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 Post subject: SKX-2 Revisit
PostPosted: Sat Dec 26, 2020 11:18 pm 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
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SKX-2 Revisit

Ed was probably sensible to omit this puzzle from his series of Revisit killers; it's quite a lot harder than the others, with a SS score of 2.35. However I felt the challenge of attempting it again; also Ruud's Special Killer-Xs were landmark puzzles, the other three only being at original Assassin level.

I spent a lot of time finding a way for my key breakthrough; of course a simpler way at that stage is R123C6 cannot be {247} because that forces R123C4 = {136} and R456C4 = {589} but 15(4) cage at R7C4 cannot be {2247}.

Here's my walkthrough for SKX-2 Revisit:
Prelims

a) R12C5 = {89}
b) R34C3 = {19/28/37/46}, no 5
c) R34C7 = {79}
d) R67C3 = {18/27/36/45}, no 9
e) R67C7 = {15/24}
f) R89C5 = {16/25/34}, no 7,8,9
g) 10(4) cage at R2C9 = {1234}
h) 27(4) cage at R6C9 = {3789/4689/5679}, no 1,2

Steps resulting from Prelims
1a. Naked pair {89} in R12C5, locked for C5 and N2
1b. Naked pair {79} in R34C7, locked for C7
1c. Naked pair {79} in R34C7, CPE no 7,9 in R4C6 using D/
1d. Naked quad {1234} in 10(4) cage at R2C9, CPE no 1,2,3,4 in R1C9
1e. 27(4) cage at R6C9 = {3789/4689/5679}, CPE no 9 in R9C9
1f. 8 in R3 only in R3C123, locked for N1

2a. 45 rule on R1234 2 innies R4C28 = 10 = {19/28/37/46}, no 5
2b. 45 rule on R6789 2 innies R6C28 = 10 = {19/28/37/46}, no 5

3a. 45 rule on N1 2 outies R4C13 = 1 innie R2C3 + 5, no 5 in R4C1 (IOU)
3b. 45 rule on N3 2 outies R4C79 = 1 innie R2C7 + 3, no 3 in R4C9 (IOU)
3c. Naked quad {1234} in 10(4) cage at R2C9, 3 locked for N3
3d. 45 rule on N7 2 outies R6C13 = 1 innie R8C3 + 4, no 4 in R6C1 (IOU)
3e. 45 rule on N9 1 outie R6C9 = 2 innies R78C7 + 4
3f. Min R78C7 = 3 -> min R6C9 = 7
3g. Max R78C7 = 5, no 5 in R7C7, no 5,6,8 in R8C7, clean-up: no 1 in R6C7

4a. 45 rule on C12 2 outies R19C3 = 14 = [59/68/95]
4b. Min R9C3 = 5 -> max R8C2 + R9C12 = 11, no 9 in R8C2 + R9C12
4c. 45 rule on C89 2 outies R19C7 = 6 = {15/24}
4d. Naked quad {1245} in R1679C7, locked for C7 -> R8C7 = 3, clean-up: no 4 in R9C5
4e. R8C7 = 3 -> R789C6 = 22 = {589/679}, 9 locked for C6 and N8
4f. 45 rule on N9 1 outie R6C9 = 1 remaining innie R7C7 + 7 -> R6C9 + R7C7 = [81/92], clean-up: no 2 in R6C7
4g. 27(4) cage at R6C9 = {4689/5679}, 6 locked for N9
4h. 45 rule on N8 1 innie R7C5 = 1 remaining outie R8C3 + 1 -> R7C5 = {23567}, R8C3 = {12456}
4i. 45 rule on N69 3 remaining innies R4C79 + R5C7 = 17, R4C7 is odd, R5C7 is even -> R4C9 must be even = {24}
4j. Naked quad {1234} in 10(4) cage at R2C9, 1 locked for N3, clean-up: no 5 in R9C7
4k. 1 in C7 only in R79C7, locked for N9
4l. 1,3 in N6 only in 15(4) cage at R4C8 = {1239/1347/1356}, no 8, clean-up: no 2 in R4C2 (step 2a), no 2 in R6C2 (step 2b)

