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someone_somewhere
Joined: 07 Aug 2005 Posts: 275 Location: Munich

Posted: Thu Sep 29, 2005 11:58 am Post subject: Example with "hidden pair", "Xwing" and 


Hello,
I found in the Sunday Time Magazine Sudoku 8:
500000000
400070006
908405000
010003090
005209600
040800020
000604503
700020004
000000009
could solved it after finding "hidden pair in row", "Xwing on column" and "XYwing". It looks to be a very nice one!
can someone do it without advanced techniques?
see u, 

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David Bryant
Joined: 29 Jul 2005 Posts: 559 Location: Denver, Colorado

Posted: Thu Sep 29, 2005 4:46 pm Post subject: What is an advanced technique? 


I have solved this puzzle. I'm not sure if I used an "advanced technique," or not.
I'll just describe the position I reached in general terms for now. After about 12 moves (I'm not sure exactly how many  I'll have to go back and retrace the early steps) I reached a position that had three noteable characteristics:
1. the only possibilities at r4c4 were {5, 7};
2. there were several columns and rows where "9" could only appear in two places; and
3. the set {2, 6, 8} already appeared in column 4.
What I noticed at this point was that placing a "5" in r4c4 made it impossible to complete the top middle 3x3 box, from which I inferred that r4c4 = 7  the rest of the puzzle was relatively straightforward. dcb 

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someone_somewhere
Joined: 07 Aug 2005 Posts: 275 Location: Munich

Posted: Thu Sep 29, 2005 4:53 pm Post subject: 


David, I suppose that you did it!
Here to compare it with a "classic" way of solving:
5 in r2c8 9 in r6c3 7 in r7c8 4 in r9c3  Unique Horizontal
2 in r9c7  Unique in 3x3 block
1 not in r7c3, it is in r7c1 or r9c1 (Column on 3x3 Block interaction)
1 not in r8c3, it is in r7c1 or r9c1 (Column on 3x3 Block interaction)
2 in r7c3  Sole Candidate
2 in r4c1  Unique Horizontal
8 not in r5c8, it is in r4c7 or r4c9 (Row on 3x3 Block interaction)
8 not in r5c9, it is in r4c7 or r4c9 (Row on 3x3 Block interaction)
1 not in r8c4, Hidden Pair 5 9 in r8c2 and r8c4 (in Row)
3 not in r8c2, Hidden Pair 5 9 in r8c2 and r8c4 (in Row)
3 not in r8c4, Hidden Pair 5 9 in r8c2 and r8c4 (in Row)
6 not in r8c2, Hidden Pair 5 9 in r8c2 and r8c4 (in Row)
8 not in r8c2, Hidden Pair 5 9 in r8c2 and r8c4 (in Row)
1 not in r8c8, Nacked Pair 1 8 in r8c6 and r8c7 (same Row)
8 not in r8c8, Nacked Pair 1 8 in r8c6 and r8c7 (same Row)
6 in r8c8  Sole Candidate
3 in r8c3  Sole Candidate
1 in r2c3  Sole Candidate
3 not in r5c2, it is in r5c1 or r6c1 (Column on 3x3 Block interaction)
8 not in r1c6, it is in r2c6 r2c7 r8c6 r8c7 (XWing on Column)
8 not in r1c7, it is in r2c6 r2c7 r8c6 r8c7 (XWing on Column)
8 not in r4c7, it is in r2c6 r2c7 r8c6 r8c7 (XWing on Column)
8 not in r9c6, it is in r2c6 r2c7 r8c6 r8c7 (XWing on Column)
8 in r4c9  Unique Horizontal
5 in r6c9  Unique Vertical
4 not in r4c5, XYWing2 X=1 Y=7 in r5c9 X=1 Z=4 in r5c5 Y=7 Z=4 in r4c7
4 in r4c7 4 in r5c5  Unique Horizontal
4 in r1c8  Unique Horizontal
8 in r1c5 8 in r2c7 8 in r8c6  Unique Horizontal
9 in r2c4 1 in r8c7 9 in r8c2  Unique Horizontal
9 in r1c7 3 in r2c2 9 in r7c5 5 in r8c4 5 in r9c2  Unique Horizontal
3 in r1c4 2 in r2c6 5 in r4c5 1 in r7c1 3 in r9c5 6 in r9c1  Unique Horizontal
6 in r4c3 8 in r7c2 8 in r9c8  Unique Horizontal
7 in r4c4 8 in r5c1 7 in r6c7 7 in r9c6  Unique Horizontal
3 in r5c8 7 in r5c2 3 in r6c1 1 in r9c4  Unique Horizontal
3 in r3c7 7 in r3c9 1 in r5c9  Unique Horizontal
1 in r1c6 7 in r1c3 1 in r3c8 2 in r3c2 1 in r6c5  Unique Horizontal
2 in r1c9 6 in r1c2 6 in r3c5 6 in r6c6  Unique Horizontal
Final SuDoku Table
5 6 7 3 8 1 9 4 2
4 3 1 9 7 2 8 5 6
9 2 8 4 6 5 3 1 7
2 1 6 7 5 3 4 9 8
8 7 5 2 4 9 6 3 1
3 4 9 8 1 6 7 2 5
1 8 2 6 9 4 5 7 3
7 9 3 5 2 8 1 6 4
6 5 4 1 3 7 2 8 9 

