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 Post subject: Assassin 311
PostPosted: Thu Mar 05, 2015 10:05 pm 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
Surpising easy once you know the way. Surprisingly hard when you don't. But at least you can nibble away rather than being completely stumped from the start. I found it very tough first time. It gets a 1.30. Later, tried changing it to get the ending harder but wasn't happy with it as the V1. Will post that version in a few days as a V1.5.

Assassin 311

Image
code: paste into solver:
3x3::k:6665:6665:8211:8211:8211:8449:8449:1546:1546:6665:4352:8211:4614:8211:8449:8449:3592:3592:6665:4352:8211:4614:4614:4614:8449:8449:4629:4352:4352:4113:4614:2829:7694:1295:1295:4629:4352:5392:4113:5899:2829:7694:7694:7694:4629:5392:5392:4113:5899:7694:7694:4372:4372:4372:3852:3852:5899:5899:6658:6658:6658:6658:6658:3852:3852:3845:3845:6151:6151:6658:5138:5138:2820:2820:3845:6151:6151:1795:1795:5138:5138:
solution:
+-------+-------+-------+
| 2 9 4 | 6 7 8 | 3 5 1 |
| 7 3 5 | 1 9 4 | 2 8 6 |
| 8 6 1 | 3 5 2 | 9 7 4 |
+-------+-------+-------+
| 5 2 3 | 7 8 6 | 1 4 9 |
| 1 8 6 | 4 3 9 | 7 2 5 |
| 9 4 7 | 2 1 5 | 6 3 8 |
+-------+-------+-------+
| 3 7 9 | 8 4 1 | 5 6 2 |
| 4 1 2 | 5 6 7 | 8 9 3 |
| 6 5 8 | 9 2 3 | 4 1 7 |
+-------+-------+-------+
Cheers
Ed


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 Post subject: Re: Assassin 311
PostPosted: Fri Mar 06, 2015 7:12 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
Thanks Ed! I found the way in quite quickly.

Walkthrough:
1. Innies r789 -> r7c34 = +17 = {89}
-> r56c4 = +6 = {15} | {24}
Innies n89 -> r78c4 = +13 = [94] | [85]
-> (45) locked for c4.

2. Innies n4 -> r4c12+r5c1 = +8 = {125} | {134}
-> r23c2 = +9 = {36} | {27}
-> r123c3 = +10 and must include a 1.
-> r1c45+r2c5 = +22 = {589} | {679}

3. 18/5(n2) must include (12) and at least one of (45)
-> r3c56 = {(1|2)(4|5)}
-> 18/5(n2) has exactly one of (45)
-> 18/5(n2) = {123((48)|(57))} (no 69)

4. Innies - Outies n1234 -> r4c4 = r3c9 + 3
-> r4c4 min 4.
But r4c4 cannot be from (456) and is max 8.
-> r4c4 from (78)
-> r3c9 from (45)
4 or 5 also in 6/2(n3)
-> 14/2(n3) = {68}
-> r123c7+r3c8 = {379(1|2)}
-> r12c6 = [8(5|4)]
-> r1c45+r2c5 = {679}
-> r123c3 = {145}

5. HS 8 in n1/r3 -> r3c1 = 8
-> HS 6 in n1/r3 -> r3c2 = 6
-> r2c2 = 3
-> 26/4(n1) = [{279}8]
Also r3c4 = 3
-> r1c7 = 3
Also r3c78 = {79}
Also 17/5@n14 = [36{125}]
-> 21(3)n4 = {489}
-> 16/3@n4 = {367}

6! HS 2 in r3 -> 2 in r3c56
-> r2c4 = 1
-> r56c4 = {24}
-> r78c4 = [85]
-> r7c3 = 9
-> r89c3 = {28}
Also r4c4 = 7
-> r3c9 = 4
-> 6/2@n3 = {15}

etc.


Rating:
Rating 0.9 - No complicated moves.


