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Louise56
Joined: 21 Sep 2005
Posts: 94
Location: El Cajon, California USA
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 Posted: Sat Nov 19, 2005 4:36 pm Post subject: Difficult newspaper puzzle 19 November |
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Here's another puzzle from the San Diego paper. It's rated 6 stars which means advanced techniques are necessary. Last week I said I wouldn't work on these anymore, but was feeling confident this morning. Alas, I couldn't solve it! What do you think the next step is?
500 009 704
000 000 600
010 800 002
037 002 000
800 605 003
000 300 260
100 003 080
009 000 000
205 400 007
I resolved it to this point and got stuck.
508 009 704
000 000 608
010 800 002
637 002 845
800 645 073
050 300 260
100 003 080
309 000 020
205 400 007 |
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David Bryant
Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado
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| Posted: Sat Nov 19, 2005 8:15 pm Post subject: This one is hard. |
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Hi, Louise!
I don't have a good answer for you, yet. I'm working on it, and none of the techniques I know about are getting any traction. I'm down to tracing out "forcing chains," which will probably work if I can figure out where to start. From the possibilities remaining, r6c3 looks like a good place to start.
Maybe someone_somewhere can apply his double-implication chains to this one. I'll let you know when I figure something out myself. dcb |
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David Bryant
Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado
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| Posted: Sat Nov 19, 2005 8:45 pm Post subject: A forcing chain will work |
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Hi again, Louise!
Here's one way to solve the puzzle. There may be a better way, but this is the one I found.
The short answer is that r6c3 = 1. Armed with that information I'm sure you can work out the rest of the puzzle.
I can prove this by following two chains rooted in r6c3. The only possibilities at r6c3 are {1, 4}. So let's assume that r6c3 = 4.
r6c3 = 4 ==> r7c3 = 6 ==> r3c3 = 3 (chain #1)
r6c3 = 4 ==> r6c1 = 9 ==> r6c9 = 1 ==> r8c9 = 6 ==> r7c9 = 9 ==> r9c8 = 3 ==> r3c7 = 3 (chain #2)
But now we have two "3"s in row 3, which is impossible. So r6c3 <> 4. dcb |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sun Nov 20, 2005 3:47 am Post subject: Difficult puzzle |
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By the way, this is David Bodycombe's puzzle, which also appears in my local, the Oakland Press, MI.
What do you call the following: From where Louise got to, you find the following possibilities:
R6C3: 1,4 R6C9: 1,9
R7C3: 4,6 R7C9: 6,9
If R7C3 is 1, then clockwise via the other cells says R8C3 must be 4.
If R7C3 is 4, then cclockwise via the other cells says R8C3 must be 6.
So, none of the other cells in C3 (other than R6 or R7) can be a 4. Correct?
If you buy this logic, then R2C3 is 2,3 and R3C3 is 3,6 and they form a triplet with R1C2 2,6. Then R2C2 is 4,9, and I am still working on it.
Keith
Waterford, MI |
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someone_somewhere
Joined: 07 Aug 2005
Posts: 275
Location: Munich
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| Posted: Sun Nov 20, 2005 10:12 am Post subject: |
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Hi,
| Code: | 500009704
000000600
010800002
037002000
800605003
000300260
100003080
009000000
205400007 |
First the classic steps, just for who is wondering how to get to our
difficult point:
6 in r4c1 - Unique Horizontal
5 in r6c2 8 in r4c7 2 in r8c8 7 in r5c8 8 in r2c9 - Unique Vertical
8 in r1c3 4 in r4c8 - Unique Vertical
5 in r4c9 3 in r8c1 - Unique in 3x3 block
1 not in r5c5, it is in r4c4 or r4c5 (Row on 3x3 Block interaction)
1 not in r6c5, it is in r4c4 or r4c5 (Row on 3x3 Block interaction)
1 not in r6c6, it is in r4c4 or r4c5 (Row on 3x3 Block interaction)
9 not in r5c5, it is in r4c4 or r4c5 (Row on 3x3 Block interaction)
9 not in r6c5, it is in r4c4 or r4c5 (Row on 3x3 Block interaction)
4 in r5c5 - Sole Candidate
7 not in r2c2, it is in r2c1 or r3c1 (Column on 3x3 Block interaction)
5 not in r3c7, it is in r2c8 or r3c8 (Column on 3x3 Block interaction)
1 not in r9c8, it is in r1c8 or r2c8 of the 3x3 (3x3 Block on Row/Column interaction)
1 not in r8c7, Hidden Pair 4 5 in r7c7 and r8c7 (in Column)
9 not in r7c7, Hidden Pair 4 5 in r7c7 and r8c7 (in Column)
Now I try the technique I called it "Double impilcation chains, starting from a pair":
I could detect a "5 star chain" starting at pair r1c2.
This is like a generalization of the Turbot Fish, where only one number was involved.
Look at the cells (like stars): r1c2, r1c5, r7c5, r3c3, r7c3
And you will find out that digit 6 can not be in r7c5!
