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 Post subject: CPFC Killer
PostPosted: Sat Apr 30, 2016 9:33 pm 
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Grand Master
Grand Master

Joined: Tue Jun 16, 2009 9:31 pm
Posts: 280
Location: California, out of London
To celebrate Crystal Palace F.C. in the FA Cup Final :cheers: - I created this. Quite tough... SS 3.6.3 Gives rating of 1.50.

Image

3x3::k:3353:3098:3098:4615:4615:4615:4615:5384:5384:3353:8961:8961:8961:3347:8452:8452:8452:5384:3353:8961:1298:1298:3347:8452:8452:8452:4361:5904:8961:8961:8961:3347:8452:4618:4361:4361:5904:5393:5393:5393:2071:4618:4618:4618:3605:5904:9475:9475:9475:2071:7426:7426:7426:3605:5904:9475:9475:1048:1048:7426:4118:4118:3605:3355:9475:4124:4124:6413:7426:7426:7426:3596:3355:3355:3355:4124:6413:6413:6413:3596:3596:

By the way - Is there a way to get SS to colour selected cages? if so, I couldn't find it.


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 Post subject: Re: CPFC Killer
PostPosted: Sat May 07, 2016 3:06 am 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
Thanks wellbeback for a nice puzzle. And congratulations to Crystal Palace for reaching the FA Cup Final!

Here is my walkthrough:
Prelims

a) R1C23 = {39/48/57}, no 1,2,6
b) R3C34 = {14/23}
c) R56C5 = {17/26/35}, no 4,8,9
d) R7C45 = {13}
e) R7C78 = {79}
f) 21(3) cage at R1C8 = {489/579/678}, no 1,2,3
g) 21(3) cage at R5C2 = {489/579/678}, no 1,2,3
h) 13(4) cage at R8C1 = {1237/1246/1345}, no 8,9
i) 37(6) cage at R6C2 = {256789/346789}, no 1
j) 29(7) cage at R6C6 = {1234568}, no 7,9

Steps resulting from Prelims
1a. R7C45 = {13}, locked for R7 and N8
1b. R7C78 = {79}, locked for R7 and N9
1c. 13(4) cage at R8C1 = {1237/1246/1345}, 1 locked for N7
1d. 9 in N7 only in R8C23, locked for R8

2. 45 rule on C1 2 outies R9C23 = 4 = {13}, locked for R9 and N7
2a. R9C23 = 4 -> R89C1 = 9 = {27/45}

3. 45 rule on R1 1 outie R2C9 = 1 innie R1C1 + 6 -> R1C1 = {123}, R2C9 = {789}

4. 45 rule on C9 3 outies R149C8 = 21 = {489/678} (cannot be {579} which clashes with R7C8), 8 locked for C8
4a. Killer pair 7,9 in R149C8 and R7C8, locked for C8

5. 45 rule on C1234 2 innies R17C4 = 5 = [23/41]

6. 45 rule on R1234 2 innies R4C17 = 13 = {49/58/67}, no 1,2,3

7. 45 rule on C5 4 innies R1789C5 = 24
7a. R7C5 = {13} -> R1789C5 = {1689/3489/3579} (cannot be {3678} which clashes with R56C5), no 2, 9 locked for C5
7b. R7C5 = {13} -> no 1,3 in R1C5
7c. R56C5 = {17/26/35}, R7C5 = {13} -> combined cage R567C5 = {17}3/{26}1/{26}3/{35}1
7d. 13(3) cage at R2C5 = {148/157/247/256} (cannot be {238/346} which clashes with R567C5), no 3
7e. 3 in C5 only in combined cage R567C5 = {17}3/{26}3/{35}1

8. 45 rule on R89 4 innies R8C2678 = 22
8a. Hidden killer pair 1,3 in R8C2678 and R8C9 for R8, 22(4) cannot contain both of 1,3 -> R8C9 = {13} and R8C2678 must contain one of 1,3
8b. R8C9 = {13} -> 14(3) cage at R8C9 = [185/365] -> R9C9 = 5, R9C8 {68}, clean-up: no 4 in R8C1 (step 2a)
8c. 21(3) cage at R1C8 = {489/678}, 8 locked for N3
8d. 25(4) cage at R8C5 = {2689/4579/4678}
8e. 5 of {4579} must be in R8C5, {2689/4678} must have one of 6,8 in R8C5 (R9C567) cannot contain both of 6,8 which would clash with 14(3) cage) -> R8C5 = {568}

9. 16(3) cage at R8C3 = {259/268/457}
9a. Double hidden killer pair 2,4 in R89C1, 16(3) cage, 25(4) cage at R8C5 and R8C2678 for R89, R89C1 contains one of 2,4, 16(3) cage contains one of 2,4, 25(4) cage contains one of 2,4 -> R8C2678 must contain one of 2,4
9b. R8C2678 (step 8) = 22 must contain one of 1,3 (step 8a) and one of 2,4 = {1489/2389/3469/3478} (only possible combinations), no 5
9c. 7,9 of R8C2678 only in R8C2 -> R8C2 = {79}