5a. 45 rule on N9 2 remaining outies R6C79 = 13 = [49/58]
5b. Consider permutations for R6C79
R6C79 = [49]
or R6C79 = [58] => 27(4) cage at R6C9 (step 4g) = 8{469}, 4 locked for N9
-> naked pair {12} in R79C7, 2 locked for C7 and N9

6a. 45 rule on N3 2 innies R23C7 = 1 outie R4C9 + 13
6b 10(4) cage at R2C9 = {1234}, R6C79 = [49/58] (step 5a)
6c. Consider placement for 4 in C7
R1C7 = 4 => 4 in 10(4) cage only in R4C9 => R23C7 = 17 = [89]
or R6C7 = 4 => R6C9 = 9 => R34C7 = [97]
-> R3C7 = 9, placed for D/, R4C7 = 7, clean-up: no 3 in R3C3, no 3 in R4C28 (step 2a), no 1 in R4C3, no 3 in R6C2 (step 2b)
6d. 15(4) cage at R4C8 (step 4l) = {1239/1356}, no 4, clean-up: no 6 in R4C2 (step 2a), no 6 in R6C2 (step 2b)
6e. Consider placements for R4C9 = {24}
R4C9 = 2 => 4 in 10(4) cage in R2C9 + R3C89, locked for N3
or R4C9 = 4 => R1C7 = 4 (hidden single in C7)
-> no 4 in R12C8

7a. 45 rule on N1 2 innies R23C3 = 1 outie R4C1 + 5
7b. R23C3 cannot total 14 (because no 5,9 in R3C3 and R234C3 = [682] + R19C3 = {59} clash with R67C3) -> no 9 in R4C1
7c. Consider placements for R4C1 = {123468}
R4C1 = {1234} => R23C3 = 6,7,8,9 => no 9 in R2C3
or R4C1 = 6 => R23C3 = 11 cannot be [92] because R234C3 = [928] clashes with R19C3
or R4C1 = 8 => R23C4 = 13 cannot be [94] because R234C3 = [946] clashes with R19C3
-> no 9 in R2C3

8. Consider combinations for R789C6 (step 4e) = {589/679}
R789C6 = {589}
or R789C6 = {679}, 6 locked for N8, 8 in N8 only in 15(4) cage at R7C4 = {1248}, CPE no 1,2,4 in R8C5 => R8C5 = 5
-> 5 in R789C6 + R8C5, locked for N8, clean-up: no 2 in R8C5, no 4 in R8C3 (step 4h)
[A key step which helped the breakthrough in step 11.]

9a. 45 rule on R12 2 outies R3C46 = 2 innies R2C19 + 7
9b. Max R3C46 = 13 => max R2C19 = 6, no 6,7,9 in R2C1
9c. Min R2C19 = 3 -> min R3C46 = 10, no 1,2 in R3C46

10a. 45 rule on R89 2 innies R8C19 = 2 outies R7C46 + 5
10b. Min R7C46 = 6 -> min R8C19 = 11, no 1 in R8C1
10c. Max R8C19 = 17 -> max R7C46 = 12, min R7C6 = 5 -> max R7C4 = 7

