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 Post subject: Assassin 283
PostPosted: Thu Jan 30, 2014 9:55 pm 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1043
Location: Sydney, Australia
An easier one than last time. Just. At least the end is easier. All standard Assassin techniques but I found that 5 are quite hidden and took some finding. Very satisfying. SudokuSolver Score, 1.35

Assassin 283

Image
code: select and paste into solver:
3x3::k:2816:2816:6145:6145:8962:8962:8962:8962:8962:2563:3332:6145:6405:4614:3847:8962:3336:3336:2563:3332:6145:6405:4614:3847:8962:777:777:4874:3332:6405:6405:4614:4614:3595:3595:3595:4874:4874:4874:6405:3852:3852:3341:3341:3341:3598:3598:1807:6405:7440:4625:4625:4625:3602:3347:1556:1807:6421:7440:7440:7440:4625:3602:3347:1556:6421:6421:6421:6421:7440:2838:2583:3347:3864:3864:3865:3865:3865:2838:2838:2583:
solution:
+-------+-------+-------+
| 3 8 2 | 6 1 5 | 4 9 7 |
| 4 1 9 | 2 3 7 | 6 5 8 |
| 6 5 7 | 4 9 8 | 3 1 2 |
+-------+-------+-------+
| 9 7 1 | 8 2 4 | 5 6 3 |
| 2 3 5 | 7 6 9 | 8 4 1 |
| 8 6 4 | 3 5 1 | 7 2 9 |
+-------+-------+-------+
| 1 4 3 | 9 7 6 | 2 8 5 |
| 5 2 8 | 1 4 3 | 9 7 6 |
| 7 9 6 | 5 8 2 | 1 3 4 |
+-------+-------+-------+
Cheers
Ed


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 Post subject: Re: Assassin 283
PostPosted: Sat Feb 01, 2014 12:52 am 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1894
Location: Lethbridge, Alberta, Canada
Thanks Ed for another nice Assassin!

I'll guess that my breakthrough step wasn't what you intended, but it was very effective.

Thanks Afmob for your comments and corrections. I've also added an alternative way for my step 11, which I realised while I was getting up the following morning.
Here is my walkthrough for Assassin 283:
Prelims

a) R1C12 = {29/38/47/56}, no 1
b) R23C1 = {19/28/37/46}, no 5
c) R23C6 = {69/78}
d) R2C89 = {49/58/67}, no 1,2,3
e) R3C89 = {12}
f) R5C56 = {69/78}
g) R6C12 = {59/68}
h) R67C3 = {16/25/34}, no 7,8,9
i) R67C9 = {59/68}
j) R78C2 = {15/24}
k) R89C9 = {19/28/37/46}, no 5
l) R9C23 = {69/78}
m) 11(3) cage at R8C8 = {128/137/146/236/245}, no 9

1. Naked pair {12} in R3C89, locked for R3 and N3, clean-up: no 8,9 in R2C1

2. 45 rule on R6789 1 innie R6C4 = 3, clean-up: no 4 in R7C3

3. 45 rule on N3 1 outie R1C56 = 6 = {15/24}
3a. 35(7) cage at R1C5 must contain at least one of 6,7
3b. R2C89 = {49/58} (cannot be {67} which clashes with 35(7) cage), no 6,7 in R2C89

4. 45 rule on N1 2(1+1) outies R1C4 + R4C2 = 13 = {49/58/67}, no 1,2,3

5. 45 rule on N7 2 innies R78C3 = 11 = [29/38/56/65], clean-up: no 6 in R6C3

6. 45 rule on N4 3 innies R4C23 + R6C3 = 12 = {129/147/246} (cannot be {156} which clashes with R6C12), no 5,8, clean-up: no 5,8 in R1C4 (step 4), no 2 in R7C3, no 9 in R8C3 (step 5)
6a. 9 of {129} must be in R4C2, no 9 in R4C3

