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Ed
 Post subject: Assassin 425
PostPosted: Sun Jan 15, 2023 5:09 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Attachment:
a425x.png
a425x.png [ 90 KiB | Viewed 2367 times ]
note: x puzzle so 1-9 cannot repeat on either diagonal.

A425
Couldn't solve this one after a really good crack at it. But it has such an interesting JSudoku log and 'only' gets 1.55 so worth getting out there. Look forward to seeing what I missed.
spoiler alert: JSudoku log with my bespoke solver order:
Techniques used:
60 Naked Single
16 Hidden Single
4 Single Innies & Outies
6 Unique Pair
3 Naked Pair
2 Hidden Pair
1 Complex Hidden Single
3 Unique Triplet
14 Intersection
1 Naked Triplet
5 Odd Pairs
7 Odd Triplets
2 Double Innies & Outies
1 Mandatory Inclusion
5 Odd Quads
1 Complex Intersection
1 Triple Innies & Outies
3 Double Outies minus Innies
4 Complex Naked Pair
1 Complex Hidden Pair
2 Conflicting Pair
2 Quadruple Innies & Outies
1 Triple Outies minus Innies
2 Pointing Pair
2 Odd Combinations
1 Swordfish
2 Locked Cages
1 Complex Naked Triplet
1 Conflicting Partial Pair
1 Turbot Fish
3 Multiple Innies & Outies
1 Multiple Overlaps
4 Cages Grouping
26 Conflicting Partial Triplet
1 Grouped Turbot Fish
3 Grouped XY-Chain up to 3 links
triple click code:
3x3:d:k:8960:8960:8960:8960:769:7170:7170:6147:3076:8960:8960:5125:4614:769:7170:6147:6147:3076:8960:5125:5125:4614:5895:7170:6147:6147:3076:5125:5125:4872:4614:5895:6665:5130:5130:4107:5132:4872:4872:5895:5895:6665:5130:5130:4107:5132:5132:5132:2317:2317:6665:2062:2062:2319:5136:5136:5136:5137:5137:6665:3346:3346:2319:5136:5139:5139:5137:3092:6665:3346:4117:2070:5139:5139:1559:1559:3092:6665:4117:4117:2070:
solution: 
+-------+-------+-------+
| 7 5 6 | 4 1 8 | 9 2 3 |
| 9 3 4 | 7 2 6 | 8 1 5 |
| 1 8 2 | 3 9 5 | 6 7 4 |
+-------+-------+-------+
| 5 1 9 | 8 3 4 | 2 6 7 |
| 2 7 3 | 6 5 1 | 4 8 9 |
| 4 6 8 | 2 7 9 | 5 3 1 |
+-------+-------+-------+
| 3 4 7 | 9 6 2 | 1 5 8 |
| 6 9 1 | 5 8 3 | 7 4 2 |
| 8 2 5 | 1 4 7 | 3 9 6 |
+-------+-------+-------+
Cheers
Ed
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wellbeback
 Post subject: Re: Assassin 425
PostPosted: Thu Jan 19, 2023 7:22 pm 
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 282
Location: California, out of London
I've never seen a puzzle with such a long easy start before hitting such a high wall! Nice one Ed!
Here's how I solved it...
Assassin 425 WT:
1. The start is straightforward so won't detail it here, just list the position.
r1c4 = 4
r2c1 = 9
20(5)n14 = [{4(28|37)}{15}]
18(3)n25 = [{37}8]
3(2)n2 = {12}
23(4)n25 = [93{56}]
c6 = [{568}419{237}]
r1c7 = 9
12(3)n3 = [6{24}] or {345} with r1c9 from (35)
19(3)n4 = {379} with 3 in r5c23
20(4)n4 = [2{468}]
9(2)n5 = [27]
20(4)n6 = [{26}{48}]
16(2)n6 = {79}
8(2)n6 = {35}
9(2)n69 = [18]
9 in n7 in r7c3 or r8c2
6(2)n78 = [51]
20(3)n8 = {569} with 9 in r78c4
12(2)n8 = {48}
7 in n9 in 13(3)n9
r9c8 = 9

2. 6 in c9 only in r189c9
Either 6 in r19c9 which puts r5c45 = [65] and r7c5 = 6
or 6 in r8c9 which puts 6 in r7c45
Either way 6 in r7 in r7c45

3! For (68) in n7 one of:
a) 6 in r8c1, 8 in 20(4)r8c2
b) 8 in r8c1, 6 in 20(4)r8c2
c) Both (68) in 20(4)r8c2

For (a) possibilities for r7c123 are:
[{14}9]
{347}
Cannot be {239} since that puts r7c456 = [{56}7] which leaves no solution for 13(3)n9

For (b) possibilities for r7c123 are:
[129] only
Cannot be {147} since that leaves no solution for 13(3)n9
Cannot be {237} since one of (237) in r7c6

For (c)
20(4)r8c2 = {6824} -> 20(4)r7c1 = {1379}

Note that in all cases each 20(4)n7 consists of two pairs of +10(2)!

