Assassin 30

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sudokuEd
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Post by sudokuEd »

Bullseye 3 tag solution continued
Caida wrote: I was getting desperate
There's always Para's WT if we want to give up :wink: .
Caida wrote:How about a move looking at c9?
Nice one. I couldn't get anything worth-while before you showed how. I think there's one little mistake, but easy to get around. And found another way to do c9 which adds a little extra. :)
Caida wrote:29b1. h23(5)c9 <> {13469}
This one still looks OK: 13(2) = {58} & 9(2) = {27}. Get around that in just a sec.

ALT step 29b2. No 7 possible in r5c9: like this. From the nice observation in step 29a3, 1 in r19c2 -> 1 in r5c7. If 7 in r5c9 -> 1 cannot be in r5c7 (combo's 18(3)) -> 1 cannot be in r19c2. Also, 7 in r5c9 -> no 2 in 9(2)c9 -> must have 2 in h23(5) = 27{149/158/356}: but each of these combo's forces 1 in r19c2 (cage sums) which is not possible when r5c9 = 7. Therefore, no 7 in r5c9 -> h23(5)c9 <> {12479) blocked as r12/89c9 = {12/47} forces 3 into both r19c8.

29e. h23(5)c9={13469} is blocked blocked same way as step 29a3. r12c9 = {13} -> r589c9 = [6]{49} ([9]{46} blocked by combo's in 14(3)) -> 1 in r9c8 (cage sum) -> 1 in r5c7: but [16] is not possible in 18(3)n6.

Add 29f. h23(5)c9 = 18{239/257/356}(no 4)
29g. no {67} possible in r89c9 -> {167} combo blocked from 14(3)n9
29h. no {37} possible in r89c9 -> {347} combo blocked from 14(3)n9

Couple more
30. no 7 in r5c9 (see alt step 29b2).
30a. 6 in {567} in 18(3)n6 must be in r5c9 -> no 6 r5c78

31. {459} is blocked from 18(3)n6 by 4s in c9. Like this.
31a. 4 in 13(2) or 9(2) in c9. If 4 is in r46c9 -> no {459} in 18(3)n6 OR 4 is in r37c9 -> r46c9 = 9/5 (cage sums)-> {459} blocked from 18(3)
31b. 18(3)n6 = {189/567}(no 4)

Code: Select all

.-------------------------------.-------------------------------.-------------------------------.
| 458       458       79        | 79        568       123       | 456       123       123       |
| 458       123679    123679    | 123468    56789     123468    | 456789    456789    123       |
| 13679     13679     123       | 123       56789     468       | 46789     456789    4679      |
&#58;-------------------------------+-------------------------------+-------------------------------&#58;
| 13479     13456789  13        | 589       2         569       | 46789     13456789  4679      |
| 12458     12458     12458     | 46        3         79        | 15789     15789     689       |
| 346789    3456789   3456789   | 478       1         458       | 23456789  23456789  234567    |
&#58;-------------------------------+-------------------------------+-------------------------------&#58;
| 123467    123456789 12345678  | 123567    46789     12356789  | 123456789 123456789 234567    |
| 124689    123456789 123456789 | 1235679   46789     12356789  | 123456789 123456789 12356789  |
| 124689    236789    234678    | 123567    46789     12356     | 12456     12345     12356789  |
'-------------------------------.-------------------------------.-------------------------------'
Caida
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Post by Caida »

Found an easy elimination (very easy - so hope my logic is right!)