7. R78C3 (step 5) = [38/56/65]
7a. Killer pair 6,8 in R78C3 and R9C23, locked for N7

8. 45 rule on R9 2 innies R9C19 = 1 outie R8C8 + 4, IOU no 4 in R9C1

9. 45 rule on C123 2 innies R48C3 = 1 outie R1C4 + 3
9a. R1C4 = {4679} -> R48C3 = 7,9,10,12 -> no 6 in R4C3 (because no 1,3,4 in R8C4)

10. 45 rule on C12 1 innie R9C2 = 1 outie R5C3 + 4, R9C2 = {6789} -> R5C3 = {2345}

[I can see a complex clash in N1 to make the following elimination, but prefer this way …]
11. Consider combinations for R4C23 + R6C3 (step 6) = {129/147/246}
R4C23 + R6C3 = {129/147}, 1 locked for C3
or R4C23 + R6C3 = {246} => R4C2 = 6, R1C4 = 7 (step 4) => 24(4) cage at R1C3 cannot be {1689} (only combination for 24(4) containing 1)
-> no 1 in R12C3)
[With hindsight, an alternative way is 24(4) cage at R1C3 cannot be {1689} because {168}9 clashes with R78C3 and {189}6 clashes with R4C23 + R6C3 = 7{14} using step 4 -> no 1 in R12C3.
I’d originally seen the first half of that, but had blocked {189}6 by the combined cage R1C12 + R23C1 which must contain at least one of 8,9 because of the interactions between R1C12 and R23C1.]


12. 1 in R1 only in R1C56 = {15} (step 3), locked for R1, N2 and 35(7) cage at R1C5, no 5 in R23C7, clean-up: no 6 in R1C12
12a. 5 in N3 only in R2C89 = {58}, locked for R2 and N3, clean-up: no 7 in R3C6
12b. 2,8 in R1 only in R1C123, locked for N1
12c. R1C12 cannot contain both of 2,8 -> R1C3 = {28} (hidden killer pair 2,8 for N1)
[Afmob pointed out that this also implies R1C12 = {29/38}.]

13. 24(4) cage at R1C3 = {2679/3489/3678/4578} (cannot be {2589} which contains both of 2,8, other combinations don’t contain 2 or 8)
13a. 24(4) cage = {2679/3678} (cannot be {3489} which clashes with R1C12, cannot be {4578} which clashes with R78C3), no 4,5, clean-up: no 9 in R4C2 (step 4)

14. R3C2 = 5 (hidden single in N1), clean-up: no 9 in R6C1, no 1 in R78C2
14a. Naked pair {24} in R78C2, locked for C2 and N7, clean-up: no 7,9 in R1C1
14b. R3C2 = 5 -> R24C2 = 8 = [17], R1C4 = 6 (step 4), clean-up: no 4 in R1C1, no 9 in R23C6, no 9 in R3C1, no 8 in R9C3
14c. 6 in N3 only in R23C7, locked for C7
14d. R23C1 = {46} (hidden pair in N1), locked for C1, clean-up: no 8 in R6C2

15. R23C6 = [78], clean-up: no 7,8 in R5C5
15a. Naked pair {69} in R5C56, locked for R5 and N5
[With hindsight, I could have done naked quad {6789} in R23C6 and R5C56, CPE no 6,7,8,9 in R6C6 earlier, but it wouldn’t have achieved much.]

16. R3C3 = 7 (hidden single in N1), clean-up: no 8 in R9C2
16a. Naked pair {69} in R9C23, locked for R9 and N7, clean-up: no 1,4 in R8C9
16b. 13(3) cage at R7C1 = {157} (only remaining combination), locked for C1 and N7 -> R6C1 = 8, R6C2 = 6, R78C3 = [38], R6C3 = 4, R5C123 = [235], R4C13 = [91], R1C1 = 3, R1C2 = 8, R12C3 = [29], R9C23 = [96], clean-up: no 6,8 in R7C9, no 2 in R9C9
16c. Naked pair {59} in R67C9, locked for C9 -> R2C89 = [58], clean-up: no 2 in R8C9, no 1 in R9C9

17. 6 in N6 only in 14(3) cage at R4C7 = {356} (only remaining combination) -> R4C7 = 5, R67C9 = [95]

18. R35C9 = [21] (hidden pair in C9), R3C8 = 1
18a. Naked pair {27} in R6C78, locked for R6, N5 and 18(4) cage at R6C6, no 2,7 in R7C8

