must have a particular value, so their common buddies cannot contain that value, as follows:
One of R4C9 and R6C7 must be <1>.
One of R6C7 and R6C5 must be <4>.
One of R6C5 and R4C6 must be <6>.
One of R4C6 and R4C9 must be <8>.
Thus we can deduce that:
R6C9  cannot contain <1> because of R4C9 and R6C7.
R5C7  cannot contain <1> because of R4C9 and R6C7.
R5C9  cannot contain <1> because of R4C9 and R6C7.
R6C1  cannot contain <4> because of R6C7 and R6C5.
R6C4  cannot contain <4> because of R6C7 and R6C5.
R6C8  cannot contain <4> because of R6C7 and R6C5.
R6C4  cannot contain <6> because of R6C5 and R4C6.
R5C4  cannot contain <6> because of R6C5 and R4C6.
(3 links were considered before finding this loop)
Deduction pass 5; 33 squares solved; 48 remaining.
* Found a 7link Simple Forcing Chain. If we assume that square R1C5 is <4> then we can make the following chain of conclusions:
R1C3 must be <2>, which means that
R1C9 must be <7>, which means that
R3C9 must be <1>, which means that
R4C9 must be <8>, which means that
R4C6 must be <6>, which means that
R6C5 must be <4>, which means that
R1C5 can't be <4>.
Since this is logically inconsistent, R1C5 cannot be <4>.
(19 links were considered before finding this chain)
Deduction pass 6; 33 squares solved; 48 remaining.
* Found a 5link Comprehensive Chain. If we assume that square R2C3 is <9> then we can make the following chain of conclusions:
R1C3 must be <2> (C3 pin), which means that
R1C9 must be <7> (force), which means that
R1C2 must be <3> (force), which means that
R3C1 must be <9> (force), which means that
R2C3 can't be <9> (buddy contradiction).
Since this is logically inconsistent, R2C3 cannot be <9>.
(30 links were considered before finding this chain)
Deduction pass 7; 33 squares solved; 48 remaining.
* Intersection of row 2 with block 3. The value <9> only appears in one or more of squares R2C7, R2C8 and R2C9 of row 2. These squares are the ones that intersect with block 3. Thus, the other (nonintersecting) squares of block 3 cannot contain this value.
R3C7  removing <9> from <1349> leaving <134>.
R3C8  removing <9> from <3479> leaving <347>.
Deduction pass 8; 33 squares solved; 48 remaining.
* Found a 4link Comprehensive Chain. If we assume that square R2C5 is <3> then we can make the following chain of conclusions:
R6C5 must be <4> (C5 pin), which means that
R4C6 must be <6> (B5 pin), which means that
R3C6 must be <8> (C6 pin), which means that
R2C5 must be <5> (B2 pin (which contradicts our original choice)).
Since this is logically inconsistent, R2C5 cannot be <3>.
(34 links were considered before finding this chain)
Deduction pass 9; 33 squares solved; 48 remaining.
* Found a 5link Simple Forcing Loop. If we assume that square R1C5 is <3> then we can make the following chain of conclusions:
R1C2 must be <7>, which means that
R2C2 must be <5>, which means that
R2C5 must be <4>, which means that
R6C5 must be <6>, which means that
R1C5 must be <3>.
This forms a inherently bidirectional loop through the puzzle that permits reductions. On each edge of the loop, one of the two squares must have a particular value, so their common buddies cannot contain that value, as follows:
One of R1C5 and R1C2 must be <3>.
One of R1C2 and R2C2 must be <7>.
One of R2C2 and R2C5 must be <5>.
One of R2C5 and R6C5 must be <4>.
One of R6C5 and R1C5 must be <6>.
Thus we can deduce that:
R1C4  cannot contain <3> because of R1C5 and R1C2.
R2C3  cannot contain <5> because of R2C2 and R2C5.
R7C5  cannot contain <6> because of R6C5 and R1C5.
(5 links were considered before finding this loop)
Deduction pass 10; 33 squares solved; 48 remaining.
* Squares R1C3 and R2C3 in column 3 form a simple naked pair. These 2 squares both contain the 2 possibilities <24>. Since each of the squares must contain one of the possibilities, they can be eliminated from the other squares in the column.
R3C3  removing <4> from <459> leaving <59>.
Deduction pass 11; 33 squares solved; 48 remaining.
