Maximum number of clues for a given SE rating

Everything about Sudoku that doesn't fit in one of the other sections

Postby susume » Tue Aug 05, 2008 12:38 pm

Thanks for the additional info and Gurth's proof (Im not sure it's a proof so much as a convincing argument, but in any case I'm convinced). In this particular puzzle, each equivalence set consists of 4 symmetrically placed candidates, each 90 degrees anti-clockwise from the last, not just 2, so I think you can get 4x as much information from each chain. Example: the three cells symmetrical to r1c7 are r3c1, r9c3, and r7c9. One set of candidates in them that must all be true together or all be false together are (4)r1c7, (1)r3c1, (2)r9c3, (3)r7c9. The set of digits 1234 are equivalent in every symmetrical set of cells in which one of them appears, and the set 6789 are equivalent in every symmetrical set of cells in which one of them appears.

I believe this allows substitution of any proposition in the equivalence set into a true proposition and it will still be true. Example: (4)r1c7=(4)r3c9 (biloc in block). The following must then also be true:
(4)r1c7=(1)r1c3
(4)r1c7=(2)r7c1
(4)r1c7=(3)r9c7
(1)r3c1=(4)r3c9
(2)r9c3=(4)r3c9
(3)r7c9=(4)r3c9
(please let me know if my notation is bad - still learning)

It may be unusual to have strong links on different digits in different cells, but the ones shown above still behave as regular strong links - i.e. exactly one of the pair must be true. I found a simple nice loop to accomplish the first elimination found by the SE solver.

Oops- Glyn I see you also stated 4-way symmetry while I was writing.... I chose anti-clockwise only because the clues in this puzzle are placed anti-clockwise, so it's easier for me to be sure I'm getting the right digits in the equivalence set.

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Re: re: arguing from symmetry

Postby tarek » Tue Aug 05, 2008 1:03 pm

Pat wrote:
    so all i had to do was,
    carefully read RW's post on the present page---
tarek, are you convinced?

Fair enough,

if the clues displayed symmetry, then it would be easy to extract this extra constraint & use to advance the puzzle:?:

I'm not sure it would be that easy if the clues didn't display symmetry. In these cases it can be extracted later on as more clues are uncovered or probably if the PM grid shows the symmetry:?::?:

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Postby RW » Tue Aug 05, 2008 1:07 pm

Glyn wrote:Actually the symmetry of the Mauricio 37 clue puzzle is even greater
Using 90 degree clockwise rotations
Code: Select all
2-->1  6-->9
^   |  ^   |
|   v  |   v
3<--4  7<--8

4 times the info?

You are absolutely right! Using this info, the puzzle is not that hard at all!:)

Code: Select all
If r2c6=9 => r6c8=8, r8c4=7, r4c2=6 (symmetry)
   => r3c6=7 => r3c8=6 => r3c9=4 => r1c2=4 => No candidate 7 in column 2
    => r2c6<>9

If r2c4=9 => r4c8=8 => r8c6=7 => r6c2=6 (symmetry)
  => r2c6=6 => r3c6=1 => r1c3=1
  => r4c2=7 => r2c3=7 => r8c3=8 => r8c4=empty cell
    => r2c4<>9


=> r2c3=9 (and r3c8=8, r8c7=7, r7c3=6), the rest is singles!:D

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Postby Glyn » Tue Aug 05, 2008 1:25 pm

Well done RW no need for me to do it now.
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Postby RW » Tue Aug 05, 2008 3:51 pm

Glyn wrote:Well done RW no need for me to do it now.

Well, that was done very quickly, found the solution on almost my first attempt. I'm sure there's still a shorter solution for you to find!:D

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Postby JPF » Tue Aug 05, 2008 4:16 pm

How does it work for this one :

Code: Select all
 1 2 3 | . 5 . | . . .
 4 . . | 7 . . | . 2 .
 7 . 9 | . . 3 | . . .
-------+-------+-------
 . 3 . | . . . | 8 . .
 5 . . | . . . | . . 1
 . . 7 | . . . | . 6 .
-------+-------+-------
 . . . | 6 . . | 9 . 8
 . 4 . | . . 8 | . . 2
 . . . | . 1 . | 6 4 5       ER=10.4


With 2 symmetries :

D symmetry :
1-1 2-4 3-7 5-5 6-8 9-9

AD symmetry :
1-5 2-2 3-8 4-4 6-7 9-9

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Postby RW » Tue Aug 05, 2008 4:32 pm

JPF wrote:How does it work for this one :

With 2 symmetries :

D symmetry :
1-1 2-4 3-7 5-5 6-8 9-9

AD symmetry :
1-5 2-2 3-8 4-4 6-7 9-9

JPF

Can't get any immediate results from the diagonal symmetries, but yes, I believe they can be used just like the rotational symmetry in the previous example.