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PostPosted: Fri Mar 06, 2015 9:31 pm 
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Grand Master
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Joined: Mon Apr 21, 2008 9:44 am
Posts: 310
Location: MV, Germany
Thanks for the new Assassin! I'm quite surprised by SudokuSolver's rating as this Killer doesn't need any complicated moves.

A311 Walkthrough:
1. R789 !
a) Innies R789 = 17(2) = {89} locked for R7+23(4)
b) Innies N89 = 13(2) = [85/94]
c) Outies R789 = 6(2) = {15/24}
d) ! Killer pair (45) locked in Outies R789 + R8C4 for C4

2. N23 !
a) ! Innies+Outies R123+N4: 3 = R4C4 - R3C9 -> R4C4 = (678), R3C9 = (345)
b) 18(5) = 12{348/357/456} -> 1,2 locked for N2; R2C4+R3C456 <> 6,7,8 since R4C4 = (678)
c) 7 locked in 33(6) @ N3 for 33(6)
d) 7 locked in Outies N14 @ N2 = 22(3) = {679} locked for N2+32(6)
e) 8 locked in 33(6) @ N2 for 33(6)
f) 8 locked in 14(2) @ N3 = {68} locked for R2+N3

3. N14
a) Innies N4 = 8(3) = 1{25/34} -> 1 locked for N4+17(5)
b) 1 locked in Innies N14 @ N1 = 10(3) = {145} locked for C3+N1
c) Hidden Single: R1C6 = 8 @ N2, R3C1 = 8 @ R3, R3C2 = 6 @ R3
d) Outie N4 = R2C2 = 3
e) Hidden Single: R1C7 = 3 @ R1

4. R123+N58
a) Killer pair (12) locked in R2C4+23(4) for C4
b) R3C4 = 3
c) Killer pair (45) locked in 18(5) + R3C9 for R3
d) R3C3 = 1
e) 18(5) = 123{48/57} -> R2C4 = 1; 2 locked for R3
f) 23(4) = {2489} -> 2,4 locked for C4+N5
g) Innies N89 = 13(2) = [85] -> R8C4 = 5, R7C4 = 8
h) 18(5) = {12357} -> R4C4 = 7; 5 locked for R3+N2
i) Outies N9 = 8(3) = {134} locked for N8
j) 7(2) = [16/34]

5. N367
a) 18(3) = {459} since R3C9 = 4 and {468} blocked by R2C9 = (68) -> R3C9 = 4; 5,9 locked for C9+N6
b) 33(6) = {234789} -> R2C6 = 4, R2C7 = 2
c) 5(2) = {14} locked for R4+N6 since R4C7 = (14)
d) 17(3) = 8{27/36} -> 8 locked for R6+N6
e) 30(6) = {125679} -> R5C8 = 2, R5C7 = 7
f) Hidden Single: R7C7 = 5 @ C7
g) 5 locked in 11(2) @ N7 = {56} -> R9C2 = 5, R9C1 = 6

6. Rest is singles.

Rating:
1.0. I used some Killer pairs.


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 Post subject: Re: Assassin 311
PostPosted: Sat Mar 07, 2015 7:03 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
Thanks Ed for your latest Assassin.

Ed wrote:
Surpising easy once you know the way. Surprisingly hard when you don't.
Unlike wellbeback and Afmob, I didn't spot the 'easy' way in but it still wasn't difficult.

Here is my walkthrough for Assassin 311:
Thanks Afmob for pointing out that my original step 10a was incorrect. I've re-worked that area, using a simpler step which I ought to have spotted when I originally solved this puzzle.

Prelims

a) R1C89 = {15/24}
b) R2C89 = {59/68}
c) R45C5 = {29/38/47/56}, no 1
d) R4C78 = {14/23}
e) R9C12 = {29/38/47/56}, no 1
f) R9C67 = {16/25/34}, no 7,8,9
g) 21(3) cage at R5C2 = {489/579/678}, no 1,2,3
h) 26(4) cage at R1C1 = {2789/3689/4589/4679/5678}, no 1
i) 17(5) cage at R2C2 = {12347/12356}, no 8,9
j) 18(5) cage at R2C4 = {12348/12357/12456}, no 9

[17(5) cage at R2C2 and 21(3) cage at R5C2 almost form a combined cage, but there’s the possibility that R23C2 and R6C1 may contain the same value, since these cells don’t ‘see’ each other.]

