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Texas Jigsaw Killer 18
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Para
Yokozuna
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Posts: 384
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PostPosted: Fri Jul 06, 2007 10:33 am    Post subject: Reply with quote

Ok a bit of overlap with Mike. Here is what he left out that i did get.

118.First some clean up: R6C2: no 5(i/o C12); R6C5: no 4(I/O c1234)

119. 11(3) in R4C4 = {128/146/236}: 2 only in R5C5 -> R5C5: no 3
119a. Needs one of {24}: has to go in R5C5 -> R4C5: no 4

120. 14(3) at R5C4 = {158/347}; R6C5 needs {3/8} so nowhere else in 14(3) -> R56C4: no 3

Para

More later
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Para
Yokozuna
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PostPosted: Fri Jul 06, 2007 11:19 am    Post subject: Reply with quote

Here's some more. I can't see that 6 opening things up but i might be overlooking something.

121. 6 in N2 locked for R1
121a. 6 in N1 locked for R4
121b. 6 in N1 locked in 11(3) at R4C3 -->> 11(3) = {164/632}: no 8; {3/4..}
121c. Clean up: R5C9: no 5

122. R3C12 = {39/48}: {8/9..}
122. Killer Triple {689} in R3C1267: locked for R3

123. 18(3) at R1C5 = {189/369/459/567}
123a. {89} only in R12C5 -> R12C5: no 1
123b. 6 only in R1C5 -> R1C5: no 7

124. 16(3) at R2C3 = {178/259/358/457}
124a. {89} only in R2C3 -> R2C3: no 1,3

125. colouring on 1's from C4: R3C5 <> 1
125a. R4C4 = 1 -> R3C9 = 1: R3C5 <> 1
125b. R5C4 = 1 -> R1/2C6 = 1: R3C5 <> 1

126. one more colouring 1's from C4: R9C1 <> 1
126a. R4C4 = 1 -> R9C5 = 1: R9C1 <> 1
126b. R5C4 = 1 -> R6C1 = 1: R9C1 <> 1

127. 1 in 12(3) at R8C2 only in R9C2 -->> R9C2: no 5

128. "45" R1234 : R5C589 = 17 = [2]{69}/[278]/[458]/[476]: R5C8: no 2,8; R5C9: no 2
128a. 12(3) at R3C9 = [219/156/138]: R4C9: no 9

Para

Time for other things right now. Maybe later tonight some more. Leave something for me Wink


Last edited by Para on Fri Jul 06, 2007 5:36 pm; edited 1 time in total
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rcbroughton
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PostPosted: Fri Jul 06, 2007 12:34 pm    Post subject: Reply with quote

A couple of quick ones

129. 2 in n5 (r3c9) only in r3c9/r6c8 - > no 2 in r3c8

130. from 125b. 18(3) r1c5 - no 1, so {189} no longer valid. No 8 r12c5

131. 8 in c5 locked at r6r8 -> CPE no 8 at r8c7

132. 10(3) ar r2c7 = {127}/{145}/{235} (6 has gone)
132a. 2 only at r2c8 -> no 3,7 r2c8

133. 12(3) r3c9 = [219]/[138]/[156] - > no 9 at r4c9

134. 18(3)r4c7 ={378}/{468}/{567} - no 9
134a. 8 only at r4c8 -> no 3 at r4c8
134b. 6 only at r5c8 -> no 5 at r4c8

135. 9 now locked at r56c9 for c9 - nowhere else in c9
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rcbroughton
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PostPosted: Fri Jul 06, 2007 1:20 pm    Post subject: Reply with quote

Just had to post this one - might just have cracked it!!

136. 45 on r4. Innies r4c456789 = 31
only combinations are:
{135679}, {145678}. {134689}
(can't place {235678}, {234679}. {234589}, {125689}. {124789})
136a. {135679}/{134689} - 9 must be at r4c6
136b. {145678} - r4c45={16} r4c6={8} r4c9={5} - so r4c78={47} but it can't be {47} as you can't make 18(3).
136c. r4c6 = 9
136d. r3c7= 6
136e. r3c6 = 8
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mhparker
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PostPosted: Fri Jul 06, 2007 2:28 pm    Post subject: Reply with quote

Thanks, Richard!

Unfortunately, you beat me to it with your last move. Here's another way of peeling the proverbial onion:

136. r4c789 cannot contain both of {45} due to r7c9
136a. only other place for {45} on r4 is 14(3)r4c1
136b. -> 14(3)r4c1 = {(4/5)..} = {248/257} (no 3,9)
136c. -> hidden single in r4 at r4c6 = 9
136d. -> r3c67 = [86]

As requested, I think we should leave the rest for Para... Wink
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Para
Yokozuna
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PostPosted: Fri Jul 06, 2007 5:24 pm    Post subject: Reply with quote

You could at least have provided me with a marks pic Mad Wink .
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Para
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PostPosted: Fri Jul 06, 2007 6:00 pm    Post subject: Reply with quote

I get to do all singles, i am so happy. I think i can handle that.