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David Bryant
Joined: 29 Jul 2005 Posts: 559 Location: Denver, Colorado

Posted: Thu Sep 29, 2005 6:52 pm Post subject: How I solved it 


I've gone back and retraced the early moves I made on this puzzle. Here's what it looked like to me.
1. r2c8 = 5 (because of "5" at r1c1, r3c6, and r7c7  sole candidate)
2. r6c3 = 9 (because of "9" at r3c1, r4c8, and r5c6  sole candidate)
3. r9c3 = 4 (because of "4" at r2c1, r6c2, and r8c9  sole candidate)
4. r9c7 = 2 (because of "2" at r6c8 and r8c5  sole candidate)
5. r7c8 = 7 (this is the only cell in row 7 that can contain a "7"  note r8c1 and r2c5)
6. r7c3 = 2 (sole candidate  note that "1" must appear at r1c3 or r2c3
7. r4c1 = 2 (this is the only cell in row 4 that can contain a "2"  note r5c4 and r6c8)
At this point the grid looked like this  I've only entered the possibilities that I had pencilled in, leaving the more complex possibilities blank. I have also marked all the spots where a "9" can possibly appear with an asterisk ("*").
Code: 
5 . . * * . * . .
4 . . * 7 . * 5 6
9 . 8 4 1/3/6 5 1/3/7 1/3 1/2/7
2 1 6/7 5/7 4/5/6 3 . 9 .
. . 5 2 1/4 9 6 . .
. 4 9 8 1/5/6 1/6/7 . 2 .
1/8 8/9* 2 6 8/9* 4 5 7 3
7 * . * 2 . 1/8 1/6/8 4
. . 4 . . . 2 . 9

At this point I noticed that if I placed a "5" at r4c4 the middle psrt of column 5 (that is, r4c5, r5c5, and r6c5, in the middle box) would be reduced to the triplet {1, 4, 6}, substantially simplifying the analysis of columns 4, 5, and 6. So I traced out the following forced moves:
8. r4c4 = 5 (hypothetical only)
9. r9c5 = 5 (because of "5" at r4c4, r3c6, and r7c7  sole candidate)
10. r8c2 = 5 (because of "5" at r5c3, r7c7, and r9c5  sole candidate)
11. r7c2 = 9 (only place it can now fit in column 2  see * above)
12. r7c5 = 8 (by elimination on {8, 9} pair)
13. r1c5 = 9 (only place it can now fit in column 5  see * above)
14. r3c5 = 3 (sole candidate , because of {1, 4, 6} triplet in center 3x3 box)
But now there are only four cells (r1c4, r1c6, r2c4, and r2c6) left open in the top middle 3x3 box, and these four cells must contain the values {1, 2, 6, 8}  since the values {2, 6, 8} already appear in column 4, this clearly cannot be done. Therefore the hypothetical move r4c4 = 5 can be eliminated, and we must have r4c4 = 7. The rest of the puzzle is easily solved from there.
There's always more than one way to skin a cat! dcb
Last edited by David Bryant on Fri Sep 30, 2005 6:28 pm; edited 1 time in total 

