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 Post subject: Assassin 379
PostPosted: Mon Jul 01, 2019 8:15 am 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1043
Location: Sydney, Australia
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a379.JPG
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An x-puzzle so 1-9 cannot repeat on either diagonal.

Assassin 379
The sudokuSolver score of 1.20 and the JSudoku log suggests an easier one. A fairly routine start but took me a long time to find the 'easy' solution after that. I'm sure I missed something. Enjoy hunting!

I have a 1.60 version I couldn't solve if anyone wants to try. Same solution as the original and the start is the same.

code:
3x3:d:k:1792:6401:6401:6401:6401:4354:2307:8964:8964:1792:5381:6150:4354:4354:4354:2307:8964:7943:5381:5381:6150:6150:6150:6150:8964:8964:7943:5381:2824:5381:4873:4873:4873:8964:8964:7943:4362:2824:3339:4873:6932:6932:2572:2572:7943:4362:2824:3339:7950:6932:6932:4111:4111:7943:7950:7950:7950:7950:6932:3088:4111:4111:5137:5394:5651:7950:5651:6932:3088:3088:5137:5137:5394:5394:5651:5651:4365:4365:4365:4365:4365:
solution:
Code:
+-------+-------+-------+
| 4 6 9 | 2 8 5 | 7 3 1 |
| 3 8 5 | 4 7 1 | 2 6 9 |
| 2 7 1 | 3 6 9 | 4 5 8 |
+-------+-------+-------+
| 1 2 3 | 6 4 8 | 9 7 5 |
| 9 4 7 | 1 5 3 | 8 2 6 |
| 8 5 6 | 9 2 7 | 1 4 3 |
+-------+-------+-------+
| 6 1 2 | 5 9 4 | 3 8 7 |
| 5 3 8 | 7 1 2 | 6 9 4 |
| 7 9 4 | 8 3 6 | 5 1 2 |
+-------+-------+-------+
Cheers
Ed


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 Post subject: Re: Assassin 379
PostPosted: Tue Jul 02, 2019 8:18 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1043
Location: Sydney, Australia
puzzle pic a379v1.60:
Attachment:
a379v1.60.JPG
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Andrew wants the harder one also! Look forward to seeing what I missed. I'll put it under cover so no chance of a spoiler.
code:
3x3:d:k:1792:6401:6401:6401:6401:4354:2307:8964:8964:1792:5381:6150:4354:4354:4354:2307:8964:7943:5381:5381:6150:6150:6150:6150:8964:8964:7943:5381:2824:5381:4873:4873:4873:8964:8964:7943:4362:2824:3339:4873:6156:3341:3341:3341:7943:4362:2824:3339:7950:6156:6156:4111:4111:7943:7950:7950:7950:7950:6156:3088:4111:4111:5137:5394:5651:7950:5651:6156:3088:3088:5137:5137:5394:5394:5651:5651:4372:4372:4372:4372:4372:
Same solution as the V1.
Cheers
Ed


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 Post subject: Re: Assassin 379
PostPosted: Thu Jul 04, 2019 3:46 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1894
Location: Lethbridge, Alberta, Canada
Thanks Ed for your latest Assassin.

I used a short forcing chain. SudokuSolver probably used more combination analysis to get the low score of 1.20.

Here is my walkthrough for Assassin 379:
Prelims

a) R12C1 = {16/25/34}, no 7,8,9
b) R12C7 = {18/27/36/45}, no 1
c) R56C1 = {89}
d) R56C3 = {49/58/67}, no 1,2,3
e) R5C78 = {19/28/37/46}, no 5
f) 11(3) cage at R4C2 = {128/137/146/236/245}, no 9
g) 20(3) cage at R7C9 = {389/479/569/578}, no 1,2
h) 21(3) cage at R8C1 = {489/579/678}, no 1,2,3
i) 17(5) cage at R9C5 = {12347/12356}, no 8,9

