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daj95376
Joined: 23 Aug 2008
Posts: 3854
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 Posted: Tue Apr 20, 2010 11:27 pm Post subject: Puzzle 10/04/20 (E) ___ Advanced |
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Chains are the easy way out.
| Code: | +-----------------------+
| 4 8 . | . . . | . . 7 |
| 9 . 3 | 4 6 2 | 1 . . |
| . 6 1 | . . 7 | . . . |
|-------+-------+-------|
| . 9 . | . 2 6 | . 1 . |
| . 3 . | 9 . . | . 7 . |
| . 2 7 | 3 . 1 | . . . |
|-------+-------+-------|
| . 5 . | . . . | 7 6 . |
| . . . | 6 9 . | 4 . . |
| 3 . . | . . . | . . . |
+-----------------------+
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Play this puzzle online at the Daily Sudoku site |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Wed Apr 21, 2010 4:50 am Post subject: |
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| Quote: | | Chains are the easy way out. |
Would someone be kind enough to explain--briefly, I'm not looking for someone to write a thesis--how chains can be looked for efficiently. Thanks. |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Wed Apr 21, 2010 8:02 am Post subject: |
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| Marty R. wrote: | | Quote: | | Chains are the easy way out. |
Would someone be kind enough to explain--briefly, I'm not looking for someone to write a thesis--how chains can be looked for efficiently. Thanks. |
For what it's worth my thinking path is like this (I use Simple Sudoku btw)
1) Filter all the bivalue
a) Very few bivalues - try something else
b) Bivalues but mostly the same pairs in same blocks - look for UR
c) Good spread of bivalues - look for chains/xywings etc
The next step is the one that made the main difference for me in finding chains...
2) Start by looking for victims that you would like to eliminate i.e not for the start/end of the chain. Good victims are obviously ones that make singles or pairs or locked sets etc.
3) If so can you find bivalue pincers that would do the elimination
a) Pick a promising pincer and start looking for xy-chains
I usually scan for xy-chains firsty because I find them a lot easier.
No xy-chains...
4) Filter through by each digit to get the lie of the land for strong links on each digit. Particularly look for strong links that have a bivalue at one end and something at the other that you can't get to with xy-chain logic
5) Go back to 3 and try to bridge stretches of xy-chain with strong links on a single digit
As I say looking for promising victims rather than methodically plodding through all the start points is the thing I would most recommend. |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Wed Apr 21, 2010 8:07 am Post subject: |
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| I now believe that my solution to this puzzle is inappropriate. If you like chains, then it may still prove interesting. |
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wapati
Joined: 10 Jun 2008
Posts: 472
Location: Brampton, Ontario, Canada.
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| Posted: Wed Apr 21, 2010 12:43 pm Post subject: |
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I am happy with this one because I found a finned swordfish.
It wasn't necessary as a finned x-wing does the same elimination.
One of the above, an xy-wing and a w-wing solve it. |
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arkietech
Joined: 31 Jul 2008
Posts: 1834
Location: Northwest Arkansas USA
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| Posted: Wed Apr 21, 2010 1:10 pm Post subject: |
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I had the same solution as wapati
finned x-wing 2
xy-wing 258
w-wing 58
did anyone find a chain?  |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Wed Apr 21, 2010 3:50 pm Post subject: |
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| Thanks Peter. |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Wed Apr 21, 2010 4:56 pm Post subject: |
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After deleting the 8 in r9c8 (I used colouring), I found a forcing chain in r1c3 which forced 8 in r8c3 ... but it didn't do any more than the xy-wing..