19. R5C4 = 7 (hidden single in R5)
19a. 25(6) cage at R2C4 = {123478} (only remaining combination) -> R234C4 = [248], R23C1 = [46], R23C5 = [39], R23C7 = [63], R5C56 = [69]

20. 11(3) cage at R8C8 = {128/137/146} (cannot be {236} which clashes with R89C9) -> R9C7 = 1, R9C4 = 5, R9C1 = 7, R78C1 = [15], clean-up: no 3 in R8C9
20a. R9C4 = 5 -> R9C56 = 10 = [82]

21. R78C4 = [91], R8C3 = 8 -> R8C56 = 7 = [43], R78C2 = [42]

22. R7C56 = [76], R7C78 = [28], R8C7 = 9, R6C5 = 5 (cage sum)

and the rest is naked singles.

Rating Comment:
I'll rate my walkthrough for A283 at Easy 1.5. I used a short forcing chain.
I'll stick with my rating, even though the alternative step 11 might possibly be Hard 1.25.


Last edited by Andrew on Mon Feb 03, 2014 3:16 am, edited 2 times in total.

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 Post subject: Re: Assassin 283
PostPosted: Sat Feb 01, 2014 9:33 am 
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Grand Master
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Joined: Mon Apr 21, 2008 9:44 am
Posts: 310
Location: MV, Germany
Thank you for this Assassin, Ed! It took me some time to find the most important move (step 1d) even though it wasn't really hidden nor technical difficult.

A283 Walkthrough:
1. N6789 !
a) Innie R6789 = R6C4 = 3
b) Outies N6789 = 10(3) = 1{27/45} -> 1 locked for R6
c) Innies+Outies R5: 2 = R4C1 - R5C4 -> R4C1 <> 1,2,5; R5C4 <> 8,9
d) ! Hidden Killer pair (13) in 13(3) @ N6 since 14(3) can only have one of them

2. C1234 !
a) Outies N1 = 13(1+1) <> 1,2,3
b) Outies C12 = 11(2): R5C3 = (2345)
c) Innies N4 = 12(3) = {129/147/246} <> 5,8 since {156} blocked by Killer pair (56) of 14(2)
d) 7(2): R7C3 <> 1,2,4
e) Innies N7 = 11(2) = [38]/{56}
f) Killer pair (68) locked in 15(2) + Innies N7 for N6
g) 10(2) <> {37} since it's a Killer pair of 13(3) @ C1
h) Innies N4 = 12(3): R4C3 <> 9 since R4C3 >= 4
i) ! 3 locked in 19(4) @ N4 = 3{259/268/457} <> 1 because 13{69/78} blocked by Killer pair (13) of 13(3) @ R5
j) 1 locked in Innies N4 @ N4 = 12(3) = 1{29/47} for C3

3. R1234
a) 3(2) = {12} locked for R3+N3
b) 1 locked in Outies N1 @ R1 = 6(2) = {15} locked for R1+N2+35(7)
c) 5 locked in 13(2) @ N3 = {58} locked for R2+N3
d) Outies N1 = 13(1+1): R1C4 <> 7,8
e) 2,8 locked in R1C123 @ R1 for N1

4. C1234
a) Killer pair (14) locked in 10(2) + 13(3) for C1
b) Killer pair (14) locked in 13(3) + 6(2) for C2
c) 11(2) = {29/38}
d) 13(3) @ C2: R23C2 <> 4,7,9 since R4C2 = (479)
e) 7 locked in 24(4) @ N1 for C3
f) Innies N4 = 12(3) = 1{29/47} -> R4C2 = (79)
g) 13(3) @ C2 = 1{39/57} -> R2C2 = 1
h) 10(2) = {46} locked for C1+N1
i) 15(2): R9C2 <> 8
j) 8 locked in R89C3 @ N7 for C3