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Victor
Joined: 29 Sep 2005 Posts: 207 Location: NI

Posted: Thu Sep 29, 2005 9:11 pm Post subject: 


This IS a tricky sudoku. You can do it by finding pairs, etc. AND by the Xwing that Someone_Somewhere gives: at one stage, 8 can go in rows 2 and 8 in the columns 6 and 7 only. So logically you can have an 8 nowhere else in these columns, which turns out to be enough to solve it all without the need for XYwing, by later finding sneakily hidden multiples. 

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Louise56
Joined: 21 Sep 2005 Posts: 94 Location: El Cajon, California USA

Posted: Fri Sep 30, 2005 6:14 pm Post subject: 


someone_somewhere wrote:  David, I suppose that you did it!
Here to compare it with a "classic" way of solving:
5 in r2c8 9 in r6c3 7 in r7c8 4 in r9c3  Unique Horizontal
2 in r9c7  Unique in 3x3 block
1 not in r7c3, it is in r7c1 or r9c1 (Column on 3x3 Block interaction)
1 not in r8c3, it is in r7c1 or r9c1 (Column on 3x3 Block interaction)
2 in r7c3  Sole Candidate
2 in r4c1  Unique Horizontal
8 not in r5c8, it is in r4c7 or r4c9 (Row on 3x3 Block interaction)
8 not in r5c9, it is in r4c7 or r4c9 (Row on 3x3 Block interaction)
1 not in r8c4, Hidden Pair 5 9 in r8c2 and r8c4 (in Row)
3 not in r8c2, Hidden Pair 5 9 in r8c2 and r8c4 (in Row)
3 not in r8c4, Hidden Pair 5 9 in r8c2 and r8c4 (in Row)
6 not in r8c2, Hidden Pair 5 9 in r8c2 and r8c4 (in Row)
8 not in r8c2, Hidden Pair 5 9 in r8c2 and r8c4 (in Row)
1 not in r8c8, Nacked Pair 1 8 in r8c6 and r8c7 (same Row)
8 not in r8c8, Nacked Pair 1 8 in r8c6 and r8c7 (same Row)
6 in r8c8  Sole Candidate
3 in r8c3  Sole Candidate
1 in r2c3  Sole Candidate
3 not in r5c2, it is in r5c1 or r6c1 (Column on 3x3 Block interaction)
8 not in r1c6, it is in r2c6 r2c7 r8c6 r8c7 (XWing on Column)
8 not in r1c7, it is in r2c6 r2c7 r8c6 r8c7 (XWing on Column)
8 not in r4c7, it is in r2c6 r2c7 r8c6 r8c7 (XWing on Column)
8 not in r9c6, it is in r2c6 r2c7 r8c6 r8c7 (XWing on Column)
8 in r4c9  Unique Horizontal
5 in r6c9  Unique Vertical
4 not in r4c5, XYWing2 X=1 Y=7 in r5c9 X=1 Z=4 in r5c5 Y=7 Z=4 in r4c7
4 in r4c7 4 in r5c5  Unique Horizontal
4 in r1c8  Unique Horizontal
8 in r1c5 8 in r2c7 8 in r8c6  Unique Horizontal
9 in r2c4 1 in r8c7 9 in r8c2  Unique Horizontal
9 in r1c7 3 in r2c2 9 in r7c5 5 in r8c4 5 in r9c2  Unique Horizontal
3 in r1c4 2 in r2c6 5 in r4c5 1 in r7c1 3 in r9c5 6 in r9c1  Unique Horizontal
6 in r4c3 8 in r7c2 8 in r9c8  Unique Horizontal
7 in r4c4 8 in r5c1 7 in r6c7 7 in r9c6  Unique Horizontal
3 in r5c8 7 in r5c2 3 in r6c1 1 in r9c4  Unique Horizontal
3 in r3c7 7 in r3c9 1 in r5c9  Unique Horizontal
1 in r1c6 7 in r1c3 1 in r3c8 2 in r3c2 1 in r6c5  Unique Horizontal
2 in r1c9 6 in r1c2 6 in r3c5 6 in r6c6  Unique Horizontal
Final SuDoku Table
5 6 7 3 8 1 9 4 2
4 3 1 9 7 2 8 5 6
9 2 8 4 6 5 3 1 7
2 1 6 7 5 3 4 9 8
8 7 5 2 4 9 6 3 1
3 4 9 8 1 6 7 2 5
1 8 2 6 9 4 5 7 3
7 9 3 5 2 8 1 6 4
6 5 4 1 3 7 2 8 9 
Hello Someone!
I decided to give your puzzle a try, and went as far as I could. In looking at your steps to solve, I don't understand what a "naked pair" is. You identified 1,8 as a naked pair in r8c6 and r8c7. Please explain this to me. Thanks! 

