The New Sudoku Players' Forum

[champagne]

now ok with blues count for the R90 ED puzzles
thanks to blue for the file pointing on the last bug
the final process lasted 1 hour for that symmetry.
The run time should be nearly the same printing the "symmetry minimal" but not globally minimal.

[eleven]

Here are 5 puzzles with 23 givens having backdoor of size 3 in singles and 180 deg rotational symmetry

These puzzles are a challenge also with symmetry. Below are my steps to bring the first one down to "normal advanced" difficulty.
Are there any other solutions available, maybe by a solver ?

Hidden Text: Show

Code: Select all

. . . . . 7 . . 6
. . 3 . 4 . 2 . .
7 . . 8 . . . 4 .
. . 1 . . . . 2 3
. . . . 1 . . . .
4 5 . . . . 1 . .
. 3 . . . 6 . . 9
. . 5 . 3 . 4 . .
8 . . 9 . . . . .
+----------------------+----------------------+----------------------+
| 1259   12489  2489   | 1235   259    7      | 3589   13589  6      |
|*1569  *1689   3      | 156    4      159    | 2      15789  1578   |
| 7      1269   269    | 8     *2569   12359  | 359    4      15     |
+----------------------+----------------------+----------------------+
|*69    *6789   1      | 4567   56789  4589   |*56789  2      3      |
| 3      26789  26789  |*2567   1      2589   | 56789  56789  4      |
| 4      5     *26789  | 2367   26789  2389   | 1     *6789   78     |
+----------------------+----------------------+----------------------+
| 12     3      247    | 12457  2578   6      | 578    1578   9      |
|*1269  *12679  5      | 127    3      128    | 4     *1678   1278   |
| 8     *12467 *2467   | 9      257    1245   |*3567  *13567  1257   |
+----------------------+----------------------+----------------------+

Note, that a 6 in r5c4 forces 6r3c7, (else x-wing, 6r9c3,r6c8, empty b9).
79r6c3/r3c7->68r3c12/r6c89->6r5c4 contr. => r6c3/r3c7 <>79

Code: Select all

+----------------------+----------------------+----------------------+
| 1259   12489  2489   | 1235   259    7      | 3589   13589  6      |
| 159    1689   3      | 156    4     a159    | 2     b15789  1578   |
| 7      1269   269    | 8      2569   12359  | 359    4      15     |
+----------------------+----------------------+----------------------+
|a69     6789   1      | 4567  b56789  4589   | 568    2      3      |
| 3      26789  26789  | 2567   1      2589   | 56789  56789  4      |
| 4      5      268    | 2367  a26789  2389   | 1      6789  b78     |
+----------------------+----------------------+----------------------+
| 12     3      247    | 12457  2578   6      | 578    1578   9      |
| 1269  a12679  5      |a127    3      128    | 4      1678   127    |
| 8      12467  2467   | 9      257    1245   | 3567   13567  1257   |
+----------------------+----------------------+----------------------+

6r4c1/8r6c9 -> 2r6c3/5r4c7
9r4c1/7r6c9->9r8c2/7r2c8->9r2c6/7r8c4->9r6c5/7r4c5->68r4c2/68r6c8 -> 2r6c3/5r4c7

Code: Select all

+----------------------+----------------------+----------------------+
| 1259   12489  489    | 1235   259    7      | 389    13589  6      |
| 1569   1689   3      | 156    4      159    | 2      15789  1578   |
| 7      1269   69     | 8      2569   12359  | 39     4      15     |
+----------------------+----------------------+----------------------+
| 69     6789   1      | 467    6789   489    | 5      2      3      |
| 3      6789   6789   | 25     1      25     | 6789   6789   4      |
| 4      5      2      | 367    6789   389    | 1      6789   78     |
+----------------------+----------------------+----------------------+
| 12     3      47     | 12457  2578   6      | 78     1578   9      |
| 1269   12679  5      | 127    3      128    | 4      1678   1278   |
| 8      12467  467    | 9      257    1245   | 367    13567  1257   |
+----------------------+----------------------+----------------------+

4r7c3 or 7r7c3/9r3c7->6r3c3->4r9c3 => r1c3<>4, r9c7<>3
Then 8r1c3 or 9r1c3/7r9c7->8r7c7 => r1c7<>8,r9c3<>6

[champagne]

Here are 5 puzzles with 23 givens having backdoor of size 3 in singles and 180 deg rotational symmetry

These puzzles are a challenge also with symmetry. Below are my steps to bring the first one down to "normal advanced" difficulty.
Are there any other solutions available, maybe by a solver ?