5. R1234
a) 8 locked in 11(2) @ N1 = {38} locked for R1+N1
b) R3C2 = 5 -> R4C2 = 7
c) 24(4) = {2679} -> R1C3 = 2, R1C4 = 6; 9 locked for C3
d) 15(2) = {78} -> R2C6 = 7, R3C6 = 8
e) 25(6) = {123478} -> R3C4 = 4, R2C4 = 2, R4C3 = 1, R4C4 = 8, R5C4 = 7
f) 18(4) = {2349} since R23C5 = (39) -> 2,4 locked for R4+N5; 3,9 locked for C5
g) 14(3) = {356} locked for R4+N7

6. R789
a) R6C3 = 4 -> R7C3 = 3
b) 15(2) = {69} -> R9C3 = 6, R9C2 = 9
c) Hidden Single: R8C3 = 8 @ N7
d) 15(3) = 5[73/82] because R9C4 = (15) -> R9C4 = 5
e) 25(5) = {13489} because R78C4 = (19) -> R8C5 = 4, R8C6 = 3; 1,9 locked for N8
f) R9C6 = 2 -> R9C5 = 8, R7C6 = 6, R7C5 = 7
g) 29(5) = {25679} -> R6C5 = 5; 2,9 locked for C7+N9
h) 14(2) = {59} -> R6C9 = 9, R7C9 = 5

7. Rest is singles.

Rating:
Hard 1.0. I used a Hidden Killer pair.

Going through Andrew's walkthrough I noticed that I could have used his step 12c to replace my steps 4a-4c.


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 Post subject: Re: Assassin 283
PostPosted: Tue Feb 04, 2014 9:43 am 
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Ok, this is going back to the old days! Here is a harder version of the original that closes up Afmob's really nice solution (I checked and SS does it the same way) and leaves open the way I solved the original. Andrew's lovely breakthrough step should still work but this time doesn't completely break it open. When Ruud used to make an Assassin that felt a bit easy, I used to enjoy trying to make them harder - exactly the opposite of HATMAN's new Messy One which makes A283 easier. Nice to have HATMAN for the next Assassin.

Assassin 283 V1.5 (SudokuSolver score 1.50)

Image
code:paste into solver:
3x3::k:2816:2816:6145:6145:8962:8962:8962:8962:8962:2563:3332:6145:6405:4614:3847:8962:3336:3336:2563:3332:6145:6405:4614:3847:8962:777:777:4874:3332:6405:6405:4614:4614:6931:6931:6931:4874:4874:4874:6405:3852:3852:6931:6931:6931:3598:3598:1807:6405:7440:4625:4625:4625:3602:7191:1556:1807:6421:7440:7440:7440:4625:3602:7191:1556:6421:6421:6421:6421:7440:2838:2573:7191:7191:7191:3851:3851:3851:2838:2838:2573:

Same solution as the original.

Cheers
Ed


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PostPosted: Tue Feb 04, 2014 6:36 pm 
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Thanks for this version, Ed! It was a bit harder to solve than V1 but it still can be solved without using chains with one advanced move.

A283 V1.5 Walkthrough:
1. R123
a) 3(2) = {12} locked for R3+N3
b) Outies N3 = 6(2) = {15/24}
c) 13(2) <> {67} since it's a Killer pair of 35(7)
d) Outies = N1 = 13(1+1) <> 1,2,3

2. N46789 !
a) Innies R6789 = R6C4 = 3
b) Outies N6789 = 10(3) = 1{27/45} -> 1 locked for R6
c) Innies N4 = 12(3) = {129/147/246} <> 5,8 since {156} blocked by Killer pair (56) of 14(2)
d) 7(2): R7C3 <> 1,2,4
e) Innies N7 = 11(2) = [38]/{56}
f) ! Innies N47 = 16(2+1): R4C2 <> 4 since [475] blocked by Killer pair (45) of 6(2)

3. R123 !
a) Outies N1 = 13(1+1): R1C4 <> 5,8,9
b) 3,6,7 locked in 35(7) @ N3 = 34567{19/28}
c) ! Killer quad (4567) locked in R1C4 + 35(7) for R1 -> 11(2) can have at most one of (4567) -> 11(2) = {29/38}
d) 24(4) <> 1 since {1689} blocked by Killer pair (89) of 11(2)
e) 1 locked in Outies N1 @ R1 = 6(2) = {15} locked for R1+N2+35(7)
f) 5 locked in 13(2) @ N3 = {58} locked for R2+N3
g) Hidden Killer pair (28) locked in 11(2) + R1C3 @ R1 for N1 -> R1C3 = (28)

