The New Sudoku Players' Forum

[denis_berthier]

.
Puzzles are known that have no (3D-)bivalue cell - [no bivalue/bilocal] - at the start and after basic moves. I don't have an example at hand but I clearly remember seeing several and I think I could find a few if I needed.

But my first question is about trivalue: does anyone know a puzzle that has no (3D-)bivalue cell AND no (3D-)trivalue cell - [no trivalue/trilocal] - also at the start and after basic moves?

In case an example exists, has anyone studied what's the simplest (say SER-wise) puzzle with this property?

Edit: corrected an error inside the first [], that made the question ambiguous.

Edit: It seems my first sentence was based on a false belief and it was true only for rc-bivalue. See http://forum.enjoysudoku.com/the-bivalue-conjecture-t38764.html
.

[coloin]

I am not sure if there are puzzles without a bivalue/bilocal set of 2 similar pm values in a box,row or column ... Edit.. is this what you mean ?

Golden Nugget SE 11.9 has ? 6 [ 24/25, 41/63, 42/53, 51/62, 54/66, 81/92] !!! [wow]

Code: Select all

+---+---+---+
|...|...|.39|
|...|..1|..5|
|..3|.5.|8..|
+---+---+---+
|..8|.9.|..6|
|.7.|..2|...|
|1..|4..|...|
+---+---+---+
|..9|.8.|.5.|
|.2.|...|6..|
|4..|7..|...|
+---+---+---+  Golden Nugget

+----------------------+----------------------+----------------------+
| 25678  14568  124567 | 268    2467   4678   | 1247   3      9      |
| 26789  4689   2467   | 23689  23467  1      | 247    2467   5      |
| 2679   1469   3      | 269    5      4679   | 8      12467  1247   |
+----------------------+----------------------+----------------------+
| 235    345    8      | 135    9      357    | 123457 1247   6      |
| 3569   7      456    | 13568  136    2      | 13459  1489   1348   |
| 1      3569   256    | 4      367    35678  | 23579  2789   2378   |
+----------------------+----------------------+----------------------+
| 367    136    9      | 1236   8      346    | 12347  5      12347  |
| 3578   2      157    | 1359   134    3459   | 6      14789  13478  |
| 4      13568  156    | 7      1236   3569   | 1239   1289   1238   |
+----------------------+----------------------+----------------------+

Here is a 20 clue puzzle I found with ED=10.4/10.4/9.7...

Code: Select all

+---+---+---+
|.1.|..2|..3|
|42.|...|...|
|..5|.6.|...|
+---+---+---+
|...|...|5..|
|.7.|..4|..2|
|..8|...|.9.|
+---+---+---+
|...|.8.|9..|
|...|.9.|.6.|
|3..|..1|..7|
+---+---+---+  20C 

+-------------------------+-------------------------+-------------------------+
| 6789    1       679     | 45789   457     2       | 4678    4578    3       |
| 4       2       3679    | 135789  1357    35789   | 1678    1578    15689   |
| 789     389     5       | 134789  6       3789    | 12478   12478   1489    |
+-------------------------+-------------------------+-------------------------+
| 1269    3469    123469  | 1236789 1237    36789   | 5       13478   1468    |
| 1569    7       1369    | 135689  135     4       | 1368    138     2       |
| 1256    3456    8       | 123567  12357   3567    | 13467   9       146     |
+-------------------------+-------------------------+-------------------------+
| 12567   456     12467   | 234567  8       3567    | 9       12345   145     |
| 12578   458     1247    | 23457   9       357     | 12348   6       1458    |
| 3       45689   2469    | 2456    245     1       | 248     2458    7       |
+-------------------------+-------------------------+-------------------------+

there are ? 5 bilocals at 29/39, 23/32, 37/38, 78/87, 92/93,

Now the question I have .....

is it possible to do the "FORCING-T&E applied to bivalue candidates" technique here ?

[denis_berthier]

I am not sure if there are puzzles without a bivalue/bilocal set of 2 similar pm values in a box,row or column ... Edit.. is this what you mean ?