Referring back to step 8 - copied here for ease of reference

8. Outties and Innie c9: r5c9 less r19c8 = 3
8a. -> min r8c19 = 3 -> min r5c9 = 6 (no 1,2,4,5)
8b. -> max r5c9 = 9; -> max r19c8 = 6 (no 6..9)

Continuation of tag for 30v2-1

32. r5c9 <> 6 here’s how:
32a. if r5c9 = 6 -> r19c8 = 3 = {12} (step 8)
32b. -> but: from step 29a3. if there is a 1 in c8 of n3 or n9 then there is a 1 in 18(3)n6 (r5c7 = 1)
32c. -> no combination possible for 18(3)n3 with 1 in r5c7 and 6 in r5c9
32d. -> r5c9 no 6

Hopefully this leads to a bunch more - I think it even places a 6 somewhere.
But - I must get back to my paying job now - deadlines looming! I'll have to explore more later.
Nasenbaer
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Post by Nasenbaer »

I hope you don't mind my intrusion. :wink:

Caida, I didn't see your step 32 when I started to write down my steps so my numbering also starts at 32.

My main work here is on 26(5) in n3, mostly with help from h16(3) in n5.


32. n236: 26(5) <> {12689|23489|23678}
32a. a combination in 26(5) with 2 in r2c6 must also include 5 because [25] in r1c67 is blocked

32b. 26(5) <> {13679}
32c. 26(5) = {13679} -> r2c6 r4c8 = {13}, r2c78 r3c8 = {679} -> r34c9 = [49], r13c7 = [68], r4c46 = [86] -> no valid placement for n236: 21(3)

32d. 26(5) <> {23579}
32e. 26(5) = {23579} -> r2c6 = 2, r4c38 = [13], r2c78 r3c8 = {579} -> r1c6 r3c9 = {46}, r3c7 = 8 -> r4c79 = {79} -> r4c46 = [86] -> r3c6 = 4, r4c4c79 = [97], r3c9 = 6, r1c67 = [34] -> r3c4 = 1 not possible

32f. no 9 in r4c8 possible
32g. 9 in r4c8:
26(5) = {13589} -> can't place it
26(5) = {14579} -> r2c6 = 1 -> r1c67 = [16] -> conflict
26(5) = {24569} -> r2c6 = 2 -> no number left for r1c7 -> conflict

33. no 8 in r4c78 possible
33a. 8 in r4c78: r4c46 = {59}, r5c46 = [67] -> can't place 18(3) in r5

34. no 5 in r4c8
34a. 5 in r4c8: r4c46 = [86], r5c46 = [49] -> cant place 18(3) in r5

35. n7: 17(3): no 8 in r9c2

36. no 5 in r4c2
36a. n124: 22(5) <> {23458} -> only 2 cells for {458} left -> conflict
36b. 22(5) = {12568} -> r4c2 = 5, r2c4 = 8, r4c46 = [86] -> conflict
36c. 22(5) = {13459} -> r4c2 = 5, r2c4 = 4, r5c4 = 6, r4c46 = [86] -> conflict
36d. 22(5) = {13567} -> r4c2 = 5, r4c46 = [86], r5c46 = [49], r6c46 = [75] -> r1c34 = [79] -> conflict because 7 is locked in 22(5) in n1

37. n5: 5 locked in r4c46 for r4 and n5
37a. -> 9 locked in r4c46 for r4 and n5
37b. -> r5c46 = [67]
37c. -> {48} locked in r6c46 for r4 and n5

38. r5: 18(3) = {189} -> 1,8,9 locked for r5 and n6
38a. 11(3) = {245} -> 2,4,5 locked for r5 and c4

39. Cleanup
39a. n14: 10(2): no 1,6 in r3c1
39b. n47: 10(2): no 2,6 in r7c1
39c. n36: 13(2): no 4 in r3c9
39d. n69: 9(2): no 5 in r7c9
39e. n236: 21(3): no 4 in r3c67
39f. -> hidden single: r3c8 = 4
39g. -> hidden single: r2c4 = 4
39h. -> hidden single: r3c5 = 5
39i. r6c46 = [84]
39j. using hidden cages h19(3) and h20(3) in n5: r4c46 = [59]
39k. n23: 7(2): no 3 in r1c6
39l. n78: 9(2): no 3,4 in r9c3