10. 16(3) cage at R8C3 (step 9) = {259/268/457}, R8C5 = {568} -> 16(3) cage + R8C5 = {259}6/{259}8/{268}5/{457}6/{457}8 (cannot be {268}6 or {268}8 because all cells of 16(3) cage ‘see’ R8C5), 5 locked for R8, clean-up: no 4 in R9C1 (step 2a)
10a. Naked pair {27} in R89C1, locked for C1 and N7 -> R8C2 = 9, clean-up: no 3 in R1C3, no 8 in R2C9 (step 3), no 6 in R4C7 (step 6)
10b. 13(3) cage at R1C1 = {139/148/346}, no 5
10c. 21(3) cage at R1C8 = {489/678}, 8 locked for R1, clean-up: no 4 in R1C23
10d. R2C9 = {79} -> no 7,9 in R1C89
10e. R47C8 = {79} (hidden pair in C8)

11. 37(6) cage at R6C2 = {256789/346789}, 7 locked for R6, clean-up: no 1 in R5C5

12. Double killer pair 2,7 in R89C1, 16(3) cage at R8C3 and 25(4) cage at R8C5, locked for R89
12a. R8C2678 (step 9b) = {1489/3469}, 4 locked for R8 and 29(7) cage at R6C6, no 4 in R6C678 + R7C6
12b. 4 in N7 only in R7C123, locked for R7
12c. Killer pair 2,4 in R1C4 and 16(3) cage at R8C3, locked for C4, clean-up: no 1,3 in R3C1
12d. Naked pair {13} in R37C4, locked for C4
12e. R8C14 = {27} (hidden pair in R8)
12f. 16(3) cage at R8C3 (step 9) = {259/268/457}
12g. R8C4 = {27} -> no 2,7 in R9C4

13. 17(3) cage at R3C9 = {179/269/278/467} (cannot be {368} because R4C8 only contains 7,9), no 3

14. 14(3) cage at R5C9 = {167/239/248} (cannot be {149/347} because R7C9 only contains 2,6,8)
14a. {167} can only be [716] -> no 1 in R5C9, no 6 in R56C9
14b. 14(3) cage = {239/248} (cannot be [716] which clashes with 14(3) cage at R8C9 (step 8b) = [185/365]), 2 locked for C9
14c. 17(3) cage at R3C9 (step 13) = {179/467}, no 8

[There may be a technically simpler way to continue, but this short forcing chain is effective …]
15. Consider combinations for 14(3) cage at R5C9 (step 14b) = {239/248}
14(3) cage = {239}, locked for C9 => R2C9 = 7
or 14(3) cage = {248}, locked for C9 => R1C9 = 6 => 21(3) cage at R1C8 (step 8c) = {678} => R2C9 = 7
-> R2C9 = 7
15a. R2C9 = 7 -> 21(3) cage at R1C8 (step 8c) = {678}, 6 locked for R1 and N3
15b. 17(3) cage at R3C9 (step 13) = {179/467} -> R4C8 = 7 -> R7C78 = [79], clean-up: no 6 in R4C1 (step 6), no 4 in R4C9
[Cracked, the rest is fairly straightforward.]

16. 33(7) cage at R2C6 must contain 6, locked for C6
16a. 6 in R7 only in R7C123, locked for N7
16b. 16(3) cage at R8C3 (step 9) = {259/268/457}
16c. R8C3 = {58} -> no 8 in R9C4

17. 2,4 in R1 only in 18(4) cage at R1C4 = {2349/2457}, no 1
17a. R1C1 = 1 (hidden single in R1) -> R23C1 = 12 = {39/48}, no 6
17b. 6 in N1 only in R2C23 + R3C2, locked for 35(7) cage at R2C2, no 6 in R2C4 + R4C234
17c. 1 in N4 only in R4C23, locked for R4, clean-up: no 9 in R3C9 (step 15b)

18. R3C34 = [23] (cannot be [41] which clashes with R3C9) -> R7C45 = [13], R1C4 = 4 (step 5), clean-up: no 9 in R2C1 (step 17a), no 5 in R56C5
18a. R8C4 = 2 (hidden single in C4) -> R89C1 = [72]
18b. R7C9 = 2 (hidden single in R7)
18c. 29(7) cage at R6C6 must contain 2, locked for R6, clean-up: no 6 in R5C5

19. 37(6) cage at R6C2 = {346789} (only remaining combination), 3 locked for R6, clean-up: no 9 in R5C9 (step 14b)
19a. 29(7) cage at R6C6 must contain 3, locked for R8 -> R8C9 = 1, R9C8 = 8 (cage sum)
19b. R8C2678 (step 12a) = {3469} (only remaining combination) -> R8C6 = 4, R8C78 = {36}, locked for N9 and 29(7) cage at R6C6, no 6 in R6C78
19c. 29(7) must contain 1, locked for R6, R6C5 = 6 -> R5C5 = 2

20. R9C4 = 6 (hidden single in N8) -> R8C3 = 8 (cage sum)
20a. Naked pair {46} in R7C23, locked for R7 and 37(6) cage at R6C2, no 4 in R6C23
20b. R7C1 = 5 -> R7C6 = 8, R8C5 = 5, clean-up: no 8 in R4C7 (step 6)
20c. Naked triple {378} in R6C234, locked for R6, 3 also locked for N4
20d. R4C5 = 4 (hidden single in C5), clean-up: no 9 in R4C17 (step 6)
20e. R2C1 = 3 (hidden single in C1) -> R3C1 = 9 (cage sum), R6C1 = 4, R6C9 = 9, R5C9 = 3 (cage sum)
20f. R23C5 = {18} (hidden pair in C5), locked for N2

and the rest is naked singles.

