DB 8.3

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DB 8.3

Postby denis_berthier » Thu Mar 18, 2021 8:57 am

Code: Select all
    . . . . . . . . .
    1 3 . . . 7 6 . .
    7 . 5 . . 9 . 8 .
    3 . . 5 . 8 . . 2
    8 . . 7 . 2 5 6 .
    5 . . 6 . 1 . . 3
    4 . 2 . . 6 . 5 .
    6 7 . . . 5 2 . .
    . . . . . . . . .

.........13...76..7.5..9.8.3..5.8..28..7.256.5..6.1..34.2..6.5.67...52........... # 95912 FNBWXYK C29.m/S2.h
29 clues, SER = 8.3
denis_berthier
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Re: DB 8.3

Postby DEFISE » Fri Mar 19, 2021 2:02 pm

I have 2 radically different solutions:
1) with 12 whips [<= 5]

Hidden Text: Show
Singles: 9r9c1, 2r1c1, 3r5c5, 2r6c2, 8r6c7, 5r9c2, 6r9c9, 2r2c8
Alignment: 9c4b8 => -9r7c5 -9r8c5
Alignment: 4b5c5 => -4r1c5 -4r2c5 -4r3c5 -4r8c5 -4r9c5
Xwing: 3r37c47 => -3r1c4 -3r1c7 -3r8c4 -3r9c4 -3r9c7

whip[4]: r2c5{n8 n5}- r1n5{c5 c9}- c9n7{r1 r7}- c9n8{r7 .} => -8r8c5
Single: 1r8c5
whip[5]: b3n9{r1c7 r2c9}- c9n5{r2 r1}- c9n7{r1 r7}- r7c5{n7 n8}- c2n8{r7 .} => -9r1c2
Alignment: 9c2b4 => -9r4c3 -9r5c3 -9r6c3
whip[3]: r9c6{n4 n3}- b2n3{r1c6 r3c4}- c4n2{r3 .} => -4r9c4
Naked triplet: 278b8p278 => -8r7c4 -8r8c4
whip[5]: r2c5{n8 n5}- r1n5{c5 c9}- c9n7{r1 r7}- r7c5{n7 n8}- c2n8{r7 .} => -8r2c3
Alignment: 8r2b2 => -8r1c4 -8r1c5
whip[4]: r4n6{c3 c2}- r3c2{n6 n4}- r3c9{n4 n1}- r5n1{c9 .} => -1r4c3
whip[4]: r8c4{n9 n4}- r8c8{n4 n3}- r1n3{c8 c6}- c6n4{r1 .} => -9r8c9
whip[5]: r3c9{n1 n4}- r3c2{n4 n6}- r1n6{c2 c5}- r1n5{c5 c9}- c9n7{r1 .} => -1r7c9
whip[3]: r7n1{c7 c2}- r4n1{c2 c8}- c9n1{r5 .} => -1r3c7
whip[5]: r3c9{n1 n4}- r3n1{c9 c4}- r1c4{n1 n4}- b1n4{r1c2 r2c3}- r5c3{n4 .} => -1r5c9
Alignment: 1r5b4 => -1r4c2
Alignment: 1c9b3 => -1r1c7 -1r1c8
whip[4]: r7c2{n1 n8}- r7c5{n8 n7}- r7c9{n7 n9}- r5n9{c9 .} => -1r5c2
Singles: 1r5c3, 1r7c2, 8r1c2
whip[5]: r5n4{c2 c9}- r8c9{n4 n8}- r7n8{c9 c5}- r2n8{c5 c4}- r2n4{c4 .} => -4r3c2
Singles: 6r3c2, 2r3c5, 6r1c5, 5r1c9, 1r1c4, 5r2c5, 8r2c4, 2r9c4, 1r3c9, 6r4c3, 7r6c3, 7r7c9, 8r7c5, 7r9c5, 8r9c3, 3r8c3, 8r8c9
Naked pair: 49r4c25 => -4r4c7 -9r4c7 -4r4c8 -9r4c8
Naked pair: 49c8r68 => -4r1c8 -9r1c8 -4r9c8
whip[3]: r9c7{n1 n4}- r3n4{c7 c4}- c6n4{r1 .} => -1r4c7
STTE


2) with 3 whips [<= 12] (I think there is no solution with less than 3 braids and a fortiori with less than 3 whips).