* Found a 4link Simple Forcing Loop. If we assume that square R2C5 is <4> then we can make the following chain of conclusions:
R1C4 must be <6>, which means that
R7C4 must be <3>, which means that
R7C5 must be <5>, which means that
R2C5 must be <4>.
This forms a inherently bidirectional loop through the puzzle that permits reductions. On each edge of the loop, one of the two squares must have a particular value, so their common buddies cannot contain that value, as follows:
One of R2C5 and R1C4 must be <4>.
One of R1C4 and R7C4 must be <6>.
One of R7C4 and R7C5 must be <3>.
One of R7C5 and R2C5 must be <5>.
Thus we can deduce that:
R2C4  cannot contain <4> because of R2C5 and R1C4.
R3C4  cannot contain <4> because of R2C5 and R1C4.
(6 links were considered before finding this loop)
Deduction pass 12; 33 squares solved; 48 remaining.
* Intersection of row 3 with block 3. The values <14> only appears in one or more of squares R3C7, R3C8 and R3C9 of row 3. These squares are the ones that intersect with block 3. Thus, the other (nonintersecting) squares of block 3 cannot contain these values.
R2C7  removing <4> from <2349> leaving <239>.
R2C8  removing <4> from <23479> leaving <2379>.
Deduction pass 13; 33 squares solved; 48 remaining.
* Found a 5link Simple Forcing Loop. If we assume that square R2C4 is <7> then we can make the following chain of conclusions:
R2C2 must be <5>, which means that
R2C5 must be <4>, which means that
R1C4 must be <6>, which means that
R7C4 must be <3>, which means that
R2C4 must be <7>.
This forms a inherently bidirectional loop through the puzzle that permits reductions. On each edge of the loop, one of the two squares must have a particular value, so their common buddies cannot contain that value, as follows:
One of R2C4 and R2C2 must be <7>.
One of R2C2 and R2C5 must be <5>.
One of R2C5 and R1C4 must be <4>.
One of R1C4 and R7C4 must be <6>.
One of R7C4 and R2C4 must be <3>.
Thus we can deduce that:
R2C8  cannot contain <7> because of R2C4 and R2C2.
R3C4  cannot contain <3> because of R7C4 and R2C4.
(10 links were considered before finding this loop)
Deduction pass 14; 33 squares solved; 48 remaining.
* Found a 6link Simple Forcing Loop. If we assume that square R3C9 is <7> then we can make the following chain of conclusions:
R3C4 must be <8>, which means that
R3C6 must be <5>, which means that
R7C6 must be <6>, which means that
R4C6 must be <8>, which means that
R4C9 must be <1>, which means that
R3C9 must be <7>.
This forms a inherently bidirectional loop through the puzzle that permits reductions. On each edge of the loop, one of the two squares must have a particular value, so their common buddies cannot contain that value, as follows:
One of R3C9 and R3C4 must be <7>.
One of R3C4 and R3C6 must be <8>.
One of R3C6 and R7C6 must be <5>.
One of R7C6 and R4C6 must be <6>.
One of R4C6 and R4C9 must be <8>.
One of R4C9 and R3C9 must be <1>.
Thus we can deduce that:
R3C8  cannot contain <7> because of R3C9 and R3C4.
(35 links were considered before finding this loop)
Deduction pass 15; 33 squares solved; 48 remaining.
* Intersection of column 8 with block 6. The values <67> 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 these values.
R6C9  removing <7> from <578> leaving <58>.
Deduction pass 16; 33 squares solved; 48 remaining.
* Found a 4link Comprehensive Chain. If we assume that square R4C8 is <4> then we can make the following chain of conclusions:
R6C8 must be <7> (C8 pin), which means that
R6C5 must be <6> (R6 pin), which means that
R6C7 must be <4> (R6 pin), which means that
R4C8 can't be <4> (buddy contradiction).
Since this is logically inconsistent, R4C8 cannot be <4>.
(17 links were considered before finding this chain)
Deduction pass 17; 33 squares solved; 48 remaining.
* Found a 7link Simple Forcing Chain. If we assume that square R2C8 is <3> then we can make the following chain of conclusions:
R2C4 must be <7>, which means that
R2C2 must be <5>, which means that
R2C5 must be <4>, which means that
R6C5 must be <6>, which means that