1. 45 rule on N9 3 outies R7C56 + R9C7 = 8 = {125/134}, 1 locked for N8, clean-up: no 1 in R9C7

2. 45 rule on R789 2 innies R7C34 = 17 = {89}, locked for R7
2a. 45 rule on R789 2 outies R56C4 = 6 = {15/24}
2b. 45 rule on N89 2 innies R78C4 = 13 = [85/94]
2c. Killer pair 4,5 in R56C4 and R8C4, locked for C4
2d. Killer pair 4,5 in R8C4 and R7C56 + R9C7, locked for N8

3. 45 rule on N7 3 innies R789C3 = 19 = {289/379/469/478/568}, no 1

4. 45 rule on N14 3 innies R123C3 = 10 = {127/136/145/235}, no 8,9
4a. 45 rule on N14 3 outies R1C45 + R2C5 = 22 = {589/679}, 9 locked for N2

5. 18(5) cage at R2C4 = {12348/12357} (cannot be {12456} which clashes with R56C4), no 6

6. 45 rule on N4 3 innies R4C12 + R5C1 = 8 = {125/134}, 1 locked for N4
6a. R4C12 + R5C1 = {125/134} -> R2C23 = {27/36}
6b. R4C12 + R5C1 must be {15}2/{25}1/{14}3 (other combinations clash with R4C78), no 3 in R4C12, no 4,5 in R5C1
6c. Killer pair 1,2 in R4C12 and R4C78, locked for R4, clean-up: no 9 in R5C5
6d. 18(5) cage at R2C4 (step 5) = {12348/12357}, 1,2 locked for N2

7. 45 rule on N3 2 outies R12C6 = 1 innie R3C9 + 8
7a. Max R12C6 = 13 (R12C6 cannot be {68/78} which clash with R1C45 + R2C5) -> max R3C9 = 5
7b. 7 in N3 only in R123C7 + R3C8, locked for 33(6) cage at R1C6, no 7 in R12C6

8. 33(6) cage at R1C6 contains 7 = {135789/234789/245679/345678} (cannot be {126789} which clashes with R1C89)
8a. 33(6) cage = {135789/234789/245679} (cannot be {345678} which would have 4,5 in R12C6 because of R2C89 = {59}, R1C89 = {24}, clashing with 18(5) cage at R2C4), 9 locked for N3, clean-up: no 5 in R2C89
8b. Naked pair {68} in R2C89, locked for R2 and N3, clean-up: no 3 in R3C2 (step 6a)
8c. Killer pair 1,2 in R1C89 and 33(6) cage, locked for N3
8d. 6,8 of 33(6) cage only in R1C6 -> R1C6 = {68}
8e. Killer pair 6,8 in R1C6 and R1C45 + R2C5, locked for N2
8f. 6,8 in N2 only in R1C456, locked for R1

9. R3C1 = 8 (hidden single in N1), clean-up: no 3 in R9C2
9a. 26(4) cage at R1C1 = {2789/4589}, no 3

[Afmob pointed out that my original step 10a, using hidden killer triple 3,4,5 in N2, was incorrect because of the possibility that R1C45 + R2C5 could contain 5, so I’ll re-work this part.]
R12C6 = R3C9 + 8 (step 7), R3C9 = {345} -> R12C6 = 11,12,13 = [83/84/85] (cannot be [65] which clashes with R1C45 + R2C5) -> R1C6 = 8, R1C45 + R2C5 (step 4a) = {679}, locked for N2 and R123C3, no 6,7 in R123C3.