137. I/O dif R123: R3C9 = 2 -> R45C9 = [19]
137a. R4C45 = [63]; R5C5 = 2; R6C5 = 8
137b. R56C4 = [15](last possible combi); R789C6 = [625]; R7C5 = 7
137c. R89C5 = [61]; R7C13 = [51](hidden 8(3) cage); R6C1 = 1(hidden)
137d. R7C7 = 2; R7C9 = 5(hidden); R9C9 = 6(hidden); R5C7 = 5(hidden)
137e. R1C6 = 8(hidden 19(3) in C7); R3C8 = 1(hidden); R7C7 = 1(hidden); R7C8 = 9
137f. R1C3 = 6; R6C2 = 6; R5C8 = 6; R4C8 = 8; R8C9 = 8(all hidden)
137g. R4C7 = 4; R8C8 = 7; R6C89 = [37]; R56C6 = [34]; R56C3 = [79]
137h. R9C8 = 4; R9C3 = 8
137i. R2C2 = 8; R5C1 = 8; R7C4 = 8(all hidden); R5C2 = 4; R7C123 = [423]
137j. R8C247 = [349]; R9C12 = [27]; R9C47 = [93]; R4C123 = [752]
137k. R3C12 = [39]; R1C12 = [91]; R12C6 = [71]; R2C78 = [72]
137l. R12C4 = [23]; R12C9 = [34]; R1C8 = 5; R123C5 = [495]; R23C3 = [54]; R3C4 = 7

And we have this:

Code:
.-------.-------.---.-------.-------.
| 9   1 | 6   2 | 4 | 7   8 | 5   3 |
:---.   :---.   |   |   .---'---.   |
| 6 | 8 | 5 | 3 | 9 | 1 | 7   2 | 4 |
|   '---:   '---:   :---'---.   :---:
| 3   9 | 4   7 | 5 | 8   6 | 1 | 2 |
:-------'---.---'---:   .---'---:   |
| 7   5   2 | 6   3 | 9 | 4   8 | 1 |
:-------.---+---.   :---'---.   |   |
| 8   4 | 7 | 1 | 2 | 3   5 | 6 | 9 |
|   .---'   |   '---:   .---+---'---:
| 1 | 6   9 | 5   8 | 4 | 2 | 3   7 |
:---'---.---'---.---'---:   '---.   |
| 4   2 | 3   8 | 7   6 | 1   9 | 5 |
|   .---:   .---+---.   :---.---'---:
| 5 | 3 | 1 | 4 | 6 | 2 | 9 | 7   8 |
:---'   :---'   |   '---:   '---.   |
| 2   7 | 8   9 | 1   5 | 3   4 | 6 |
'-------'-------'-------'-------'---'



Para
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mhparker
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PostPosted: Fri Jul 06, 2007 7:42 pm    Post subject: Reply with quote

Para wrote:
Leave something for me Wink

Mike wrote:
As requested, I think we should leave the rest for Para... Wink

Para wrote:
I get to do all singles, i am so happy. I think i can handle that.

Good to see that you're taking that with a sense of humor!

BTW, I don't remember seeing a puzzle crack as quickly as this one. The word explosion might be more appropriate! If we were to optimize the walkthrough, we could probably go from the first placement to all singles within 5 moves. But getting that first placement was an absolute nightmare!

I can only speak for myself, but I must say that I really enjoyed doing this puzzle. It was a real team effort, requiring all of our different approaches to solve it. I'm glad that I was proved wrong in saying it couldn't be done. But I was right in one sense: we only managed to crack it because we were successfully able to turn our attention to the top-left corner of the grid.

Now on to Ed's A57V2X...
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mhparker
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PostPosted: Fri Jul 06, 2007 8:12 pm    Post subject: Reply with quote

Just a quick post to pick up on one of the moves Ed made in the walkthrough...

sudokuEd wrote:
107. weak links on 1s in r7 & 12(3)r6c7
107a. -> no 1 r8c1 (1 in r6c7 -> 1 in r7 in n6r4c3 -> no 1 in 8c1: 1 in r7c1 -> no 1 in r8c1))

Well spotted, Ed! Of course, what you found here was (to use the official jargon) an Alternating Inference Chain (AIC), consisting of both strong and weak links. To be more specific, it is a special case of an AIC referred to as a "grouped x-cycle" (grouped because at least one of the nodes in the chain consists of multiple cells, and x-cycle because the loop is based on the same digit ("x") all the way round).

This becomes more apparent if one includes a couple of nodes you omitted:

r6c1<>1 -> r6c7=1 -> r7c78<>1 -> r7c124=1

So either r6c1=1 or r7c124=1 (or both), allowing us to eliminate 1 from any common peer of both of these end nodes (only r8c1 in this case).

In Eureka notation, we would express this loop (and the associated elimination) as:

(1): r6c1=r6c7-r7c78=r7c124 => r8c1<>1

Because (being an x-cycle) the same digit is used throughout, it is only listed once at the beginning of the expression.
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