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Assassin 192X


Afmob thank you for reminding me - total memory failure.

SS gives it a 1.07 - but it feels easier to me. After you have solved it you may wish to try without the N5 20(3) cage.



Solution:
284376195
965241837
173985246
326857914
547193628
891624753
738419562
612538479
459762381
Puzzle (sumocue PS):
3x3:d:k:2816:5633:2:3:4868:5:6:5639:2568:5641:2816:5633:12:4868:14:5639:2568:5137:18:5641:2816:5633:4868:5639:2568:5137:26:27:28:5641:5150:5150:5150:5137:34:35:4132:4132:4132:39:40:41:4138:4138:4138:45:46:2607:48:49:50:5683:52:53:54:2607:3384:2873:2618:5435:3388:5683:62:2607:3384:2873:66:2618:68:5435:3388:5683:3384:2873:74:75:2618:77:78:5435:3388:
Without the 20(3) cage:
3x3:d:k:2816:5633:2:3:4868:5:6:5639:2568:5641:2816:5633:12:4868:14:5639:2568:5137:18:5641:2816:5633:4868:5639:2568:5137:26:27:28:5641:30:31:32:5137:34:35:4132:4132:4132:39:40:41:4138:4138:4138:45:46:2607:48:49:50:5683:52:53:54:2607:3384:2873:2618:5435:3388:5683:62:2607:3384:2873:66:2618:68:5435:3388:5683:3384:2873:74:75:2618:77:78:5435:3388:

Thanks HATMAN for a fun Assassin.

My solving path seems fairly long for this puzzle so there may well be quicker ways to solve it or I may have taken too long to spot the key steps.


Here is my walkthrough for A192X.

Prelims

a) 11(3) cage at R1C1 = {128/137/146/236/245}, no 9
b) 22(3) cage at R1C2 = {589/679}
c) 19(3) cage at R1C5 = {289/379/469/478/568}, no 1
d) 22(3) cage at R1C8 = {589/679}
e) 10(3) cage at R1C9 = {127/136/145/235}, no 8,9
f) 22(3) cage at R2C1 = {589/679}
g) 20(3) cage at R2C9 = {389/479/569/578}, no 1,2
h) 20(3) cage at R4C4 = {389/479/569/578}, no 1,2
i) 10(3) cage at R6C3 = {127/136/145/235}, no 8,9
j) 22(3) cage at R6C7 = {589/679}
k) 11(3) cage at R7C4 = {128/137/146/236/245}, no 9
l) 10(3) cage at R7C5 = {127/136/145/235}, no 8,9
m) 21(3) cage at R7C6 = {489/579/678}, no 1,2,3

1. 45 rule on D/ 3 innies R4C6 + R5C5 + R6C4 = 22 = {589/679}, 9 locked for N5 and D/

2. 45 rule on D\ 3 innies R4C4 + R5C5 + R6C6 = 21 = {489/678} (cannot be {579} which clashes with R4C6 + R5C5 + R6C4), no 1,2,3,5, 8 locked for N5 and D\
2a. 8 of {678} must be in R5C5 (permutations with 6 or 7 in R5C5 clash with R4C6 + R5C5 + R6C4) -> no 6,7 in R5C5

3. 20(3) cage at R4C4 = {479/569/578} (cannot be {389} which clashes with R5C5), no 3

4. R4C4 + R5C5 + R6C6 (step 2) = {489/678}
4a. 4 of {489} must be in R6C6 -> no 4 in R4C4

5. R5C46 + R6C5 = {123} (hidden triple in N5)

6. 19(3) cage at R1C5 = {379/469/478/568} (cannot be {289} which clashes with R5C5), no 2

7. 45 rule on R5 3 innies R5C456 = 13 = {139/238}, 3 locked for R5 and N5

8. 10(3) cage at R7C5 = {136/145/235} (cannot be {127} which clashes with R6C5), no 7

9. 19(3) cage at R1C5 (step 6) = {379/469/478} (cannot be {568} which clashes with 10(3) cage at R7C5), no 5
9a. Killer pair 3,4 in 19(3) cage at R1C5 and 10(3) cage at R7C5, locked for C5

10. R6C6 = 4 (hidden single in N5), placed for D\
10a. R4C4 + R5C5 + R6C6 (step 2) = 21 -> R4C4 + R5C5 = 17 = [89], 9 placed for D\