4. C123
a) Outies C12 = 11(2) <> 1
b) 1 locked in Innies N4 @ C3 = 12(3) = 1{29/47} for N4
c) 13(3) @ N1 = 1{39/57} since R4C2 = (79) -> R2C2 = 1; R3C2 = (35)
d) Outies N1 = 13(1+1): R1C4 <> 7
e) 24(4) = 7{269/368/458} because R1C3 = (28), R1C4 = (46) and {3489} blocked by Killer pair (89) of 11(2) -> 7 locked for C3+N1
f) 10(2) = {46} locked for C1+N1
g) Outies C12 = 11(2) <> 4
h) 1,4 locked in Innies N4 @ C3 = 12(3) = {147} for N4 -> R4C2 = 7
i) Cage sum: R3C2 = 5
j) Outie N1 = R1C4 = 6

5. N2678
a) 15(2) = {78} -> R2C6 = 7, R3C6 = 8
b) 25(6) = {123478} -> R3C4 = 4, R2C4 = 2, R4C3 = 1, R4C4 = 8, R5C4 = 7
c) 18(4) = {2349} since R23C5 = (39) -> 3,9 locked for C5; R4C56 = (24) locked for R4+N5
d) R6C3 = 4 -> R7C4 = 3
e) Innie N7 = R8C3 = 8
f) 28(5) = 679{15/24} -> 6 locked for R9
g) 6 locked in R4C789 @ R4 for N6
h) 25(5) = 18{259/349/457} <> 6 since R78C4 = (159) -> 1 locked for N8

6. N89
a) 15(3) = 5[73/82] because R9C4 = (59) and {249} blocked by Killer pair (24) of 25(5) -> R9C4 = 5
b) 25(5) = {13489} -> R8C6 = 3, R8C5 = 4
c) R9C6 = 2 -> R9C5 = 8
d) Hidden Single: R7C5 = 7 @ N8, R7C6 = 6 @ N8
e) 29(5) = {25679} -> R6C5 = 5; 2,9 locked for C7+N9
f) 14(2) = {59} -> R6C9 = 9, R7C9 = 5

7. Rest is singles.

Rating:
1.25. I used a Killer quad.


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 Post subject: Re: Assassin 283
PostPosted: Fri Feb 07, 2014 11:41 pm 
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Thanks Ed for this version.

Thanks Afmob for pointing out that I forgot to change the cages in N6 when I changed my diagram from A283 to A283 V1.5. It's something that can easily happen when one sets up one's own diagrams, especially when changing them from an original puzzle to a variant with combined cages. I've reworded my step 17 to allow for that. Combining the cages in N6 made no difference to this puzzle (that applies to both Afmob's solving path and my one); the V1.5 was made by combining cages in N7.


Ed wrote:
Here is a harder version of the original that closes up Afmob's really nice solution (I checked and SS does it the same way) and leaves open the way I solved the original. Andrew's lovely breakthrough step should still work but this time doesn't completely break it open.
My way wasn't really any harder than my solving path for Assassin 283; I used many of the same steps.

Here is my walkthrough for Assassin 283 V1.5:
Prelims

a) R1C12 = {29/38/47/56}, no 1
b) R23C1 = {19/28/37/46}, no 5
c) R23C6 = {69/78}
d) R2C89 = {49/58/67}, no 1,2,3
e) R3C89 = {12}
f) R5C56 = {69/78}
g) R6C12 = {59/68}
h) R67C3 = {16/25/34}, no 7,8,9
i) R67C9 = {59/68}
j) R78C2 = {15/24}
k) R89C9 = {19/28/37/46}, no 5
l) 11(3) cage at R8C8 = {128/137/146/236/245}, no 9

