JeJ wrote:Also, when you post your puzzles, it would get you more help if you enclose your candidates using the CODE button

will do, thanks

24 posts
• Page **2** of **2** • 1, **2**

JeJ wrote:Also, when you post your puzzles, it would get you more help if you enclose your candidates using the CODE button

will do, thanks

- Maria
**Posts:**30**Joined:**19 March 2011

JeJ wrote:EDIT

I hope you did not pay too much attention to my previous advice, because I didn't check well and it wasn't completely right.

What you need to find is:

Naked pair on column 9

Locked candidates (2) in column 3

Naked triple in box 2

hopefully it was late and I could not take much time, here is also difficult because I have traced the three issues but do not know how to act, I have to improve

- Maria
**Posts:**30**Joined:**19 March 2011

JeJ wrote:EDIT

I hope you did not pay too much attention to my previous advice, because I didn't check well and it wasn't completely right.

What you need to find is:

Naked pair on column 9

Locked candidates (2) in column 3

Naked triple in box 2

thank you for yr usual help, I can find them but do not know what to do with them, maybe I will have to follow what someone else suggests, zoo puzzles which may mean first degree...... thanks a lot

- Maria
**Posts:**30**Joined:**19 March 2011

PIsaacson wrote:Maria,

SSTS gets your puzzles quickly to the following PM grid, at which point Simple Sudoku can make no further progress.A W-Wing at r8c3.<27> and r9c8.<27> with a stong link on 2 between col 3 and col 8 produces r8c89 <> 7, and that reduces the puzzle to singles from there on.

- Code: Select all
`*-----------------------------------------------------------*`

| 3 126 19 | 69 8 14 | 7 5 24 |

| 5 12 8 | 7 3 14 | 6 9 24 |

| 679 67 4 | 569 2 569 | 1 3 8 |

|-------------------+-------------------+-------------------|

| 69 8 3 | 1 69 7 | 2 4 5 |

| 2 4 79 | 369 5 369 | 8 1 67 |

| 167 167 5 | 8 4 2 | 3 67 9 |

|-------------------+-------------------+-------------------|

| 17 9 127 | 4 67 68 | 5 2678 3 |

| 4 5 27 | 236 1 368 | 9 2678 67 |

| 8 3 6 | 25 79 59 | 4 27 1 |

*-----------------------------------------------------------*

Cheers,

Paul

I got it following yr instructions but it will be difficult to find out next time, I am sure - thanks a lot

- Maria
**Posts:**30**Joined:**19 March 2011

Maria wrote: excellently presented, thanks!!!

Sorry, of course i did not expect, that you would immediately understand my post. I guess, that not more than 20 active users here understood. But i always like it, when i can find a nice pattern in a puzzle, which was neither intended by the creator nor is found by most solvers, at least not as early as it can be spot easily by manual solvers (much easier than say a naked triple). This is a kind of uniqueness technique, maybe you have heard about "unique rectangles". E.g. when you have a pattern like this (a "rectangle" in 2 boxes), where none of the "x" is a given,

- Code: Select all
`x . . | x . .`

x . . | x . .

. . . | . . .

it is not possible in a unique (one solution) puzzle, that you have only 2 digits in the 4 cells. This can be used in various ways. The most common is, when you have candidates like this,

- Code: Select all
`12 . . | 12 . .`

12 . . | x . .

. . . | . . .

you know that x cannot contain 1 or 2.

Another one is, that you know, that for each digit pair at least one of the digits must be elsewhere in the 2 boxes, the 2 rows and the 2 columns.

E.g. in this case

- Code: Select all
`x . 3 | x 5 6 | . . .`

x . 4 | x . 7 | . . .

. . . | . # . | 1 2 .

----------------

. 1 . | . . . | . . .

. 2 . | . 1 . | . . .

you can immediately say, that the cell # must be 2, because there is no other possibility for 1 and 2 in the 2 boxes.

How to spot that: Obviously you have a hidden pair 12 in r12c1. In such a case i always look, if 12 can form a DP (Deadly Pattern). This can be done in seconds. You only have to look at boxes 2 or 3 (if the pair is in the same line, you look at the vertical boxes). When you already have one of the digits there, forget it (box 3 here). So r12c4 are the only candidates. Now you can look, if it is useful in some way (in most cases it will not be).

Its very similar with the 6 cell pattern in your puzzle (again no given in the cells). Here a unique solution is not possible with only 3 digits in the cells.

If you want to know, why, suppose you have a solution like this:

- Code: Select all
`1 . . | 2 . . | 3 . .`

3 . . | 1 . . | 2 . .