10. 18(5) cage at R2C4 (step 5) = {12348/12357}
10a. 7,8 only in R4C4 -> R4C4 = {78}
10b. Deleted
10c. 18(5) cage = {12348/12357}, 3 locked for N2

11. Deleted
11a. R12C6 = R3C9 + 8 (step 7), R1C6 = 8 -> R2C6 = R3C9 -> R3C9 = {45}

12. 1 in N1 only in R123C3 (step 4) = {145} (only remaining combination), locked for C3 and N1
12a. R23C2 = [36] (hidden pair in N1), locked for 17(5) cage at R2C2 -> R4C12 + R5C1 (step 6) = {125}, locked for N4, 5 also locked for R4, clean-up: no 6 in R5C5, no 5 in R9C1
12b. 2 in C3 only in R789C3 (step 3) = {289} (only possible combination), locked for C3 and N7, clean-up: no 3 in R9C1
12c. Naked triple {367} in 16(3) cage at R4C3, locked for N4
12d. Deleted
[So 17(5) cage at R2C2 and 21(3) cage at R5C2 do happen to form a combined cage, but this didn’t necessarily follow from the starting position.]

13. 18(3) cage at R3C9 = {459/567} (cannot be {189/279/369/378} because R3C9 only contains 4,5, cannot be {468} which clashes with R2C9), no 1,2,3,8, 5 locked for C9, clean-up: no 1 in R1C8

14. R1C7 = 3 (hidden single in R1), clean-up: no 2 in R4C8, no 4 in R9C6

15. Killer pair 1,2 in R2C4 and R56C4, locked for C4 -> R3C4 = 3

16. 2 in R7 only in 26(6) cage at R7C5 = {123479/123569/123578/124568}
16a. 8,9 only in R8C7 -> R8C7 = {89}
16b. 26(6) cage = {123479/123569/123578/124568}, 1 locked for R7

17. Naked quad {2679} in R1C1245, locked for R1, clean-up: no 4 in R1C89
17a. R1C89 = [51], R1C3 = 4
17b. R3C9 = 4 -> 18(3) cage at R3C9 (step 13) = {459} -> R45C9 = [95], clean-up: no 6 in R4C5, no 2 in R5C5, no 1 in R6C4 (step 2a)
17c. R2C6 = 4 (hidden single in N2)
17d. R2C3 = 5 (hidden single in R2), R3C3 = 1
17e. Naked pair {25} in R3C56, locked for R3 and N2 -> R2C4 = 1, clean-up: no 5 in R6C4 (step 2a)
17f. Naked pair {24} in R56C4, locked for C4 and N5 -> R8C4 = 5, R89C3 = 10 = {28}, locked for C3 -> R7C34 = [98], R4C4 = 7, clean-up: no 2 in R9C7

18. Naked pair {38} in R45C5, locked for C5 and N5
18a. R4C6 = 6, R5C6 + R6C56 = {159} = 15 -> R5C78 = 9 = {27}, locked for R5 and N6, clean-up: no 3 in R4C8
18b. Naked pair {14} in R4C78, locked for R4 and N6
18c. R2C7 = 2 (hidden single in N3) -> R5C78 = [72], R3C78 = [97], R8C7 = 8, R6C7 = 6, clean-up: no 1 in R9C6

19. R9C8 = 1 (hidden single in R9)
19a. Naked pair {45} in R79C7, locked for C7 and N9
19b. R8C8 = 9 (hidden single in N9)
19c. R89C8 = [91] = 10 -> R89C9 = 10 = {37}

20. 6,7 in N8 only in 24(4) cage at R8C5 = {2679}, locked for N8 -> R9C6 = 3, R9C7 = 4

and the rest is naked singles.

Rating Comment:
I'll rate my walkthrough for A311 at either Easy 1.25 or 1.25; it's hard to know what rating to give my step 8a. With the 'easy' step, which I missed, this puzzle is definitely no more than 1.0.

I'm not sure what the maximum number of nonets SudokuSolver is programmed to look at but I'll guess three, although it's possible that it might be programmed to look at a 2x2 block. If it was programmed to look at N1234 then arguably it would also have to be programmed to look at N1235.


