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Clement
Joined: 24 Apr 2006
Posts: 1113
Location: Dar es Salaam Tanzania
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 Posted: Sat Aug 30, 2014 9:02 pm Post subject: Daily Telegraph August 29, 2014 |
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| Code: |
+-------+-------+-------+
| 8 . . | 5 . 2 | . . 7 |
| . 7 . | 1 . 6 | 3 . . |
| 5 . . | . . . | 2 . . |
+-------+-------+-------+
| . . 5 | . 4 . | 1 . . |
| . . . | 2 . . | . . . |
| . . 3 | . 1 . | 4 . . |
+-------+-------+-------+
| . . 4 | . . . | . . 3 |
| . . 7 | 8 . 3 | . 9 . |
| 3 . . | 7 . 4 | . . 6 |
+-------+-------+-------+
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Play this puzzle online at the Daily Sudoku site |
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JC Van Hay
Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium
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| Posted: Sat Aug 30, 2014 9:37 pm Post subject: |
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| Hint : analysis of the puzzle from the 2 solutions of B3. |
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arkietech
Joined: 31 Jul 2008
Posts: 1834
Location: Northwest Arkansas USA
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| Posted: Sun Aug 31, 2014 11:45 am Post subject: |
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| Code: | *-----------------------------------------------------------*
| 8 349 1 | 5 c39 2 | 69 46 7 |
| 249 7 a29 | 1 8 6 | 3 45 a59 |
| 5 39 6 | 4 c379 79 | 2 8 1 |
|-------------------+-------------------+-------------------|
| 2679 269 5 | 3 4 89 | 1 67 28 |
| 14 14 8 | 2 b67 57 |b69 3 b59 |
| 2679 269 3 | 69 1 589 | 4 567 28 |
|-------------------+-------------------+-------------------|
| 69 8 4 | 69 5 1 | 7 2 3 |
| 126 126 7 | 8 26 3 | 5 9 4 |
| 3 5 9-2 | 7 c29 4 | 8 1 6 |
*-----------------------------------------------------------*
(2=5)r2c39-(5=7)r5c579-(7=2)r139c5 => -2r9c3; ste |
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Clement
Joined: 24 Apr 2006
Posts: 1113
Location: Dar es Salaam Tanzania
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| Posted: Sun Aug 31, 2014 2:31 pm Post subject: Daily Telegraph August 29, 2014 |
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| Code: |
+-------------+------------+-----------+
| 8 349 1 | 5 39 2 | c69 d46 7 |
| 249 7 f29 | 1 8 6 | 3 e45 59 |
| 5 39 6 | 4 379 79 | 2 8 1 |
+-------------+------------+-----------+
| 2679 269 5 | 3 4 89 | 1 67 28 |
| 14 14 8 | 2 aj7-6 57 | b69 3 59 |
| 2679 269 3 | 69 1 589 | 4 567 28 |
+-------------+------------+-----------+
| 69 8 4 | 69 5 1 | 7 2 3 |
| 126 126 7 | 8 i26 3 | 5 9 4 |
| 3 5 g29 | 7 h29 4 | 8 1 6 |
+-------------+------------+-----------+
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(7=6)r5c5-(6=9)r5c7-(9=6)r1c7-(6=4)r1c8-(4=5)r2c8-(5=9)r2c9-(9=2)r2c3-(2=9)r9c3-(9=2)r9c5-(2=6)r8c5-(6=7)r5c5 (contradiction) => -6r5c5; stte. |
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dongrave
Joined: 06 Mar 2014
Posts: 572
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| Posted: Thu Sep 04, 2014 1:09 am Post subject: |
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This puzzle seems like it's perfect to use to ask a few questions that I have about Sudoku chains and Eureka. Not too long ago I discovered what I've been referring to as 'Van Hay' chains. I've been calling them this because I learned about them from him and I don't know what they're really called. The solutions go as follows: A or B; if A then C; if B then C; therefore C. When I recently sent one of these solutions to Marty and asked him to represent it in Eureka, I was surprised to find that he transformed this 'double path' chain to one continuous one. Since then I've read that many Sudoku experts prefer chains that work in both directions over single direction chains and if that's true, then I'm thinking that they also must prefer single chains over these 'double path' ones.
