The New Sudoku Players' Forum

[edddy]

I have this puzzle
http://sudokuhelper.kevinpluck.net/?g=9092a7765707d9e676a57e4a222b1603

I can resolve it using trial and error, but I know that there must be a way to solve it with a logical move.

can you help me?

Please be kind in the explanation, I'm not a native english speaker.

[JL]

Skyscraper in 1's. *'s show the skyscraper, -'s what 1's are eliminated. Puzzle finishes easily after that.

Code: Select all

.------------------.------------------.------------------.
| 9     8     3    | 7     6     5    | 4     2     1    |
| 2     16    16   | 4     3     9    | 7     8     5    |
| 7     5     4    | 2     1     8    | 6     3     9    |
:------------------+------------------+------------------:
| 6     -19    2    | 5     89    7    | *18    4     3    |
| 5     4     7    | 1     89    3    | 28    69    26   |
| *18    3     189  | 6     4     2    | 5     -19    7    |
:------------------+------------------+------------------:
| *18    169   1689 | 3     5     4    | *12    7     26   |
| 4     7     16   | 9     2     16   | 3     5     8    |
| 3     2     5    | 8     7     16   | 9     16    4    |
'------------------'------------------'------------------'

[Cec]

"....Please be kind in the explanation, I'm not a native english speaker."

Hi Edddy and welcome to the forum,

As an alternative to the "Skyscraper" technique you may prefer the "colouring" solving method. Thanks to Angusj and Sped's clear explanation of the "colouring" technique, which I have kept in my "treasure box", you can read about this by clicking on HERE

Allowing for a naked pair [16] in column 3 your grid now looks like this. I have marked the conjugate pairs - coincidently also candidates [16] - using a "+" and "-" symbol for the candidate 1's in lieu of using colors.

Code: Select all

 *-----------------------------------------------------------*
 | 9     8     3     | 7     6     5     | 4     2     1     |
 | 2    +16   -16    | 4     3     9     | 7     8     5     |
 | 7     5     4     | 2     1     8     | 6     3     9     |
 |-------------------+-------------------+-------------------|
 | 6    +19    2     | 5     89    7     |-18    4     3     |
 | 5     4     7     | 1     89    3     | 28    69    26    |
 |-18    3     89    | 6     4     2     | 5    +19    7     |
 |-------------------+-------------------+-------------------|
 |+18    169   89    | 3     5     4     |+12    7     26    |
 | 4     7    +16    | 9     2    -16    | 3     5     8     |
 | 3     2     5     | 8     7    +16    | 9    -16    4     |
 *-----------------------------------------------------------*

Considering the conjugate pairs [16] in row2 (Box 1) then if digit 1 was correct in cell r2c2 (marked with a "+" sign) then digit 1 cannot be correct for cell r2c3 (shown by a "-" sign). Continuing this sequence then candidate1 in cell r8c3 must be "+" and therefore candidate1 must be "-" in r8c6. This sequence is continued for all the conjugate pairs [16]. Because candidate1 in r7c1 "sees" both a "-" sign in cell r6c1 and a "+" sign in r8c3 then candidate1 can be excluded from cell r7c1 leaving 8 as the correct candidate in that cell. The remaining puzzle now reduces to all singles.

Cec

[edddy]

Thanks JL, still trying to understand the Skyscraper thing

[daj95376]

Although the Skyscraper is the better solution, if you don't understand it, then it doesn't help. There is another solution, but I'm not sure if it will help, either.

You missed a Naked Pair for <16> in [c3]. This reduces the puzzle to where there is only one cell with more than two candidates -- [r7c2]=<169>. This is called a BUG+1. If you look at those three candidates -- and the row, column, and box containing [r7c2] -- then you will see that <1> is the only candidate that occurs three times in each. Setting [r7c2]=1 reduces the puzzle to singles afterwards.