a central symmetry can be shared with exocets, this should be considered

[David P Bird]

a central symmetry can be shared with exocets, this should be considered

Yes, applying a Double JExocet before applying symmetry considerations the first puzzle reduces to

Hidden Text: Show

Code: Select all

 *-----------------*-----------------*-----------------*
 | 125  124  489   | 1235 259  <7>   | 389  135  <6>   |
 | 1569 689  <3>   | 156  <4>  159   | <2>  789  1578  |
 | <7>  12   69    | <8>  2569 1235  | 39   <4>  15    |
 *-----------------*-----------------*-----------------*
 | 69   6789 <1>   | 467  6789 489   | 5    <2>  <3>   |
 | 3    6789 6789  | 25   <1>  25    | 6789 6789 4     |
 | <4>  <5>  2     | 367  6789 389   | <1>  6789 78    |
 *-----------------*-----------------*-----------------*
 | 12   <3>  47    | 1245 2578 <6>   | 78   15   <9>   |
 | 1269 679  <5>   | 127  <3>  128   | <4>  678  1278  |
 | <8>  124  467   | <9>  257  1245  | 367  135  125   |
 *-----------------*-----------------*-----------------*

[champagne]

Yes, applying a Double JExocet before applying symmetry considerations the first puzzle reduces to

and in that position, using the symmetry of given, you have r1c1=5 ..... to the end

[eleven]

a central symmetry can be shared with exocets, this should be considered

Yes, applying a Double JExocet before applying symmetry considerations the first puzzle reduces to ...

Oh nice, and so much extra eliminations.

... and in that position, using the symmetry of given, you have r1c1=5 ..... to the end

For example, yes. I need 2 steps for that, and something after it, before i can finish without chains.

[champagne]

Yes, applying a Double JExocet before applying symmetry considerations the first puzzle reduces to ...

Oh nice, and so much extra eliminations.

if I red correctly the print, this works for the 5 puzzles

... and in that position, using the symmetry of given, you have r1c1=5 ..... to the end

For example, yes. I need 2 steps for that, and something after it, before i can finish without chains.

I looked at the first solution proposed by my solver. I don't know what is the best path, but I did not see any "hard" elimination.
A typical elimination in that position is as that one

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125  124  489  |1235 259  7    |389  135  6   
1569 689  3    |156  4    159  |2    789  1578
7    12   69   |8    2569 1235 |39   4    15   
----------------------------------------------
69   6789 1    |467  6789 489  |5    2    3   
3    6789 6789 |25   1    25   |6789 6789 4   
4    5    2    |367  6789 389  |1    6789 78   
----------------------------------------------
12   3    47   |1245 2578 6    |78   15   9   
1269 679  5    |127  3    128  |4    678  1278
8    124  467  |9    257  1245 |367  135  125 

2r1c1 - 2r7c1 = 1r7c1 - {1r7c8;1r3c2} = 2r3c2 - 2r1c1

somewhere in the chain, you replace the candidate (1r7c8) by the "given symmetric" here (1r3c2), '1' is its own symmetric

[Leren]

David P Bird Wrote :Yes, applying a Double JExocet before applying symmetry considerations the first puzzle reduces to

I think that if you apply the symmetry considerations first you can find the first Exocet r4c1 r4c2 r6c5 r5c7 6789 and then deduce that there must be a second Exocet r6c8 r6c9 r5c3 r4c5 6789 from symmetry ie without having to look for it in the normal way. So the symmetry gives you two Exocets (in fact a double Exocet) for the price of one. That looks like a bargain to me .

Leren

[David P Bird]

David P Bird Wrote :Yes, applying a Double JExocet before applying symmetry considerations the first puzzle reduces to

I think that if you apply the symmetry considerations first you can find the first Exocet r4c1 r4c2 r6c5 r5c7 6789 and then deduce that there must be a second Exocet r6c8 r6c9 r5c3 r4c5 6789 from symmetry ie without having to look for it in the normal way. So the symmetry gives you two Exocets (in fact a double Exocet) for the price of one. That looks like a bargain to me .

I imagine it's a question of which is the quicker to conduct; testing for symmetrical givens or testing for Exocets. Testing for JExocets alone should be much faster than a full search and will detect 95%+ of all of Exocets (I forget the figure Champagne reported). For a manual player once the double JExocet is processed the symmetry becomes glaring.