Hidden Text: Show
Singles: 9r9c1, 2r1c1, 3r5c5, 2r6c2, 8r6c7, 5r9c2, 6r9c9, 2r2c8
Alignment: 9c4b8 => -9r7c5 -9r8c5
Alignment: 4b5c5 => -4r1c5 -4r2c5 -4r3c5 -4r8c5 -4r9c5
Xwing: 3r37c47 => -3r1c4 -3r1c7 -3r8c4 -3r9c4 -3r9c7

whip[12]: r7c2{n8 n1}- r7c5{n1 n7}- c9n7{r7 r1}- r1n5{c9 c5}- c5n6{r1 r3}- r3c2{n6 n4}- r3c9{n4 n1}- r3c7{n1 n3}- r7c7{n3 n9}- r8c9{n9 n4}- r5c9{n4 n9}- r5c2{n9 .} => -8L7C9
Singles: 8r8c9, 1r8c5, 3r8c3

whip[11]: r3c9{n4 n1}- r3c7{n1 n3}- r3c4{n3 n2}- c5n2{r3 r9}- c5n7{r9 r7}- r7c9{n7 n9}- r2n9{c9 c3}- r5n9{c3 c2}- r5n1{c2 c3}- c2n1{r4 r7}- r7c7{n1 .} => -4L3C2
Singles: 6r3c2, 2r3c5, 6r1c5, 5r1c9, 5r2c5, 6r4c3, 2r9c4, 7r6c3, 7r7c9, 8r7c5, 1r7c2, 8r9c3, 7r9c5, 8r2c4, 8r1c2, 1r5c3, 1r3c9, 1r1c4
Naked pair: 49r4c25 => -4r4c7 -9r4c7 -4r4c8 -9r4c8
Naked pair: 49c8r68 => -4r1c8 -9r1c8 -4r9c8

whip[2]: r3n4{c4 c7}- r9n4{c7 .} => -4L8C4
STTE
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Re: DB 8.3

Postby denis_berthier » Sat Mar 20, 2021 4:25 am

DEFISE wrote:I have 2 radically different solutions:
1) with 12 whips [<= 5]

