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daj95376
Joined: 23 Aug 2008
Posts: 3854
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 Posted: Wed Jun 15, 2011 6:32 am Post subject: Puzzle 11/06/15: ~ XY (BBDB) |
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| Code: | +-----------------------+
| 4 . . | . . 3 | 2 . 7 |
| . . 7 | . 6 . | 1 . . |
| . 5 . | 2 1 7 | . . . |
|-------+-------+-------|
| . . 3 | 5 . . | . . . |
| . 2 4 | . 3 6 | . . . |
| 7 . 5 | . 2 4 | 6 . . |
|-------+-------+-------|
| 5 6 . | . . 2 | 7 . . |
| . . . | . . . | . 6 . |
| 9 . . | . . . | . . 5 |
+-----------------------+
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Play this puzzle online at the Daily Sudoku site |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Wed Jun 15, 2011 8:13 am Post subject: |
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Two moves - I think the first is really clear example of its kind!
| Quote: | Looking at r4 and c4 and the potential 89 pairs it's visually pretty clear that whatever happens with the 1/7 in b5 you are going to have two 89 pairs.. doubly linked als-xz. As a loop...
als(89=7)r4c25 - als(7=89)r15c4 - als(89=1)r16c4 - als(1=89)r4c26 - loop ; r4c78<>89, r8c4<>89
remotepair(89) r1c4,r5c4,r5c9,r6c8 ; r1c8<>89=5 |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Wed Jun 15, 2011 1:34 pm Post subject: |
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Another two step............
| Quote: | anp(89=7)r5c98-als(7=1)r516c4-als(1=89)b6q86; r4c78<>89
RP(89)r2c9,r5c9,r5c4,r4c6; r2c6<>89=5
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Ted |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Wed Jun 15, 2011 1:47 pm Post subject: |
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Once again Peter and I have the same solution but I approached it from a different perspective!
Question/Issue
Here is my grid after the first move:
| Code: | *--------------------------------------------------*
| 4 1 6 | 89 58 3 | 2 589 7 |
| 2 3 7 | 4 6 589 | 1 589 89 |
| 8 5 9 | 2 1 7 | 3 4 6 |
|----------------+----------------+----------------|
| 6 89 3 | 5 7 89 | 4 2 1 |
| 1 2 4 | 89 3 6 | 5 7 89 |
| 7 89 5 | 1 2 4 | 6 89 3 |
|----------------+----------------+----------------|
| 5 6 8 | 3 9 2 | 7 1 4 |
| 3 4 1 | 7 58 58 | 9 6 2 |
| 9 7 2 | 6 4 1 | 8 3 5 |
*--------------------------------------------------* |
I first viewed this as a BUG+2. If so, r2c6 must be a (8), but what is r2c8? My simple rule of three occurrences in the row, column and box fails; both digit (8) and (9) meet the condition.
Ted
[Edited to correct value of r2c6; was 5, should be 8.]
Last edited by tlanglet on Wed Jun 15, 2011 9:20 pm; edited 1 time in total |
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ronk
Joined: 07 May 2006
Posts: 398
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| Posted: Wed Jun 15, 2011 3:03 pm Post subject: |
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| tlanglet wrote: | Question/Issue
Here is my grid after the first move:
| Code: | *--------------------------------------------------*
| 4 1 6 | 89 58 3 | 2 59+8 7 |
| 2 3 7 | 4 6 59+8 | 1 58+9 89 |
| 8 5 9 | 2 1 7 | 3 4 6 |
|----------------+----------------+----------------|
| 6 89 3 | 5 7 89 | 4 2 1 |
| 1 2 4 | 89 3 6 | 5 7 89 |
| 7 89 5 | 1 2 4 | 6 89 3 |
|----------------+----------------+----------------|
| 5 6 8 | 3 9 2 | 7 1 4 |
| 3 4 1 | 7 58 58 | 9 6 2 |
| 9 7 2 | 6 4 1 | 8 3 5 |
*--------------------------------------------------* |
I first viewed this as a BUG+2. If so, r2c6 must be a (5), but what is r2c8? My simple rule of three occurrences in the row, column and box fails; both digit ( and (9) meet the condition. |
It's actually BUG+3. I added the notation for the extra candidates to your pencilmarks above. An obvious deduction is r2c8<>8 ... which [edit: doesn't lead] to cascading singles.
Last edited by ronk on Wed Jun 15, 2011 4:17 pm; edited 1 time in total |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Wed Jun 15, 2011 3:29 pm Post subject: |
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Ted:
I missed Ron's layout of the BUG+3 as well. Upon review of his markings, the following seems to explain his results.