4. Innies n1 cannot be {347} since that puts r1c89 = [73], r9c1 = 7 (HS 7 on D/) and 3 in r78c1 which contradicts conclusion from Step 3
-> Innies n1 = {248} with r3c3 from (24)

5. Since (19) are in same cage in n7, and 9 in D/ in r7c3 or r8c2 -> 1 not in r7c3 or r8c2
-> 1 in D/ in r2c8 or r3c7

6. If 8(2)n9 is {26} that puts 16(3)n9 = [{34}9] and 4 in n7 in r7 which puts r8c1 = 6. (By Step 3)
-> 8(2)n9 cannot be [62]
-> 6 in c9 in r19c9
-> r5c45 = [65]
-> r7c5 = 6 and r78c4 = {59}

7. Either 8(2)n9 = [26], 16(3)n9 = [{34}9], 13(3)n9 = {157}
Or 8(2)n9 = [53], 16(3)n9 = [169], 13(3)n9 = {247}
-> 6 in n9 in r9c79
For (24) on D\ one is in r3c3 and the other must be in n9
But trying r3c3 = 4 puts r7c7 = 2 and r9c7 = 6 contradicting r4c78 = {26}
-> r3c3 = 2

8. Also 4 in D\ in r7c7 or r8c8
-> Either 8(2)n9 = [26], 16(3)n9 = [439], 13(3)n9 = {157}
Or 8(2)n9 = [53], 16(3)n9 = [169], 13(3)n9 = [427] (cannot be [472] since 2 would already be in r4c7)
I.e., r9c79 = {36}

9. Since 2 already in c13 and D/ -> 2 in n7 in r79c2
Trying 2 in r7c2 puts r8c1 = 8 (By Step 3), 12(2)n8 = [48] which puts 13(3)n9 = [427] (Two 2's on r7)
-> r9c2 = 2
-> r9c6 = 7
Also (Since 4 already on D/) (HS 4 in r9) 12(2)n8 = [84]
-> r9c1 = 8

10! 6 in c9 in r19c9
4 in n7 either in r7c12 which puts r8c1 = 6 (Step 3)
Or 4 in r8c3 which puts r8c2 = 6 (Step 3)
But the latter puts 13(3)n9 = [427] and 16(3)n9 = [169] which leaves no place for 6 in c9
-> 4 in r7c12 and r8c1 = 6

11. Finishing up
Also r8c8 = 4 which puts 16(3)n9 = [439], 8(2)n9 = [26], and 13(3)n9 = {157}
Also 12(3)n3 = [3{45}
Also r78c6 = [23]
-> 3 in n7 in r7
-> 20(4)r7c1 = [{347}6]
-> r8c23 = [91]
-> 13(3)n9 = [157]
Also r5c78 = [48]
-> r2c7 = 8
-> r3c2 = 8 and r2c3 = 4
-> r6c123 = [468]
-> r7c123 = [347]
Also r2c8 = 1
-> 3(2)n2 = [12]
-> r3c1 = 1
etc.
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Andrew
 Post subject: Re: Assassin 425
PostPosted: Tue Jan 31, 2023 6:04 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Thanks Ed for probably the most challenging Assassin we've had for a while. As wellbeback said an easier than usual start, then one had to 'change gear' and think much harder. Finished just in time before the next puzzle.

Here's how I solved Assassin 425:
Prelims

a) R12C5 = {12}
b) R45C9 = {79}
c) R6C45 = {18/27/36/45}, no 9
d) R6C78 = {17/26/35}, no 4,8,9
e) R67C9 = {18/27/36/45}, no 9
f) R89C5 = {39/48/57}, no 1,2,6
g) R89C9 = {17/26/35}, no 4,8,9
h) R9C34 = {15/24}
i) 19(3) cage at R4C3 = {289/379/469/478/568}, no 1
j) 20(3) cage at R7C4 = {389/479/569/578}, no 1,2
k) 28(4) cage at R1C6 = {4789/5689}, no 1,2,3