Rating 5 can also be obtained using only reversible chains (bivalue-chains and a single z-chain):
Hidden Text: Show
Starting after Singles ans Whips[1]:
x-wing-in-rows: n3{r3 r7}{c4 c7} ==> r9c7 ≠ 3, r9c4 ≠ 3, r8c4 ≠ 3, r1c7 ≠ 3, r1c4 ≠ 3
finned-x-wing-in-rows: n9{r2 r5}{c9 c3} ==> r6c3 ≠ 9, r4c3 ≠ 9
biv-chain[3]: r2c4{n8 n4} - c6n4{r1 r9} - b8n3{r9c6 r7c4} ==> r7c4 ≠ 8
biv-chain[3]: r7c2{n1 n8} - b9n8{r7c9 r8c9} - r8c5{n8 n1} ==> r8c3 ≠ 1, r7c4 ≠ 1, r7c5 ≠ 1
biv-chain[3]: r7c5{n8 n7} - c9n7{r7 r1} - r1n5{c9 c5} ==> r1c5 ≠ 8
biv-chain[3]: r9c6{n4 n3} - b2n3{r1c6 r3c4} - c4n2{r3 r9} ==> r9c4 ≠ 4
hidden-triplets-in-a-block: b8{n3 n4 n9}{r7c4 r9c6 r8c4} ==> r8c4 ≠ 8, r8c4 ≠ 1
biv-chain[4]: r2c5{n8 n5} - b3n5{r2c9 r1c9} - c9n7{r1 r7} - b9n8{r7c9 r8c9} ==> r8c5 ≠ 8
naked-single ==> r8c5 = 1
biv-chain[4]: r8n8{c9 c3} - r8n3{c3 c8} - r7n3{c7 c4} - b8n9{r7c4 r8c4} ==> r8c9 ≠ 9
biv-chain[4]: c4n1{r1 r3} - b2n2{r3c4 r3c5} - c5n6{r3 r1} - r1n5{c5 c9} ==> r1c9 ≠ 1
biv-chain[4]: r2c5{n8 n5} - b3n5{r2c9 r1c9} - c9n7{r1 r7} - r7c5{n7 n8} ==> r9c5 ≠ 8
finned-x-wing-in-columns: n8{c5 c2}{r7 r2} ==> r2c3 ≠ 8
whip[1]: r2n8{c5 .} ==> r1c4 ≠ 8
biv-chain[4]: r2n9{c3 c9} - r2n5{c9 c5} - c5n8{r2 r7} - c2n8{r7 r1} ==> r1c2 ≠ 9
whip[1]: c2n9{r5 .} ==> r5c3 ≠ 9
biv-chain[4]: c9n7{r1 r7} - r7c5{n7 n8} - r2c5{n8 n5} - b3n5{r2c9 r1c9} ==> r1c9 ≠ 4, r1c9 ≠ 9
biv-chain[4]: r7c5{n7 n8} - r2c5{n8 n5} - b3n5{r2c9 r1c9} - c9n7{r1 r7} ==> r7c7 ≠ 7
z-chain[4]: c3n6{r4 r1} - r3c2{n6 n4} - r3c9{n4 n1} - r5n1{c9 .} ==> r4c3 ≠ 1
biv-chain[5]: r3c2{n4 n6} - r3c5{n6 n2} - r9c5{n2 n7} - r7c5{n7 n8} - c2n8{r7 r1} ==> r1c2 ≠ 4
biv-chain[4]: r3c9{n1 n4} - r3c2{n4 n6} - r1c2{n6 n8} - r7c2{n8 n1} ==> r7c9 ≠ 1
biv-chain[3]: r7n1{c7 c2} - c3n1{r9 r5} - c9n1{r5 r3} ==> r1c7 ≠ 1, r3c7 ≠ 1
biv-chain[3]: r3c7{n4 n3} - r1n3{c8 c6} - c6n4{r1 r9} ==> r9c7 ≠ 4
biv-chain[4]: c9n1{r3 r5} - c3n1{r5 r9} - r9n8{c3 c4} - c4n2{r9 r3} ==> r3c4 ≠ 1
singles ==> r1c4 = 1, r3c9 = 1
whip[1]: r5n1{c3 .} ==> r4c2 ≠ 1
biv-chain[3]: r5c9{n4 n9} - r6n9{c8 c5} - b5n4{r6c5 r4c5} ==> r4c7 ≠ 4, r4c8 ≠ 4
whip[1]: c7n4{r3 .} ==> r1c8 ≠ 4, r2c9 ≠ 4
biv-chain[3]: b3n4{r1c7 r3c7} - r3n3{c7 c4} - r1c6{n3 n4} ==> r1c3 ≠ 4
biv-chain[3]: r1n4{c7 c6} - r2n4{c4 c3} - b1n9{r2c3 r1c3} ==> r1c7 ≠ 9
biv-chain[3]: c2n9{r5 r4} - c7n9{r4 r7} - r7n1{c7 c2} ==> r5c2 ≠ 1
singles ==> r5c3 = 1, r7c2 = 1, r1c2 = 8
naked-pairs-in-a-row: r7{c4 c7}{n3 n9} ==> r7c9 ≠ 9
biv-chain[3]: b1n4{r2c3 r3c2} - r5n4{c2 c9} - c9n9{r5 r2} ==> r2c3 ≠ 9
stte



DEFISE wrote:2) with 3 whips [<= 12] (I think there is no solution with less than 3 braids and a fortiori with less than 3 whips).

