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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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 Posted: Fri Dec 23, 2011 8:24 pm Post subject: Free Press December 23, 2011 |
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Not yet started:
| Code: | Puzzle: FP122311
+-------+-------+-------+
| . . . | . 7 . | . . 9 |
| . . 6 | . . 2 | . 8 . |
| . 9 . | . . . | . . 1 |
+-------+-------+-------+
| . . . | . 3 5 | 8 9 . |
| 9 . . | . . . | . . 7 |
| . 3 4 | . . . | 2 . . |
+-------+-------+-------+
| 8 . . | 1 . . | . 7 . |
| . 4 . | 6 . . | 5 . . |
| 7 . . | . 5 . | . . . |
+-------+-------+-------+
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Play this puzzle online at the Daily Sudoku site
Keith |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat Dec 24, 2011 12:20 am Post subject: |
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Looking at the potential 35 DP in boxes 12, one of r13c1=4 or r3c4=8. Common outcome; r1c3, r2c2, r3c3=172.
| Code: |
+-----------+------------+----------+
| 345 8 12 | 35 7 146 | 46 25 9 |
| 45 17 6 | 59 19 2 | 47 8 3 |
| 345 9 27 | 358 68 468 | 467 25 1 |
+-----------+------------+----------+
| 2 17 17 | 4 3 5 | 8 9 6 |
| 9 5 8 | 2 16 16 | 3 4 7 |
| 6 3 4 | 7 89 89 | 2 1 5 |
+-----------+------------+----------+
| 8 6 5 | 1 4 3 | 9 7 2 |
| 1 4 9 | 6 2 7 | 5 3 8 |
| 7 2 3 | 89 5 89 | 1 6 4 |
+-----------+------------+----------+
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sat Dec 24, 2011 5:56 am Post subject: |
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| Marty R. wrote: | Looking at the potential 35 DP in boxes 12, one of r13c1=4 or r3c4=8. Common outcome; r1c3, r2c2, r3c3=172.
| Code: |
+-----------+------------+----------+
| 345 8 12 | 35 7 146 | 46 25 9 |
| 45 17 6 | 59 19 2 | 47 8 3 |
| 345 9 27 | 358 68 468 | 467 25 1 |
+-----------+------------+----------+
| 2 17 17 | 4 3 5 | 8 9 6 |
| 9 5 8 | 2 16 16 | 3 4 7 |
| 6 3 4 | 7 89 89 | 2 1 5 |
+-----------+------------+----------+
| 8 6 5 | 1 4 3 | 9 7 2 |
| 1 4 9 | 6 2 7 | 5 3 8 |
| 7 2 3 | 89 5 89 | 1 6 4 |
+-----------+------------+----------+
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Play this puzzle online at the Daily Sudoku site |
Marty,
I don't like your solution, if you don;t mind me saying so. When I run R3C4=8, I get different results in the three cells you mention.
Now, certainly, R13C4=4 and/or R3C4=8. After any number of steps, and by any path, if you can show (for example) that R1C3=1, then indeed R3C1=1.
However, you can also assume R3C4=8 and show R3C1=2.
Let me put is in a different way. I don't like your solution for the same reason I believe you would not like the Almost Naked Pair 35 in R13C4.
Since I do also not like chains that are not patterns, I now have a problem.
Bah! Humbug!
Keith |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat Dec 24, 2011 6:18 am Post subject: |
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| Quote: | | When I run R3C4=8, I get different results in the three cells you mention. |
Keith, I stand by my solution. An 8 in r3c4 yields a 7 in r3c7 and 2 in r3c3.
| Quote: | | I don't like your solution, if you don;t mind me saying so. |
I don't mind you saying so at all. And if my statement above is incorrect, I'd want to know about it. I always want input if it's not an insult and has the potential to improve my game.
I'd have probably been happier with some other solution, but I have to do what I have to do in order to solve puzzles.
But just for the sake of argument, a couple of years ago I asked you about a UR you used, claiming that it looked like a Forcing Chain which you were led to by a potential UR. You replied that you saw nothing wrong with spotting a pattern and then examining the ramifications of that pattern. Is that or is that not what I did?
At any rate, I'll say the same thing you said, that I hope you don't mind me saying what I said. I'm heading to bed but look forward to your reply Saturday.