Problem of language, here. I don't understand either what you mean: "two similar pm values.."
So let me say what I meant:
rc-bivalue: a (rc-)cell that contains only two candidates
rn-bivalue: a rn-cell that contains only two candidates; i.e. two candidates conjugated in a row
cn-bivalue: a cn-cell that contains only two candidates; i.e. two candidates conjugated in a column
bn-bivalue: a bn-cell that contains only two candidates; i.e. two candidates conjugated in a block
3D-bivalue = rc-bivalue or rn-bivalue or bn-bivalue or cn-bivalue
Replace bivalue everywhere by trivalue and two by three and you get the corresponding definitions.

Golden Nugget SE 11.9 has ? 6 [ 24/25, 41/63, 42/53, 51/62, 54/66, 81/92] !!! [wow]

Code: Select all

+---+---+---+
|...|...|.39|
|...|..1|..5|
|..3|.5.|8..|
+---+---+---+
|..8|.9.|..6|
|.7.|..2|...|
|1..|4..|...|
+---+---+---+
|..9|.8.|.5.|
|.2.|...|6..|
|4..|7..|...|
+---+---+---+  Golden Nugget

+----------------------+----------------------+----------------------+
| 25678  14568  124567 | 268    2467   4678   | 1247   3      9      |
| 26789  4689   2467   | 23689  23467  1      | 247    2467   5      |
| 2679   1469   3      | 269    5      4679   | 8      12467  1247   |
+----------------------+----------------------+----------------------+
| 235    345    8      | 135    9      357    | 123457 1247   6      |
| 3569   7      456    | 13568  136    2      | 13459  1489   1348   |
| 1      3569   256    | 4      367    35678  | 23579  2789   2378   |
+----------------------+----------------------+----------------------+
| 367    136    9      | 1236   8      346    | 12347  5      12347  |
| 3578   2      157    | 1359   134    3459   | 6      14789  13478  |
| 4      13568  156    | 7      1236   3569   | 1239   1289   1238   |
+----------------------+----------------------+----------------------+

So, Golden Nugget has
- no rc-bivalue cell
- rn bivalue cells: r2n3,

but a lot of:
rc-trivalue cells: r1c4, r2c7, r3c4...
rn-trivalue: r1n1, r1n5, r2n8,..
cn-trivalue: c1n8, c2n8, c2n9,...
bn-trivalue: b1n1, b1n5...

So, Golden Nugget has both 3D-bivalue and 3D-trivalue cells.
What I'm looking for is a puzzle that has none of them. There are puzzles that have no 3D-bivalue cell (I think I've seen some already) but my (weak) conjecture is: there is no puzzle that has no 3D-bivalue cell AND no 3D-trivalue cell.

Here is a 20 clue puzzle I found with ED=10.4/10.4/9.7...

Code: Select all

+---+---+---+
|.1.|..2|..3|
|42.|...|...|
|..5|.6.|...|
+---+---+---+
|...|...|5..|
|.7.|..4|..2|
|..8|...|.9.|
+---+---+---+
|...|.8.|9..|
|...|.9.|.6.|
|3..|..1|..7|
+---+---+---+  20C 

+-------------------------+-------------------------+-------------------------+
| 6789    1       679     | 45789   457     2       | 4678    4578    3       |
| 4       2       3679    | 135789  1357    35789   | 1678    1578    15689   |
| 789     389     5       | 134789  6       3789    | 12478   12478   1489    |
+-------------------------+-------------------------+-------------------------+
| 1269    3469    123469  | 1236789 1237    36789   | 5       13478   1468    |
| 1569    7       1369    | 135689  135     4       | 1368    138     2       |
| 1256    3456    8       | 123567  12357   3567    | 13467   9       146     |
+-------------------------+-------------------------+-------------------------+
| 12567   456     12467   | 234567  8       3567    | 9       12345   145     |
| 12578   458     1247    | 23457   9       357     | 12348   6       1458    |
| 3       45689   2469    | 2456    245     1       | 248     2458    7       |
+-------------------------+-------------------------+-------------------------+

there are ? 5 bilocals at 29/39, 23/32, 37/38, 78/87, 92/93,

ok, I see.
What you call bilocal at 29/39 is what I call a cn-bivalue cell: c9n9 (for values r2, r3)
What you call bilocal at 23/32 is what I call a bn-bivalue cell: b1n3 (for values r2c3, r3c2 - or s6, s8)
...
SudoRules finds more, because some pairs can be bivalue in two different ways:

Hidden Text: Show

f-10764 (bivalue 0 237 238 232)
f-10766 (bivalue 0 992 993 299)
f-10768 (bivalue 0 415 495 354)
f-10770 (bivalue 0 929 939 399)
f-10772 (bivalue 0 323 332 413)
f-10774 (bivalue 0 929 939 439)
f-10776 (bivalue 0 237 238 432)
f-10778 (bivalue 0 748 767 467)
f-10780 (bivalue 0 378 387 493)
f-10782 (bivalue 0 992 993 479)

Here, candidates n1r3c7 and n1r3c8 are both rn-bivalue in r3n1 and bn-bivalue in b3n1. Note that, which way they are bivalue is irrelevant to Forcing-T&E (but may be relevant in other patterns).

Finally, this puzzle has both 3D-bivalue and 3D-trivalue cells (r1c3, r1c5...). I'm looking for the opposite.

Now the question I have .....
is it possible to do the "FORCING-T&E applied to bivalue candidates" technique here ?

Yes, it is possible (and SudoRules will try only 7 pairs, irrespective of the ways in which they are bivalue).
But FORCING-T&E doesn't give anything in the present case. This should not be a surprise: no technique is guaranteed to solve any puzzle. And as I wrote previously FORCING-T&E is equivalent (elimination-wise) to forcing-braids - which, in turn, don't seem to be more powerful than braids.

However, you are perfectly right to allude to FORCING-T&E in this thread, because it was the motivation for my original question.
It is easy for me to extend FORCING-T&E to both 3D-bivalue and 3D-trivalue cells. But I wondered if I would also need to extend it to 3D-quadrivalue cells.

[coloin]

I understand and can see the bi values [ pencil marks] in a cell or box or row or column......

I dont think that I have seen a puzzle which doesnt have at least one incidence of a bi value .... [ and the above ultra hard puzzles have 5 and 6] - but I will look some more ...

f you take enough of these proposition pairs - more than 2 pairs - or forcing T and E ... surely all valid puzzles will succumb !

Despite trying hard I still cant visualize the bivalues that SudoRules finds or the 3D ones either !!!

[denis_berthier]

Hi Coloin

I understand and can see the bi values [ pencil marks] in a cell or box or row or column......
I dont think that I have seen a puzzle which doesnt have at least one incidence of a bi value .... [ and the above ultra hard puzzles have 5 and 6] - but I will look some more ...

I thought I had seen one long ago, but I'm unable to find it. Unfortunately, I've been busy with other things and I haven't had time to explore this question.

If you take enough of these proposition pairs - more than 2 pairs - or forcing T and E ... surely all valid puzzles will succumb !

If you mean that all the puzzles can be solved by repeatedly applying Forcing-T&E to all the bivalue cells, then no. For this, I have many counter-examples, starting from EasterMonster.
Moreover, I also have examples (including EasterMonter) for which Forcing-T&E starting from trivalue cells is not enough either.

Despite trying hard I still cant visualize the bivalues that SudoRules finds or the 3D ones either !!!

???
They are the same thing as in your first line.