1. Naked pair {12} in R3C89, locked for R3 and N3, clean-up: no 8,9 in R2C1

2. 45 rule on R6789 1 innie R6C4 = 3, clean-up: no 4 in R7C3

3. 45 rule on N3 1 outie R1C56 = 6 = {15/24}
3a. 3 in N2 only in R23C5, locked for C5

4. 35(7) cage at R1C5 must contain at least one of 6,7
4a. R2C89 = {49/58} (cannot be {67} which clashes with 35(7) cage), no 6,7 in R2C89

5. 45 rule on N1 2(1+1) outies R1C4 + R4C2 = 13 = {49/58/67}, no 1,2,3

6. 45 rule on N7 2 innies R78C3 = 11 = [29/38/56/65], clean-up: no 6 in R6C3

7. 45 rule on N4 3 innies R4C23 + R6C3 = 12 = {129/147/246} (cannot be {156} which clashes with R6C12), no 5,8, clean-up: no 5,8 in R1C4 (step 5), no 2 in R7C3, no 9 in R8C3 (step 6)
7a. 9 of {129} must be in R4C2, no 9 in R4C3

8. 45 rule on C12 2 outies R59C3 = 11 = {29/38/47/56}, no 1

[Using the hindsight way for the step I used in A283 …]
9. R4C23 + R6C3 (step 7) = {129/147/246}
9a. 24(4) cage at R1C3 cannot be {1689} because {168}9 clashes with R78C3 and {189}6 clashes with R4C23 + R6C3 = 7{14} using step 5 -> no 1 in R12C3
9b. 1 in C3 only in R46C3, locked for N4
9c. R4C23 + R6C3 = {129/147}, no 6, clean-up: no 7 in R1C4 (step 5)

10. 1 in R1 only in R1C56 = {15} (step 3), locked for R1, N2 and 35(7) cage at R1C5, no 5 in R23C7, clean-up: no 6 in R1C12
10a. 5 in N3 only in R2C89 = {58}, locked for R2 and N3, clean-up: no 7 in R3C6
10b. Killer pair 5,8 in R2C9 and R67C9, locked for C9, clean-up: no 2 in R89C9

11. 2,8 in R1 only in R1C123, locked for N1
11a. Hidden killer pair 2,8 in R1C12 and R1C3 for R1, R1C12 cannot contain both of 2,8 -> R1C12 = {29/38}, R1C3 = {28}
11b. 7 in R1 only in R1C789, locked for N3

12. 24(4) cage at R1C3 = {2679/3489/3678/4578} (cannot be {2589} which contains both of 2,8, other combinations don’t contain 2 or 8)
12a. 24(4) cage = {2679/3678} (cannot be {3489} which clashes with R1C12, cannot be {4578} which clashes with R78C3), no 4,5, 7 locked for C3 and N1, clean-up: no 3 in R23C1, no 9 in R4C2 (step 5), no 4 in R59C3 (step 8)

13. R4C23 + R6C3 (step 9c) = {147} (only remaining combination) -> R4C2 = 7, R46C3 = {14}, locked for N4, R1C4 = 6 (step 5), clean-up: no 9 in R23C6, no 5 in R7C3, no 6 in R8C3 (step 6)
13a. 6 in N3 only in R23C7, locked for C7

14. R4C2 = 7, R3C2 = 5 (hidden single in N1) -> R2C2 = 1 (cage sum), clean-up: no 9 in R3C1, no 9 in R6C1
14a. Naked pair {46} in R23C1, locked for C1, clean-up: no 8 in R6C2
14b. Naked pair {24} in R78C2, locked for C2 and N7, clean-up: no 9 in R1C1

15. R23C6 = [78], clean-up: no 7,8 in R5C5
15a. Naked pair {69} in R5C56, locked for R5 and N5, clean-up: no 5 in R9C3 (step 8)

16. 25(6) cage at R2C4 = {123478} (only remaining combination) -> R23C4 = [24], R4C3 = 1, R45C4 = [87], R6C3 = 4, R7C3 = 3, R8C3 = 8 (step 6), R1C3 = 2, R5C3 = 5, R6C1 = 8, R6C2 = 6, R5C12 = [23], R4C1 = 9