1. pair: 2 & 6 in r1c2
(a1) r1c2 <> 6
(a2) r3c3 <> 6
(B1) r1c5 = 6
(B2) r7c3 = 6
(b1) r7c5 <> 6
(b2) r7c5 <> 6
*** Pair Contradiction of small letters in cell r7c5
This has a similariry with "Forcing chains" techniques that states:
"Forcing chains is a technique that allows you to deduce with certainty the content of a cell from considering the implications resulting from the placement of each of another cell's candidates"
Here we "deduce with certainity that a digit can't be in a cell from ...." instead of that a digit is in a cell.
| Code: | 5 26 8 12 1236 9 7 13 4
. Aa . Ba babB . . b. .
. aA . .. ...a . . .. .
479 249 234 1257 12357 147 6 1359 8
... a.b a.. b... b.... b.. . .... .
... ... ... .... ..... ... . .... .
479 1 346 8 3567 467 39 359 2
... . bbB . ..b. .b. .. ... .
... . ..a . .... ... .. ... .
6 3 7 19 19 2 8 4 5
. . . b. .. . . . .
. . . .. .. . . . .
8 29 12 6 4 5 19 7 3
. aB bB . . . .b . .
. .. .. . . . .. . .
49 5 14 3 78 78 2 6 19
.b . .. . .. .. . . ..
.. . .. . .. .. . . ..
1 467 46 2579 25679 3 45 8 69
. ... .b .... ..b.. . .. . ..
. .a. bB .... ..b.. . .. . ..
3 4678 9 157 15678 1678 45 2 16
. .... . b.. ..b.. .... .. . ..
. .a.. . ... ..... .... .. . ..
2 68 5 4 1689 168 139 39 7
. .. . . .b.. ... ... .. .
. aB . . .... ... ... .. . |
But all the efforts to find out that digit 6 is not in r7c5 was for nothing.
It warmed me up. Now I was ready to cruck the nut !!!
Starting from digit 2 in r5c2:
(a1) r5c2 <> 9 r2c2 <> 2
(B1) r2c3 = 2 r5c7 = 9 r5c3 = 1
(b1) r2c3 <> 3 r5c7 <> 1 r3c7 <> 9
(C1) r3c3 = 3 r3c7 = 3
*** Contradiction on row ***
Excluded 2 from r5c2
| Code: | 5 26 8 12 1236 9 7 13 4
. .. . .. .... . . .. .
479 249 234 1257 12357 147 6 1359 8
... ... Bbb .... ..... ... . .... .
479 1 346 8 3567 467 39 359 2
... . C.. . .... ... Cb ... .
6 3 7 19 19 2 8 4 5
. . . .. .. . . . .
8 29 12 6 4 5 19 7 3
. Aa Ba . . . bB . .
49 5 14 3 78 78 2 6 19
.. . .. . .. .. . . ..
1 467 46 2579 2579 3 45 8 69
. ... .. .... .... . .. . ..
3 4678 9 157 15678 1678 45 2 16
. .... . ... ..... .... .. . ..
2 68 5 4 1689 168 139 39 7
. .. . . .... ... ... .. . |
This is a "6 star constellation" !
And excluding 2 from r5c2 is really crucking this nut.
see u,
I am nuts to crack a nut |
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someone_somewhere
Joined: 07 Aug 2005
Posts: 275
Location: Munich
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| Posted: Sun Nov 20, 2005 10:18 am Post subject: |
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Hi,
Just for the recond, the rest of the solution:
9 in r5c2 - Sole Candidate
1 in r5c7 4 in r6c1 - Sole Candidate
2 in r5c3 1 in r6c3 9 in r6c9 - Sole Candidate
6 in r7c9 - Sole Candidate
4 in r7c3 1 in r8c9 - Sole Candidate
3 in r2c3 7 in r7c2 5 in r7c7 - Sole Candidate
6 in r3c3 4 in r8c7 - Sole Candidate
2 in r1c2 - Sole Candidate
1 in r1c4 4 in r2c2 - Sole Candidate
3 in r1c8 7 in r2c6 9 in r4c4 - Sole Candidate
6 in r1c5 9 in r2c1 4 in r3c6 9 in r3c7 1 in r4c5 8 in r6c6 2 in r7c4 9 in r9c8 - Sole Candidate
5 in r2c4 7 in r3c1 5 in r3c8 7 in r6c5 9 in r7c5 6 in r8c6 8 in r9c5 3 in r9c7 - Sole Candidate
2 in r2c5 1 in r2c8 3 in r3c5 8 in r8c2 7 in r8c4 5 in r8c5 6 in r9c2 1 in r9c6 - Sole Candidate
And the final result:
| Code: | 5 2 8 1 6 9 7 3 4
9 4 3 5 2 7 6 1 8
7 1 6 8 3 4 9 5 2
6 3 7 9 1 2 8 4 5
8 9 2 6 4 5 1 7 3
4 5 1 3 7 8 2 6 9
1 7 4 2 9 3 5 8 6
3 8 9 7 5 6 4 2 1
2 6 5 4 8 1 3 9 7 |
see u, |
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