11a. 45 rule on N8 4 innies R789C4 + R7C5 = 16 = {1267/1348/2347}
11b. 3 of {1348} must be in R7C5, 3 of {2347} must be in R79C4 (cannot be in R7C5 because 15(4) cage at R7C4 cannot be {247}2) -> R789C4 must contain one of 1,3
11c. 17(4) cage at R1C4 = {1367/1457/2357/2456}
11d. 19(4) cage at R1C6 = {1378/1468/2368/2458/2467} (cannot be {1567} = {157}6 which clashes with R789C6, cannot be {3457} because R2C8 only contains 6,8)
11e. 19(4) cage = {1378/1468/2368/2458} (cannot be {2467} = {247}6 which clashes with 17(4) cage = {1457/2357/2456} while 17(4) cage = {1367} with R123C4 = {136} clashes with R789C4)
[Fairly straightforward from here.]
11f. 19(4) cage = {1378/1468/2368/2458} -> R2C7 = 8, R12C5 = [89], R5C7 = 6, R6C9 = 8 (hidden single in N6), R6C7 = 5 (step 5a) -> R7C7 = 1, placed for D\, R1C7 = 4, R9C7 = 2
11g. Naked triple 1,2,3 in R2C9 + R3C89, 2 locked for N3 and 10(4) cage at R2C9 -> R4C9 = 4
11h. R6C9 = 8 -> 27(4) cage at R6C9 (step 4g) = {4689} -> R7C8 = 4, R78C9 = {69}, locked for C9, 9 locked for N9
11i. Naked pair {57} in R19C9, CPE no 5,7 in R1C1 + R5C5 + R9C1 using diagonals
Clean-ups: no 6 in R3C3, no 9 in R4C3, no 4 in R6C2, no 2 in R6C8 (both step 2b), no 5 in R8C5

12a. 5 in N8 only in R789C6 (step 4e) = {589}, 5,8 locked for C6, 8 locked for N8
12b. 2,7 in N8 only in R789C4 + R7C5, CPE no 2,7 in R6C4
12c. R45C4 = {89} (hidden pair in N5)
12d. R4C5 = 5 (hidden single in N5) -> 19(4) cage at R3C5 = {1459/2359/2458} (cannot be {1567/3457} because R4C4 only contains 8,9), no 6,7
12e. 4 of {1459} must be in R3C5 -> no 1 in R3C5
12f. 45 rule on N2 1 remaining outie R2C3 = 1 innie R3C5 = {234}
12g. R89C5 = {16} (cannot be [43] which clashes with R35C5, ALS block), locked for C5 and N8
12h. 4 in N8 only in R89C4, locked for C4
12i. 6 in N5 only in R6C46 -> 19(4) cage at R6C4 = {2467} (only remaining combination), no 1,3, 4 locked for R6 and N5
12j. R6C4 = 6, placed for D/
Clean-ups: no 3 in R6C3, no 3,5 in R7C3, no 2,5 in R8C3 (step 4h)

13a. R5C7 = 6 -> 26(5) cage at R5C3 = {13679/14678/23678/24569/34568} (cannot be {12689} = {89}[216] which clashes with R5C89, ALS block)
13b. R5C4 = {89} -> no 8,9 in R5C3
13c. 9 on C3 only in R19C3 (step 4a) = {59}, 5 locked for C3
13d. 45 rule on C123 3 innies R258C3 = 12 = {147/246} (cannot be {237} because R8C3 only contains 1,6), no 3
13e. R4C3 = 3 (hidden single in C3) -> R3C3 = 7, placed for D\, R9C9 = 5, R8C8 = 8, both placed for D\
13f. R1C9 = 7, R2C8 = 5, both placed for D/
13g. R7C3 = 8 (hidden single in C3) -> R6C3 = 1, clean-up: no 9 in R6C28 (step 2b)
13h. R8C3 = 6 -> R7C5 = 7 (step 4h)
13i. R9C3 = 9 -> 16(4) cage at R8C2 = {1249} -> R8C2 = 2, placed for D/, R4C6 = 1, placed for D/ -> R9C12 = [41], 4 placed for D/
13j. R4C4 = 9, R5C5 = 3, both placed for D\
13k. R1C8 = 6 -> R1C1 = 2, placed for D\

and the rest is naked singles, without using the diagonals.