1a. Naked pair {12} in R12C5, locked for C5 and N2, clean-up: no 7,8 in R6C4
1b. Naked pair {79} in R45C9, locked for C9 and N6
1c. 45 rule on N6 1 innie R6C9 = 1 -> R7C9 = 8, clean-up: no 8 in R6C5
1d. 45 rule on N3 1 innie R1C7 = 9
1e. 45 rule on N7 1 innie R9C3 = 5 -> R9C4 = 1, clean-up: no 7 in R8C5, no 3 in R8C9
1f. 45 rule on N8 3 remaining innies R789C6 = 12, must contain 2 for N8 = {237/246}, no 5,8,9
1g. 45 rule on C6 using R123C6 = 19, 3 remaining innies R456C6 = 14, must contain 1 for C6 = {149/158} (cannot be {167} which clashes with R789C6), no 2,3,6,7
[Oops! I was overlooking that R123C6 must contain both of 1,9 for C6! Fortunately this didn’t make much difference and was OK after step 1n.]
1h. R6C45 = [27] (cannot be {36} which clashes with R6C78, cannot be {45} which clashes with R456C6), 2 placed for D/, clean-up: no 6 in R6C78, no 5 in R8C5
1i. Naked pair {35} in R6C78, locked for R6 and N6
1j. 4 in C9 only in 12(3) cage at R1C9, locked for N3
1k. 2,3 in C6 only in R789C6 = {237}, 3,7 locked for N8, 7 locked for C6, clean-up: no 9 in R89C5
1l. Naked pair {48} in R89C5, locked for C5 and N8
1m. R123C6 = 19 = {568} (only remaining combination), locked for N2, 5,8 locked for C6
1n. Naked triple {149} in R456C6, 4,9 locked for N5
1o. 45 rule on N58 1 outie R3C5 = 1 innie R4C4 + 1 -> R3C5 = 9, R4C4 = 8, placed for D\
1p. 6,8 in R6 only in R6C123 -> 20(4) cage at R5C1 = 2{468} (cannot be {1568} because 1,5 only in R5C1), locked for N4
1q. R6C6 = 9 (hidden single in R6), placed for D\
1r. 45 rule on N4 2 innies R4C12 = 6 = {15}, locked for R4 and 20(5) cage at R2C3, 5 locked for N4
1s. R45C6 = [41], 4 placed for D/
1t. Naked pair {26} in R4C78, 6 locked for R4 and N6 -> R4C5 = 3
1u. R4C4 = 8 -> R23C4 = 10 = {37}, locked for N2 -> R1C4 = 4
1v. 20(5) cage at R2C3 contains 4 for N1 = {248}{15}/{347}{15}, no 6,9
1w. R2C1 = 9 (hidden single in N1)
1x. 9 on D/ only in R7C3 + R8C2, locked for N7
1y. R9C8 = 9 (hidden single in R9) -> R8C8 + R9C7 = 7 = [16/34/43] (cannot be [52] which clashes with R89C9), no 2,5,7, no 6 in R8C8
1z. Killer pair 3,6 in R8C8 + R9C7 and R89C9, locked for N9
1aa. 12(3) cage at R1C9 = {246/345}
1ab. 6 of {246} must be in R1C9 -> no 6 in R23C9
1ac. Naked pair {56} in R57C5, CPE no 6 in R7C3 using D/, no 5 in R7C7 using D\

[17 placements after step 1 and a lot of cages reduced to one or two combinations, which may be the reason why the SS score is only 1.55.]

2a. 20(5) cage at R2C3 (step 1v) = {248}{15}/{347}{15}
2b. R2C3 + R3C23 = {248}/{37}4 (R23C3 cannot be {37} which clashes with R45C3 ALS block, R3C23 cannot be {37} which clashes with R3C4) -> no 3,7 in R3C3
2c. 13(3) cage at R7C7 = {157/247}
2d. Consider combinations for R8C8 + R9C7 = [16]/{34}
R8C8 + R9C7 = [16] => R4C7 = 2
or R8C8 + R9C7 = {34}, 4 locked for N9 => 13(3) cage = {157}
-> no 2 in R78C7
2e. 2 of {247} must be in R7C8 -> no 4 in R7C8