R6C8 must be <7>, which means that
R4C8 must be <3>, which means that
R2C8 can't be <3>.
Since this is logically inconsistent, R2C8 cannot be <3>.
(46 links were considered before finding this chain)
Deduction pass 18; 33 squares solved; 48 remaining.
* Squares R2C8 and R7C8 in column 8 form a simple naked pair. These 2 squares both contain the 2 possibilities <29>. Since each of the squares must contain one of the possibilities, they can be eliminated from the other squares in the column.
R5C8  removing <29> from <2469> leaving <46>.
Deduction pass 19; 33 squares solved; 48 remaining.
* Intersection of block 6 with column 7. The value <2> only appears in one or more of squares R4C7, R5C7 and R6C7 of block 6. These squares are the ones that intersect with column 7. Thus, the other (nonintersecting) squares of column 7 cannot contain this value.
R2C7  removing <2> from <239> leaving <39>.
Deduction pass 20; 33 squares solved; 48 remaining.
* Squares R5C3<16>, R5C4<14> and R5C8<46> in row 5 form a comprehensive naked triplet. These 3 squares can only contain the 3 possibilities <146>. Since each of the squares must contain one of the possibilities, they can be eliminated from the other squares in the row.
R5C1  removing <4> from <245> leaving <25>.
R5C7  removing <4> from <249> leaving <29>.
Deduction pass 21; 33 squares solved; 48 remaining.
* Intersection of block 4 with row 4. The value <4> only appears in one or more of squares R4C1, R4C2 and R4C3 of block 4. These squares are the ones that intersect with row 4. Thus, the other (nonintersecting) squares of row 4 cannot contain this value.
R4C7  removing <4> from <234> leaving <23>.
Deduction pass 22; 33 squares solved; 48 remaining.
* A set of 2 squares form a simple hidden pair. R3C7 and R6C7 all contain the 2 possibilities <14>. No other squares in column 7 have those possibilities. Since the 2 squares are the only possible locations for 2 possible values, any additional possibilities these squares have (if any) can be eliminated. These squares now become a simple naked pair.
R3C7  removing <3> from <134> leaving <14>.
Deduction pass 23; 33 squares solved; 48 remaining.
* Found a 6link Simple Forcing Loop. If we assume that square R7C1 is <2> then we can make the following chain of conclusions:
R5C1 must be <5>, which means that
R5C9 must be <9>, which means that
R9C9 must be <3>, which means that
R8C9 must be <2>, which means that
R7C8 must be <9>, which means that
R7C1 must be <2>.
This forms a inherently bidirectional loop through the puzzle that permits reductions. On each edge of the loop, one of the two squares must have a particular value, so their common buddies cannot contain that value, as follows:
One of R7C1 and R5C1 must be <2>.
One of R5C1 and R5C9 must be <5>.
One of R5C9 and R9C9 must be <9>.
One of R9C9 and R8C9 must be <3>.
One of R8C9 and R7C8 must be <2>.
One of R7C8 and R7C1 must be <9>.
Thus we can deduce that:
R4C1  cannot contain <2> because of R7C1 and R5C1.
R8C1  cannot contain <2> because of R7C1 and R5C1.
(23 links were considered before finding this loop)
Deduction pass 24; 33 squares solved; 48 remaining.
Heuristics used:
4 x Comprehensive Forcing Chains
2 x Simple Forcing Chains
7 x Simple Forcing Loops
2 x Comprehensive Naked Sets
5 x Intersection Removal
1 x Simple Hidden Sets
2 x Simple Naked Sets
Deduction completed...
The current puzzle is invalid; it has multiple solutions. Deductive methods may make some pointless progress, but some of them may take a very long time to run (in particular the more complex ones). Remember that you can hold down cmd. to cancel deducing at any time.
++++
 1 37 24  46 36 9  5 8 27 
 8 57 24  37 45 1  39 29 6 
 39 6 59  78 2 58  14 34 17 
++++
 47 24 16  5 9 68  23 37 18 
 25 8 16  14 7 3  29 46 59 
 57 9 3  18 46 2  14 67 58 
++++
 29 1 7  36 35 56  8 29 4 
 34 24 8  9 1 7  6 5 23 
 6 35 59  2 8 4  7 1 39 
++++
* None of the currently selected rules can be applied to the puzzle in order to provide a hint.