Steps resulting from Prelims
1a. Naked pair {89} in R56C1, locked for C1 and N4, clean-up: no 4,5 in R56C3
1b. Naked pair {67} in R56C3, locked for C3 and N4
1c. 11(3) cage at R4C2 = {245} (only remaining combination), locked for C2 and N4
1d. Naked pair {13} in R4C13, locked for R4 and 21(5) cage at R2C2
1e. 17(5) cage at R9C5 = {12347/12356}, 1,2,3 locked for R9
1f. 8,9 in R9 only in R9C234 -> no 8,9 in R8C2 (CPE)

2a. 21(3) cage at R8C1 = {579/678} (cannot be {489} because 8,9 only in R9C2), no 4
2b. 8,9 only in R9C2 -> R9C2 = {89}
2c. R89C1 = {57/67}, 7 locked for C1 and N7

3a. R4C13 = {13} = 4 -> R2C2 + R3C12 = 17 = {269/278/467} (cannot be {458} because 4,5 only in R3C1)
3b. 2,4 only in R3C1 -> R3C1 = {24}
[At this stage I saw combined cage R12C1 + R2C2 + R3C12 must contain 2 in R123C1 but decided to leave that for later if necessary, which it wasn’t.]

4. 45 rule on N7 4 outies R6789C4 = 29 = {5789}, locked for C4

5. 45 rule on C9 1 outie R8C8 = 2 innies R19C9 + 6
5a. Min R19C9 = 3 -> R8C8 = 9, R19C9 = {12}, 9 placed for D\, clean-up: no 1 in R5C7
5b. Naked pair {12} in R19C9, locked for C9
5c. Naked pair {12} in R19C9, CPE no 1,2 in R1C1 + R5C5 using D\, clean-up: no 5,6 in R2C1

6a. 45 rule on N36 2 innies R6C78 = 5 = {14/23}
6b. 45 rule on N36 2 outies R7C78 = 11 = {38/47/56}, no 1,2

7. Hidden killer pair 8,9 in R9C2 and R9C34 for R9, R9C2 = {89} -> R9C34 must contain one of 8,9
7a. 22(4) cage at R8C2 = {1489/1579/3478} (cannot be {4567} which doesn’t contain 8 or 9, cannot be {1678/3469/3568} because 1,3,6 only in R8C2), no 6
7b. 4 of {1489/3478} must be in R9C3 -> no 8 in R9C3
7c. 45 rule on R9 4 innies R9C1234 = 28 = {4789/5689}
7d. Consider placement for 7 in N7
R8C1 = 7 => R9C12 = 14 = [59] (cannot be [68] because R9C34 cannot be {59} without 7 in R8C4)
or R9C1 = 7
-> R9C1 = {57}
7e. R9C1234 = {4789} (only remaining combination), no 5 -> R9C1 = 7, placed for D/, R9C3 = 4, R9C24 = {89}
7f. 22(4) cage contains 4 = {1489/3478}, no 5
7g. 5 in C4 only in R67C4, locked for 31(6) cage at R6C6
7h. R8C1 = 5 (hidden single in N7) -> R9C2 = 9 (cage sum), R89C4 = [78] -> R8C2 = 3 (cage sum), placed for D/, clean-up: no 2 in R2C1
7i. Killer pair 1,3 in R12C1 and R4C1, locked for C1
7j. R2C2 + R3C12 (step 3a) = {278} (only remaining combination, cannot be {467} which clashes with R12C1) -> R3C1 = 2, R23C2 = {78}, locked for N1, 8 locked for C2 -> R7C12 = [61], clean-up: no 1 in R2C1, no 5 in R7C78 (step 6b)
7k. Naked pair {34} in R12C1, 3 locked for C1 and N1
7l. R1C2 = 6 -> R1C345 = 19 = {289/379} (cannot be {478} because no 4,7,8 in R1C3) -> R1C3 = 9, R1C45 = [28/37], clean-up: no 3 in R2C7
7m. Naked pair {15} in R23C3, locked for 24(5) cage at R2C3
7n. R23C3 = {15} = 6 -> R3C456 = 18 = {369/468} (cannot be {378} which clashes with R3C2), no 7, 6 locked for R3 and N2

