The New Sudoku Players' Forum

[MrHamilton]

My travelers can only travel so far, it seems.
What's the old saying about travelers' tales?
Anyway, if the method (or something based upon it) helps you, use it; and if not, use another.
It has helped me occasionally.
I thank Ravel and Jeff etc. for their thoughts and corrections.
As a validity check for other techniques or guesses, it works well enough.

[ravel]

Looking at Tareks puzzle, i think that more than box elimination can be done with travelers:

Code: Select all

*-----------------------------------------------------------------------*
| 348   349    3489       | 5     29     6      | 289    7       1      |   
| 1345  2      134569     | 8     19     7      | 569    3469    34569  |   
| 158   7      15689      | 4     129    3      | 25689  2689    25689  | 
|-----------------------------------------------------------------------|   
| 7     134    12348      | 139   6      1489   | 1289   5       2489   |   
| 1458  1456   14568      | 2     1489   14589  | 3      14689   7      |   
| 9     13456  1234568    | 7     1348   1458   | 1268   12468   2468   |   
|-----------------------------------------------------------------------|   
| 6     13459  123459     | 139   7      12489  | 12589  12389   23589  |   
| 1235  8      7          | 1369  1239   129    | 4      12369   23569  |   
| 1234  1349   12349      | 1369  5      12489  | 7      123689  23689  |     
*-----------------------------------------------------------------------*


Possible partners are:
1: 29
2: 19
3: 67
4: 8
5:
6: 3
7: 3
8: 4
9: 12

Possible traveling pairs are 12 or 19 or 29, 36 or 37, and 48
Looking at r789c5: Since 5 is alone r8c5 must be 3 (the only possible partner of 7)
Correct ?
[Edit:]
Ah,no, 129 can be box eliminated.
But lets continue:
r4c4 must be 3.
The 48 pair of box 7 must be in r79c6 -> eliminate 129 from r79c6
Same for col 5 in box 4: eliminate 19 from r5c5 and 13 from r6c5 and 48 from r456c6
19 must be a pair then (5 is alone), 2 is alone, r8c6=2.
Funny.
Its a combination of box elimination and (normal) pairs.

[aeb]

...i think that more than box elimination can be done with travelers: ...Possible partners are: 1: 29 ...
Its a combination of box elimination and (normal) pairs.

Let us see. You are talking about the middle vertical band, columns 456.
The condensed picture is

Code: Select all

458   129    367
12379 134689 14589
1369  123579 12489


5 is going down left, alone, since 4 and 8 are going down right. It follows that three digits go down left, alone, and six digits go down right, in pairs. 2 is going down left, hence is alone. 6 is going down left, hence is alone. We see that the three pairs are 48, 19, 37, and the picture becomes

Code: Select all

458  129  367
237  468  159
169  357  248


That excluded 13 digits in the condensed picture, a big success.
Now by line-box interactions. In box 8 digits 4,8 live in column 6,
so there is no 4 or 8 in box 5 in column 6. In box 2 digits 1,2,9 live in column 5, so there are no 1,2,9 in column 5 in box 5 or 8. This means that column 5 in box 8 has digits 3,5,7, and there are no digits 3 in column 5, box 5, or in column 4 box 8. Now in boxes 5,8 two of the three columns are known, and the third follows.

Looking at the condensed picture is the same as looking at box-line intersections.

[MrHamilton]

Anecdotally, I've found that even when the candidates individually cannot be deduced locally, you can rule out more of the possible permutations of them that pencilmarks alone allow for.
So you get more of these situations where something is either a certain number OR a certain color and you know that solving that dilemma may well solve the whole puzzle.

[MrHamilton]

Any way of determining whether triple triples can exist in a solution? And the frequency of the different solution types?
You may have a solution with either no triples, or 1 - 6 triple-chutes.
Seven possible types--which are possible, which not?
I have witnessed double triples in a puzzle, but I can't recall seeing a higher incidence.

[MrHamilton]

Returning again to Eppstein's puzzle:

Code: Select all

 *-----------------------------------------------------------*
 | 5     6     279   | 47    39    1     | 249   349   8     |
 | 2479  248   1279  | 5     39    478   | 6     1349  1249  |
 | 349   348   139   | 48    6     2     | 5     7     149   |
 |-------------------+-------------------+-------------------|
 | 367   9     37    | 2     47    5     | 1     8     467   |
 | 27    25    4     | 68    1     68    | 3     59    579   |
 | 16    15    8     | 3     47    9     | 47    2     56    |
 |-------------------+-------------------+-------------------|
 | 1234  7     6     | 9     8     34    | 24    145   1245  |
 | 1349  134   5     | 467   2     3467  | 8     1469  1479  |
 | 8     24    29    | 1     5     467   | 2479  469   3     |
 *-----------------------------------------------------------*


Consider the middle box of the leftmost stack.
There are five bivalue cells and one with three candidates.
The "extra" candidate in c1r4 is 7.
The "extra" candidate is also wrong and makes the puzzle unsolvable (it sets off a chain of deductions that ends in a contradictory double 9 in column 7!)
Coincidence, or a clue to something?