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 Post subject: Re: Assassin 311
PostPosted: Mon Mar 09, 2015 1:44 am 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
Thanks guys for your WTs. You all did way better than me. I checked the SudokuSolver solution and it misses Afmob's step 2a (end), as did I. Giving it that step when it is first available gives 30 less steps which is a lot for a weighted step count method of scoring.

Here's the V1.5. Please don't feel the need to post a complete WT. Am just interested in how you guys get started. I used an ugly chain but hoping there is a nice way. The end is very straightforward so boring. I originally thought it was a nice hard ending but it wasn't on the second solve. Missed a hidden single first time. Hence, it never made it as the V1. It gets a 1.60

A311 V1.5

Image
code: paste into solver:
3x3::k:6665:6665:8195:8195:8195:8449:8449:1546:1546:6665:4352:8195:4614:8195:8449:8449:3592:3592:6665:4352:8195:4614:4614:4614:8449:8449:4620:4352:4352:4113:4614:2829:7694:1295:1295:4620:4352:5392:4113:5899:2829:7694:7694:7694:4620:5392:5392:4113:5899:7694:7694:4357:4357:4357:0000:0000:5899:5899:6658:6658:6658:6658:6658:0000:0000:0000:0000:6151:6151:6658:5138:5138:2820:2820:0000:6151:6151:0000:0000:5138:5138:
. Same solution.

Cheers
Ed


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PostPosted: Tue Mar 10, 2015 7:19 pm 
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Grand Master
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Joined: Mon Apr 21, 2008 9:44 am
Posts: 310
Location: MV, Germany
Since I didn't have to post the full walkthrough, it made optimizing the start so much faster! But nonetheless, I took me two days to solve this Killer since I tried to avoid chainy moves but in the end I had to use one (step 2e).

A311 V1.5 Walkthrough start:
1. N123456
a) Innies N123456 = 6(2) = {15/24}
b) Innies N4 = 8(3) = 1{25/34} -> 1 locked for N4+17(5)
c) 1 locked in Innies N14 @ N4 = 10(3) = 1{27/36/45} <> 8,9 for C3+32(6)
d) Outies N14 = 22(3) = 9{58/67} -> 9 locked for N2
e) Innies+Outies N1234: 3 = R4C4 - R3C9 -> R4C4 <> 1,2,3; R3C9 = (12345)
f) 18(5) = 12{348/357/456} -> 1,2 locked for N2
g) 7 locked in 33(6) @ N3 for 33(6)
h) Using step 1h: 33(6) = 7{13589/23489/24569/34568} since {126789} blocked by Killer pair (12) of 6(2)
i) R2C6 <> 6,8 since it sees all 6,8 of N3

2. N1234 !
a) Innies N1234 = 15(4+1) -> R2C4 <> 6,7,8
b) Innies+Outies N3: 8 = R12C6 - R3C9: R1C6 <> 3 since R2C6 <= 5
c) Innies+Outies N3: 8 = R12C6 - R3C9: R1C6 <> 4 since [45]-1 blocked by Killer pair (14) of 6(2)
d) Innies+Outies N3: 8 = R12C6 - R3C9 = {36/45}-1 / {46}-2 / {38}-3 / {48}-4 / {58}-5 since {56}-3 blocked by Killer pair (56) of Outies N14
e) ! 33(6) = 378{159/249/456} since {245679} <> 3 forces R3C9 = 3 (HS @ N3) which forces R2C6 = 3 @ 33(6) (step 2d)
f) Hidden Killer pair (69) in 33(6) @ N3 since 14(2) can only have one of them -> R1C6 <> 6 since 33(6) can only have one of (69)
g) Outies N14 = 22(3) = {679} locked for N2+32(6) since {589} blocked by R1C6 = (58)
h) Innies N14 = 10(3) = {145} locked for C3+N1
i) 18(5): R4C4 <> 5 since 6,7 only possible there