Rating Comment:
I'll rate my walkthrough at Easy 1.5 because I used a short forcing chain to crack the puzzle. Until then my steps were in the 1.25 range.


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 Post subject: Re: CPFC Killer
PostPosted: Sat May 28, 2016 6:30 pm 
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Grand Master
Grand Master

Joined: Tue Jun 16, 2009 9:31 pm
Posts: 280
Location: California, out of London
Thanks for your WT Andrew. I've not noticed double killer pairs before! Nice.

Here is mine. It shares some features with Andrew's WT, but is mostly quite different.

Typos fixed thanks to Andrew.

CPFC Killer WT:
1. Outies c1 -> r9c23 = +4(2) = {13}
4(2)n8 = {13}
-> (13) in n9/r8c789
Innies r89 = r8c2678 = +22(4) -> cannot contain both (13)
-> r8c9 from (13)
16(2)n9 = {79}
-> 14(3)n9 from [1{58}] or [3{56}]
Outies c9 = r149c8 = +21(3)
Since r7c8 from (79) -> H21(3) not {579} - i.e., cannot contain a 5
-> 14(3)n9 from [185] or [365]. I.e., r9c9 = 5

2. 29(7)r6c6 = {1234568} (Not 79)
-> (79) in n8 in r8c45 and r9c456.

16(3)r8c3 does not contain 1 or 3 -> must be from {259}, {268}, {457}
25(4)r8c5 also does not contain 1 or 3 -> must be from {4678}, {4579}, {2689}

-> Either (a) (79) both in 25(4) -> 25(4) = [5{79}4] -> 16(3) = {268}
or (b) One of (79) in r89c4
Either way r8c3 cannot be 7 or 9
-> HS 9 in n7 -> r8c2 = 9
-> HS 7 in n7 -> r89c1 = {27}

3! Innies - Outies n8 -> r7c6 + r8c6 = r8c3 + r9c7
-> Whatever is in r7c6 cannot go in n9 in r9c7 (would put r8c3 = r8c6)
Also r7c6 cannot be (13) -> cannot be the same as r8c9
-> Whatever is in r7c6 is in n9 in r9c89 -> from {568}
-> Whatever it is must be in n7 in r8c3
-> r8c6 = r9c7
Since remaining innies r89 = r8c678 = +13(3)
-> r8c78 + r9c7 = +13(3)
-> r7c9 = 2

4. -> 2 in 29(7) in r6c678 (cannot be in r8c6 since r8c6 = r9c7)
-> 2 not in 37(7)
-> 37(7) = {346789}
-> 3 in 37(7) in r6c234
-> 3 in 29(7) in r8c78
-> r8c9 = 1
-> r9c8 = 8

5. Since H21(3) r149c8 cannot be {579} and since r9c9 = 5 -> 21(3)r1c8 cannot be {579}
-> 21(3)r1c8 must contain an 8 in r12c9
-> 14(3)r5c9 = [392]

6. Innies - Outies r1 -> r2c9 = r1c1 + 6
Since 2 already in c1 and 9 already in c9
-> r1c1 = 1 and r2c9 = 7
-> 21(3)r1c8 = [687]
-> Remaining Outies c9 -> r4c8 = 7
-> 16(2)n9 = [79]
Also r34c9 = [46]
Also 33(7)r2c6 must contain a 6 in r23c6
-> r8c6 not 6
-> r9c7 not 6
-> HS 6 in n9 -> r8c7 = 6
-> r8c8 = 3
-> r9c7 = 4
-> r8c6 = 4

7. -> 2 in n8 only in r89c4
-> Innies c1234 = r17c4 = +5(2) = [41]
-> r7c5 = 3
-> 5(2)r3c3 = [23]
Also HS 4 in n5/c5 -> r4c5 = 4
Also HS 3 in n5/c6 -> r4c6 = 3
-> HS 3 in n3 -> r1c7 = 3
-> r23c78 = {1259}
-> r23c6 = [67]
-> r23c5 = {18}
-> 8(2)n5 = [26]
-> 18(4)r1 = [4923]
-> 25(4)r8c5 = [5794]
-> 16(3)r8c3 = [826]
-> r7c6 = 8
etc.

Rating (on feel):
1.25 - I found it easier the second time through.

Now for Assassin 335 X fives!


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