8. 1 on D/ only in R1C9 + R2C8 + R3C7, locked for N3, clean-up: no 8 in R12C7
8a. 35(7) cage at R1C8 contains 1 = {1235789/1245689/1345679}, 9 locked for C7, clean-up: no 1 in R5C8
8b. 1 in N6 only in R6C78 = {14} (step 6a), locked for R6, 4 locked for N6 and 16(4) cage at R6C7, clean-up: no 6 in R5C78, no 7 in R7C78 (step 6b)
8c. Naked pair {38} in R7C78, locked for R7 and N9 -> R78C3 = [28], 2 placed for D/
8d. R1C9 = 1 -> R9C9 = 2, placed for D\
8e. R7C9 = 7 (hidden single in R7 and N9) -> R8C9 = 4 (cage sum)
8f. R3C3 = 1 (hidden single in R3 and on D\) -> R2C3 = 5, clean-up: no 4 in R1C7
8g. 5 on D\ only in R5C5 + R6C6, locked for N5 and 27(6) cage at R5C5
8h. R6C4 = 9, placed for D/, R7C4 = 5

9. 19(4) cage at R4C4 = {1468/2368/2467} (cannot be {1378} because R4C4 only contains 4,6), 6 locked for N5
9a. R4C4 = 6 (hidden single on D\)
9b. R2C8 = 6 (hidden single on D/), clean-up: no 3 in R1C7
9c. 35(7) cage at R1C8 contains 1,6 = {1245689/1345679}, 4 locked for N3, clean-up: no 5 in R1C7
9d. Naked pair {27} in R12C7, locked for C7 and N3
9e. Naked pair {38} in R57C7, locked for C7
9f. Naked pair {38} in R57C7, CPE no 8 in R5C5
9g. R45C8 = {27} (hidden pair in C8 and N6)
9h. R4C6 = 8 (hidden single on D/)
9i. R4C46 = [68] = 14 -> R4C5 + R5C4 = 5 = [23/41]
9j. R4C8 = 7 (hidden single in R4) -> 35(7) cage = {1345679}, 3 locked for C8 and N3
9k. R7C8 = 8 -> R7C7 = 3, placed for D\
9l. R1C1 = 4, placed for D\, R2C1 = 3
9m. R5C5 = 5, placed for D/, R6C6 = 7, placed for D\
9n. R3C7 = 4, R3C4 = 3, R3C8 = 5 -> R9C8 = 1
9o. R8C7 = 6 -> R78C6 = 6 = [42]

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

Rating Comment:
I'll rate my walkthrough for A379 at Easy 1.5; I used a short forcing chain.


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 Post subject: Re: Assassin 379
PostPosted: Sat Jul 06, 2019 3:57 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1894
Location: Lethbridge, Alberta, Canada
Thanks Ed for posting the harder version. Definitely more challenging to find some of the key steps; however I found it more enjoyable and a better version of this puzzle. I loved the way that I was able to use the diagonals and CPEs.

Thanks Ed for several detail corrections to my walkthrough.

Here is my walkthrough for Assassin 379 V1.60:
I’ve used my starting steps for A379 up to step 8a, with a few detail omissions.

Prelims

a) R12C1 = {16/25/34}, no 7,8,9
b) R12C7 = {18/27/36/45}, no 1
c) R56C1 = {89}
d) R56C3 = {49/58/67}, no 1,2,3
e) Not applicable for V1.60
f) 11(3) cage at R4C2 = {128/137/146/236/245}, no 9
g) 20(3) cage at R7C9 = {389/479/569/578}, no 1,2
h) 21(3) cage at R8C1 = {489/579/678}, no 1,2,3
i) 17(5) cage at R9C5 = {12347/12356}, no 8,9

Steps resulting from Prelims
1a. Naked pair {89} in R56C1, locked for C1 and N4, clean-up: no 4,5 in R56C3
1b. Naked pair {67} in R56C3, locked for C3 and N4
1c. 11(3) cage at R4C2 = {245} (only remaining combination), locked for C2 and N4
1d. Naked pair {13} in R4C13, locked for R4 and 21(5) cage at R2C2
1e. 17(5) cage at R9C5 = {12347/12356}, 1,2,3 locked for R9
1f. 8,9 in R9 only in R9C234 -> no 8,9 in R8C2 (CPE)