11. 20(3) cage at R4C4 (step 3) = {578} (only remaining combination), 5,7 locked for R4 and N5 -> R6C4 = 6, R4C4 = 7 (step 1), both placed for D/, R4C5 = 5

12. 19(3) cage at R1C5 (step 9) = {478} (only remaining combination), locked for C5 and N2

13. R5C456 (step 7) = {139} (only remaining combination), 1 locked for R5 and N5 -> R6C5 = 2

14. Naked triple {136} in 10(3) cage at R7C5, locked for N8

15. 10(3) cage at R1C9 = {145/235}, 5 locked for D/ and N3

16. 8 on D/ only in 13(3) cage at R7C3, locked for N7

17. 11(3) cage at R1C1 = {137/236}, no 5, 3 locked for D\ and N1

18. 5 on D\ only in 13(3) cage at R7C7, locked for N9
18a. 22(3) cage at R6C7 = {589/679}
18b. 5 of {589} must be in R6C7 -> no 8 in R6C7

19. 45 rule on N7 2 innies R7C1 + R9C3 = 2(1+1) outies R6C3 + R7C4 + 11
19a. Min R6C3 + R7C4 = 3 -> min R7C1 + R9C3 = 14, no 1,2,3,4
19b. Max R7C1 + R9C3 = 16 -> max R6C3 + R7C4 = 5, no 5,7

20. 13(3) cage at R7C7 must contain 5 (step 18) = {157/256}
20a. 22(3) cage at R6C7 = {589/679}
20b. 5 of {589} must be in R6C7, 7 of {679} must be in R6C7 (R7C8 + R8C9 cannot be {67} which clashes with 13(3) cage) -> no 9 in R6C7, no 7 in R7C8 + R8C9

21. 45 rule on N9 2(1+1) outies R6C7 + R7C6 = 2 innies R7C9 + R9C7 + 11
21a. Max R6C7 + R7C6 = 16 -> max R7C9 + R9C7 = 5, no 6,7,8,9

22. 21(3) cage at R7C6 = {489/579} (cannot be {678} which clashes with 13(3) cage at R7C7), no 6

23. Hidden killer pair 1,2 in 11(3) cage at R1C1 and R1C3 + R3C1 for N1, 11(3) cage contains one of 1,2 -> R1C3 + R3C1 must contain one of 1,2
23a. 4 in N1 only in R1C3 + R3C1 -> R1C3 + R3C1 = {14/24}
23b. 45 rule on N1 2(1+1) outies R3C4 + R4C3 = 2 innies R1C3 + R3C1 + 10
23c. R1C3 + R3C1 can only be {14/24} = 5,6, R3C4 + R4C3 can only be [59/96/99] = 14,15,18 -> R1C3 + R3C1 = 5 = {14}, locked for N1, R3C4 + R4C3 = 15 = [96], clean-up: no 7 in 11(3) cage at R1C1 (step 17)

24. Naked triple {236} in 11(3) cage at R1C1, locked for D\ and N1
24a. Naked triple {157} in 13(3) cage at R7C7, locked for N9, clean-up: no 5 in R6C6 (step 22)

25. 21(3) cage at R7C6 = {489} (only remaining combination), 4 locked for N9

26. R6C7 + R7C6 = R7C9 + R9C7 + 11 (step 21)
26a. R7C9 + R9C7 = {23} = 5 -> R6C7 + R7C6 = 16 = [79]

27. Naked pair {48} in R8C7 + R9C8, locked for N9 -> R7C8 = 6, R8C9 = 9

28. 7 in R5 only in R5C123 -> 16(3) cage at R5C1 = {457} (only remaining combination), locked for R5 and N4
28a. Naked triple {268} in 16(3) cage at R5C7, locked for N6

29. R2C1 = 9 (hidden single in N1), R3C2 = 7 (cage sum)
29a. R9C3 = 9 (hidden single in N7)

30. 22(3) cage at R1C8 = {589/679} -> R1C8 = 9, R3C6 = 5, R2C7 = 8

31. Naked pair {28} in R89C6, locked for C6 and N8 -> R7C4 = 4

32. 11(3) cage at R7C4 = {146/245}, no 3,7
32a. 6 of {146} must be in R9C2 -> no 1 in R9C2

33. R7C3 = 8 (hidden single in R7)
33a. 13(3) cage at R7C3 = {148/238}
33b. Killer pair 1,2 in 13(3) cage at R7C3 and 11(3) cage at R7C4, locked for N7

34. 10(3) cage at R6C3 = {136/145}, no 7 -> R6C3 = 1, clean-up: no 6 in R9C2 (step 32)
34a. Naked pair {25} in R8C3 + R9C2, locked for N7 -> R7C2 = 3, R8C1 = 6
34b. R8C7 = 4, R8C2 = 1, R9C1 = 4, both placed for D/

and the rest is naked singles without using the diagonals.