40. n2: {68} locked in r1c5 and r4c6 for n2

41. n236: 26(5) = 4{1579|1678|2569} (other combinations blocked by r2c1) -> no 3 in r2c6 and r4c8

42. n2: {12} locked in r12c6 for c6 and n2
42a. r3c4 = 3, r34c3 = [21]
42b. -> r34c1 = [73]
42c. r1c34 = [97]
42d. r12c5 = [69], r1c67 = [25], r34c6 = [18]

The rest should be easy.
sudokuEd
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Post by sudokuEd »

Here is a simplified version of the Bullseye 3 tag solution. It uses 1 hypothetical so should be rated a 2.0. However, Para's solution is much better (just leave out step 27 since it's not needed) because he found a "45" that we missed. Great work Para :D . He uses a 'trick' chain move to unlock it - so I would give a 2.0 rating for Para's WT as well.

Great to have this one solved properly.

Bullseye 3 simplified tag WT
Preliminaries
0. r5c5 = 3

a) 6(3)n124 and n3 = {123} (no 4..9) -> {123} locked for n3
b) 16(2)n12 = {79} (no 1..6,8) -> {79} locked for r1
c) 10(2)n14 and n47 = {19/28/37/46} (no 5)
d) 20(3)n2 = {479/569/578} (no 1,2)
e) 7(2)n23 = [16/25/34] (no 7,8) -> r1c6 no 4,5,6; -> {123} locked for r1 in c689
e1) 7(2)n89 = {16/25/34} (no 7..9)
f) 21(3)n236 = {489/579/678} (no 1..3)
g) 13(2) n36 = {49/58/67} (no 1..3)
h) 11(3)n4 = {128/146/245} (no 7,9)
i) 9(2)n46 and n78 = {18/27/36/45} (no 9)
j) 19(3)n8 = {289/469/478/568} (no 1)

1. Innies c5: r46c5 = 3(2) = {12} (no 4..9)
1a. -> {12} locked for n5 and c5

2. 20(3)n2 = {569/578}(no 4) (combo {479} blocked by r1c4)
2a. -> 5 locked in 20(3)n2 for c5 & n2
2b. -> 4 locked in 19(3)n8 for c5 & n8
2c. cleanup: 7(2)n89: r9c7 no 3
2d. cleanup: 9(2)n78: r9c3 no 5

3. killer pair {79} locked for n2 in 20(3) and r1c4

4. 17(3)n1
4a. min r1c12 = 9 -> max r2c1 = 8 (no 9)
4b. max r1c12 = 14 -> min r2c1 = 3 (no 1,2)
4c. -> r2c1 no 6 (no valid permutations)

5. Innies r1234: r4c456 = h16(3) & must have 1/2 = {169/178/259/268}
5a. -> r4c46 no 4

6. Innies r6789: r6c456 = h13(3) = {148/157/247/256}
6a. -> r6c46 no 9

7. Innies c1234: r456c4 = h19(3) = {469/478/568}
7a. Innies c6789: r456c6 = h20(3) = {479/569/578}

8. "45" r5: r5c46 = h13(2) = {49/58/67} = [4/6/8..]
8a. {468} blocked from 18(3)n6

9. from steps 1 & 8: In n5 a Very hidden cage Vh26(4)r46c46 = {4589/4679/5678}
9a. Each combo must have exactly 2 numbers overlap with the 2 hidden cages in c4 & 6
i. {4589} -> h19(3)c4 & h20(3)c6 = {568-479} -> the 2 left-over candidates must be in r5c46 = 13 = [67]
ii. ........-> h19(3)c4 & h20(3)c6 = {469-578} -> the 2 left-over candidates must be in r5c46 = 13 = [67]
iii.........-> h19(3)c4 & h20(3)c6 = {478-569} -> the 2 left-over candidates must be in r5c46 = 13 = [76]
iv. {4679} -> h19(3)c4 & h20(3)c6 = {478-569} -> the 2 left-over candidates must be in r5c46 = 13 = [85]
v. {5678} -> h19(3)c4 & h20(3)c6 = {478-569} -> the 2 left-over candidates must be in r5c46 = 13 = [49]