But your puzzle does have 180 rotational symmetry: 1-5 2-4 3-6 7-8 9-9

This solves r5c5=9, which brings ER down to 7.3!:)
(Actually, r5c5 could also be solved by looking at the diagonal symmetries alone.)

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Postby Glyn » Tue Aug 05, 2008 4:41 pm

JPF I'm sure there are many ways. Here's one

Box 7 AD symmetry 2 locked in r7c3 or r9c1 => r7c1=3 and r9c3=8
Box 9 D symmetry r8c8=1
Box 1 D symmetry r2c2=5
Singles to finish
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Postby JPF » Tue Aug 05, 2008 6:33 pm

Excellent Glyn:!:

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Postby ttt » Tue Aug 05, 2008 11:24 pm

Hi All,
Very interesting and impressive!
I have one dumb question : how can I define a puzzle symmetry? Only see givens then conclude a puzzle is symmetry?

Thanks
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Postby susume » Wed Aug 06, 2008 1:04 am

Well done RW and Glyn!

It seems symmetrical givens can only produce symmetrical eliminations and placements. So once givens + solved cells exhibits symmetry, there is no way asymmetry can re-enter the solution.

Anyone care to offer a puzzle with a symmetrical solution, where the givens are not symmetrical (or only in placement - symmetry of placement does not match symmetry of solution) and the symmetry of the solution only appears part way through the solving?

Is there a good term to distinguish digit-location symmetry from the usual location-only symmetry of the givens?

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Postby udosuk » Wed Aug 06, 2008 5:56 am

susume wrote:Anyone care to offer a puzzle with a symmetrical solution, where the givens are not symmetrical (or only in placement - symmetry of placement does not match symmetry of solution) and the symmetry of the solution only appears part way through the solving?

Here is a "subpuzzle" of the 11.2 one on the previous pages (with digits permuted):
Code: Select all
7.......1
.6..2..8.
..8..54..
..5.7...4
.4.9...6.
....385..
..65..2..
.2..8..4.
9....2..3

Once you establish symmetry by filling all the singles you can start applying symmetrical techniques.

susume wrote:Is there a good term to distinguish digit-location symmetry from the usual location-only symmetry of the givens?

Yes, the term is Emerald.
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Postby RW » Wed Aug 06, 2008 6:40 am

susume wrote:It seems symmetrical givens can only produce symmetrical eliminations and placements. So once givens + solved cells exhibits symmetry, there is no way asymmetry can re-enter the solution.

That is correct! Actually, not even all givens + solved cells need to display symmetry, as long as all givens and their symmetrically opposite cells display symmetry. But beware, as long as there is one single given whose symmetrically opposite cell is not solved, the solution might still not be symmetrical, no matter how promising the rest of the grid looks. An example:
Code: Select all
 *-----------*
 |721|368|459|
 |45.|917|.32|
 |39.|245|1..|
 |---+---+---|
 |.34|176|9.5|
 |...|493|...|
 |6.9|582|34.|
 |---+---+---|
 |..2|631|.94|
 |14.|829|763|
 |963|754|218|
 *-----------*

62 solved cells, all solved symetrically opposite cells form the pairs 1-2 3-4 5-6 7-8 9-9. Only one given (r8c7) whose opposite cell is not yet solved. The solution is not symmetrical.

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Postby 999_Springs » Thu Aug 07, 2008 5:29 pm

Intrigued by the failure of Bob Hanson's solver to remove a single candidate and the fact that SE needs 12 sub-chains in some enormous first step, I ask these questions:
1. How far does Bob Hanson's ++depth actually go?
2. Do all puzzles of this sort of difficulty stump Bob Hanson's solver, or just this one?
3. Why choose r1c7#3 in Mauricio's puzzle? Is this the simplest elimination? (I can hardly believe this is.)

Back on topic.
I found this puzzle lurking in the corners of these forums.
Pat wrote:
Code: Select all
 9 7 . | . 4 8 | 3 1 6
 3 6 8 | 1 . 7 | . 4 .
 1 4 . | . 3 6 | 7 . 8
-------+-------+------
 2 1 9 | 3 7 5 | 8 6 4
 8 3 6 | 4 1 9 | 2 5 7
 7 5 4 | 8 6 2 | 1 3 9
-------+-------+------
 5 . 7 | 6 . 3 | 4 8 1
 4 8 3 | 7 . 1 | 6 . .
 6 . 1 | . 8 4 | . 7 3

ER=1.7 [edit: 2.6], 65 clues
Last edited by l$ on Fri Aug 08, 2008 1:47 pm, edited 1 time in total.
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Postby JPF » Thu Aug 07, 2008 7:27 pm

ER=2.6:?:
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