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David Bryant
Joined: 29 Jul 2005 Posts: 559 Location: Denver, Colorado

Posted: Fri Sep 30, 2005 6:57 pm Post subject: Naked pairs 


Louise56 wrote:  Hello Someone!
I decided to give your puzzle a try, and went as far as I could. In looking at your steps to solve, I don't understand what a "naked pair" is. You identified 1,8 as a naked pair in r8c6 and r8c7. Please explain this to me. Thanks! 
A "naked pair" is a set of two numbers that must appear in two cells  either in a row, or in a column, or in some 3x3 box. Here's an example.
Code: 
1 2 3/7 4 5 6 3/7 8 9 
In this hypothetical example, we already know how to place 7 digits in a row, but the position of the other two numbers {3, 7} is still indeterminate.
This can be a useful piece of information, as in the following example.
Code: 
1 2 3/7 4 ? 6 3/7 8 9 
The cell marked "?" is not yet known  we can immediately see that it must be a "5" because of the "naked pair" {3, 7}.
This kind of setup is called a "naked pair" to distinguish it from a "hidden pair"  here's an illustration of a "hidden pair":
Code: 
1 2/4/5/6 3 2/4/6/9 5/6/9 5/6/9 7 8 5/6/9 
In this example the "hidden pair" appears in the second and fourth positions. We can see it there because the two digits {2, 4} cannot appear in any other spots in the row, except the second and fourth positions. But it's also "hidden" because there are apparently some other values that _might_ appear in the 2nd and 4th positions.
Another way to spot the "hidden pair" is to notice the "naked triplet" {5, 6, 9} appearing in the 5th, 6th, and 9th positions. Since those three cells must contain the 3 digits {5, 6, 9}, we can eliminate {5, 6, 9} from the 2nd and 4th positions, exposing the "hidden pair."
I hope that helps! dcb 

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Louise56
Joined: 21 Sep 2005 Posts: 94 Location: El Cajon, California USA

Posted: Fri Sep 30, 2005 7:47 pm Post subject: 


Thanks, David for explaining that to me! I don't see the naked pair in row 8. I have a 1,8 pair in row 8 in r8c6, r8c7 and r8c8. I've been following SomeoneSomewhere's solution and he must have put a 6 in r8c8 in order to get the naked pair. I don't see how the 6 gets there. In box 9 I have a 6 in r8c8 and r9c8.
Thanks again for your response! 