10. 6 in Vh26(4) in {4679/5678} must be in c6 (see 9iv & v) -> no 6 in r46c4
10a. 7 in Vh26(4) in {4679/5678} must be in c4 (see 9iv & v) -> no 7 in r46c6
10b. r5c4 = {4678}(no 59)(step 9a.i..v)
10c. r5c6 = {5679}(no 48)(h13(2)r5c46)

11. Same thing with r46 in n5. Very hidden Vh26(4)r46c46 = {4589/4679/5678}
11a. Each combo must have exactly 2 numbers overlap with the 2 hidden cages in r4 & r6
i. {4589} -> h16(3)r4 & h13(3)r6 = {259-148} = {5[2]9-4[1]8}
ii. {4679} Blocked: Like this: h16(3)r4 & h13(3)r6 = {169-247} = [916-724] but r46c4 = [79] clashes with r1c4
iii. {5678} -> h16(3)r4 & h13(3)r6 = {178-256} = [718-526] BUT BLOCKED: h19(3)c4 cannot have [75]
iv.....................................= {268-157} = [826-715]

12. In summary, h16(3)r4 & h13(3)r6 = {259-148/268-157} = 8{..}
12a. r46c5 = [21]
12b. 8 locked for n5 -> h13(2)r5c46 = {49/67}(no 5)
12c. h16(3)r4 = [2]{59}/[826](no 7)(no 8 in r4c6)
12d. h13(3)r6 = 1{48/57}(no 6) = {4[1]8}/[715](no 5 in r6c4)
12e. clean-ups:no 8 in r3c1
12f. no 9 in r7c1
12g. no 8 in r7c9
12h. 2 in 6(3)n1 must be in r3: 2 locked for r3
12i. no 8 in r4c1

13. 21(3)n236 = {489/678}(no 5) ({579} blocked by no digits in r3c6)

14. Outties and Innie c9: r5c9 less r19c8 = 3
14a. min r19c8 = 3 -> min r5c9 = 6 (no 1,2,4,5)
14b. max r5c9 = 9 -> max r19c8 = 6 (no 6..9)

Now the trick that Caida spotted. Maybe this is one for Sudoku SolverV3
15. Outies c9: r159c8 + r5c7 = 15 AND empty rectange/generalized X-Wing on 1 for c89 between 6(3)n3 & r9c8 IF r19c8 = 1 -> min r19c8 = 5. Like this.
15a. If r19c8 = [12/13] = 3/4 -> 1 in c9 must be in n9 -> 1 in n6 must be in r5c7 which is in outies of c9. But since outies of c9 = 15 and 3/4 + 1 = 4/5 ->

r5c8 would have to be 10/11! therefore r19c8 must be min of 5 = [14]/{23}
15b. If r19c8 = [21/31] = 3/4 -> generalized X-Wing on 1 with 6(3)n3 -> 1 in n6 must be in r5c7. But since outies c9 = 15, this would force r5c8 = 10/11.

Therefore, r19c8 cannot be [21/31]
15b. -> min r19c8 = 5 -> max r5c78 = 10 (outies c9)
15c. -> min r5c9 = 8 (cage sum) (no 6,7)
15d. min r19c8 = 5 -> min r9c8 = 2

16. 18(3)n6 must have 8/9 = {189/459}(no 2,6,7) ({279} blocked by h13(2)r5c46)
16a. = 9{18/45}: 9 locked for r5 & n6
16b. no 4 in r5c4 (h13(2)r5c46) or r3c9
16c. 9 in {489} in 21(3)n2 must be in r3c7 -> no 4 in r3c7

