The New Sudoku Players' Forum

[Mauriès Robert]

Hi Denis, P.O.,
Since the one-step solution of this puzzle relies on the almost closed set 23p578b3, the simplest one-step solution seems to me to be the following, with a relatively short sequence (TDP track) :
P(4r3c7) : [ [ 4r3c7->3r6c7->3r4c5->5r4c4->4r4c3 ]->4r1c2->7r1c6 ]->7r5c7->2r5c8->... 8b6 empty => -4r3c7, lclste.


Robert

[denis_berthier]

.
@P.O.
You don't FIND the 3 chains independently.

[P.O.]

hi Robert, effectively removing n4 from r3c7 gives the 2 triplets whose eliminations are enough to solve the puzzle; to set n4 in r3c7 to prove a contradiction is a good way to do it; so supposing n4 in r3c7 here your track in chain notation:
c7n3{r3 r6} - b5n3{r6c6 r4c5} - r4n5{c5 c4} - r4n4{c4 c3} - c2n4{r5 r1} - r1c6{n4 n7} - c7n7{r1 r5} - r5n2{c7 c8} = b6 empty of 8 => r3c7 <> 4
the difficulty is to start the chain, what reason to assert n4 in r3c7?; i like these kinds of finds but it seems that a logical step is missing at the start of the resolution process.

[Mauriès Robert]

Hi P.O.,

the difficulty is to start the chain, what reason to assert n4 in r3c7?; i like these kinds of finds but it seems that a logical step is missing at the start of the resolution process.

Perhaps you prefer the logic of the interaction of two conjugate tracks that eliminates the 4r3c7, like this starting with the 3r6 pair :
P(3r6c7) : 3r6c7->[ 179p239b6->7r1c7>4r1c6 and 3r4c5->5r4c4->4r4c3 ]->4r3c2->...
P(3r6c6) : 3r6c6->3r3c7->...
=> -4r3c7 => closed set 236p578b3 => lclste.

puzzle: Show

Note that the track P(3r6c7) leads to a contradiction if we continue its development while the track P(3r6c6) leads to the solution without having to use the closed set.
Robert

[P.O.]

hi Robert,
yes i actually prefer this solution, although more complicated the beginning does not have the arbitrary character of the previous one, as either n3r6c6 or n3r6c7 is true being able to obtain the same conclusion, the elimination of n4 in r3c7, from both of them is in accordance with the rules we have set for ourselves to play this game.

[Mauriès Robert]

Hi P.O.,

yes i actually prefer this solution, although more complicated the beginning does not have the arbitrary character of the previous one, as either n3r6c6 or n3r6c7 is true being able to obtain the same conclusion, the elimination of n4 in r3c7, from both of them is in accordance with the rules we have set for ourselves to play this game.

If I understand correctly, what bothers you is to use the contradiction to eliminate 4r3c7 as in my first resolution. You can avoid the contradiction like this with an anti-track:
P'(3r3c7) : (-3r3c7)->[ 3r6c7->179p239b6->7r1c7->4r1c6 and 3r4c5->5r4c4->4r4c3 ]->4r3c2->... => -4r3c7.

puzzle: Show

That said, Denis Berthier's whips and braids are chains of contradiction and this is not a problem.
Robert

[P.O.]

hi Robert,
in fact the question i was asking myself is why you choose to put 4 in r3c7 in your first resolution.