Beginning with the diagram of either of the two solutions posted above, I discovered the following 'Van Hay' solution: r2c3 is 2 or 9; if it's 2, r9c3=2 so r9c5=2 so r13c5=9 so r3c6=7; if it's 9, then r2c9=5 so r5c9=9 so r5c6=5 so r3c6=7; therefore r3c6=7.
After looking at it for a while and thinking about how Marty represented my previous 'Van Hay' solution as a single chain, I realized that this one could also be represented as a single chain like: (7=9)r3c6-r13c5=r9c5-(9=2)r9c3-(2-9)r2c3-(9=5)r2c9-r5c9=r5c6 => r3c6=7 (contradiction).
Could one of you Eureka experts let me know if this expression is correct? You'll probably also notice that I shifted the starting point of this chain for
this chain. The reason was that I wasn't exactly sure how to represent the nodes at the very end of the this expression when they appeared in the middle of the chain. Could someone also show me how this chain would look if you shift the starting point to some other cell (like maybe r2c3?) Thanks in advance, Don. |
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arkietech
Joined: 31 Jul 2008
Posts: 1834
Location: Northwest Arkansas USA
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| Posted: Thu Sep 04, 2014 1:10 pm Post subject: |
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| dongrave wrote: | This puzzle seems like it's perfect to use to ask a few questions that I have about Sudoku chains and Eureka. Not too long ago I discovered what I've been referring to as 'Van Hay' chains. I've been calling them this because I learned about them from him and I don't know what they're really called. The solutions go as follows: A or B; if A then C; if B then C; therefore C. When I recently sent one of these solutions to Marty and asked him to represent it in Eureka, I was surprised to find that he transformed this 'double path' chain to one continuous one. Since then I've read that many Sudoku experts prefer chains that work in both directions over single direction chains and if that's true, then I'm thinking that they also must prefer single chains over these 'double path' ones.
Beginning with the diagram of either of the two solutions posted above, I discovered the following 'Van Hay' solution: r2c3 is 2 or 9; if it's 2, r9c3=2 so r9c5=2 so r13c5=9 so r3c6=7; if it's 9, then r2c9=5 so r5c9=9 so r5c6=5 so r3c6=7; therefore r3c6=7.
After looking at it for a while and thinking about how Marty represented my previous 'Van Hay' solution as a single chain, I realized that this one could also be represented as a single chain like: (7=9)r3c6-r13c5=r9c5-(9=2)r9c3-(2-9)r2c3-(9=5)r2c9-r5c9=r5c6 => r3c6=7 (contradiction).
Could one of you Eureka experts let me know if this expression is correct? You'll probably also notice that I shifted the starting point of this chain for
this chain. The reason was that I wasn't exactly sure how to represent the nodes at the very end of the this expression when they appeared in the middle of the chain. Could someone also show me how this chain would look if you shift the starting point to some other cell (like maybe r2c3?) Thanks in advance, Don. |
| Code: | *-----------------------------------------------------------*
| 8 349 1 | 5 39 2 | 69 46 7 |
| 249 7 29 | 1 8 6 | 3 45 59 |
| 5 39 6 | 4 379 79 | 2 8 1 |
|-------------------+-------------------+-------------------|
| 2679 269 5 | 3 4 89 | 1 67 28 |
| 14 14 8 | 2 67 57 | 69 3 59 |
| 2679 269 3 | 69 1 589 | 4 567 28 |
|-------------------+-------------------+-------------------|
| 69 8 4 | 69 5 1 | 7 2 3 |
| 126 126 7 | 8 26 3 | 5 9 4 |
| 3 5 92 | 7 29 4 | 8 1 6 |
*-----------------------------------------------------------*
(7=9)r3c6-r13c5=r9c5-(9=2)r9c3-(2-9)r2c3-(9=5)r2c9-r5c9=r5c6 => r3c6=7 (contradiction).