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Wed Apr 21, 2010 5:39 pm Post subject: |
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Here's the solution that caught my attention.
| Code: | after basics and an XY-Wing
strong-link eliminations leave a BUG+2
+------------------------------------------------------+
| 4 8 2 | 5 1 9 | 6 3 7 |
| 9 7 3 | 4 6 2 | 1 58 58 |
| 5 6 1 | 8 3 7 | 9 24 24 |
|-----------------+------------------+-----------------|
| 8 9 45 | 7 2 6 | 3 1 45 |
| 1 3 45 | 9 *48+5 *48 | 2 7 6 |
| 6 2 7 | 3 45 1 | 58 458 9 |
|-----------------+------------------+-----------------|
| 2 5 9 | 1 *48 *48+3 | 7 6 38 |
| 7 1 8 | 6 9 35 | 4 25 235 |
| 3 4 6 | 2 7 58 | 58 9 1 |
+------------------------------------------------------+
# 24 eliminations remain
r57c56 <48> UR via s-link <> 4 r5c5
r57c56 <48> UR via s-link <> 8 r7c6
BUG+2 (r6c8=5 or r8c9=5) <> 5 r8c8
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Wed Apr 21, 2010 10:07 pm Post subject: |
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If you don't like the UR s-links approach, then it can always be turned into a BUG+4 by adding:
| Code: | (4)r5c5 - (4=5)r6c5 - r6c78 = r4c9 - r2c9 = (5)r2c8 => r8c8<>5
(8)r7c6 - (8=4)r5c6 - (4=5)r6c5 - r6c78 = r4c9 - r2c9 = (5)r2c8 => r8c8<>5
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Luke451
Joined: 20 Apr 2008
Posts: 310
Location: Southern Northern California
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| Posted: Thu Apr 22, 2010 12:02 am Post subject: |
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| arkietech wrote: | | did anyone find a chain? |
There's always the ole hidden pair trick.
| Code: | *--------------------------------------------------------------------*
| 4 8 25 | 15 135 9 | 6 23 7 |
| 9 7 3 | 4 6 2 | 1 58 58 |
| 25 6 1 | 58 358 7 | 39 2349 2349 |
|----------------------+----------------------+----------------------|
| 58 9 458 | 7 2 6 | 358 1 3458 |
| 1 3 458 | 9 458 458 | 2 7 6 |
| 6 2 7 | 3 458 1 | 589 4589 4589 |
|----------------------+----------------------+----------------------|
| 28 5 9 | 128 148 348 | 7 6 1238 |
| 7 1 28 | 6 9 358 | 4 2358 2358 |
| 3 4 6 | 1258 7 58 | 589 2589 12589 |
*--------------------------------------------------------------------*
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| Code: | 1. hp(12)r9c49=(2)r9c8-(2=3)r1c8-(3=9)r3c7-(9=58)als:r9c67
=>r9c49 can't be 58
2. (5=2)r3c1-(2=8)r7c1-(8=12)als:r79c4-(1=5)r1c4
=>r3c4 can't be 5=8
3. (8=4)r5c6-(4=5)r6c5-(5=8)r6c7-(8=5)r9c7-(5=8)r9c6-continuous
=>r5c5<>4,r6c8<>5, r7c6<>8
Drat, still needs a BUG+1 |
Must be something better chainwise, but break's over... |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Thu Apr 22, 2010 10:26 pm Post subject: |
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| Looking at the 58 UR in boxes 39, either DP killer (2 or 3) forces the same solution in c1. I used an M-Wing (58) to finish. |
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oaxen
Joined: 10 Jul 2006
Posts: 96
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| Posted: Fri Apr 23, 2010 10:25 am Post subject: |
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Thank you Dan.
This puzzle was a good one as one chain is not enough. I needed to open up a secondary chain, means it was a two stepper.
Lars |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Fri Apr 23, 2010 1:54 pm Post subject: |
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| Lars, can you share? |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Fri Apr 23, 2010 4:12 pm Post subject: |
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| peterj wrote: | | Lars, can you share? |
Although Lars'/oaxen's approach solves puzzles (for himself and others -- including my aunt), it doesn't lend itself to being shared with others because there isn't a pattern involved and, thus, there's nothing the rest of us can do to recreate his solution. |
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oaxen
Joined: 10 Jul 2006
Posts: 96
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| Posted: Fri Apr 23, 2010 6:08 pm Post subject: |
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| peterj wrote: |
Lars, can you share? |
| daj95376 wrote: |
Although Lars'/oaxen's approach solves puzzles (for himself and others -- including my aunt), it doesn't lend itself to being shared with others because there isn't a pattern involved and, thus, there's nothing the rest of us can do to recreate his solution. |
There is a pattern. First look after promising bivalues. Preferably bivalues which occurs double in a line, column or block. With yours eyes only test that both of the candidates can be followed for a long while. That is because if the candidate you chose is wrong you must go on with the other (obvious the right one) and you don't like to come to a sudden end.