2a. 21(3) cage at R8C1 = {579/678} (cannot be {489} because 8,9 only in R9C2), no 4
2b. 8,9 only in R9C2 -> R9C2 = {89}
2c. R89C1 = {57/67}, 7 locked for C1 and N7

3a. R4C13 = {13} = 4 -> R2C2 + R3C12 = 17 = {269/278/467} (cannot be {458} because 4,5 only in R3C1)
3b. 2,4 only in R3C1 -> R3C1 = {24}
[At this stage I saw combined cage R12C1 + R2C2 + R3C12 must contain 2 in R123C1 but decided to leave that for later if necessary, which it wasn’t.]

4. 45 rule on N7 4 outies R6789C4 = 29 = {5789}, locked for C4

5. 45 rule on C9 1 outie R8C8 = 2 innies R19C9 + 6
5a. Min R19C9 = 3 -> R8C8 = 9, R19C9 = {12}, 9 placed for D\
5b. Naked pair {12} in R19C9, locked for C9
5c. Naked pair {12} in R19C9, CPE no 1,2 in R1C1 + R5C5 using D\, clean-up: no 5,6 in R2C1

6a and 6b. Not applicable for V1.60

7. Hidden killer pair 8,9 in R9C2 and R9C34 for R9, R9C2 = {89} -> R9C34 must contain one of 8,9
7a. 22(4) cage at R8C2 = {1489/1579/3478} (cannot be {4567} which doesn’t contain 8 or 9, cannot be {1678/3469/3568} because 1,3,6 only in R8C2), no 6
7b. 4 of {1489/3478} must be in R9C3 -> no 8 in R9C3
7c. 45 rule on R9 4 innies R9C1234 = 28 = {4789/5689}
7d. Consider placement for 7 in N7
R8C1 = 7 => R9C12 = 14 = [59] (cannot be [68] because R9C34 cannot be {59} without 7 in R8C4)
or R9C1 = 7
-> R9C1 = {57}
7e. R9C1234 = {4789} (only remaining combination), no 5 -> R9C1 = 7, placed for D/, R9C3 = 4, R9C24 = {89}
7f. 22(4) cage contains 4 = {1489/3478}, no 5
7g. 5 in C4 only in R67C4, locked for 31(6) cage at R6C6
7h. R8C1 = 5 (hidden single in N7) -> R9C2 = 9 (cage sum), R89C4 = [78] -> R8C2 = 3 (cage sum), placed for D/, clean-up: no 2 in R2C1
7i. Killer pair 1,3 in R12C1 and R4C1, locked for C1
7j. R2C2 + R3C12 (step 3a) = {278} (only remaining combination, cannot be {467} which clashes with R12C1) -> R3C1 = 2, R23C2 = {78}, locked for N1, 8 locked for C2 -> R7C12 = [61], clean-up: no 1 in R2C1
7k. Naked pair {34} in R12C1, 3 locked for C1 and N1
7l. R1C2 = 6 -> R1C345 = 19 = {289/379} (cannot be {478} because no 4,7,8 in R1C3) -> R1C3 = 9, R1C45 = [28/37], clean-up: no 3 in R2C7
7m. Naked pair {15} in R23C3, locked for C3 and 24(5) cage at R2C3
7n. R23C3 = {15} = 6 -> R3C456 = 18 = {369/468} (cannot be {378} which clashes with R3C2), no 7, 6 locked for R3 and N2

8. 1 on D/ only in R1C9 + R2C8 + R3C7, locked for N3, clean-up: no 8 in R12C7
8a. 35(7) cage at R1C8 contains 1 = {1235789/1245689/1345679}, 9 locked for C7
8b. 9 in C7 only in R34C7, CPE no 9 in R4C6
[That was also available for A379 but I didn’t spot it then.]