17. NP {67} in r5c46: both locked for r5 & n5

18. h16(3)r4c456 = [2]{59}: all locked for r4 & n5
18a. no 1 r3c1
18b. no 4 or 8 in r3c9

19. NP {48} in r6c46: both locked for r6
19a. no 2 or 6 in r7c1
19b. no 1 or 5 in r7c9

20. 2 locked for r5 in 11(3)n4 = 2{18/45}
20a. 2 locked for n4
20b. no 8 in r7c1
20c. min r6c3 = 3 -> max r7c34 = 9 (no 9)

21. h19(3)r456c4 = {469/568}
21a. r5c4 = 6
21b. r5c6 = 7
21c. no 3 in r9c3

22. deleted

23. Outies and Innie r1: r2c19 – r1c5 = 1
23a. min r1c5 = 5 -> min r2c19 = 6 -> r2c1 no 3

24. 17(3)n1 = {458/467} -> 4 locked in n1, not elsewhere in n1
24a. r4c1 no 6

Para calls this next move a killer Y-wing in his WT
25. r456c4 must contain either 8 or 9 – if they contain a 9 then r123c5 must contain 9 and r789c5 contains an 8; otherwise r456c4 contains an 8
25b. -> r789c4 no 8 (either 8 is in r456c4 or it is in r789c5)
25c. -> r9c3 no 1
(note: Para's "45" on n236 eliminates 8 from r2c4 which would have now left hidden single 8 in r6c4 for us)

Now a very productive hypothetical - which means we should rate this puzzle as a 2.0
26. "45" i/o c9: r19c8 = 5/6 = {23}/[14/15/24] -> r1289 = 14/15 (cage sums = 20) AND must have 1 for c9
26a. r19c8 = [23] -> r1289c9 = 15: r12c9 = {13} = 4 -> r89c9 = 11 = {29/56} ({47} blocked by [1/3/4/7] needed by 9(2)c9)
26b. r19c8 = [32] -> r1289c9 = 15: r12c9 = {12} = 3 -> r89c9 = 12 = {39/57} ({48} blocked by [4/8] needed by 9(2) & 13(2)c9)
26c. r19c8 = [14] -> r1289c9 = 15: r12c9 = {23} = 5 -> r89c9 = 10 = {19} ({46} blocked - r89 must have 1 for c9)
26d. r19c8 = [15] -> r1289c9 = 14: r12c9 = {23} = 5 -> r89c9 = 9 = {18} ({45} blocked - 1 must be in r89 for c9)
26e. r19c8 = [24] -> r1289c9 = 14: r12c9 = {13} = 4 -> r89c9 = 10 = {28} ({46} blocked by 4 in r9c8)

27. In summary, r1289c9 = 1{356/239/238/257}(no 4)
27a. 14(3)n9 = {149/158/239/248/257/356} ({347} blocked by no {37} possible in r89c9 steps 26a..e.)

28. {459} is blocked from 18(3)n6 by 4s in c9. Like this.
28a. 4 in r7c9 -> 5 in r6c9 -> {459} blocked from 18(3)
28b. 4 in r4c9 -> {459} blocked from 18(3)
28c. 18(3)n6 = {189}: all locked for r5 & n6
28d. no 5 or 8 in r3c9
28e. 8 must be in 21(3)n2: only in r3: 8 locked for r3

29. 11(3)n4 = {245}: all locked for n4
29a. no 6 in r3c1
29b. no 8 in r7c1

30. r4c2 = 8 (hsingle n4)

31. r6c4 = 8 (hsingle c4)

32. 8 in n1 only in 17(3) = {458}: all locked for n1
32a. 8 locked for c1
32b. 17(3)n7: no 8 = {179/269/359/467}

33. r2c4 = 4 (hsingle c4)
33b. 4 in n1 only in r1: 4 locked for r1
33c. no 3 r1c6
33d. 3 in r1 only in c89: no 3 in r2c9