Notice that your solution is not really 3-step, as there are Subsets.
There are 330 W1-anti-backdoor pairs. None of them gives rise to a solution in W13. I didn't try with longer whips due to increasing computation times and to what I consider absurd for a puzzle solvable in Z5.

What's noticeable in many of the examples I've tried is how a relatively easy puzzle can become intractable with an additional condition. Single-step solutions are not too hard to find for puzzles that have one but, for those that don't, 2-step solutions tend to be beyond human possibilities.
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Re: DB 8.3

Postby DEFISE » Sat Mar 20, 2021 8:12 pm

denis_berthier wrote:Notice that your solution is not really 3-step, as there are Subsets.

Yes if you want, but note that I have deliberately avoided talking about “steps”.
I simply said that I had used 3 whips [<= 12]. However I should have specified whips of length >= 2 and <= 12 because my alignments are whips [1].

denis_berthier wrote:There are 330 W1-anti-backdoor pairs. None of them gives rise to a solution in W13. I didn't try with longer whips due to increasing computation times and to what I consider absurd for a puzzle solvable in Z5.
What's noticeable in many of the examples I've tried is how a relatively easy puzzle can become intractable with an additional condition. Single-step solutions are not too hard to find for puzzles that have one but, for those that don't, 2-step solutions tend to be beyond human possibilities.

In this example it’s faster to check that:
for each of the 330 W1-anti-backdoor pairs (Ai,Bi) we never have (1) or (2):
(1) [T&E(BRT, Ai, RS) => error] and [T&E(BRT, Bi, RS-{Ai}) => error]
(2) [T&E(BRT, Bi, RS) => error] and [T&E(BRT, Ai, RS-{Bi}) => error]

According to your “T&E vs braids” theorem, this is equivalent to say that we never have (1’) or (2’):
(1’) Ai target of a braid and Bi target of a braid, after deleting Ai
(2’) Bi target of a braid and Ai target of a braid, after deleting Bi

and finally we never have (1’’) or (2”):
(1’’) Ai target of a whip and Bi target of a whip, after deleting Ai
(2’’) Bi target of a whip and Ai target of a whip, after deleting Bi
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Location: France

Re: DB 8.3

Postby denis_berthier » Sun Mar 21, 2021 2:54 am

DEFISE wrote:
denis_berthier wrote:There are 330 W1-anti-backdoor pairs. None of them gives rise to a solution in W13. I didn't try with longer whips due to increasing computation times and to what I consider absurd for a puzzle solvable in Z5.
What's noticeable in many of the examples I've tried is how a relatively easy puzzle can become intractable with an additional condition. Single-step solutions are not too hard to find for puzzles that have one but, for those that don't, 2-step solutions tend to be beyond human possibilities.

In this example it’s faster to check that:
for each of the 330 W1-anti-backdoor pairs (Ai,Bi) we never have (1) or (2):
(1) [T&E(BRT, Ai, RS) => error] and [T&E(BRT, Bi, RS-{Ai}) => error]
(2) [T&E(BRT, Bi, RS) => error] and [T&E(BRT, Ai, RS-{Bi}) => error]

Hi François,
You are right, but my main purpose in such cases is not to prove there's no 2-step solution. It is to find one with the smallest whips. Of course, I don't check all the pairs; I first eliminate all those that don't contain a candidate that can be eliminated in the given RS with the chosen rules.
In the present case, as you had done the testing up to W12, I started with W13 (which anyway would have given absurdly long whips).

What I want to remember from this example is how such arbitrary requirements on the number of steps lead to situations that are totally disconnected from the real difficulty of the puzzle.
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