9. 6 on D\ only in R4C4 + R5C5 + R6C6, locked for N5
9a. 6 on D/ only in R2C8 + R5C5, CPE no 6 in R5C8

10. Killer pair 3,8 in R1C45 and R3C456, locked for N2
10a. 17(4) cage at R1C6 = {1259/1457}
10b. R2C456 cannot contain both of 1,5, which would clash with R2C3 -> R1C6 = {15}
10c. Killer pair 1,5 in R2C3 and R2C456, locked for R2, clean-up: no 4 in R1C7

11. 19(4) cage at R4C4 = {1468/1567/2368/2458/2467/3457} (cannot be {1279} because 7,9 only in R4C5, cannot be {1369/1378} because 1,3 only in R5C4, cannot be {1459/2359} which clash with R6C4), no 9
11a. 9 in R4 only in R4C79, locked for N6

12. 45 rule on N3689 3 innies R7C45 + R8C5 = 1 outie R5C6 + 12
12a. Max R7C45 + R6C5 = 20 -> max R5C6 = 8
12b. R5C1 = 9 (hidden single in R5) -> R6C1 = 8

13. 2 in N3 only in R1C789 + R2C78, CPE no 2 in R4C7
[Another one that I didn’t spot for A379; it was available after step 5.]

14. 17(4) cage at R1C6 (step 10a) = {1259/1457}
14a. Consider permutations for R1C45 (step 7l) = [28/37]
R1C45 = [28] => 17(4) cage = {1457}, 4 locked for R2 => R2C1 = 3
or R1C45 = [35] => R1C1 = 4
-> R12C1 = [43], 4 placed for D\

15. 19(4) cage at R4C4 (step 11) = {1468/1567/2368/2458/2467} (cannot be {3457} because R4C4 only contains 2,6)
15a. Consider placements for R1C9
R1C9 = 1 => R9C9 = 2, placed for D\ => R4C4 = 6 => 19(4) cage = {1468/1567/2368/2467}
or R1C9 = 2 => R1C6 = 1 (hidden single in R1) => R5C4 = 1 (hidden single in C4) => 19(4) cage = {1468/1567}
-> 19(4) cage = {1468/1567/2368/2467} -> R4C4 = 6
15b. R2C8 = 6 (hidden single on D/), clean-up: no 3 in R1C7
15c. 4 on D/ only in R3C7 + R4C6, CPE no 4 in R3C6 + R4C78
15d. 35(7) cage at R1C8 contains 1 and 6 = {1245689/1345679}, 4 locked for N3, clean-up: no 5 in R1C7
[Ed pointed out that 4 is also locked for R3. This would immediately place R3C4 = 3, R1C45 = [28], R5C5 = 5 etc. which would simplify some of my later steps.]
15e. Naked pair {27} in R12C7, locked for C7 and N3
15f. 35(7) cage at R1C8 = {1245689/1345679} contains one of 2,7 -> R4C8 = {27}

16a. R1C9 = 1, R9C9 = 2, placed for D\, R1C6 = 5
16b. R3C3 = 1 (hidden single in R3), placed for D\
16c. 5 on D\ only in R5C5 + R7C7, CPE no 5 in R35C7 + R7C5
16d. 5 on D/ only in R5C5 + R6C4, locked for N5
16e. Consider placement for 5 on D\
R5C5 = 5 => R67C4 = [95]
or R7C7 = 5
-> 5 in R7C47, locked for R7
16f. R8C8 = 9 -> R78C9 = 11 = [38/74]
16g. 6 in N9 only in R89C7, locked for C7

17. Caged X-Wing for 5 in 35(7) cage at R1C8 and 31(5) cage at R1C8, no other 5 in N36

18. 19(4) cage at R4C4 (step 15a) = {1468/2368/2467}
18a. Consider permutations for R1C45 (step 7l) = [28/37]
R1C45 = [28] => 19(4) cage = {1468/2368} (cannot be {2467} = [6724] which clashes with R4C8)
or R1C45 = [37] => 19(4) cage = {1468/2368}
-> 19(4) cage = {1468/2368}, no 7, 8 locked for R4 and N5
18b. 1,3 of 19(4) cage only in R5C4 -> R5C4 = {13}
[Now I’m back where I would have been if I hadn’t overlooked 4 locked for R3 in step 15d.]
18c. R5C5 = 5 -> R67C4 = [95], 9 placed for D/
18d. R4C7 = 9 (hidden single in C7)
18e. 7 in R4 only in R4C89, locked for N6
18f. 3 on D\ only in R6C6 + R7C7, CPE no 3 in R6C78 + R7C56