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Louise56
Joined: 21 Sep 2005 Posts: 94 Location: El Cajon, California USA

Posted: Fri Sep 30, 2005 9:02 pm Post subject: 


David,
I found the naked pair! I had put a 3 in r8c6 which was a mistake since column 6 already had a 3 in it! It's amazing that one mistake can throw off the whole puzzle! I'll see if I can finish this bad boy, otherwise you may hear from me again! I appreciate your patience and excellent teaching! 

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someone_somewhere
Joined: 07 Aug 2005 Posts: 275 Location: Munich

Posted: Sat Oct 01, 2005 6:50 am Post subject: 


Hi David,
Do you have an example that can be solved with "quadruples" but not with "pairs" or "triples"? Could be in paralel with Xwing, XYwing, coloring or some more advanced than "quadruples".
Thank you in advance,
see u, 

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David Bryant
Joined: 29 Jul 2005 Posts: 559 Location: Denver, Colorado


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Louise56
Joined: 21 Sep 2005 Posts: 94 Location: El Cajon, California USA

Posted: Sat Oct 01, 2005 8:32 pm Post subject: 


someone_somewhere wrote:  Hi David,
Do you have an example that can be solved with "quadruples" but not with "pairs" or "triples"? Could be in paralel with Xwing, XYwing, coloring or some more advanced than "quadruples".
Thank you in advance,
see u, 
I did this puzzle yesterday from USA Today. I needed to use the X Wing technique to solve it. I'm not a whiz kid like you and David are so perhaps this one can be solved without advanced techniques!
[/url]http://puzzles.usatoday.com/sudoku/ 

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someone_somewhere
Joined: 07 Aug 2005 Posts: 275 Location: Munich

Posted: Sun Oct 02, 2005 8:47 am Post subject: 


Hi David,
Thank you for your efforts.
The first one you mentioned, can be solved without "quadruples".
"Hidden pair" is enough:
 6    5   
5    7  9  
7 4 3 1 6 9 8  
  4  8 1   7
2    5  1  
 1      8 
1   6 3 4 2 9 8
4  6  9    1
   5 1   4 
2 not in r1c8, it is in r3c8 or r3c9 (Row on 3x3 Block interaction)
2 not in r1c9, it is in r3c8 or r3c9 (Row on 3x3 Block interaction)
2 not in r2c8, it is in r3c8 or r3c9 (Row on 3x3 Block interaction)
2 not in r2c9, it is in r3c8 or r3c9 (Row on 3x3 Block interaction)
5 not in r8c2, it is in r7c2 or r7c3 (Row on 3x3 Block interaction)
7 not in r8c2, it is in r7c2 or r7c3 (Row on 3x3 Block interaction)
7 not in r9c2, it is in r7c2 or r7c3 (Row on 3x3 Block interaction)
7 not in r9c3, it is in r7c2 or r7c3 (Row on 3x3 Block interaction)
3 not in r6c9, Hidden Pair 2 5 in r3c9 and r6c9 (in Column)
4 not in r6c9, Hidden Pair 2 5 in r3c9 and r6c9 (in Column)
6 not in r6c9, Hidden Pair 2 5 in r3c9 and r6c9 (in Column)
9 not in r6c9, Hidden Pair 2 5 in r3c9 and r6c9 (in Column)
9 in r5c9  Unique Vertical
... (and the rest is no problem)
Now, I will take a look at the second one.
see u, 

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someone_somewhere
Joined: 07 Aug 2005 Posts: 275 Location: Munich

Posted: Sun Oct 02, 2005 8:50 am Post subject: 


Hi,
Second one, I have already analysed. No need for "quadruples".
See the discussion at that point.
So, still I don't have an example (hidden or nacked qadruples) that I am looking for.
Still, thank you for your efforts, and keep your eyes open ...
see u, 

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someone_somewhere
Joined: 07 Aug 2005 Posts: 275 Location: Munich

Posted: Sun Oct 02, 2005 9:21 am Post subject: 