3. N1234 !
a) ! Innies N1234 = 15(4+1): R2C4 <> 2,3 since 145{2/3} blocked by R3C3 = (145)
b) ! 18(5): R4C4 <> 4 since 1{238}4 blocked by Killer pair (14) of Innies N123456
c) 18(5) can only have one of (678) -> R3C456 <> 8
d) Hidden Single: R1C6 = 8 @ N2, R3C1 = 8 @ R3

4. Puzzle is cracked. But you might want to use:

5. R789
a) Outies N123456 = 17(2) = {89} locked for R7
b) Grouped X-Wing: 1 locked in R78C12 + 26(6) for R78
c) Using Innies C12: 1 locked in Outies R1234567 @ R8 = 13(3) = 1{39/48/57} <> 2,6

Rating:
1.5. I used a small contradiction chain.


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 Post subject: Re: Assassin 311
PostPosted: Wed Mar 11, 2015 11:54 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
Here's mine for V1.5. Not too different from V1.0. Thanks Ed.

Hidden Text:
1. Outies n123456 -> r7c34 = {89}
-> r56c4 = {15} | {24}

2. Innies n4 -> r4c12+r5c1 = +8 = {125} | {134}
-> r23c2 = {36} | {27}
-> r123c3 = +10 and includes a 1
(89) in n1 in 26(4)
r1c45+r2c5 = +22 = {589} | {679}

3. 18(5)n2 = (12) + two of (345) + one of (678) (no 9)
Innies - Outies n1234 -> r4c4 = r3c9 + 3
-> r3c9 from (12345), r4c4 from (45678).

a) Try r56c4 = {15} - Puts 1 in r3c56 and not 5 in r4c4 -> not (12) in r3c9
b) Try r56c4 = {24} - Puts 2 in r3c56 and not 4 in r4c4 -> not (12) in r3c9

-> r3c9 from (345), r4c4 from (678)
-> one of (12) in 33(6) in n3
-> 9 must be in 33(6) in n3
-> 14(2)n3 = {68}

4. Try 3 not in 33(6)
Puts r3c9 = 3, r2c4 = 3, r4c4 = 6
But 18(5) cannot contain both 3 and 6
-> 3 in 33(6)
-> 8 in 33(6)
-> r1c6 = 8
-> HS 8 in r3/n1 -> r1c3 = 8

5. Also 6 in r1c45
-> H22(3)n2 = {679}
-> r123c3 = {145}
-> HS 6 in r3/n1 -> r1c2 = 6
-> r2c2 = 3
-> NS 3 in r1 -> r1c7 = 3
-> 26(4)n1 = {2789} with (7|9) in r2c1, 2 in r1c12 and 8 in r3c1
-> 6(2)n3 = {15}
-> r3c9 = 4, r2c6 = 4, r1c3 = 4, r4c4 = 7

Also NP (79) in r3/n3 -> r3c78 = {79}
-> r2c7 = 2

6. NP r2c34 = {15}
-> r56c4 = {24}

Also H8(3)n4 = {125}
Also r45c9 = {59}
-> 5 in r6 in r6c56
-> NP 11(2)n5 = {38}

etc.

Later on an important step is:
2 is in r89c3
9 is in r89c7
-> 9 not in 28(6)
-> Remaining two values not in 28(6) = +10
Since 2 in r7 in 26(6) -> 8 in 26(6) -> r8c7 = 8.


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 Post subject: Re: Assassin 311
PostPosted: Mon Mar 16, 2015 3:38 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 Ed for this variant. I was busy with other things for several days, then took some time to find my breakthrough which felt like a 'human solvable' step. It's in the same area as one of wellbeback's steps, but looked at the other way round.