3a. 20(4) cages in N7 = {1289/1379/1469/1478/2369/2378/2468/3467}
3b. 20(4) cage at R7C1 = {1289/1379/1469/2369/3467} (cannot be {147}8 which clashes with R7C7, cannot be {237}8 which clashes with R7C6, cannot be {2468} because R7C3 only contains 1,3,7,9)
3c. 9 of {1289/1379/1469} must be in R7C3 -> no 1 in R7C3
3d. Naked triple {379} in R457C3, locked for C3
3e. R2C3 + R3C23 (step 2b) = {248} (only remaining combination), 2,8 locked for N1
3f. 20(4) cage at R8C2 = {1289/1478/2378/2468/3467} (cannot be {1379/1469/2369} which clash with 20(4) cage at R7C1)
3g. 9 of {1289} must be in R8C2 -> no 1 in R8C2
3h. 1 on D/ only in R2C8 + R3C7, locked for N3
3i. Variable hidden killer pair 1,4 in R7C12 and R7C78 for R7, R7C78 cannot contain both of 1,4 (step 2c) -> R7C12 must contain at least one of 1,4
3j. 20(4) cage at R7C1 = {1289/1379/1469/3467} (cannot be {2369} which doesn’t contain 1 or 4)
3k. 20(4) cage at R8C2 = {1289/2378/2468/3467} (cannot be {1478} which clashes with 20(4) cage at R7C1)
3l. Consider combinations for 13(3) cage at R7C7 (step 2c) = {157/247}
13(3) cage = {157} => 4 in R7 only in R7C12 => 20(4) cage at R7C1 = {1469/3467}
or 13(3) cage = {247} => R7C8 = 2 => R89C9 = {53}, R1C9 = 6, placed for D/ => 20(4) cage at R8C2 cannot be {2468} (because 2,4,6 then only in R8C3 + R9C2) => 20(4) cage at R7C1 cannot be {1379}
-> 20(4) cage at R7C1 = {1289/1469/3467}
3m. 8 of {1289} must be in R8C1, 6 of {1469} must be in R8C1 (R7C123 cannot be {169/469} which clash with R7C45, ALS block) -> no 1 in R8C1
3n. 20(4) cage at R8C2 = {1289/2378/3467} (cannot be {2468} which clashes with 20(4) cage at R7C1)
3o. Consider combinations for 20(4) cage at R7C1 = {1289/1469/3467}
20(4) cage at R7C1 = {1289}, 2 locked for R7 => 13(3) cage = {157}
or 20(4) cage at R7C1 = {1469} = {149}6, 4 locked for R7
or 20(4) cage at R7C1 = {3467} => 1 in R7 only in R7C78 => 13(3) cage = {157}
-> no 4 in R7C7, 13(3) cage = {157}/[724]
3p. 4 on D\ only in R3C3 + R8C8, CPE no 4 in R8C3

[At last I can use triangles.]
4a. 13(3) cage at R7C7 = {157}/[724], R8C8 + R9C7 = [16]{34}, R89C9 = {26}/[53]
4b. Consider placements for R1C1
R1C1 = 1, placed for D\ => R8C8 + R9C9 = {34}, 4 locked for N9
or R1C1 = 3, placed for D\ => R89C9 = {26}, 2 locked for N9
or R1C1 = {56} => naked pair {56} in R1C1 + R5C5 => R1C9 = 3 (CPE using diagonals) => R89C9 = {26}, 2 locked for N9
or R1C1 = 7, placed for D\ => R7C7 = 1
-> 13(3) cage = {157}, 1,5 locked for N9, clean-up: no 3 in R9C9
4c. Naked pair {26} in R89C9, locked for C9, 6 locked for N9
4d. Naked triple {345} in R123C9, 3,5 locked for N3
4e. 4 in R7 only in R7C12, locked for N7
4f. 4 in R7C12 -> 20(4) cage at R7C1 (step 3n) = {1469/3467}, no 2,8, 6 locked for N7
4g. 5 on D/ only in R1C9 + R5C5, CPE no 5 in R1C1 using diagonals
4h. 20(4) cage at R7C1 = {3467} (cannot be {1469} = {149}6 because of killer ALS clash with R7C45 = {56} and R7C78 = {157}), locked for N7

5a. 20(4) cage at R8C2 = {1289} = [9182], 8 placed for D/
5b. R9C9 = 6, placed for D\ -> R5C5 = 5, placed for both diagonals -> R1C9 = 3, placed for D/ -> R7C7 = 7, placed for D/ -> R7C7 = 1, place for D\
5c. R9C5 = 4 -> R9C7 = 3, R8C8 = 4, placed for D\

and the rest is naked singles, without using the diagonals.
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Andrew
 Post subject: Re: Assassin 425
PostPosted: Wed Feb 01, 2023 7:35 pm 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
I've now had time to go through wellbeback's WT. His step 2, which I really ought to have spotted, led to the powerful step 3, the result of which was used to great effect in later steps. So while we had to work in the same area, our reasons and order of eliminations in that area were quite different, maybe more different than usual.
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