The New Sudoku Players' Forum

[champagne]

Here four puzzles of the current pattern game (236)
The serate rating is in the range 10.2 to 11.1
They all have common "exotic properties"
which ones and what is the best path to solve them

Code: Select all

.....1....2..3..4...15....63.7....8.2...8...3.8....4.25....69...7..2..3....9.....
.....1....2..3..4...15....62.7....8.3...8...2.8....4.35....69...7..2..3....9.....
.....1....2..3..4...56....72.8....3.3...8...4.4....8.96....71...3..4..9....5.....
.....1....2..3..4...56....72.8....9.3...9...2.9....4.36....71...8..2..3....5.....

[pjb]

All have a double exocet. The first has 1569 at r4c7, r4c9, r5c2 and r6c5 => -4 r5c2, -7 r6c5; then 1695 at r6c1, r6c3, r4c5 and r5c8 => -4 r4c5, -7 r5c8. Then the double exocet configuration allows removal of 1569 from r6c8 and r4c2. Following a naked quad of 1569 at r5c2378, then is a nice little continuous loop: (1=3)r7c2-(3=9)r3c2-(9=2)r3c8-(2=1)r7c8 with multiple eliminations, after which the puzzle is easy. The other 3 are similar.

Phil

[champagne]

All have a double exocet.
Phil

They have more that a double exocet, they have a symmetry of given.

I did not check, but this should make the solution even simpler.

[pjb]

champagne wrote:

They have more that a double exocet, they have a symmetry of given.
I did not check, but this should make the solution even simpler.

I'd be grateful if you could demonstrate how the symmetry is used to make eliminations
Thanks, Phil

[champagne]

I'd be grateful if you could demonstrate how the symmetry is used to make eliminations
Thanks, Phil

here is the first puzzle after Phil moves plus the naked pair 47 in column 5

Code: Select all

478   569   348    |2478  69     1     |2378   59    78 
6789  2     5689   |678   3      789   |1578   4     15789
478   39    1      |5     47     2478  |2378   29    6     
---------------------------------------------------------
3     4     7      |126   1569   259   |156    8     159   
2     1569  569    |47    8      47    |156    1569  3     
169   8     569    |136   1569   359   |4      7     2     
---------------------------------------------------------
5     13    2348   |3478  47     6     |9      12    1478 
14689 7     4689   |148   2      458   |1568   3     1458 
48    16    2348   |9     15     3478  |278    156   478

At that point the easiest seems to use the symmetry constraint 9r1c5<->1r9c5
We can then either prove r1c6=6 or r9c5=5

1r9c5 =>9r1c5 - 9r1c8 =5r1c8 - 5r9c8 = 5r9c5 - 1r9c5 <> 1r9c5

after r1c5 = 6 and r9c5 = 5

r2c6=9 r7c4=1 r4c9=9 r6c1=1 ....


EDIT generally speaking, in my experience, once the limitations for the digits self mirror applied (in a central symmetry, this is only affecting r5c5, here given) the symmetry is best used in small contradiction chains skipping once or twice from a candidate to the mirror candidate.

Each contradiction chain as a mirror one, but the eliminations can also be duplicated using the symmetry.

[David P Bird]

Champagne,

I'm curious, have you a method to detect various types of symmetry even when the row and column orders are scrambled?

David

[champagne]

Champagne,

I'm curious, have you a method to detect various types of symmetry even when the row and column orders are scrambled?

David

right, but the worst case for a player.

The code try all possible starts for a symmetry of given (also just for a symmetry upon request) and looks for a possible re morphing of the puzzle.

AFAIK, gsf's code has functions performing better than my own code.