Here is my start for Assassin 311 V1.5:
Prelims

a) R1C89 = {15/24}
b) R2C89 = {59/68}
c) R45C5 = {29/38/47/56}, no 1
d) R4C78 = {14/23}
e) R9C12 = {29/38/47/56}, no 1
f) 21(3) cage at R5C2 = {489/579/678}, no 1,2,3
g) 26(4) cage at R1C1 = {2789/3689/4589/4679/5678}, no 1
h) 17(5) cage at R2C2 = {12347/12356}, no 8,9
i) 18(5) cage at R2C4 = {12348/12357/12456}, no 9

1. 45 rule on R123456 2 outies R7C34 = 17 = {89}, locked for R7
1a. 45 rule on R123456 2 innies R56C4 = 6 = {15/24}

2. 45 rule on N4 3 innies R4C12 + R5C1 = 8 = {125/134}, 1 locked for N4
2a. R4C12 + R5C1 = {125/134} -> R2C23 = {27/36}
2b. R4C12 + R5C1 must be {15}2/{25}1/{14}3 (other combinations clash with R4C78), no 3 in R4C12, no 4,5 in R5C1
2c. Killer pair 1,2 in R4C12 and R4C78, locked for R4, clean-up: no 9 in R5C5
2d. 18(5) cage at R2C4 = {12348/12357/12456}, 1,2 locked for N2

3. 1 in N1 only in R123C3, locked for C3
3a. 45 rule on N14 3 innies R123C3 = 10 = {127/136/145}, no 8,9
3b. 45 rule on N14 3 outies R1C45 + R2C5 = 22 = {589/679}, 9 locked for N2

[The step I missed when solving Assassin 311]
4. 45 rule on N1234 1 outie R4C4 = 1 innie R3C9 + 3 -> R3C9 = {12345}, R4C4 = {45678}
4a. 7 in N3 only in R123C7 + R3C8, locked for 33(6) cage at R1C6, no 7 in R12C6

[It took me a long time to find this next step]
5. R3C9 + R4C4 (step 4) = [36/47/58] (cannot be [14/25] because R24C4 + R3C9 = [141/252], where R2C4 is hidden single in N2, clash with R56C4), no 1,2 in R3C9, no 4,5 in R4C4

6. 18(5) cage at R2C4 = {12348/12357/12456}
6a. R4C4 ={678} -> no 6,7,8 in R2C4 + R3C456

7. 7 in N2 only in R1C45 + R2C5 (step 3a) = {679}, locked for N2 and 32(6) cage at R1C3, no 6,7 in R123C3
7a. R123C3 (step 3) = {145}, locked for C3 and N1

8. 8 in N2 only in R12C6, locked for 33(6) cage at R1C6
8a. 8 in N3 only in R2C89 = {68}, locked for R2 and N3, clean-up: no 3 in R3C2 (step 2a)
8b. R1C6 = 8 (hidden single in N2) -> R3C1 = 8 (hidden single in N1), clean-up: no 3 in R9C2

9. 16(3) cage at R4C3 = {268/367}, no 9, 6 locked for N4 and C3
9a. 9 in C3 only in R789C3, locked for N7, clean-up: no 2 in R9C12

10. Naked quint {12345} in R3C34569, locked for R3
10a. R3C2 = 6 (hidden single in R3), R2C2 = 3 (step 2a), clean-up: no 5 in R9C1

11. R4C12 + R5C1 = 8 = {125} (only remaining combination), locked for N4, 5 also locked for R4
11a. 16(3) cage at R4C3 (step 9) = {367}, locked for N4 and C3

12. Naked quad {2679} in R1C1245, locked for R1, clean-up: no 4 in R1C89

and the rest is a lot easier.

Rating Comment:
Possibly a little low, but I'll rate my start for A311 V1.5 at 1.25.

As I commented to Ed, it was only necessary to remove two cages from the A311 grid. When removing cages, at least two cages need to be removed as removing one would just leave a hidden cage. Removing the 15(4) cage at R7C1 makes no difference; it's still there as a hidden cage in C12.


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 Post subject: Re: Assassin 311
PostPosted: Mon Mar 16, 2015 6:56 am 
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Grand Master
Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
Great solution Andrew! Easily the easiest to follow. Really enjoyed the other WTs too. wellbeback used the same really neat key step then Afmob found a really tough to see 4+1 innies. All infinitely better than my way. Won't bore you with the details. Thanks guys!


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