Cheers! |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sat Dec 24, 2011 7:20 am Post subject: |
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Hmmm!!! First off, my solver yields a more complex grid after basics.
| Code: | after basics
+-----------------------------------------------------+
| *35+4 8 12 | *35 7 146 | 46 25 9 |
| 45 17 6 | 59 149 2 | 47 8 3 |
| *35+4 9 27 | *35+8 468 468 | 467 25 1 |
|-----------------+-----------------+-----------------|
| 2 17 17 | 4 3 5 | 8 9 6 |
| 9 5 8 | 2 16 16 | 3 4 7 |
| 6 3 4 | 7 89 89 | 2 1 5 |
|-----------------+-----------------+-----------------|
| 8 6 5 | 1 24 3 | 9 7 24 |
| 1 4 9 | 6 28 7 | 5 3 28 |
| 7 2 3 | 89 5 489 | 1 6 48 |
+-----------------------------------------------------+
# 40 eliminations remain
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You can use Marty's sequence:
(8)r3c4 - (8=467)r3c567 - (7=2)r3c3
And then derive:
(4)r13c1 - (4=5=9*=8=2=4)r2c1,r2c4,r9c4,r8c5,r7c5 - (*94=1)r2c5 - (1=7=2)r2c2,r3c3
Or you can use the contradiction to Marty's sequence:
(8)r3c4 - (8=461)r3c56,r1c6 - (1=2)r1c3
And create a single chain:
(8=461)r3c56,r1c6 - (1=2=7)r1c3,r3c3 - (746=8)r3c756 => r3c4<>8
The remaining UR Type 2 cracks the puzzle. |
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SudoQ
Joined: 02 Aug 2011
Posts: 127
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| Posted: Sat Dec 24, 2011 8:41 am Post subject: |
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If you like chains, it is easy to show that
r2c4=5 leads to a contradiction in r1(c7).
/SudoQ |
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JC Van Hay
Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium
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| Posted: Sat Dec 24, 2011 8:48 am Post subject: |
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A chain : "Wing" : ANP(9=45)r2c14-AHP(4r1c1=46r1c67)-1r1c6=1r2c5 => -9r2c5;stte
or
A pattern : ALS-XZ Rule : (X=1,Z=4) : ANP(4=17)r2c27-ANQ(1=2345)r1c1348 => -4r1c7, -4r2c1;stte |
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arkietech
Joined: 31 Jul 2008
Posts: 1834
Location: Northwest Arkansas USA
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| Posted: Sat Dec 24, 2011 4:40 pm Post subject: |
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| JC Van Hay wrote: | | A pattern : ALS-XZ Rule : (X=1,Z=4) : ANP(4=17)r2c27-ANQ(1=2345)r1c1348 => -4r1c7, -4r2c1;stte |
Thanks JC,
I have struggled with this for a long time. You turned lights on for me.
| Code: | *--------------------------------------------------*
|a345 8 a12 |a35 7 146 | 6-4 a25 9 |
| 5-4 b17 6 | 59 149 2 |b47 8 3 |
| 345 9 27 | 358 468 468 | 467 25 1 |
|----------------+----------------+----------------|
| 2 17 17 | 4 3 5 | 8 9 6 |
| 9 5 8 | 2 16 16 | 3 4 7 |
| 6 3 4 | 7 89 89 | 2 1 5 |
|----------------+----------------+----------------|
| 8 6 5 | 1 24 3 | 9 7 24 |
| 1 4 9 | 6 28 7 | 5 3 28 |
| 7 2 3 | 89 5 489 | 1 6 48 |
*--------------------------------------------------*
als-xz(x=1,z=4):anq(1=2345)r1c1348-anp(4=17)r2c27 => r1c7,r2c1<>4
(1)a=(4)b;r2c7=4;
(1)b=(4)a;r1c1=4; => r1c7,r2c1<>4
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat Dec 24, 2011 4:51 pm Post subject: |
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| Quote: | | Hmmm!!! First off, my solver yields a more complex grid after basics. |
Danny, I did this three times (once with pencil and paper and twice on Draw/Play) and got three different post-basics grids. However, on my last attempt (on Draw/Play) it agreed with yours.  |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sat Dec 24, 2011 5:02 pm Post subject: |
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Use of an atypical UR strong inference.
