The New Sudoku Players' Forum

Website · by **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

by **DEFISE** » Fri Mar 19, 2021 2:02 pm

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

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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

Website · by **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

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.

by **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

Website · by **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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