Hi Louise,
The Sudoku from USATODAY is extreme interesting.
I am wondering how you solved it with the Xwing technique.
I could not do it!
So here are my efforts:
Initial Sudoku Table:
123045000
000000000
600700200
070080030
040020060
050010090
009003008
000000000
000290714
3 in r3c5  Sole Candidate
6 in r2c5  Sole Candidate
2 in r2c6  Unique Horizontal
7 not in r2c8, it is in r1c8 or r1c9 (Row on 3x3 Block interaction)
7 not in r2c9, it is in r1c8 or r1c9 (Row on 3x3 Block interaction)
3 not in r8c1, it is in r9c1 or r9c2 (Row on 3x3 Block interaction)
3 not in r8c2, it is in r9c1 or r9c2 (Row on 3x3 Block interaction)
5 not in r7c1, it is in r9c1 or r9c3 (Row on 3x3 Block interaction)
5 not in r8c1, it is in r9c1 or r9c3 (Row on 3x3 Block interaction)
5 not in r8c3, it is in r9c1 or r9c3 (Row on 3x3 Block interaction)
3 in r9c2 7 in r1c8  Unique Vertical
1 not in r8c3, it is in r7c2 or r8c2 (Column on 3x3 Block interaction)
6 not in r8c3, it is in r7c2 or r8c2 (Column on 3x3 Block interaction)
6 not in r9c3, it is in r7c2 or r8c2 (Column on 3x3 Block interaction)
9 not in r2c1, it is in r2c2 or r3c2 (Column on 3x3 Block interaction)
5 not in r7c4, it is in r7c5 or r8c5 (Column on 3x3 Block interaction)
5 not in r8c4, it is in r7c5 or r8c5 (Column on 3x3 Block interaction)
7 not in r8c6, it is in r7c5 or r8c5 (Column on 3x3 Block interaction)
2 not in r8c9, it is in r7c8 or r8c8 (Column on 3x3 Block interaction)
4 not in r2c7, it is in r2c8 or r3c8 (Column on 3x3 Block interaction)
8 not in r1c7, it is in r2c8 or r3c8 (Column on 3x3 Block interaction)
8 not in r2c7, it is in r2c8 or r3c8 (Column on 3x3 Block interaction)
8 in r1c4 6 in r9c6  Unique Horizontal
8 in r8c6  Unique Vertical
1 in r3c6  Unique Vertical
9 in r2c4  Unique in 3x3 block
9 in r3c2  Unique in 3x3 block
8 in r2c2 5 in r3c9  Sole Candidate
4 in r2c8 4 in r3c3  Sole Candidate
8 in r3c8  Sole Candidate
4 not in r4c4, it is in r4c6 or r6c6 (Column on 3x3 Block interaction)
4 not in r6c4, it is in r4c6 or r6c6 (Column on 3x3 Block interaction)
5 not in r7c7, it is in r7c8 or r8c8 (Column on 3x3 Block interaction)
5 not in r8c7, it is in r7c8 or r8c8 (Column on 3x3 Block interaction)
6 in r7c7  Sole Candidate
9 in r1c7 1 in r7c2  Sole Candidate
6 in r1c9 4 in r7c4 6 in r8c2 3 in r8c7  Sole Candidate
1 in r2c7 1 in r8c4 9 in r8c9  Sole Candidate
3 in r2c9  Sole Candidate
4 in r8c1  Unique Horizontal
And now I am stucked at the position:
Final SuDoku Table
1 2 3 8 4 5 9 7 6
 8  9 6 2 1 4 3
6 9 4 7 3 1 2 8 5
 7   8   3 
 4   2   6 
 5   1   9 
 1 9 4  3 6  8
4 6  1  8 3  9
 3  2 9 6 7 1 4
or with the exclude table:
1 2 3 8 4 5 9 7 6
57 8 57 9 6 2 1 4 3
6 9 4 7 3 1 2 8 5
29 7 126 56 8 49 45 3 12
389 4 18 35 2 79 58 6 17
238 5 268 36 1 47 48 9 27
27 1 9 4 57 3 6 25 8
4 6 27 1 57 8 3 25 9
58 3 58 2 9 6 7 1 4
From here only some coloring, multicoloring, nisho, etc can help.
Or someone else . . . ?
see u,
Thank you in advance for further hits. 