Code: Select all

+--------------------------------------------------------------------+
|  9      8      3     |  7      6      5     |  4      2      1     |
|  2      16     16    |  4      3      9     |  7      8      5     |
|  7      5      4     |  2      1      8     |  6      3      9     |
|----------------------+----------------------+----------------------|
|  6      19     2     |  5      89     7     |  18     4      3     |
|  5      4      7     |  1      89     3     |  28     69     26    |
|  18     3      89-1  |  6      4      2     |  5      19     7     |
|----------------------+----------------------+----------------------|
|  18     169    89-16 |  3      5      4     |  12     7      26    |
|  4      7      16    |  9      2      16    |  3      5      8     |
|  3      2      5     |  8      7      16    |  9      16     4     |
+--------------------------------------------------------------------+


Code: Select all

Skyscraper:
1) Strong links on <1> in  [r6c1] to [r7c1] and [r4c7] to [r7c7]
2) Weak   link  on <1> for           [r7c1] to            [r7c7]
3) Any <1> jointly seen by [r6c1] and           [r4c7] can be eliminated
+-----------------------------------+
|  .  .  .  |  .  .  .  |  .  .  1  |
|  .  1  1  |  .  .  .  |  .  .  .  |
|  .  .  .  |  .  1  .  |  .  .  .  |
|-----------+-----------+-----------|
|  . -1  .  |  .  .  .  | *1  .  .  |
|  .  .  .  |  1  .  .  |  .  .  .  |
| *1  .  1  |  .  .  .  |  . -1  .  |
|-----------+-----------+-----------|
| *1  1  1  |  .  .  .  | *1  .  .  |
|  .  .  1  |  .  .  1  |  .  .  .  |
|  .  .  .  |  .  .  1  |  .  1  .  |
+-----------------------------------+


[edddy]

Thanks Cec also. I'm looking into it (trying not to make my head explode)

[edddy]

Thanks daj95376 for taking the time to teach me.

I understand that I missed the Naked Pair for <16> in [c3], but

If you look at those three candidates -- and the row, column, and box containing [r7c2] -- then you will see that <1> is the only candidate that occurs three times in each. Setting [r7c2]=1

Why you can deduce that [r7c2]=1? I mean, why this cell and not other?

Sorry if I am a little dorky.

[daj95376]

Thanks daj95376 for taking the time to teach me.

I understand that I missed the Naked Pair for <16> in [c3], but

If you look at those three candidates -- and the row, column, and box containing [r7c2] -- then you will see that <1> is the only candidate that occurs three times in each. Setting [r7c2]=1

Why you can deduce that [r7c2]=1? I mean, why this cell and not other?

Sorry if I am a little dorky.

I'm not sure where I lost you. After the Naked Pair, there is only one cell left with more than two candidates. There is an elaborate explanation, i.e. BUG+1, for constraints on the remaining cell with three candidates. The bottom line, for finding the correct candidate in that cell, is the conditions I listed above.

[edddy]

I mean that, i.e. if [r7c2]=6 (instead of 1) there is not direct implications in the row, col and group (regardless that the puzzle can't be resolved afterwards)

Maybe I have to look into the Bug+1 technique (first time I heard that such thing exists)

[Cec]

"...Maybe I have to look into the Bug+1 technique (first time I heard that such thing exists)"

Thanks to tso, an explanation of the BUG principle is given HERE though I'm not quite sure what the "Bug+1" technique means. I suspect there is a detailed explanation somewhere as to why you can immediately place the missing digit for the only three-candidate cell in a grid but I'm happy to accept this condition is true rather than concluding it is true by the more tedious alternative of trial and error.

As I see it, the alternative techniques to solve this puzzle are either "coloring" or the "Bug" technique and it's a matter of personal choice in both understanding and detecting either (or both) of these situations.

Thanks Cec also. I'm looking into it (trying not to make my head explode)

Whilst not wishing to overload you it seems the "Coloring" technique may still not be clear. To understand Coloring it is necessary to understand what "Conjugate pairs" mean.