The 2 and 9 inr2c3 must be a strong link (2=9)r2c3
I like to put a value on the last link =5r5c6
(7=9)r3c6-r13c5=r9c5-(9=2)r9c3-(2=9)r2c3-(9=5)r2c9-r5c9=5r5c6
=> no 7 in c6 |
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dongrave
Joined: 06 Mar 2014
Posts: 572
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| Posted: Fri Sep 05, 2014 12:34 am Post subject: |
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Whoops! I knew that was strong link. I really need to be more careful. I see now that I also had a typo in my Van Hay solution ('if it's 2, r9c3=2' s.b. 'r9c3=9'). Thanks for the input Arkitech! (I think Marty has referred to you as Dan in previous posts.) I was really surprised at how little I was able to find on Eureka notation when I searched the internet. One link that I did find though had a great example of the type of chain that I'm not used to seeing. It was the following example that Marty posted a couple of years ago. He said that he got it from someone named Ted and I noticed that the discussion also included input from you, keith, and van hay.
| Code: |
*-----------------------------------------------------------*
| 6 1 59 | 8 45 349 | 7 2 34 |
| 78 4 3 | 19 17 2 | 189 5 6 |
| 2 59 78 | 6 1457 349 | 189 49 134 |
|-------------------+-------------------+-------------------|
| 9 6 2 | 7 3 5 | 4 1 8 |
| 57 3 57 | 4 8 1 | 69 69 2 |
| 1 8 4 | 2 9 6 | 3 7 5 |
|-------------------+-------------------+-------------------|
| 458 25 158 | 3 1246 7 | 1256 46 9 |
| 3 7 19 | 5 26 49 | 26 8 14 |
| 45 259 6 | 19 124 8 | 125 3 7 |
*-----------------------------------------------------------*
9r9c2=9r3c2-(9=4)r3c8-r7c8=(4-1)r8c9=(1-9)r8c3=9r9c2 => 9r9c2
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Interesting! I'm so used to seeing strong links strung together that when I first saw an expression like this, it had me scratching my head! I see that I need to learn more about these chains with weak inferences. I'd appreciate any more examples that would help me better understand this notation! Thanks, Don. |
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arkietech
Joined: 31 Jul 2008
Posts: 1834
Location: Northwest Arkansas USA
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| Posted: Fri Sep 05, 2014 1:16 am Post subject: |
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| dongrave wrote: | | I'd appreciate any more examples that would help me better understand this notation! Thanks, Don. |
I post a daily puzzle on http://forum.enjoysudoku.com/ there are many examples.
dan |
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dongrave
Joined: 06 Mar 2014
Posts: 572
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| Posted: Sat Sep 06, 2014 1:11 am Post subject: |
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| arkietech wrote: | | dongrave wrote: | | I'd appreciate any more examples that would help me better understand this notation! Thanks, Don. |
I post a daily puzzle on http://forum.enjoysudoku.com/ there are many examples.
dan |
Great! Thanks for the link Dan. I'll check it out this weekend. Am I right to assume that most or all of these puzzles will require some sort of chain to solve them? Or is it possible to solve some of them with just basics and standard advanced moves? Thanks, Don. |
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arkietech
Joined: 31 Jul 2008
Posts: 1834
Location: Northwest Arkansas USA
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| Posted: Sat Sep 06, 2014 11:35 am Post subject: |
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| dongrave wrote: | | Am I right to assume that most or all of these puzzles will require some sort of chain to solve them? Or is it possible to solve some of them with just basics and standard advanced moves? Thanks, Don. |
standard advanced moves?   
To solve them with a single step will require something greater than a w-wing. |
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dongrave
Joined: 06 Mar 2014
Posts: 572
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| Posted: Sat Sep 06, 2014 2:51 pm Post subject: |
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| arkietech wrote: | | dongrave wrote: | | Am I right to assume that most or all of these puzzles will require some sort of chain to solve them? Or is it possible to solve some of them with just basics and standard advanced moves? Thanks, Don. |
standard advanced moves?
To solve them with a single step will require something greater than a w-wing. |
Hi Dan, Marty refers to 'standard advanced moves' as the group of moves necessary to solve the VHs on this site. They include X-wing, XY-wing, and XYZ-wing. I was surprised to hear that URs were not even included so I tried to coax him into adding them (with no luck). Thanks, Don. |
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