Best wishes,
Keith 

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keith
Joined: 19 Sep 2005 Posts: 3223 Location: near Detroit, Michigan, USA

Posted: Mon May 22, 2006 11:00 pm Post subject: Really 


Angel wrote:  This site regularly features puzzles that have exactly two or three solutions.
Look for the special puzzles. 
Angel,
I have not yet looked at the site. But, I am surprised at your statement. All I have seen so far is that nonunique puzzles have only zero or an even number of solutions.
Maybe I am wrong: The puzzles to which Uniquess solution methods apply, are those which otherwise would have zero or an even number of solutions.
We learn every day!
Best wishes,
Keith 

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David Bryant
Joined: 29 Jul 2005 Posts: 559 Location: Denver, Colorado

Posted: Mon May 22, 2006 11:33 pm Post subject: Here's one with 3 solutions 


Keith wrote:  All I have seen so far is that nonunique puzzles have only zero or an even number of solutions. 
An example I posted during our discussion (in February) has exactly three solutions.
Code:  . . . . 1 . . 4 8
9 . . . . . . 2 .
. 5 3 7 . . . . .
. . . . . 5 9 6 .
3 . . . 9 . . . 1
. 8 9 4 . . . . .
. 3 . . . 6 8 5 .
. . 2 . . . . . 6
5 4 . . 7 . . . . 
After making all the obvious moves you will reach this position.
Code:  6 2 7 35 1 9 35 4 8
9 1 8 6 45 34 357 2 57
4 5 3 7 8 2 6 1 9
1 7 4 8 3 5 9 6 2
3 6 5 2 9 7 4 8 1
2 8 9 4 6 1 57 3 57
7 3 1 9 2 6 8 5 4
8 9 2 35 45 34 1 7 6
5 4 6 1 7 8 2 9 3 
Here, if you apply "uniqueness" to the r26c79 rectangle you will conclude that r2c7 = 3, and you can proceed to a solution.
But that solution is not unique! You can also set r1c7 = 3, and find two more solutions (there are two more because of the possible permutation of the digits "5" and "7" around the corners of the rectangle). dcb
PS The puzzle we recently kicked around with zoltag over on the Nightmare forum had five solutions, as I recall. 

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ravel
Joined: 21 Apr 2006 Posts: 536

Posted: Tue May 23, 2006 7:19 am Post subject: Re: Here's one with 3 solutions 


David Bryant wrote: 
Here, if you apply "uniqueness" to the r26c79 rectangle you will conclude that r2c7 = 3, and you can proceed to a solution.

You can also use the BUG+1 (only 1 cell with 3 candidates) and set r2c7=5 (there are 2 more 5's in the same row and column). So in this case 2 uniqueness methods give different solutions. 

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Angel
Joined: 26 Mar 2006 Posts: 31

Posted: Tue May 23, 2006 7:39 am Post subject: Re: Really 


keith wrote: 
Angel,
I have not yet looked at the site. But, I am surprised at your statement. 
Keith,
I should mention here that I have not tried these puzzles myself, I've merely noticed they exist on the site, and pass on the information to you guys who are interested. Maybe I've misunderstood what they are offering compared to what you are discussing. 

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