19. 13(3) cage at R5C6 = {148/238} (cannot be {247} = [742] which clashes with R6C78, ALS block, but simpler is {247} clashes with R5C2), no 7, 8 locked for N6
19a. R5C3 = 7 (hidden single in R5) -> R6C3 = 6
19b. R5C9 = 6 (hidden single in R5)

20a. Naked quad {3478} in R7C789 + R8C9, locked for N9
20b. Deleted after an earlier correction
20c. R789C9 + R8C9 = {3478} = 22, R78C9 (step 16f) = 11 -> R7C78 = 11 = {38}, locked for N9, 8 also locked for R7 -> R78C9 = [74], R46C9 = [53], R7C3 = 2, placed for D/, R6C6 = 7, placed for D\, R2C2 = 8, placed for D\
20d. R7C78 = 11 -> R6C78 = 5 = {14}, locked for R6 and N6
20e. R5C78 = [82] -> R5C6 = 3 (cage sum)
20f. R36C7 = [41], R8C7 = 6 -> R78C6 = 6 = [42]

and the rest is naked singles.

Rating Comment:
I'll rate my walkthrough for A379 V1.60 at 1.5. I used several forcing chains.


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 Post subject: Re: Assassin 379
PostPosted: Sun Jul 07, 2019 6:02 pm 
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Posts: 281
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Thanks Ed. Here's how I did the harder version.
I noticed after I finished writing the WT that I ended it before I got to the area with the differences between the two versions!
Assassin 379(2) WT:
1. IOD c9 -> r19c9 = {12} and r8c8 = 9

2. 17(2)n4 = {89}
-> 13(2)n4 = {67}
-> 11(3)n4 = {245}
-> r4c13 = {13}

3. 21(3)n7 = [{57}9] or [{67}8]
Innies r9 -> r9c1234 = +28(4) = {(47|56)89}
Outies n7 -> r6789c4 = +29(4) = {5789}
Outies r9 -> r8c124 = +15(3)
Since Min r8c14 = [65] -> Max r8c2 = 4
Since (24) already in c2 -> r8c2 from (13)

4. 7 in r8c1 or r9c1
Trying r8c1 = 7 puts outies r9 = r8c124 = [735] which leaves no solution for 22(4)
-> r9c1 = 7
-> r9c3 = 4, r9c24 = {89}
Outies r9 -> r8c4 > 5
-> 5 in r67c4
-> 5 in n7 only in 21(3)
-> 21(3)n7 = [579]
-> r9c4 = 8
-> r8c24 can only be [37]
-> r67c4 = {59}

5. 7(2)n1 from {16} or {34} and 9 already in c2
-> Outies n4 = r2c2, r3c1, r3c2 = +17(3) can only be r3c1 = 2, r23c2 = {78}
-> HS 4 in c1 -> 7(2)n1 = {34}
-> r4c13 = [13]
-> r7c12 = [61]
-> r78c3 = {28}
Also r1c2 = 6
-> r123c3 = {159}

6. 25(4)r1 has a 6 in r1c2 -> cannot also have a 1 or a 5 in it.
-> r1c3 = 9 and r23c3 = {15}
-> r1c45 from [28] or [37]
-> (15) in n2 both in 17(4)n2 with one of (15) in r1c6
-> n2 from:
(A) [28], {1457} with (47) in r2, {369}
(B) [37], {1259} with (29) in r2, {468}
Both (A) and (B) -> 7(2)n1 = [43]

7. 6 in D\ only in n5
6 in D/ in n3 or n5
6 in n3 in r2
-> 6 in D/ only in r2c8 or r5c5
Also note that r4c4 only from (26)

8. 35(7)n3 is missing a 10(2)
Those two values cannot both go in 9(2)n3
-> One of the values in 9(2)n3 also goes in 35(7) at r4c8 and in n9 in r789c9
Also whatever goes in r4c7 must go in n3 in r23c9
-> For the 'missing' 10(2) - one of the values goes in r12c7 and the other goes in r23c9

Enough prep work. Now the cracker!

