dailysudoku.com

[immpy]

Enjoy the puzzle.....

[JC Van Hay]

{5C58} -> -{5r8c47, 5r7c7}

[ZeroAssoluto]

Hi everyone,

after basic

[JC Van Hay]

[ZeroAssoluto]

Hi JC,

I'm not trying to prove he has no solutions: or has one or more than one
If the schema has only one solution, can not present a situation with the same pairs of numbers at the 4 vertices of the rectangle;

[ZeroAssoluto]

It's not a real technique, but almost a trick you can use when you're sure that the schema has only one solution.

Ciao Gianni

[immpy]

I'm going to defend the Unique Rectangle as a solving technique, based on the fact that a TRUE Sudoku (which I understand roughly translates as "place one number") puzzle has only one possible solution. We can use this fact to help us in our solving. Why not? We eliminate candidates in rows columns and boxes based on the fact that a number can only appear one time in each respective area.
And the Unique Rectangle technique has more than one way of appearing. I am aware of six, depending on where any extra candidates may be located.

In this very puzzle, one could also identify the Unique Rectangle that occurs with (1,4) in rows 1 and 9, columns 5 and 6. This time there occur two extra candidates (7 and 8) in both of the row 9 ends of the rectangle. One could make the case that 7 and 8 are now a strong pair which will fill these two grid spots. Any grid spot that can be seen by both of these two, can have the 7 and the 8 eliminated from them. This means r8c6<>7, and r9c9<>7. And the puzzle quickly solves from there.

The UR in rows 2 and 3, columns 3 and 8 is much preferred here, as it is simpler, with only one extra candidate (6).

I have learned something else along these lines. The UR must be in only 2 boxes. If it occurs in 4 boxes it becomes technically a Unique Square, and any extra candidates become eliminations in such a case, which again becomes a very valuable solving technique.

In my next posted puzzle I will try to include a Unique Square.

immpy

[Ajò Dimonios]

Hi everyone

Solutions with two step after basics,fish and fins fish.

1) [(3)R4C6-(3=1)R5C4 and (3-4)R4C6=(4)R4C7-(4=1)R5C9 =>contradiction two 1 in line 5=>-3R4C6
2) [(4)R5C8-(4)R23C8=(4)R3C7-(4=1)R7C7 and (4-5)R5C8=(5)R5C5-(5=1)R7C5=>contradiction two 1 in line 7=>-4R5C8=>solution stte.

Ciao a tutti
Paolo

[Marty R.]

[immpy]

Indeed yes it is. Just wanted to reaffirm this after ZeroAssoluto's post.

[JC Van Hay]

UR(14)r23c38 -> 0 solution because, for example :

[Ajò Dimonios]

Hi everyone

[ZeroAssoluto]

[Marty R.]

[ZeroAssoluto]

Hi Marty
I think you wont say that you haven't seen case A (more than one solution).

I did not mean to say that there are so many cases with more than one solution and even less those who then have an UR.

But to say that, just because in most cases the schemes have one solution, that if you are in a situation with 4 pairs equal to the 4 vertices of a rectangle, you do not have to go to the end; You know at this point that something is wrong.

Only this

Ciao Gianni

[Ajò Dimonios]

Hi everyone

Another solutions with one step. One elimination

[JC Van Hay]

[Ajò Dimonios]

Hi JC


JC Van Hay wrote:

[JC Van Hay]

Hi Paolo,

Typo : read -5r8c7 instead of -5r8c4

[Ajò Dimonios]

Hi JC

Perhaps the logic you described to me is also included in the following AIC, which first eliminates 1 in R5C456 and consequently also eliminates 5 in R8C7 and R6C4.

AIC
(1)R5C9=(1)R9C9-(1)R7C78=(1-5)R7C5=(5)R8C4-(5=1)R6C4=>-1R5C456=>R6C4=1;-5R6C4;R6C7=5 and -5R8C7=>solution stte.

Ciao a tutti
Paolo