It also surprises me that you seem to need to conduct a second search to discover if a single JExocet is half of a double. Again testing manually it becomes immediately obvious, and in that case the gain in the available eliminations is enormous. From what you say you're suggesting finding a single Exocet first, then testing for symmetry, then finding the other single JExocet (which was available from the start) in a repeat search which seems rather peculiar.

[champagne]

I think that if you apply the symmetry considerations first you can find the first Exocet r4c1 r4c2 r6c5 r5c7 6789 and then deduce that there must be a second Exocet r6c8 r6c9 r5c3 r4c5 6789 from symmetry ie without having to look for it in the normal way. So the symmetry gives you two Exocets (in fact a double Exocet) for the price of one.
Leren

Perfect "if the exocet is in the middle band or middle stack".
What if ( and is it possible ?) the exocet is in an external band/stack.

Then IMO, you'll have surely 2 exocets, but not a double exocet.

[eleven]

If a symmetry is not in normal form, it is probably easier to find an exocet, than to see the symmetry.
But as soon as you know, that a puzzle is digit symmetric (e.g. because it was announced as such like here), and you find an Exocet, you know, there is a second one. As champagne noted, if it's not a double exocet, you can forget it - symmetry makes the rest. (btw it would be interesting to see a JExocet in a say diagonal symmetric puzzle.)

David, i had a look at the extra eliminations now. Some are straightforward, others not (1r2c2, 5r2c8, if i remember right). Do you have a systematical way to find them manually?
Just testing "suspicious" candidates can be very boring, because it can take long to find out, that they don't contradict to the exocet conditions.

[Leren]

As it turned out my solver found two other Exocets r4c2 r5c2 r1c3 r9c3 6789 and r5c8 r6c8 r1c7 r9c7 6789.

These both have aligned target cells, are in different stacks and don't form a double when combined. Nevertheless I could have deduced the existence of the fourth Exocet (via symmetry) having found the third.

From a programming perspective I don't see this as a big issue. My solver finds a list of Exocets and then looks at each pair to see if they from a double when combined. Nevertheless the symmetry would allow you to "double" the Exocet list without actually finding each "second" Exocet in the normal way.

Leren

[David P Bird]

If a symmetry is not in normal form, it is probably easier to find an exocet, than to see the symmetry.
But as soon as you know, that a puzzle is digit symmetric (e.g. because it was announced as such like here), and you find an Exocet, you know, there is a second one. As champagne noted, if it's not a double exocet, you can forget it - symmetry makes the rest. (btw it would be interesting to see a JExocet in a say diagonal symmetric puzzle.)

Good points but there could be symmetric single JExocets say in tiers 1 & 3 (as per Leren's post I now see).

David, i had a look at the extra eliminations now. Some are straightforward, others not (1r2c2, 5r2c8, if i remember right). Do you have a systematical way to find them manually?
Just testing "suspicious" candidates can be very boring, because it can take long to find out, that they don't contradict to the exocet conditions.

These eliminations come from naked triples (124)r139c2 & (135)r179c8

As I reduce the grid my spread sheet highlights tuples, line/box intersections, and simple fish - as you say boring, time consuming tasks. Anything more advanced I have to find for myself with the assistance of various colouring options.

[eleven]

These eliminations come from naked triples (124)r139c2 & (135)r179c8

What am i missing ? I thought, the triples were a result of the eliminations.

Code: Select all

+----------------------+----------------------+----------------------+
| 125-9  124-89 489    | 1235   259    7      | 389    135-89 6      |
| 1569   689-1  3      | 156    4      159    | 2      789-15 1578   |
| 7      12-69  69     | 8      2569   1235-9 | 39     4      15     |
+----------------------+----------------------+----------------------+
| 69     6789   1      | 467    6789   489    | 5      2      3      |
| 3      6789   6789   | 25     1      25     | 6789   6789   4      |
| 4      5      2      | 367    6789   389    | 1      6789   78     |
+----------------------+----------------------+----------------------+
| 12     3      47     | 1245-7 2578   6      | 78     15-78  9      |
| 1269   679-12 5      | 127    3      128    | 4      678-1  1278   |
| 8      124-67 467    | 9      257    1245   | 367    135-67 125-7  |
+----------------------+----------------------+----------------------+

[David P Bird]

These eliminations come from naked triples (124)r139c2 & (135)r179c8

What am i missing ? I thought, the triples were a result of the eliminations.

As there are 4 target cells, the base digits 6789 are all true in the two single JExocets so can be eliminated in the non-'S' cells in their respective cover rows; (6) in r3 & r9, (7) in r8 & r9, (8) in r1 & r7, and (9) in r1 & r3.