17. R5C789 = {148}, 3,6 in R4 only in R4C789 -> 27(6) cage at R4C7 = {356}{148} (only possible combination), 5 locked for R4 and N6 -> R6C9 = 9, R7C9 = 5, clean-up: no 1 in R89C9

18. R35C9 = [21] (hidden pair in C9), R3C8 = 1
18a. Naked pair {27} in R6C78, locked for R6 and 18(4) cage at R6C6, no 2,7 in R7C8
18b. R6C78 = {27} = 9 -> R6C6 + R7C8 = 9 = [18/54]
18c. Naked pair {48} in R57C8, locked for C8

19. 11(3) cage at R8C8 = {137} (only remaining combination, cannot be {128/146} because 1,4,8 only in R9C7, cannot be {236} which clashes with R89C9) -> R9C7 = 1, R89C8 = {37}, locked for C8 and N9
19a. Naked pair {46} in R89C9, locked for C9 and N9 -> R7C8 = 8, R6C6 = 1 (step 18b)

20. Deleted. Afmob pointed out that I’d reached naked singles after step 19.

and the rest is naked singles.

Rating Comment:
Rating Comment. I'll rate my walkthrough for A283 V1.5 at Hard 1.25, because I used the hindsight version of my original breakthrough step for A283.


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 Post subject: Re: Assassin 283
PostPosted: Sun Feb 09, 2014 4:42 am 
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Grand Master
Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 1043
Location: Sydney, Australia
Thanks for your WTs. I much prefer the v1.5 and wish I'd found that one for the original. Worked out fine in the end just put me behind in keeping up with this forum. Get so little time...
Andrew wrote:
Combining the cages in N6 made no difference to this puzzle (that applies to both Afmob's solving path and my one)
Don't think this is true for Afmob's way. His key steps 1d ->2i in the original don't work for the v1.5 and this is what forced him to use a different solution for the v1.5. And a really neat way he found too! Really had to think hard about his 3e. I still find that description of "killer quad" very confusing but once I see the eliminations it is really cool! Also, loved finding what Afmob has put as 2f. This is why I wanted a v1.5 puzzle that would allow that elimination to be more influential.

Here's the way I worked in the key area.

Alternate cracker for A183v1.5
(also works for a183)
2 more steps:
Candidates at end of Afmob's v1.5 WT step 3a; paste into A283v1.5 in SudokuSolver:
.-------------------------------.-------------------------------.-------------------------------.
| 23456789 23456789 123456789 | 467 1245 1245 | 3456789 3456789 3456789 |
| 12346789 123456789 123456789 | 12456789 123456789 6789 | 3456789 4589 4589 |
| 346789 3456789 3456789 | 456789 3456789 6789 | 3456789 12 12 |
:-------------------------------+-------------------------------+-------------------------------:
| 123456789 679 124679 | 12456789 12456789 12456789 | 123456789 123456789 123456789 |
| 123456789 123456789 123456789 | 12456789 6789 6789 | 123456789 123456789 123456789 |
| 5689 5689 124 | 3 12457 12457 | 2456789 2456789 5689 |
:-------------------------------+-------------------------------+-------------------------------:
| 123456789 1245 356 | 12456789 123456789 123456789 | 123456789 123456789 5689 |
| 123456789 1245 568 | 12456789 123456789 123456789 | 123456789 12345678 12346789 |
| 123456789 123456789 123456789 | 12456789 123456789 123456789 | 12345678 12345678 12346789 |
'-------------------------------.-------------------------------.-------------------------------'

End of Afmob's 3a then...
4. 1 in r1 in h6(2)r1c56 = {15} or in 24(4)r1c3 = {1689}: ie must have 5 or 6
4a. -> {56} blocked from 11(2)n1
4b. 11(2) = {29/28/47}(no 5,6)

5. 24(4)r1c3: {1689} as {189}6 only forces 11(2) = {47} but {4789} in n1 blocks all combos for 10(2)
5a. -> {1689} blocked from 24(4): only combo with 1 -> no 1 in r12c3

Much easier from there as Afmob and Andrew's Wts show.
Cheers
Ed


Attachments:
A283 v1.5 Afmob WT end step 3a.ssv [107.19 KiB]
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