| Code: | after basics
+-----------------------------------------------------+
| *35+4 8 12 | *35 7 146 | 6-4 25 9 |
| 5-4 17 6 | 59 149 2 | 47 8 3 |
| *35+4 9 27 | *35+8 468 468 | 467 25 1 |
|-----------------+-----------------+-----------------|
| 2 17 17 | 4 3 5 | 8 9 6 |
| 9 5 8 | 2 16 16 | 3 4 7 |
| 6 3 4 | 7 89 89 | 2 1 5 |
|-----------------+-----------------+-----------------|
| 8 6 5 | 1 24 3 | 9 7 24 |
| 1 4 9 | 6 28 7 | 5 3 28 |
| 7 2 3 | 89 5 489 | 1 6 48 |
+-----------------------------------------------------+
# 40 eliminations remain
(4=7)r2c7 - (7=468)r3c567 - (48)r3c14 =UR= (4)r1c1 => r1c7,r2c1<>4
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arkietech
Joined: 31 Jul 2008
Posts: 1834
Location: Northwest Arkansas USA
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| Posted: Sat Dec 24, 2011 6:27 pm Post subject: |
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More als views
als xy=wing | Code: | *--------------------------------------------------*
|a345 8 a12 |a35 7 146 | 6-4 a25 9 |
| 5-4 c17 6 | 59 149 2 |b47 8 3 |
| 345 9 27 | 358 468 468 | 467 25 1 |
|----------------+----------------+----------------|
| 2 17 17 | 4 3 5 | 8 9 6 |
| 9 5 8 | 2 16 16 | 3 4 7 |
| 6 3 4 | 7 89 89 | 2 1 5 |
|----------------+----------------+----------------|
| 8 6 5 | 1 24 3 | 9 7 24 |
| 1 4 9 | 6 28 7 | 5 3 28 |
| 7 2 3 | 89 5 489 | 1 6 48 |
*--------------------------------------------------*
als-xy-wing(xy=17):anq(1=2345)r1c1348-(4=7)r2c7-(7=1)r2c2 => r1c7,r2c1<>4
(1)a=(7)c=(4)b;r2c7=4;
(7)b=(1)c=(4)a;r1c1=4; => r1c7,r2c1<>4 |
And then an als-chain | Code: | *--------------------------------------------------*
|a345 8 a12 |b35 7 146 | 6-4 b25 9 |
| 5-4 d17 6 | 59 149 2 |c47 8 3 |
| 345 9 27 | 358 468 468 | 467 25 1 |
|----------------+----------------+----------------|
| 2 17 17 | 4 3 5 | 8 9 6 |
| 9 5 8 | 2 16 16 | 3 4 7 |
| 6 3 4 | 7 89 89 | 2 1 5 |
|----------------+----------------+----------------|
| 8 6 5 | 1 24 3 | 9 7 24 |
| 1 4 9 | 6 28 7 | 5 3 28 |
| 7 2 3 | 89 5 489 | 1 6 48 |
*--------------------------------------------------*
als-chain:ant(4=35)r1c14-ant(5=12)r1c38-(1=7)r2c2-(7=4)r2c7 => r1c7,r2c1<>4
I like better
(4=35)r1c14-(5=12)r1c38-(1=7)r2c2-(7=4)r2c7 => r1c7,r2c1<>4
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edit typo
Last edited by arkietech on Sat Dec 24, 2011 9:11 pm; edited 1 time in total |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sat Dec 24, 2011 8:50 pm Post subject: |
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For what it's worth. After basics: | Code: | +-------------+-------------+-------------+
| 345 8 D12 | 35 7 146 | 46 C25 9 |
|c45 e17 6 |b59 149 2 |d47 8 3 |
|A345 9 27 |*358 468 468 |467 B25 1 |
+-------------+-------------+-------------+
| 2 17 17 | 4 3 5 | 8 9 6 |
| 9 5 8 | 2 16 16 | 3 4 7 |
| 6 3 4 | 7 89 89 | 2 1 5 |
+-------------+-------------+-------------+
| 8 6 5 | 1 24 3 | 9 7 24 |
| 1 4 9 | 6 28 7 | 5 3 28 |
| 7 2 3 |a89 5 489 | 1 6 48 |
+-------------+-------------+-------------+ |
If R3C4=8:
1) Cell a=9; b=5; c=4; d=7; e=1.
and
2) Cell A=3; B=5; C=2; D=1.
D and e cannot both be 1. Thus, R3C4<>8, and the puzzle is solved.
What follows is simply my opinion:
There's nothing wrong with this solution, I just don't like it. And, the longer the implication chains, the more I dislike it.
Sure, you can justify looking at whether R3C4=8 because of the UR, or because of the almost pair 35 in C4, but remember: Every bivalue cell is a finned single, or an "almost" single. Yet I would regard an almost single as simple trial and error.
Keith |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat Dec 24, 2011 10:49 pm Post subject: |
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Keith,
If you have a Type 3, say, AB-AB-ABC-ABD, do you not look at the implications of C and D being true? How is that different from the UR here?
By the way, I also found that r3c4=8 leads to an invalidity. But here I've fallen under the influence of Ronk who has posited that common outcomes are more satisfying than invalidities. Thus, I didn't use the invalidity, rather I extended the chain to achieve a common outcome.
I agree that shorter chains are more satisfying than longer ones, but I had no choice. All the other posters are using ALS's or various Eureka-notated solutions which I don't understand. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sat Dec 24, 2011 11:28 pm Post subject: |
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Marty,
Go watch the Lions. This is pretty impressive. (I did not watch the Bills.)
| Quote: | | do you not look at the implications of C and D being true? |
Well, actually, it's C or D, though C and D are permitted.
It's the length of the implication chain I find less than satisfying. (Not that there's anything wrong with that.)
Keith |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat Dec 24, 2011 11:51 pm Post subject: |
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| The Lions weren't on here and the Bills were blacked out so we were offered the Evil Empire (Pats) at 1:00 and the Eagles-Cowboys at 4:00. |
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