Solution:
+-------+-------+-------+
| 2 9 5 | 1 8 3 | 4 6 7 |
| 1 6 4 | 7 9 2 | 8 5 3 |
| 8 3 7 | 5 4 6 | 9 1 2 |
+-------+-------+-------+
| 6 8 3 | 9 5 1 | 7 2 4 |
| 5 4 2 | 8 3 7 | 6 9 1 |
| 9 7 1 | 6 2 4 | 5 3 8 |
+-------+-------+-------+
| 3 5 8 | 2 7 9 | 1 4 6 |
| 7 2 6 | 4 1 5 | 3 8 9 |
| 4 1 9 | 3 6 8 | 2 7 5 |
+-------+-------+-------+


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 Post subject: Re: SKX-2 Revisit
PostPosted: Sun Jan 17, 2021 9:39 pm 
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Grand Master
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 280
Location: California, out of London
I'm 2 puzzles behind now! Here's my WT for this one. Haven't had a chance to look at Andrew's yet.
SKX-2 WT:
1. 16(2)r3c7 = {79}
Outies c89 = r19c7 = +6(2) = {15} or {24}
6(2)r6c7 = {24} or {15}
-> Innies c789 = r258c7 = {368}

2. IOD n9 -> r6c79 = r8c7 + 10
Since Max r6c79 = [59] = +14 -> Max r8c7 = 4
-> r8c7 = 3 and r25c7 = {68}
-> r6c79 = +13(2) = [58] or [49]

3. Remaining Innies n69 = r4c79 + r5c7 = +17(3) = [746], [728], or [926]
But that last case puts r6c79 = [58] and 4 in n3 in c89 and 27(4)r6c9 = [8{469}] ...
... which leaves no place for 4 in c7.
-> 16(2)r3c7 = [97]

4. 17(2)n2 = {89}
Since r8c7 = 3 -> r789c6 = {589} or {679}
-> 9 in n5 in r45c4
Also 8 in n1 only in r3c123

5. Innies r1234 = r4c28 = +10(2) (No 5)
IOD n1 -> r23c3 = r4c1 + 5
10(2)r3c3 -> r23c3 cannot be +10(2)
-> r4c1 cannot be 5
Innies c6789 = r456c6 + r5c7 = +18(4)
Since r5c7 from (68) -> r456c6 = +12(3) or +10(3)
-> r4c6 cannot be 5 since that leaves no place for 8 in c6
-> 5 in n5/r4 in r4c45

6. -> 5 not in r3c5
Also since 7 already in r4 -> 6 cannot be in r3c5. (No solution for 19(4)r3c5)

7! Can r25c7 be [68]?
This puts r456c6 = +10(3) (No 8)
-> r789c6 = {589}
-> (56) in n2 in r123c4
-> Innies n2 = r123c4 + r3c5 = +15(4) = {1356}
But no solution for 17(4)r1c4 in that case
-> r25c7 = [86]

8. -> Remaining Outie n3 = r4c9 = 4
-> r6c79 = [58]
-> 27(4)r6c9 = [84{69}]
Also r7c7 = 1
Also Outies c89 = r19c7 = [42]
Also 15(4)n6 = {1239}
Also 22(4)n3 = [4{567}] with 6 in r12c8
Also 22(4)n9 = {2578} with 2 in r9c7 and 8 in r89c8

9! Innies r5 = r5c1289 = +19(4) -> Cannot contain both (89)
Innies r1234 = r4c28 = +10(2)
Since 6 already in r5 -> r4c2 cannot be 1 since that would put remaining cells in 24(4)n4 as [{89}6].
-> r4c8 not 9.
Similarly innies r6789 = r6c28 = +10(2) -> 1 not in r6c2 -> 9 not in r6c8
-> (HS 9 in n6) r5c8 = 9

10. -> (HS 9 in c4) r4c4 = 9
-> (HS 5 in r4) r4c5 = 5
Remaining IOD n2 -> r2c3 = r3c5 -> also goes in r1c789 -> be from (4567)
-> 19(4)r3c5 = [4951] and r2c3 = 4

11. Innies r1234 = r4c28 = +10(2) can only be [82]
-> (Since 10(2)r3c3 cannot be [46]) r4c13 = [63]
-> r3c3 = 7
-> 22(4)n9 = [8275]
Also 1 in n2 in c4 -> r123c4 = {157}
-> r123c6 = {236}
-> r789c6 = {589}
-> r56c6 = [74]
-> r7c5 = 7
-> (IOD n8) r8c3 = 6
-> 7(2)n8 = [16]
etc.


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