[coloin]

Ah yes ... there are more .... i just needed to look harder !!! Thanks

The 3D ones are the ones which maybe occur in a box and a row or col simultaneously - although there is no real difference ,,

so the bivalues can occur

in a single box

in one clue postion with 2 different clues

or 2 clue positions with a single clue

- diagonal or off set positions
- in line in 2 positions of a minirow or minicolumn [?3D]

over 2 boxes - in a row or column and this would be a single clue

Fata Morgana

Code: Select all

........3..1..56...9..4..7......9.5.7.......8.5.4.2....8..2..9...35..1..6........#Fata Morgana

Code: Select all

+----------------------+----------------------+----------------------+
| 2458   2467   245678 | 126789 16789  1678   | 24589  1248   3      |
| 2348   2347   1      | 23789  3789   5      | 6      248    249    |
| 2358   9      2568   | 12368  4      1368   | 258    7      125    |
+----------------------+----------------------+----------------------+
| 12348  12346  2468   | 13678  13678  9      | 2347   5      12467  |
| 7      12346  2469   | 136    5      136    | 2349   12346  8      |
| 1389   5      689    | 4      13678  2      | 379    136    1679   |
+----------------------+----------------------+----------------------+
| 145    8      457    | 1367   2      13467  | 3457   9      4567   |
| 249    247    3      | 5      6789   4678   | 1      2468   2467   |
| 6      1247   24579  | 13789  13789  13478  | 234578 2348   2457   |
+----------------------+----------------------+----------------------+


maybe this not so [very] hard puzzle from way back is the one which might have the least....
I only see 4 , with the clues 1 and 9 in box 3 and box 7

[denis_berthier]

Fata Morgana

Code: Select all

........3..1..56...9..4..7......9.5.7.......8.5.4.2....8..2..9...35..1..6........#Fata Morgana

Code: Select all

+----------------------+----------------------+----------------------+
| 2458   2467   245678 | 126789 16789  1678   | 24589  1248   3      |
| 2348   2347   1      | 23789  3789   5      | 6      248    249    |
| 2358   9      2568   | 12368  4      1368   | 258    7      125    |
+----------------------+----------------------+----------------------+
| 12348  12346  2468   | 13678  13678  9      | 2347   5      12467  |
| 7      12346  2469   | 136    5      136    | 2349   12346  8      |
| 1389   5      689    | 4      13678  2      | 379    136    1679   |
+----------------------+----------------------+----------------------+
| 145    8      457    | 1367   2      13467  | 3457   9      4567   |
| 249    247    3      | 5      6789   4678   | 1      2468   2467   |
| 6      1247   24579  | 13789  13789  13478  | 234578 2348   2457   |
+----------------------+----------------------+----------------------+


I only see 4 , with the clues 1 and 9 in box 3 and box 7

SudoRules finds 8 bivalue relations:

Code: Select all

f-9999  (bivalue 0 917 929 439)
f-10001 (bivalue 0 118 139 431)
f-10003 (bivalue 0 929 969 399)
f-10005 (bivalue 0 953 957 259)
f-10007 (bivalue 0 961 981 319)
f-10009 (bivalue 0 171 192 471)
f-10011 (bivalue 0 981 993 479)
f-10013 (bivalue 0 981 985 289)

In this internal representation (forget about the 0), the last term in the relation is the CSP-Variable, starting with 1 for rc (none in this puzzle), 2 for rn, 3 for cn and 4 for bn
So, for instance, (bivalue 0 917 929 439) means: candidates 917 and 929 are bivalue in block (the 4 in 439) number 3 (the 3 in 439) for digit 9 (the 9 in 439).
I'm sure you can now see all of them.

In some cases, a pair of candidates can be bivalue wrt 2 different CSP-Variables, but this is not the case for this puzzle.


[Edit]: my "souvenir" that, long ago, I had seen a puzzle with no bivalues may be wrong. What I have seen may have been only puzzles with no rc-bivalues, such as Fata Morgana.
I've therefore tried a collection of 25,000 hard puzzles and found none having no (rc, rn, cn or bn) bivalues (after application of Singles).

Anyway, my original question looses some of its practical interest wrt Forcing-T&E: contrary to what I thought, a puzzle can have bivalue candidates with none of its bivalue pairs giving rise to any elimination by forcing-T&E. The same is true for trivalues and Forcing{3}-T&E.

What remains of possible theoretical interest is:
does any puzzle not solvable by Singles necessarily have (rc, rn, cn or bn) bivalue candidates (after Singles have been applied) and if true, why?