I think it seems appropriate to call the variant without the 20(3) cage in N5 A192X V1.5.



It started almost as easily as A192X and for quite a time I was finding it a more enjoyable puzzle. Then my solving path almost ground to a halt and I found myself using a couple of much harder steps to achieve the key breakthrough. HATMAN was right in his choice of which puzzle was the main Assassin and which one the variant.


Here is my walkthrough for A192X V1.5

Prelims

a) 11(3) cage at R1C1 = {128/137/146/236/245}, no 9
b) 22(3) cage at R1C2 = {589/679}
c) 19(3) cage at R1C5 = {289/379/469/478/568}, no 1
d) 22(3) cage at R1C8 = {589/679}
e) 10(3) cage at R1C9 = {127/136/145/235}, no 8,9
f) 22(3) cage at R2C1 = {589/679}
g) 20(3) cage at R2C9 = {389/479/569/578}, no 1,2
h) 10(3) cage at R6C3 = {127/136/145/235}, no 8,9
i) 22(3) cage at R6C7 = {589/679}
j) 11(3) cage at R7C4 = {128/137/146/236/245}, no 9
k) 10(3) cage at R7C5 = {127/136/145/235}, no 8,9
l) 21(3) cage at R7C6 = {489/579/678}, no 1,2,3

1. 45 rule on D/ 3 innies R4C6 + R5C5 + R6C4 = 22 = {589/679}, 9 locked for N5 and D/

2. 45 rule on D\ 3 innies R4C4 + R5C5 + R6C6 = 21 = {489/678} (cannot be {579} which clashes with R4C6 + R5C5 + R6C4), no 1,2,3,5, 8 locked for N5 and D\
2a. 8 of {678} must be in R5C5 (permutations with 6 or 7 in R5C5 clash with R4C6 + R5C5 + R6C4) -> no 6,7 in R5C5

3. 19(3) cage at R1C5 = {379/469/478/568} (cannot be {289} which clashes with R5C5), no 2

4. 45 rule on R5 3 innies R5C456 = 13 = {139/148/238} (only combinations containing 8 or 9 for R5C5), no 5,6,7

5. Hidden killer pair 1,2 in R46C5 and R5C46 for N5, R5C46 contains one of 1,2 -> R46C5 must contain one of 1,2
5a. 10(3) cage at R7C5 = {136/145/235} (cannot be {127} which clashes with R46C5), no 7

6. R46C5 contains one of 1,2 (step 5)
6a. 45 rule on C5 3 innies R456C5 = 16 = {169/178/259/268} (only combinations which contain 1 or 2), no 3,4

7. 3 in N5 only in R5C46, locked for R5
7a. R5C456 (step 4) = {139/238}, no 4

8. 4 in N5 only in R4C4 + R6C6 -> R4C4 + R5C5 + R6C6 (step 2) = {489} (only remaining combination) -> R5C5 = 9, placed for D\, R4C4 + R6C6 = {48}, 4 locked for D\
8a. R4C6 + R5C5 + R6C4 (step 1) = {679} (only remaining combination), 6,7 locked for N5 and D/
8b. 5 in N5 only in R46C5, locked for C5

9. 10(3) cage at R7C5 (step 5a) = {136} (only remaining combination), locked for C5 and N8
9a. Naked triple {478} in 19(3) cage at R1C5, locked for N2
9b. R5C46 = {13} (hidden pair in N5), 1 locked for R5

10. 10(3) cage at R1C9 = {145/235}, 5 locked for N3 and D/
10a. 8 on D/ only in 13(3) cage at R7C3, locked for N8

11. Hidden killer pair 1,2 in 11(3) cage at R1C1 and R1C3 + R3C1 for N1, 11(3) cage contains one of 1,2 -> R1C3 + R3C1 must contain one of 1,2
11a. 4 in N1 only in R1C3 + R3C1 -> R1C3 + R3C1 = {14/24}
11b. 45 rule on N1 2(1+1) outies R3C4 + R4C3 = 2 innies R1C3 + R3C1 + 10
11c. R1C3 + R3C1 = {14/24} = 5,6 -> R3C4 + R4C3 = 15,16 = [69/96/97], no 5,8

