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X-treme 23 October - stuck
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emm
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Joined: 11 Oct 2006
Posts: 34

PostPosted: Wed Nov 01, 2006 8:23 pm    Post subject: Reply with quote

I totally agree with you, Ed. Because there are more than two 7s in the sets, cancelling one out from the diagonal does not lock the remaining cells in the set.

I am not an expert on this - in fact I only tried ALS when I read Ruud's suggestion that you should throw in the towel! The ALS I suggested was just a copy of the one you gave as a hint from ? SudoCue so I am unable to comment on it's validity - except to say that it seems to work! Very Happy

Shall we wait for help from above?
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emm
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Joined: 11 Oct 2006
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PostPosted: Wed Nov 01, 2006 10:56 pm    Post subject: Reply with quote

No need to wait!

I know I said your turn, Ed - sorry! I canít resist the temptation.

A = r234568c4
B = r6c7 r9c79
X = 3 diagonal
Y = 4
=> r9c4 <> 4

which solves it Very Happy

Edit: I jumped too soon - I see this has an xtra 3 in set B - but it works so well it's just got to be right. Doesn't it?
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sudokuEd
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Joined: 19 Jun 2006
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Location: Sydney Australia

PostPosted: Thu Nov 02, 2006 4:00 am    Post subject: Reply with quote

Havn't looked at what you've done yet emm.

Found this way to solve the puzzle - took this long working with loooong contradiction chains.

Hopefully your way works - looks much more elegant and worthy of the one 'from above' (who will hopefully, on the weekend, descend from his lofty position to help out some poor stuck mortals with their Special Killer X Smile )

Step 1
4 in r2c9 -> 4 in D\ must be in r6c6.

4 elsewhere in n3 must be in c7
-> no 4 in r6c7

Step 2
A loooong contradiction chain eliminates 4 from r9c7.
The (too?) implicit version of this move doesn't look bad so will use that.

4 in r9c7 ->3 must be in r7c7

4 in r9c7 -> 4 in r1c2 -> 6 in r1c3 -> 6 in r7c7.

But this means both 3 and 6 in r7c7 -> no 4 in r9c7

Step 3
A contradiction chain now eliminates 6 from r1c3 - (key steps only shown)

6 in r1c3 -> 6 in r7c7 -> 3 in r9c7 -> 3 in r7c1
But leaves no 3 for N1 -> r1c3 <> 6 -> r1c2 = 6

This unlocks the puzzle - the rest is no sweat.

BTW - am very glad Ruud didn't post a new X-treme yesterday. WAY behind already. What about just 1 of these 'unfair' X-tremes a week Ruud?

[edit: hmm - you're right emm - doesn't work. 3 in r4c4 means the rest of A could be {15678} and r9c79 could be {35}, which doesn't eliminate 4 from r9c4. Darn. Would have been so fitting for a dragon like this to be slain by ALS's xz sword - rather than the unweildy goblin smashing chains I used. But at least its dead Very Happy ]
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emm
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Joined: 11 Oct 2006
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PostPosted: Fri Nov 03, 2006 1:33 am    Post subject: Reply with quote

Iím not sure about our weapons, Ed! The goblinsmasher seems to have destroyed some of the evidence and the flaming sword may be a fake. Sad

Now this may be a wild goose chase, but I found this on the other forum - ALSxz Mk II - an ALS with more than one X candidate in some obscure mathematical situation. Maybe this is the sort of thing that applies to Sudoku X ALS.
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Ruud
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PostPosted: Fri Nov 03, 2006 11:10 pm    Post subject: Reply with quote

I ran out of X-Files, but the stock has now been resupplied. Sorry, Ed. There is still a daily X-Treme, but I've changed the selection criteria, so there will atmost be 1 terrible move like ALS, XY-Chain, or 3D Medusa in each puzzle. Maybe this will help you catch up. Otherwise, there's always the archive...

This week's Assassin may keep you busy for a while, though Twisted Evil

cheers,
Ruud
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sudokuEd
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PostPosted: Mon Nov 06, 2006 12:17 pm    Post subject: Reply with quote

Ruud wrote:
Sorry, Ed. There is still a daily X-Treme, but I've changed the selection criteria, so there will atmost be 1 terrible move like ALS, XY-Chain, or 3D Medusa in each puzzle. Maybe this will help you catch up.
Sounds great Ruud - I actually managed to finish todays 'unfair'. Not going to try and catch up though - seem to have become a Monday X-tremist.

emm wrote:
The goblinsmasher seems to have destroyed some of the evidence
Not destroyed - just well hidden behind a veil of embarrassment Embarassed Made a daring raid and retrieved the evidence - so here are the heavy duty versions

Step 1
4 in r2c9 -> 4 in D\ must be in r6c6.

4 elsewhere in n3 must be in c7
-> no 4 in r6c7

Step 2
A loooong contradiction chain eliminates 4 from r9c7.

4 in r9c7 ->3 must be in r7c7

4 in r9c7 -> r2c9=4 -> 4 in D\ must be in r6c6 -> 4 in D/ in r7c3 -> 4 in N4 in c1 -> 4 in r1c2 -> r1c3= 6 -> r6c2=6 -> r4c8 = 6 -> r7c7=6

But this means both 3 and 6 in r7c7 -> no 4 in r9c7

Step 3
A contradiction chain now eliminates 6 from r1c3

6 in r1c3 -> 6 in r6c2 -> 6 in r4c8 -> 6 in r7c7 -> 3 in r9c7 -> 3 in r8c5 -> 3 in r7c1
But this leaves no 3 for N1 -> r1c3 <> 6 -> r1c2 = 6

This unlocks the puzzle - the rest is no sweat.[edit: actually - it does take some doing - just feels easy after all the other headaches]
emm wrote:
I found this on the other forum - ALSxz Mk II - an ALS with more than one X candidate in some obscure mathematical situation. Maybe this is the sort of thing that applies to Sudoku X ALS.
I 'studied' that thread and was very relieved that bennys seemed to withdraw his theory as a version of ALS in his(?) last post. I couldn't see it as ALS - though could understand the elimination was logical. Also, it didn't seem to be about X puzzles.

PS - sorry to take so long to reply about these things - been assimilated elsewhere Twisted Evil . Very Happy
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emm
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Joined: 11 Oct 2006
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PostPosted: Tue Nov 07, 2006 8:22 pm    Post subject: Reply with quote

You're right Ed, bennys example was not about Sudoku-X. I was just wondering if his reasoning could be applied to it.

I think in the end he was saying that it was possible to reduce the 2X-candidate ALS to a simpler form therefore his reasoning wasnít needed.

Does this mean his original reasoning was wrong? I might know if I could figure out what was meant by this

bennys wrote:
In the ALS XZ rule we need that the x is restricted common.
Now when we have that any other common must appear in the union of A and B.
However if x is common but not restricted it does not work but if lets say in B the only place that is consistent with appearance also in A is the only place that other candidate lets say y can appear in B then we get.
If x not appear in A then z appear in A
If x not appear in B then z appear in B
if x appear in BOTH A and B then y will not appear in B and then we get that z appear in B.

Ruud said that his XY ALS was more appropriate than the 2XY example in your thread. Maybe yours is the adult version. Very Happy
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