Then you can change the digits in the solution e.g. like this: [edited, thanks to Paul]

- Code: Select all
`3 . . | 1 . . | 2 . .`

1 . . | 2 . . | 3 . .

What you get is a solution again, because after the change you still have exactly one 1, 2, and 3 in all the 3 boxes, 2 rows and 3 colums, which are affected by the change. But this would mean, that the original puzzle would have at least 2 solutions.

Last edited by eleven on Wed Mar 23, 2011 10:50 am, edited 1 time in total.

- eleven
**Posts:**1876**Joined:**10 February 2008

Maria wrote:JeJ wrote:EDIT

I hope you did not pay too much attention to my previous advice, because I didn't check well and it wasn't completely right.

What you need to find is:

Naked pair on column 9

Locked candidates (2) in column 3

Naked triple in box 2

hopefully it was late and I could not take much time, here is also difficult because I have traced the three issues but do not know how to act, I have to improve

What you need to do with those clues is making small eliminations and work little by little until the puzzle starts looking simpler.

- JeJ
**Posts:**76**Joined:**06 January 2011

Eleven,

I'm not one of those gifted 20, but isn't the DP something more like:

I'm confused by your example. It looks like you just substituted digits 1 and 3, and that's a valid transformation, isn't it???

Cheers,

Paul

eleven wrote:If you want to know, why, suppose you have a solution like this:

- Code: Select all
`1 . . | 2 . . | 3 . .`

3 . . | 1 . . | 2 . .

Then you can change the digits in the solution e.g. like this:

- Code: Select all
`3 . . | 2 . . | 1 . .`

1 . . | 3 . . | 2 . .

What you get is a solution again, because after the change you still have exactly one 1, 2, and 3 in all the 3 boxes, 2 rows and 3 colums, which are affected by the change. But this would mean, that the original puzzle would have at least 2 solutions.

I'm not one of those gifted 20, but isn't the DP something more like:

- Code: Select all
`13 . . | 12 . . | 23 . .`

13 . . | 12 . . | 23 . .

- Code: Select all
`1 . . | 2 . . | 3 . .`

3 . . | 1 . . | 2 . .

3 . . | 1 . . | 2 . .

1 . . | 2 . . | 3 . .

I'm confused by your example. It looks like you just substituted digits 1 and 3, and that's a valid transformation, isn't it???

Cheers,

Paul

- PIsaacson
**Posts:**249**Joined:**02 July 2008

Oops, thanks,

in my sample column 4 and 7 have to be interchanged. Of course in the solution you only can swap the digits in the same column/box, otherwise you would NOT still have exactly one 1, 2, and 3 in it. I correct it above.

Note that also

is a deadly pattern. However you place the digits, you can swap the column digits that way for a second solution.

in my sample column 4 and 7 have to be interchanged. Of course in the solution you only can swap the digits in the same column/box, otherwise you would NOT still have exactly one 1, 2, and 3 in it. I correct it above.

Note that also

- Code: Select all
`123 . . | 123 . . | 123 . .`

123 . . | 123 . . | 123 . .

is a deadly pattern. However you place the digits, you can swap the column digits that way for a second solution.

- eleven
**Posts:**1876**Joined:**10 February 2008

eleven wrote:

- Code: Select all
`+-------+-------+-------+`

| 3 x . | . 8 x | 7 5 x |

| 5 x 8 | 7 3 x | 6 9 x |

| . .*4 | .*2 . | 1 3 8 |

+-------+-------+-------+

| . 8 3 |*1 . 7 | 2 . 5 |

| 2 4 . | . 5 . | 8 1 . |

| . . 5 | 8 . 2 | 3 . 9 |

+-------+-------+-------+

| . 9 . |*4 . . | 5 . 3 |

| 4 5 . | . 1 . | 9 . . |

| 8 3 6 | . . . | 4 . 1 |

+-------+-------+-------+

There is a nice DP in the first 2 lines. To avoid 124 in the cells marked x, one of the 3 digits must be in r1c34.

Only possible are 12 in r1c3. So one of the digits must be there - and a 9 cant be there -> r5c3=9.

Eleven

A nice way of looking at it.

More obvious (maybe) less direct is to look at the set of additional candidates in the DP {6r1c26; 9r1c6} knowing that one of those must be true. Then noticing that since r1c4 is limited to 69, the true candidate, whichever it is, will necessarily combine with r1c4 to form a Naked Pair 69 (aka Quantum Naked Pair) thus excluding 9 from r1c3, and hence placing it at r5c3.

However even better is to notice that r12c6 is the Hidden pair 14, so to avoid the DP r1c2=6.

- aran
**Posts:**334**Joined:**02 March 2007

24 posts
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