9! All to do with 9(2)n3
9a) Either r1c45 = [37] -> 9(2)n3 not {36}
Or r1c45 = [28] -> r4c4 = 6 -> r2c8 = 6 -> 9(2)n3 not {36}

9b) 1 in r19c9 -> 1 not in r5c5
-> 1 in D/ only in n3
-> 9(2)n3 not {18}

9c) 5 must be in 35(7)
-> Trying 9(2)n3 = {45} puts 5 in r4c8 and 6 in r2c9 (Since 35(7) would be missing (46))
-> Leaves no place for 5 in c9/n9
-> 9(2)n3 not {45}

-> 9(2)n3 = {27}!

Easy from here

10. Continuing...
One of (27) in r4c8
Also HS 7 in r3 -> r23c2 = [87]
Also r19c9 = [12]
-> 1 in n2 in r2
-> r23c3 = [51]
-> r1c6 = 5
Also NS (D\, c4, r4) r4c4 = 6
-> HS 6 in D/ -> r2c8 = 6
Also NS (D\, D/) -> r5c5 = 5
-> NS 3 in D\ -> r7c7 = 3
-> NS in D\ r6c6 = 7
Also r67c4 = [95]

11. ... and more ...
Also since 1 in 35(7) -> 9 must also be in 35(7)
-> HS 9 in 35(7) -> r4c7 = 9
Also since 6 in 35(7) -> 4 must also be in 35(7)
-> 4 in 35(7) in r3c78
-> r3c456 = [3{69}]
-> r1c45 = [28]
-> r2c456 = <174>
-> 9(2)n3 = [72]
Also -> r1c8 = 3
-> Since 3 in 35(7) -> 7 must also be in 35(7) and can only go in r4c8
-> r3c789 = [458]
Also r78c9 = [74]

12. ...
Remaining innies n5 = r5c6 + r6c5 = +5(2)
r5c4 only from (14)
-> 19(4)n5 can only be [6481]
etc.


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 Post subject: Re: Assassin 379
PostPosted: Thu Jul 11, 2019 8:19 am 
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Grand Master
Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 1043
Location: Sydney, Australia
Really enjoyed both your WTs for the harder version guys. I was a fair way from solving it so glad I went for the easier one. Interesting that you both prefer the harder version so will keep that in mind for the future. Just means I will occasionally post a puzzle I haven't been able to solve. We all found different ways to do the start.

Anyway, here's my WT for the V1. I found what I missed from Andrew's v1.60 WT. Have included a note at that spot at step 18. Thanks Andrew for some clarifications and corrections.
a379 WT:
Preliminaries courtesy of SudokuSolver
Cage 17(2) n4 - cells ={89}
Cage 7(2) n1 - cells do not use 789
Cage 13(2) n4 - cells do not use 123
Cage 9(2) n3 - cells do not use 9
Cage 10(2) n6 - cells do not use 5
Cage 21(3) n7 - cells do not use 123
Cage 20(3) n9 - cells do not use 12
Cage 11(3) n4 - cells do not use 9
Cage 17(5) n89 - cells do not use 89


clean-up not done unless stated.
1. "45" on c9: 1 outie r8c8 - 6 = 2 innies r19c9
1a. min. 2 innies = 3 -> r8c8 = 9 (placed for d\)
1b. -> r19c9 = 3 = {12} only: both locked for c9 nor in r5c5 and r1c1 since they see both r19c9 through d\ (Common Peer Elimination CPE)

2. 17(2)n4 = {89}: both locked for n4 and c1

3. 13(2)n4 = {67} only: both locked for c3 and n4

4. 21(3)n7: {489} blocked by only one 8,9 available
4a. = {579/678}(no 4)
4b. must have 7: locked for n7
4c. must have 8 or 9 which are only in r9c2 -> r9c2 = (89)
4d. 7 locked for c1

5. "45" on n12: 2 outies r4c13 = 4 = {13}: both locked for r4 and n4, and locked for 21(5)r2c2

6. naked triple {245} in r456c2: all locked for c2
6a. "45" on n7: 4 outies r6789c6 = 29 = {5789} only: all locked for c4

