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Paul Site Admin
Joined: 01 Jan 1970 Posts: 93

Posted: Sun May 03, 2009 5:20 pm Post subject: One tough outlaw 


Here's the highestrated puzzle in the new batch I've just generated. Most outlaws aren't rated all that high, because they just need one or two techniques that SDXP doesn't support (so which don't count towards their rating). This one needs 1,477's worth of SDXP solving steps and a series of nonSDXP steps as well. Scanraid's solver (the Boss, in my opinion) had to use just about every trick in the book, including a couple of ALSs and some grouped inference chains.
Anyone fancy a go?
Code: 
. 8 .  . 4 .  . . .
5 . .  2 . .  . 6 .
. 9 .  . . .  . . 1
++
. 3 .  . . 7  . . 9
. . 4  . . .  2 . .
6 . .  5 . .  . 8 .
++
7 . .  . . .  . 5 .
. 1 .  . . 3  . . 7
. . .  . 8 .  . 4 .

_________________  Paul. 

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GreenLantern
Joined: 23 Nov 2006 Posts: 98

Posted: Tue May 05, 2009 6:49 am Post subject: 


Fun puzzle, Paul! Thanks for sharing. Here is how I solved it:
1. 2String Kite in 2's: r4c35,r69c6 => r9c3<>2
Code: 
++++
 123 8 12367  1369 4 169  3579 2379 235 
 5 4 137  2 1379 19  3789 6 38 
 23 9 2367  368 3567 568  4 237 1 
++++
 8 3 25  4 26 7  56 1 9 
 19 57 4  13689 1369 1689  2 37 56 
 6 27 19  5 1239 129  37 8 4 
++++
 7 26 2389  169 1269 4  13689 5 2368 
 4 1 2589  69 2569 3  689 29 7 
 239 256 359  7 8 12569  1369 4 236 
++++ 
2. (6=2)r7c2(2=7)r6c2(7=5)r5c2(5=2)r4c3(2=6)r4c5r4c7=r5c9 => r7c59<>6
3. [r9c6]=2=[r6c6]2[r6c2]7[r5c2]5[r5c9]=5=[r1c9]=2=[r79c9]2[r8c8]9[r8c4]6[r9c6] => r9c6<>6
4. [r8c5]=5=[r8c3]5[r4c3]2[r4c5]6[r8c5] => r8c5<>6
5. [r8c4]=6=[r8c7]6[r4c7]5[r5c9]=5=[r1c9]=2=[r79c9]2[r8c8]9[r8c4] => r8c4<>9
6. [r9c9]=6=[r5c9]=5=[r1c9]=2=[r13c8]2[r8c8]9[r8c7]8[r8c3]=8=[r7c3]=3=[r9c13]3[r9c9] => r9c9<>3
7. [r9c2]2[r6c2]7[r5c2]5[r5c9]6[r9c9]2[r9c2] => r9c2<>2
8. Multicoloring in 2's: r4c35,r67c2,r69c6 => r7c5/r9c1<>2
9. Unique Rectangle (38): r27c79 => r2c7<>8
10. [r9c6]=5=[r9c2]5[r5c2]=5=[r4c3]=2=[r4c5]2[r6c6]=2=[r9c6] => r6c5<>2
11. [r1c7]=5=[r4c7]5[r4c3]2[r6c2]7[r6c7]3[r1c7] => r1c7<>3
12. [r5c6]=8=[r3c6]=5=[r9c6]=2=[r9c9]=6=[r5c9]6[r5c6] => r5c6<>6
13. [r3c5]=5=[r8c5]5[r8c3]=5=[r4c3]5[r5c2]7[r5c8]3[r6c7]=3=[r6c5]3[r3c5] => r3c5<>3
14. [r2c5]=7=[r3c5]=5=[r3c6]5[r9c6]2[r9c9]6[r5c9]5[r5c2]7[r6c2]=7=[r6c7]=3=[r6c5]3[r2c5] => r2c5<>3
15. [r3c8]7[r3c5]5[r8c5]2[r4c5]=2=[r4c3]=5=[r5c2]=7=[r5c8]7[r3c8] => r3c8<>7
16. [r2c7]7[r2c5]=7=[r3c5]=5=[r8c5]=2=[r4c5]2[r4c3]=2=[r6c2]=7=[r6c7]7[r2c7] => r2c7<>7
17. XYwing (239): r2c7,r38c8 => r1c8/r8c7<>9 

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Paul Site Admin
Joined: 01 Jan 1970 Posts: 93