Also possible to follow both of the candidates simulaneously which gives you some hits which must be correct.
When the puzles is solved I just for fun test if there are more bivalues which can lead to a solution. If many, I suppose the puzzle is easy whatever technique you use.
The hard puzzles are the ones which not permit a one-stepper. They are rare, but i like them most. That is why I earlier asked Dan if he can test that before publishing.
Anyhow, if so, you have to start a secondary chain and in the end your paper is full of pencil marks. |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Sun Apr 25, 2010 10:33 pm Post subject: |
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| Luke451 wrote: | | arkietech wrote: | | did anyone find a chain? |
There's always the ole hidden pair trick.
| Code: | *--------------------------------------------------------------------*
| 4 8 25 | 15 135 9 | 6 23 7 |
| 9 7 3 | 4 6 2 | 1 58 58 |
| 25 6 1 | 58 358 7 | 39 2349 2349 |
|----------------------+----------------------+----------------------|
| 58 9 458 | 7 2 6 | 358 1 3458 |
| 1 3 458 | 9 458 458 | 2 7 6 |
| 6 2 7 | 3 458 1 | 589 4589 4589 |
|----------------------+----------------------+----------------------|
| 28 5 9 | 128 148 348 | 7 6 1238 |
| 7 1 28 | 6 9 358 | 4 2358 2358 |
| 3 4 6 | 1258 7 58 | 589 2589 12589 |
*--------------------------------------------------------------------*
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| Code: | 1. hp(12)r9c49=(2)r9c8-(2=3)r1c8-(3=9)r3c7-(9=58)als:r9c67
=>r9c49 can't be 58
2. (5=2)r3c1-(2=8)r7c1-(8=12)als:r79c4-(1=5)r1c4
=>r3c4 can't be 5=8
3. (8=4)r5c6-(4=5)r6c5-(5=8)r6c7-(8=5)r9c7-(5=8)r9c6-continuous
=>r5c5<>4,r6c8<>5, r7c6<>8
Drat, still needs a BUG+1 |
Must be something better chainwise, but break's over... |
Luke,
Great use of ALSs within an AIC! I am envious.
Ted |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Mon Apr 26, 2010 11:06 am Post subject: |
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| Sorry Luke my turn to be slow witted but how can the HP be an HP when there is another 2 in r9c8? I think I'm misreading you. I can see that it can be eliminated by colouring (follow the 2's) but ?? |
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Luke451
Joined: 20 Apr 2008
Posts: 310
Location: Southern Northern California
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| Posted: Mon Apr 26, 2010 2:15 pm Post subject: |
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| Mogulmeister wrote: | | how can the HP be an HP when there is another 2 in r9c8? |
There is indeed another (2).....but is there another (1)?
| Code: | There's always the ole hidden pair trick.
*--------------------------------------------------------------------*
| 4 8 25 | 15 135 9 | 6 23 7 |
| 9 7 3 | 4 6 2 | 1 58 58 |
| 25 6 1 | 58 358 7 | 39 2349 2349 |
|----------------------+----------------------+----------------------|
| 58 9 458 | 7 2 6 | 358 1 3458 |
| 1 3 458 | 9 458 458 | 2 7 6 |
| 6 2 7 | 3 458 1 | 589 4589 4589 |
|----------------------+----------------------+----------------------|
| 28 5 9 | 128 148 348 | 7 6 1238 |
| 7 1 28 | 6 9 358 | 4 2358 2358 |
| 3 4 6 | 12 58 7 58 | 589 2 589 12 589 |
*--------------------------------------------------------------------* |
hp(12)r9c49=(2)r9c8 means both the (12) naked pair and (2)r9c8 cannot be false.
Hidden pairs used like this will always have an extra(s) of one value (which "hides" the naked pair) and the other value will be locked (conjugate.) |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Mon Apr 26, 2010 3:11 pm Post subject: |
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| Thanks. |
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