Comments...
Sunday 2-May-2010
... by: Wiking 48
Hi Harmen Dijkstra,I'm quite new on this.
why do you bother with such a non-defined sudoku?
There are many more than 17 sulotions! Check for yourself, you see that 24, 42 in the left of your grid can be excanged.
Is there any idea to fight against non-defined?
Upper left corner can also be 1, 2 or 3 which gives a lot of solutions..
Down left corner in your example is 7 in one example which is quite common among the sulotions, 1 is more rare but possible...
Friday 5-Mar-2010
... by: Harmen Dijkstra
I made this sudoku:. 9 . | 5 . . | 6 4 7
6 . 5 | . . 7 | . 9 3
. 7 . | . . . | 2 5 8
-----------------------
. 5 6 | . 7 . | 9 . .
. . 7 |9 . 5| 3 8 6
. . 3 |8 . 2| 4 7 5
-----------------------
. . . | . 5 .| 8 3 9
5 6 . | . . .| 7 2 .
. . . | . . 8| 5 6 4
If you filling this sudoku in the sudoku solver, the solution Count will say that there are 17 solutions. But if you try TAKE STEP then you will get only this sudoku (with Unique Rectangles):
2 9 1 | 5 8 3 | 6 4 7
6 8 5 | 2 4 7 | 1 9 3
3 7 4 | 1 9 6 | 2 5 8
-----------------------
8 5 6 | 3 7 4 | 9 1 2
4 2 7 | 9 1 5 | 3 8 6
9 1 3 | 8 6 2 | 4 7 5
-----------------------
7 4 2 | 6 5 1 | 8 3 9
5 6 8 | 4 3 9 | 7 2 1
1 3 9 | 7 2 8 | 5 6 4
But there are more solutions, for example:
1 9 2 | 5 8 3 | 6 4 7
6 8 5 | 2 4 7 | 1 9 3
3 7 4 | 6 9 1 | 2 5 8
-----------------------
8 5 6 | 3 7 4 | 9 1 2
2 4 7 | 9 1 5 | 3 8 6
9 1 3 | 8 6 2 | 4 7 5
-----------------------
4 2 1 | 7 5 6 | 8 3 9
5 6 8 | 4 3 9 | 7 2 1
7 3 9 | 1 2 8 | 5 6 4
The sudoku solver said that (because of Unique Rectangles) R7C4 has to be a 6, but in this example you see that you get an other solution with a 7.
Is this a fault in the sudoku solver?
Wednesday 3-Mar-2010
... by: CS Vidyasagar
1. You have given very simple and lucid explanation of difficult concept of Unique Rectangles of four types. One comes across such situations in number of Sudoku puzzles of varying degrees of difficulty. The clear understanding of unique rectangles helps one to get out of log jam one is subjected to frequently in these interesting puzzles. I do not know whether any more explanation can be given to what you have done. Only one has to read your article may be twice but keep the most important lesson you brought out loud and clear that " DO NOT HAVE DEADLY PATTERN OF HAVING FOUR CONJUGATE PAIRS" as that leads to multple solutions. Sudoku permits only unique solution.2. One has to understand the logic and solution becomes quite clear. You have seleted many good illustrations which have explained the logic clearly and unambiguosly.
3. Looking forward to more such illustrations.
Kudos for wonderful work you are doing.
with regards,
Vidyasagar
Tuesday 23-Feb-2010
... by: Lea Hayes
Just to clarify my question:Rectangle Cells:
A(1,5,6) B(1,5,6)
C(1,5) D(1,5)
Why remove 1 from both A and B and not either of the following:
A(5,6) B(1,5)
C(1,5) D(1,5)
or
A(1,5) B(5,6)
C(1,5) D(1,5)
Tuesday 23-Feb-2010
... by: Lea Hayes
Re: Type 4 Unique Rectangles - Cracking the Rectangle with Conjugate PairsI do not understand how you came to the conclusion that 1 can be removed? Why not remove 6 instead?
Friday 22-Jan-2010
... by: John_Ha
AndiI too was puzzling over that but I think the answer is simple. See Fig 2. (5 and 7) are unsolved in Col 4 otherwise they wouldn't be candidates in R2C4. Similarly (5 and 7) are unsolved in Col 6. Similarly (5 and 7) are unsolved in Row 5. Because the cell at R5C4 is unsolved for 5 and 7, then the centre 3x3 must be unsolved for (5 and 7).
There is therefore no way before now that the cell at R5C6 could have had either 5 or 7 eliminated as candidates because there are no solved (5 or 7) that it can see. The 5 and 7 at R5C6 could not have been eliminated by something happening on R5, nor on C6, nor in its 3x3.
So, the cell at R5C6 MUST have BOTH 5 and 7 as candidates (as well as possibly others, in this case 3 and 6).
Now assume R5C6 is 5 - we get the deadly pattern. So R5C6 cannot be 5. Now assume R5C6 is 7 - we get the deadly pattern. So R5C6 cannot be 7.
So R5C6 cannot be 5 or 7 and both can be eliminated.
Friday 22-Jan-2010
... by: Andi
I do not understand this strategy. In the first example for Type 1 UR, I understand that if 3 or 6 would be removed from R5C6, you would end up with a deadly pattern. That's fine so far.But I do not see why you can eliminate 5 AND 7 for this reason from R5C6. IMHO, if in R5C6 3,5,6 would be possible, well then this would just imply R5C4 and R2C6 being 7 and R2C4 being 5. An "ordinary" solution. OTOH, if in R5C6 3,6,7 would be possible, R5C4 and R2C6 would be 5 and R2C4 would be 7. Quite straightforward, too. My point is: both possibilities would be valid. But I cannot spot a "deadly pattern" here because to remove the ambiguity in this sitatuation, all you can postulate is that *either* 5 *or* 7 must be removed from the possibilities in R5C6. With either number removed, the ambiguity is removed, thus no deadly pattern anymore. But I just don't see why you eliminate *both* 5 and 7 in R5C6?
Am I missing something?
Tuesday 5-May-2009
... by: Dennis Daft
Shouldn't your Type 4B example be "solved" using the Type 2A method? Using this method the 6 in R1C3 and R3C5 would be eliminated, forcing one of the "roof" cells to be a 6. I believe the end result is the same. This would be consistant with only using Type 4 when no other solution is possible.Dennis