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someone_somewhere
Joined: 07 Aug 2005 Posts: 275 Location: Munich

Posted: Sun Oct 02, 2005 10:18 am Post subject: 


Hi,
It's me again. From the previous position, I could apply Nishio (limited form of trial and error) and solve it.
I made the assumtion:
2 in r4c1
this leads to 1 in r4c9.
with this 2 followed 2 chains:
 7 in r5c9, 9 in r5c6, 4 in r4c6 and
 6 in r4c3, 5 in r4c4, 4 in r4c7
and now I have two "4"s in r4: one in c6 and one in c7.
This implies that my assuption "2 in r4c1" is leading to a conradiction.
The rest is now obvious: 9 in r4c1 !
Has someone a simpler solution?
see u, 

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David Bryant
Joined: 29 Jul 2005 Posts: 559 Location: Denver, Colorado

Posted: Sun Oct 02, 2005 5:51 pm Post subject: USA Today puzzles 


I haven't tried the puzzle you posted yet, Someone. I visited the USA Today site and got a different one  must have been the wrong day. Here it is.
Code: 
1 2 3 . 4 5 . . .
. . . . . . . . .
6 . . 7 . . 2 . .
. 7 . . 8 . . 3 .
. 4 . . 2 . . 6 .
. 5 . . 1 . . 9 .
. . 9 . . 3 . . 8
. . . . . . . . .
. . . 2 9 . 7 1 4 
After 27 moves I arrived at a position where the only method I could apply was to make an outright guess. I tried using a couple of different computer programs on it then, and those also had to resort to guessing.
Can anyone spot a logical way through this puzzle? Does USA Today typically publish puzzles that require trial and error? dcb 

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Guest

Posted: Sun Oct 02, 2005 6:07 pm Post subject: 


Hi,
You should have tried my one!!!
It is identical to your one!!!
As I have posted, I was able to solve it ONLY with a little bit of Nishio.
Just take a look again at my previous postings.
see u, 

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David Bryant
Joined: 29 Jul 2005 Posts: 559 Location: Denver, Colorado

Posted: Sun Oct 02, 2005 7:07 pm Post subject: Oops! 


You're right! I guess all the zeroes threw me off ... I didn't notice the three columns filled up in the three middle 3x3 boxes with all those zeroes in there!
I also tried to apply "Nishio", but started from the {1, 8} possibility in r5c3, just because it looked odd. Why did you start from r4c1? dcb
PS So you and I were working on the same puzzle, but are we sure it's the one Louise56 was talking about? Is this the one you were talking about, Louise? 

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Louise56
Joined: 21 Sep 2005 Posts: 94 Location: El Cajon, California USA

Posted: Sun Oct 02, 2005 11:00 pm Post subject: Re: Oops! 


David Bryant wrote:  You're right! I guess all the zeroes threw me off ... I didn't notice the three columns filled up in the three middle 3x3 boxes with all those zeroes in there!
I also tried to apply "Nishio", but started from the {1, 8} possibility in r5c3, just because it looked odd. Why did you start from r4c1? dcb
PS So you and I were working on the same puzzle, but are we sure it's the one Louise56 was talking about? Is this the one you were talking about, Louise? 
Hi David,
The one I did from USA today was the Sept 30 one, which is the same one you and Someone worked on. I can't remember all my steps, I'll have to go back. I did an X wing three times as I remember. The X Wing numbers were all four fairly close to each other, but that helped me elimate others. I'll go through it again and write a solution. Can you tell me what a "Nishio" is? 

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