If a BUG is to exist, then the following must be true:
| Code: | 8's are all that work in [r1] so => r1c8=59+8
8's are all that work in [b2] and [c6] so => r2c6=59+8
*--------------------------------------------------*
| 4 1 6 | 89 58 3 | 2 59+8 7 |
| 2 3 7 | 4 6 59+8 | 1 589 89 |
| 8 5 9 | 2 1 7 | 3 4 6 |
|----------------+----------------+----------------|
| 6 89 3 | 5 7 89 | 4 2 1 |
| 1 2 4 | 89 3 6 | 5 7 89 |
| 7 89 5 | 1 2 4 | 6 89 3 |
|----------------+----------------+----------------|
| 5 6 8 | 3 9 2 | 7 1 4 |
| 3 4 1 | 7 58 58 | 9 6 2 |
| 9 7 2 | 6 4 1 | 8 3 5 |
*--------------------------------------------------*
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If you now assume that "+8" isn't present in these two cells, then you're left with r2c8=58+9 to describe a BUG and the last candidate to prevent it.
| Code: | *--------------------------------------------------*
| 4 1 6 | 89 58 3 | 2 59__ 7 |
| 2 3 7 | 4 6 59__ | 1 58+9 89 |
| 8 5 9 | 2 1 7 | 3 4 6 |
|----------------+----------------+----------------|
| 6 89 3 | 5 7 89 | 4 2 1 |
| 1 2 4 | 89 3 6 | 5 7 89 |
| 7 89 5 | 1 2 4 | 6 89 3 |
|----------------+----------------+----------------|
| 5 6 8 | 3 9 2 | 7 1 4 |
| 3 4 1 | 7 58 58 | 9 6 2 |
| 9 7 2 | 6 4 1 | 8 3 5 |
*--------------------------------------------------*
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It's a recursive process on the logic!
| Code: | Ron's results
*--------------------------------------------------*
| 4 1 6 | 89 58 3 | 2 59+8 7 |
| 2 3 7 | 4 6 59+8 | 1 58+9 89 |
| 8 5 9 | 2 1 7 | 3 4 6 |
|----------------+----------------+----------------|
| 6 89 3 | 5 7 89 | 4 2 1 |
| 1 2 4 | 89 3 6 | 5 7 89 |
| 7 89 5 | 1 2 4 | 6 89 3 |
|----------------+----------------+----------------|
| 5 6 8 | 3 9 2 | 7 1 4 |
| 3 4 1 | 7 58 58 | 9 6 2 |
| 9 7 2 | 6 4 1 | 8 3 5 |
*--------------------------------------------------*
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Regards, Danny |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Wed Jun 15, 2011 5:01 pm Post subject: |
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Peter: It's also a perfect example of what I call a "bivalue wraparound loop with grouping".
| Code: | after basics
+--------------------------------------------------------------------------------+
| 4 1 6 | 89 589 3 | 2 589 7 |
| 2 3 7 | 4 6 589 | 1 589 89 |
| 8 5 9 | 2 1 7 | 34 34 6 |
|--------------------------+--------------------------+--------------------------|
| 6 89 3 | 5 789 189 | 489 124789 124 |
| 1 2 4 | 789 3 6 | 5 789 89 |
| 7 89 5 | 189 2 4 | 6 189 3 |
|--------------------------+--------------------------+--------------------------|
| 5 6 18 | 3 489 2 | 7 1489 14 |
| 3 47 128 | 1789 45789 1589 | 489 6 124 |
| 9 47 128 | 6 478 18 | 348 12348 5 |
+--------------------------------------------------------------------------------+
# 72 eliminations remain
(8=9)r1c4 - r56c4 = r4c56 - (9=8)r4c2 - r4c56 = (8)r56c4 - loop => r4c78,r8c4<>89
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Wed Jun 15, 2011 9:35 pm Post subject: |
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First, I apologize for the typo in my original post; the value I had was 5 but should be 8. And yes, I agree it is a BUG+3.
Danny, I follow you interpretation of Ron's results, but I have a problem assuming that the two (+8s) are not present in order to resolve cell r2c8. Normally all implications are considered independently to form a SIS.
Hopefully Ron will provide some more info...........
Ted |
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ronk
Joined: 07 May 2006
Posts: 398
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| Posted: Thu Jun 16, 2011 11:42 am Post subject: |
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| tlanglet wrote: | | b]Danny[/b], I follow you interpretation of Ron's results, but I have a problem assuming that the two (+8s) are not present in order to resolve cell r2c8. Normally all implications are considered independently to form a SIS. |
Yes, of course. I think Danny was describing a method of identifying the extra candidates.
Whenever one or more units (sectors, housed) contain more than one poly-valued cell, avoid addressing those units. IOW do as much as can to determine the extra candidate(s) in units with only one polyvalued cell. Then pretend those candidates don't exist and repeat. |
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Luke451
Joined: 20 Apr 2008
Posts: 310
Location: Southern Northern California
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| Posted: Thu Jun 16, 2011 1:30 pm Post subject: |
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| peterj wrote: | Two moves - I think the first is really clear example of its kind!
| Quote: | Looking at r4 and c4 and the potential 89 pairs it's visually pretty clear that whatever happens with the 1/7 in b5 you are going to have two 89 pairs.. doubly linked als-xz. As a loop...
als(89=7)r4c25 - als(7=89)r15c4 - als(89=1)r16c4 - als(1=89)r4c26 - loop ; r4c78<>89, r8c4<>89
remotepair(89) r1c4,r5c4,r5c9,r6c8 ; r1c8<>89=5 |
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Peter, that's pretty cool to identify that as a doubly-linked als-xz. A great thread on the topic survives at Players, as you may know.
I agree it's a clear example, but it's the first I've seen with identical extra candidates in both sets.
An underdeveloped aspect of these critters is that the restricted common candidates do not have to directly see one another like this for the structure to work. One, the other, or both can be "developed" with an AIC or any other means as long as the RCCs are weakly linked. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Fri Jun 17, 2011 5:03 am Post subject: |
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| This is just to let you know that I did the puzzle, hacking my way to a six-move solution. |
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