[ravel]

Let us see...

I see, thanks

I have witnessed double triples in a puzzle, but I can't recall seeing a higher incidence.

This is [Edit: NOT] called the canonical grid:

Code: Select all

+-------+-------+-------+
| 1 4 7 | 3 9 6 | 5 2 8 |
| 6 3 9 | 2 5 8 | 4 1 7 |
| 5 2 8 | 4 1 7 | 3 6 9 |
+-------+-------+-------+
| 4 7 1 | 6 3 9 | 2 8 5 |
| 3 9 6 | 5 8 2 | 1 7 4 |
| 2 8 5 | 1 7 4 | 6 9 3 |
+-------+-------+-------+
| 7 1 4 | 9 6 3 | 8 5 2 |
| 9 6 3 | 8 2 5 | 7 4 1 |
| 8 5 2 | 7 4 1 | 9 3 6 |
+-------+-------+-------+

[MrHamilton]

Let us see...

I see, thanks

I have witnessed double triples in a puzzle, but I can't recall seeing a higher incidence.

This is called the canonical grid:

Code: Select all

+-------+-------+-------+
| 1 4 7 | 3 9 6 | 5 2 8 |
| 6 3 9 | 2 5 8 | 4 1 7 |
| 5 2 8 | 4 1 7 | 3 6 9 |
+-------+-------+-------+
| 4 7 1 | 6 3 9 | 2 8 5 |
| 3 9 6 | 5 8 2 | 1 7 4 |
| 2 8 5 | 1 7 4 | 6 9 3 |
+-------+-------+-------+
| 7 1 4 | 9 6 3 | 8 5 2 |
| 9 6 3 | 8 2 5 | 7 4 1 |
| 8 5 2 | 7 4 1 | 9 3 6 |
+-------+-------+-------+

5 out of 6 possible, impressive!

[ravel]

Sorry, i copied the wrong one - this is the canonical grid:

Code: Select all

+-------+-------+-------+
| 1 2 3 | 4 5 6 | 7 8 9 |
| 7 8 9 | 1 2 3 | 4 5 6 |
| 4 5 6 | 7 8 9 | 1 2 3 |
+-------+-------+-------+
| 2 3 1 | 5 6 4 | 8 9 7 |
| 5 6 4 | 8 9 7 | 2 3 1 |
| 8 9 7 | 2 3 1 | 5 6 4 |
+-------+-------+-------+
| 3 1 2 | 6 4 5 | 9 7 8 |
| 6 4 5 | 9 7 8 | 3 1 2 |
| 9 7 8 | 3 1 2 | 6 4 5 |
+-------+-------+-------+

[r.e.s.]

For reference ...

'MrHamilton, you are invited to look at the Braid Analysis thread here: http://www.sudoku.org.uk/cgi-bin/discus/show.cgi?tpc=29&post=4735#POST4735 which builds on the Travelling Pairs concept'

[doduff]

[edit] I posted this before going through the whole thread, so the solution to the puzzle linked below has already been discussed I think, as the "canonical grid" The puzzle is interesting, nonetheless.[/edit]

Here is an example of a puzzle with only traveling triples in the solution.

I constructed it by making the solution and then deleting numbers until I couldn't delete any more without destroying uniqueness. I didn't check all of them, so there still might be a few that can be deleted and still preserve uniqueness.

The traveling pair/triplet is an interesting technique. I actually used it to make so eliminations and solved some numbers in a puzzle. I should have noted the example and posted it here.

[Condor]

Here's another (possibly original) technique I like to use involving groups.

This is the same technique as I posted here. This has just been brought to my attention by Mike Barker (Thanks).

Any way of determining whether triple triples can exist in a solution? And the frequency of the different solution types?
You may have a solution with either no triples, or 1 - 6 triple-chutes.
Seven possible types--which are possible, which not?
I have witnessed double triples in a puzzle, but I can't recall seeing a higher incidence.

In the second half of an even earlier post here, I showed how the basis for this technique could be used for calculating the number of ways to fill the top band (the first three rows) of the sudoku grid.

Using the same calculation there is 27 times as many pairs (and singles) as triples. I have seen no reason for traveling pairs or triples in 1 chute has any effect on another chute, that is, the probablity of a chute having pairs or triples does not effect the probalility in any other chute.

If that can be proved true then the probability of there being 1 triple is 1/28, 2 triples 1/28 * 1/28, 3 triples 1/28 * 1/28 * 1/28, etc.



This technique was also the basis of a checking method I posted here. By following the numbers as they travel in pairs, triples or singles across the sudoku grid you can determine if the finished puzzle is correct.

[MrHamilton]

Thanks, was unaware of the earlier post, as I suppose David Bird in the other forum was unaware of either of us.

[Condor]

Thanks, was unaware of the earlier post, as I suppose David Bird in the other forum was unaware of either of us.

It seems that this method was "discovered" by several people.

Thanks for helping to make it better known in the sudoku community. I have found it a bit of an unweildy beast to explain to people. Reading other people explanations will help me explain them easier.