34. r4c4 = 5 (cage sum h19(3)c4)
34a. no 4 in r9c3

35. r46c6 = [94]

36. 22(5)n1 must have 4 & 8 (& no 5) = 48{127/136}(no 9) = 1{..} = [2/3..]
36a. 1 required in 22(5)n1 only in n1: 1 locked for n1

37. Killer pair 2/3 between 22(5)n1 & r3c3: no 3 in r3c1
37a. no 7 in r4c1

38. NP {13} in r4c13: both locked for r4 and n4
38a. no 7 in r7c1

39. NTriple {467} in r4c789: all locked for n6
39a. no 2,3 in r7c9

40. r3c8 = 4 (hsingle r3)

41. NP {79} in n1 in r3c1 & r1c3: 7 locked for n1

42. 22(5)n1 = 48{136}(no 2): 3 locked for n1

43. r3c3 = 2
43a. no 7 in r9c4

44. 2 in n2 only in c6: locked for c6
44a. no 5 in r9c7

45. hidden pair {13} in r3c24:no 6 in r3c2
45a. 6 in n1 only in r2: 6 locked for r2

46. hidden triple {123} in n2: no 8 in r2c6

47. r3c5 = 5 (hsingle r3)

48. NP {68} in n2: no 8 in r2c5

49. "45" c1: 4 outies r159c2 & r5c3 = 20
49. max r15c2 + r5c3 = [545] = 14 -> min r9c2 = 6

50. 17(3)n7 must have 1 of 6/7/9 for c1 and 1 of 6/7/9 in r9c2 -> must have 2 of {679}
50a. = {179/269/467}(no 3,5)

51. 3 in c1 in 1 of 10(2) cages -> 7 locked for c1

No way to go - time for some creativity from Peter
52. 26(5)n236 must contain a 4 and only 1 of {123}
52a. 26(5)n236 = {14678/24569/24578}(no 3)
i. {14579} blocked: r2c6 = 1 -> r1c67 = [25] -> can’t place 5 in 26(5)
ii. {34568} blocked: r2c6 = 3 -> r3c4 = 1 -> r1c67 = [25] -> can’t place 5 in 26(5)
52b. r2c6 no 3
52c. single: r3c4 = 3: no 6 in r9c3
52d. r4c3 = 1
52e. r4c1 = 3: no 7 in r6c1
52f. r3c1 = 7: no 3 in r7c1, no 6 in r4c9
52g. r1c34 = [97]
52h. r12c5 = [69](cage sum)
52i. r3c6 = 8
52j. r1c7 = 5
52k. r1c6 = 2
52l. r1c2 = 4
52m. r1c1 = 8
52n. r2c1 = 5
52o. r2c6 = 1: no 6 in r9c7
52p. r2c9 = 2: no 7 in r7c9
52q. r3c2 = 1

53. Naked pair {78} locked for 26(5)n236 in r2c78

54. r3c8 = 6

55. r8c4 = 9 (hsingle c4)

56. r6c1 = 9 (hsingle n4)

57. r7c1 = 1
57a. r7c4 = 2
57b. r9c34 = [81]:no 6 in r9c6

58. r67c3 = [73/64] = [3/6..]
58a. r7c3 = {34}

59. Killer pair {36} in r267c3: both locked for c3

60. 31(5)n478: = [9]{4567}(no 2,3) (only permutation without 8)

61. r7c3 = 3 (hsingle n7), r6c3 = 7(cage sum), r6c2 = 6, r2c23 = [36], NP {57} at r78c2: both locked for c2 & n7
r59c2 = [29], r5c13 = [45], r8c3 = 4

62. 16(3)n689 = [259/268/358/367]
r7c7 = {789}

63. 14(3)n9 = {257/356}(no 1,8)
63a. = 5{..}: 5 locked for n9

64. r1c9 = 1 (hsingle c9)

Rest is singles. Whew
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