7. 17(4)r9c5 = {12347/12356}
7a. 1,2,3 all locked for r9

8. "45" on r9: 3 outies r8c124 = 15 = {168/357}
8a. must have 1 or 3 which are only in r8c2 -> r8c2 = (13)

9. 22(4)r8c2: must have 1 or 3 for r8c2 = {1489/1579/3478}
9a. note: can't have both 3 and 5
9b. no eliminations yet

10. h15(3)r8c124 = {168/357}
10a. but [735] blocked by 22(4) can't have both 3 & 5
10b. -> no 7 in r8c1
10c. -> r9c1 = 7 (hsingle n7): Placed for d/
10d. r8c124 = [618/537](no 5 in r8c4)

11. 17(4)r9c5 = {12356} only (no 4): 5 locked for r9

12. r9c3 = 4 (hsingle r9)
12a. 22(4)r8c2 = {1489/3478}: must have 8, locked for c4 and n8

13. 5 in c4 only in r67c4 in 31(6): 5 locked for 31(6)

14. r8c1 = 5 (hsingle n7)
14a. r9c2 = 9 (cage sum), r89c4 = [78], r8c2 = 3 (cage sum), placed for d/

15. 6 in n7 only in r7: locked for r7

16. r4c13 = 4 -> r3c1 + r23c2 = 17 = {278/467}
16a. note: if it has 4 must also have 6
16b. must have 7: locked for n1

17. 4 in c1 in 7(2) = {34} or r3c1 (-> 6 in r23c2, step 16a)
17a. -> {16} blocked from 7(2) since it would leave no 4 for n1
17b. 7(2) = {34} only: both locked for n1, 3 for c1
17c. r3c1 + r23c2 = [2]{78}: 8 locked for c2 and n1
17d. r4c13 = [13], r7c12 = [61], r1c2 = 6
17e. r1c2 = 6 -> r1c345 = 19 = [9][28/37]


Key step. I see what I missed from Andrew's WT to make step 18b much simpler
(1 on D/ only in n3 in 35(7) -> it must have 9 -> no 9 in r4c6 (CPE) -> {1279} blocked from 19(4)n5 since 7 & 9 are only in r4c5)
18. from 17e. r1c45 = 10 = [28/37] = 2 in r1c4 or 7 in r1c5
18a. 19(4)n5: all combinations with 9 are
18b. {1279} is [2791] only permutation: blocked by 2 or 7 in r1c45
18c. {1369}: blocked by 1,3 only in r5c4
18d. {1459/2359} blocked by r6c4 = (59)
18e. -> no 9 in 19(4)

19. 9 in r4 only in n6: locked for n6
19a. no 1 in 10(2)n6

20. "45" on n36: 2 innies r6c78 = 5 and must have 1 for n6 = {14} only: both locked for r6, n6 and 16(4)
20a. r6c78 = 5 -> r7c78 = 11 = {38} only: both locked for r7 and n9

21. r7c3 = 2, placed for d/, r19c9 = [12]: both placed for their diagonal

22. naked triple {459} in r7c456: 4 & 5 locked for r7 and n8
22a. r78c9 = [74](cage sum)

23. hidden single 1 on d\ -> r3c3 = 1, r2c3 = 5

24. r1c6 = 5 (hsingle n2)

This one took a long time to see
25. r5c5 = 5 (hsingle d\)(Placed for d/)
25a. r67c4 = [95](9 placed for d/), r56c1 = [98]

26. r4c7 = 9 (hsingle c7)

27. 12(3)r7c6 = {129/246}
27a. must have 2 -> r8c6 = 2

28. 6 on d\ only in n5: 6 locked for n5

29. 6 on d/ only in n3: 6 locked for n3

30. 9(2)n3 = {27} only: both locked for c7 and n3

31. 10(2)n6 = [37/82]
31a. hidden pair 2,7 in n6 -> r45c8 = {27}

32. naked pair {38} in r57c7: 8 locked for c7

33. r23c3 = 6 -> r3c456 = 18
33a. but {468} blocked by r3c7 = (46)
33b. {378} blocked by r3c2 = (78)
33c. = {369} only: all locked for n2 and r3

cracked.
Cheers
Ed


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