12. 3 in N1 only in 11(3) cage at R1C1, locked for D\
12a. 11(3) cage at R1C1 = {137/236}, no 5
12b. 5 on D\ only in 13(3) cage at R7C7, locked for N9

13. 22(3) cage at R1C2 = {589/679}
13a. 6 of {679} must be in R3C4 (R1C2 + R2C3 cannot be {67} which clashes with 11(3) cage at R1C1), no 6 in R1C2 + R2C3

14. 45 rule on N7 2 innies R7C1 + R9C3 = 2(1+1) outies R6C3 + R7C4 + 11
14a. Min R6C3 + R7C4 = 3 -> min R7C1 + R9C3 = 14, no 1,2,3,4
14b. Max R7C1 + R9C3 = 16 -> max R6C3 + R7C4 = 5, no 5,6,7,8, no 4 in R6C3

15. 11(3) cage at R7C4 = {146/236/245} (cannot be {137} because R7C4 only contains 2,4), no 7

16. Hidden killer pair 1,2 in 13(3) cage at R7C7 and R7C9 + R9C7 for N9, 13(3) cage contains one of 1,2 -> R7C9 + R9C7 must contain one of 1,2
16a. 3 in N9 only in R7C9 + R9C7 -> R7C9 + R9C7 = {13/23}
16b. 45 rule on N9 2(1+1) outies R6C7 + R7C6 = 2 innies R7C9 + R9C7 + 11
16c. R7C9 + R9C7 = {13/23} = 4,5 -> R6C7 + R7C6 = 15,16, no 4,5

17. 13(3) cage at R7C7 = {157/256}
17a. 22(3) cage at R6C7 = {679} (only remaining combination), CPE no 6,7 in R7C7
17b. 9 of 22(3) cage must be in R7C8 + R8C9 (R7C8 + R8C9 cannot be {67} which clashes with 13(3) cage), 9 locked for N9, no 9 in R6C7
17c. Naked pair {67} in R6C47, locked for R6
17d. Killer pair 6,7 in 22(3) cage at R6C7 and 13(3) cage at R7C7, locked for N9
17e. Min R6C7 + R7C6 = 15 (step 16c) -> no 7 in R7C7

18. Naked pair {48} in R8C7 + R9C8, locked for 21(3) cage at R7C6 -> R7C6 = 9

19. R8C9 = 9 (hidden single in N9)
19a. R9C3 = 9 (hidden single in R9)
19b. Naked pair {67} in R4C36, locked for R4

20. 22(3) cage at R2C1 = {679} (cannot be {589} because R4C3 only contains 6,7), no 5,8, 9 locked for N1, CPE no 6,7 in R3C3

21. 5,8 in N1 only in 22(3) cage at R1C2 = {589} (only remaining combination) -> R1C2 + R2C3 = {58}, R3C4 = 9
21a. R2C1 = 9 (hidden single in N1)

22. 22(3) cage at R1C8 = {589/679} -> R1C8 = 9
22a. 7,8 only in R2C7 -> R2C7 = {78}

23. R4C7 = 9 (hidden single in C7)
23a. 20(3) cage at R2C9 = {389/479}, no 6
23b. Killer pair 3,4 in 10(3) cage at R1C9 and 20(3) cage at R2C9, locked for N3
23c. Killer pair 7,8 in R2C7 and 20(3) cage at R2C9), locked for N3
23d. R6C2 = 9 (hidden single in R6)

24. R7C3 = 8 (hidden single in R7), R2C3 = 5, R1C2 = 8

25. 4 in R7 only in R7C24, CPE no 4 in R8C3 + R9C2
25a. 11(3) cage at R7C4 (step 15) = {146/236/245}
25b. 2 of {236} must be in R7C4, 5 of {245} must be in R9C2 -> no 2 in R9C2

26. 7 in C3 only in R45C3, locked for N4
26a. 16(3) cage at R5C1 = {268/457}
26b. 8 of {268} must be in R5C1 -> no 2,6 in R5C1
26c. 7 of {457} must be in R5C3 -> no 4 in R5C3