As explained by angusj in his excellent coloring presentation, "Conjugate cells" exist where an individual candidate (meaning the same candidate) occurs in only two cells of a given group (ie. either in a row, column or box). As such, only one of these cells in a given group contains the true value for that particular candidate which means that same candidate cannot exist in the other conjugate cell of that same group.
Lets look again at your puzzle:

Code: Select all

 *-----------------------------------------------------------*
 | 9     8     3     | 7     6     5     | 4     2     1     |
 | 2     16    16    | 4     3     9     | 7     8     5     |
 | 7     5     4     | 2     1     8     | 6     3     9     |
 |-------------------+-------------------+-------------------|
 | 6     19    2     | 5     89    7     | 18    4     3     |
 | 5     4     7     | 1     89    3     | 28    69    26    |
 | 18    3     89    | 6     4     2     | 5     19    7     |
 |-------------------+-------------------+-------------------|
 | 18    169   89    | 3     5     4     | 12    7     26    |
 | 4     7     16    | 9     2     16    | 3     5     8     |
 | 3     2     5     | 8     7     16    | 9     16    4     |
 *-----------------------------------------------------------*

It can be seen that candidate 1 forms a conjugate pair in various groups of this grid (Boxes 1, 4, 6, 8 and 9. Note that candidate 1 does NOT form a conjugate pair in Box7 because there are three candidate 1's in that group.
Now, in row 2 (Box1), as candidate 1 can only be true for either of cells r2c2 or r2c3 - it cannot be true for both - then one of these cells (it doesn't matter which one) is marked in some way, either say "+" or "true" or colored blue or green or whatever. If you like you can mark them "apples" and "oranges" the result will still be the same. The other conjugate cell in that same group is marked "-" or "false" or "oranges" etc. I decided to mark candidate 1 in cell r2c2 as "+" and "-" for r2c3.

You will see in column3 that there is another conjugate pair for the candidate 1's and with cell r2c3 already marked "-" then cell r8c3 must be marked "+" to maintain the "true/false" sequence within the same group. Note that you cannot mark the other cells containing candidate 1's in column 2 because this column contains three 1's which do not form conjugate cells for this particular group (two cells and only two cells are the criteria for conjugate pairs within a particular group).

Continuing this "conjugate pair" marking sequence for the candidate 1's will place a "-" for r8c6, a "+" sign for r9c6, a "-"for r9c8, a "+" for r6c8, a"-" for r4c7, a "+" for r4c2, a "-" for r6c1, and a "+" for r7c1.

With these congugate pairs marked as above your grid now looks like this:

Code: Select all

 *-----------------------------------------------------------*
 | 9     8     3     | 7     6     5     | 4     2     1     |
 | 2    +16   -16    | 4     3     9     | 7     8     5     |
 | 7     5     4     | 2     1     8     | 6     3     9     |
 |-------------------+-------------------+-------------------|
 | 6    +19    2     | 5     89    7     |-18    4     3     |
 | 5     4     7     | 1     89    3     | 28    69    26    |
 |-18    3     89    | 6     4     2     | 5    +19    7     |
 |-------------------+-------------------+-------------------|
 |+18    169   89    | 3     5     4     |+12    7     26    |
 | 4     7    +16    | 9     2    -16    | 3     5     8     |
 | 3     2     5     | 8     7    +16    | 9    -16    4     |
 *-----------------------------------------------------------*

Sorry for the repetition but you will see that candidate 1 in cell r7c1 "sees' a "-" cell in the same column1 (r6c1) but also "sees" a "+" cell r8c3 in the same group (Box 7).This constitutes a contradiction meaning candidate 1 is false for cell r7c1. You can easily prove this by trying candidate 1 in this cell which will yield two 1's in Box7.

I hope this makes things clearer on coloring.

Edited by Cec 1am. "Block(s)" altered to read "Box(s)"

Cec

[edddy]

Cec: thanks a lot for the detailed explanation. I think I'm ready for more difficult Sudoku!