Posted: Thu May 07, 2009 3:57 pm Post subject: 


Hi GreenLantern  I have to admit that when I said "Anyone fancy a go?" I did mean you! And only you could describe this mindboggling grid as a "fun puzzle"!
I hadn't heard of a 2string kite before, but Googled it and discovered it's a neat way of spotting a discontinuous nice loop as a static pattern.
The 2string kite consists of two conjugate pairs (on the same candidate), one in a row, the other in a column, with two of the squares (one from each pair) in the same box. The candidate can be eliminated from any square that's related to the other two squares in the pattern (i.e. the ones that aren't in the same box as each other).
Here's the puzzle at the start, before the kite elimination:
Code: 
++++
 123 8 12367  1369 4 1569  3579 2379 235 
 5 4 137  2 1379 189  3789 6 38 
 23 9 2367  368 3567 568  4 237 1 
++++
 8 3 25  4 26 7  56 1 9 
 19 57 4  13689 1369 1689  2 37 356 
 6 27 1279  5 1239 129  37 8 4 
++++
 7 26 23689  169 1269 4  13689 5 2368 
 4 1 25689  69 2569 3  689 29 7 
 239 256 23569  7 8 12569  1369 4 236 
++++

I've coloured the two conjugate pairs red (r4c35) and blue (r69c6). They're joined by r4c5 (red) and r6c7 (blue) being in the same box (this box is the 'kite'). R9c3 is related to the other ends of both conjugate pairs (the 'ends of the kite strings'), so cannot contain candidate 2.
This simple Nice Loop proves the elimination:
[r9c3]2[r9c6]=2=[r6c6]2[r4c5]=2=[r4c3][r9c3]
If r9c3 is 2 then r9c6 can't be, so r6c6 must be (because it's a conjugate pair), so r4c5 can't be, so r4c3 must be, so r9c3 can't be  a contradiction, or discontinuity.
Conjugate pairs (where a candidate  2 in this example  occurs in exactly two squares in a row, column or box) are relatively easy to spot, and always worth looking out for, as they're the key to many solving techniques. Here they've given us a 'bargain' Nice Loop, which is well worth having.
Now GreenLantern, can you explain what
(6=2)r7c2(2=7)r6c2(7=5)r5c2(5=2)r4c3(2=6)r4c5r4c7=r5c9 => r7c59<>6
means, as I've never moved beyond basic NL notation! _________________  Paul. 

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GreenLantern
Joined: 23 Nov 2006 Posts: 98

Posted: Thu May 07, 2009 6:22 pm Post subject: 


I consider 2String Kites and Skyscrapers to be simple forms of Multicoloring. For example, if there was no '2' in r6c5 in the starting candidates grid, I would have described the Step #1 deduction as a Simple Coloring in 2's. I believe SDXP would describe that as a Chain of Conjugate Pairs.
Step #2 is written in Eureka AIC (Alternating Inference Chains) notation. I am now using AICs in addition to Nice Loops because I believe they are an excellent way of notating certain patterns like XYchains. The AIC I wrote is actually 2 AICs in one. You can think of it as an extended XYchain.
If we leave the last two nodes off of Step #2, we have the following AIC:
(6=2)r7c2(2=7)r6c2(7=5)r5c2(5=2)r4c3(2=6)r4c5
This is how you notate an XYchain. '6' is the unlinked digit and r7c2 and r4c5 are the endpoints of the chain. The way you read an AIC is to first look at the endpoints. Either r7c2=6 or r4c5=6 since AICs are bidirectional like Nice Loops. Therefore, any cell that can see both endpoints can have '6' eliminated from its candidate list.
Reading left to right, r7c2=2 => (r7c2<>6 and r6c2<>2) => r6c2=7 => r5c2<>7 => r5c2=5 => r4c3<>5 => r4c3=2 => r4c5<>2 => r4c5=6. So, r7c2<>6 => r4c5=6. If we followed the AIC from right to left, then we would conclude that r4c5<>6 => r7c2=6.
Now, let's look at the whole AIC in Step #2:
(6=2)r7c2(2=7)r6c2(7=5)r5c2(5=2)r4c3(2=6)r4c5r4c7=r5c9
It is understood that when you leave off the parenthesis before a node that the digit involved will be the same as in the previous node. So, r4c5=6 => r4c7<>6 => r5c9=6. So overall, r7c2<>6 => (r4c5=6 AND r5c9=6). This is an XYchain which allows one to conclude r7c59<>6. 

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