27. R1C3 = 4 (hidden single in C3), R1C5 = 7

28. 11(3) cage at R1C1 (step 12a) = {137/236}
28a. 7 of {137} must be in R2C2 -> no 1 in R2C2
28b. 7 in N1 only in R23C3, locked for C2

29. 10(3) cage at R6C3 = {127/136/145/235}
29a. 1 of {127/136/145} must be in R6C3 (R7C2 + R8C1 = {16/17} => R8C2 + R9C1 = {23} and cannot place 4 in N7) -> no 1 in R7C2 + R8C1
29b. 2 of {235} cannot be in R6C3, here’s how
R6C3 = 2 => R7C2 + R8C1 = {35} => R8C2 + R9C1 = {14} => only even numbers available for 11(3) cage at R7C4
-> no 2 in R6C3

30. 10(3) cage at R6C3 cannot be {127}, here’s how
10(3) cage = {127} => R7C2 = 2, R8C2 + R9C1 = {14}, R8C3 + R9C2 = {36} => R7C4 = 2 clashes with R7C2
30a. -> 10(3) cage at R6C3 = {136/145/235}, no 7

31. R7C1 = 7 (hidden single in N7), R7C8 = 6, R6C7 = 7, R6C4 = 6, R4C6 = 7, R4C3 = 6, R3C2 = 7, R2C7 = 8, R3C6 = 5 (step 22), R8C7 = 4, R9C8 = 8, R23C5 = [48]

32. R1C9 = 5, R2C9 = 7, R3C8 = 4 (hidden singles in N3)

33. 10(3) cage at R1C9 (step 10) = {235} (only remaining combination), 2,3 locked for N3 and D/ -> R8C2 = 1, R9C1 = 4

34. R7C7 = 5, R8C8 = 7 (hidden singles on D\), R9C6 = 2, R9C9 = 1, placed for D\

35. R5C3 = 7 (hidden single in C3), R5C12 = [54] (step 26a)

and the rest is naked singles without using the diagonals.

Andrew

I went through your 1.5 in detail. It followed the general path I intended. However 29 and 30 are very subtle, perhaps this is simpler:

29. from 14. R7C1 + R9C3 = R6C3 + R7C4 + 11
r9c3=9, r7c1=5/6/7, r7c4=2/4, r6c3=1/2/3
r7c1<>6 no 13
r7c1=5 -> r6c3&r7c4 =[12]
r7c1=7 -> r6c3&r7c4 =[14/32]
-> r6c3 = 1/3

30. ifr7c1=5 r6c3&r7c4 =[12] r7c2<>4 (as needs 145) no 4 on R7
-> r7c1 = 7

SS gave it 2.07, again I thought that a bit high, views?

Edit: Ignore 29 above. Andrew points out that it is junk (he used nicer words), as I forgot to allow for [22] in the outies.
I must make this mistake at least once a month.

I have a paper solvable that I will post: once I manage to solve it on paper.

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Thanks for another top-notch Zero-X, Maurice!

My approach for the v1.5 was broadly similar to Andrew's approach in the early stages (which Andrew specified very methodically and succinctly in his WT ). However, after roughly the equivalent of his step 24, I took a different path:

_________________
Cheers,
Mike

Thanks for saying I put that politely. I've got into the habit of always specifying when outies don't see each other, there were several 2(1+1) examples in this puzzle, to try to avoid that error. I list 2(1+1) even for cases where the total is an odd number, when the two cells can't have the same value.

For completeness an alternative to step 29, using the same approach as HATMAN's step 30.
if r7c1 = 6 r6c3&r7c4 = [22] 10(3) cage at r6c3 cannot be [244] no 4 on r7
Combining this with step 30 -> r7c1 = 7

I look forward to that but I can't promise that I won't use elimination solving on it. However I use a form of insertion solving on my other sudoku site so I'll try it that way.

Thanks Mike for a neat alternative breakthrough. Technically much simpler than my way or HATMAN's way.

Mike

Very nice

Your comments on sameness to Ed: I like the zero-X structure as it allows smaller cages, however JSudoku finds them much easier to solve than SudokuSolver does.

I'll put up a sketch WT of 1.0 in a while, which finishes a bit quicker than Andrew's but with a very similar start.

Maurice

While HATMAN battles away with his paper version, here's one that should take out the easy opening to the V1.5. The middle and ends are the same as the V1.5 (unfortunately).

A192X V2
NOTE: 1-9 cannot repeat on the diagonals.


Solution: same as original puzzle

Cheers
Ed

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