Assassin 48

rcbroughton · Post by **rcbroughton** » Mon Apr 30, 2007 8:12 pm

Building from Mike's position - but didn't manage to reuse any of my other moves . . .

76. 24(5)n3={12489}/{12579}/{13479}/{14568}/{23469}/{23478}
76a. {14568} - 5 must be at r2c8
76b. {23469} - 3 must be at r1c9 -> 4 must be at r2c8 because of r1c7=5, r1c3=4
76c. no 6 at r2c8

77. 14(3)n14 - r3c2 can't have 4,7

78. 14(3)n47 - r7c3 can't have 8

79. innies in n1=19(4)
79a. r23c1=12,11,9,6
79b. combos in 19(4) with 6 are: 4{18}6, 4{27}6 - no 6 at r3c1

80. 45 on n1 - r1c3+r3c2=r4c1+6
80a. r1c3+r3c2=7,8,10,13
80b. r1c3=4,5
80c. r3c2=3,6,8,9 - no 5

81. 45 on n1 - innies r1c3+r2c12+r3c3=19(4)
81a. r23c1=6,9,11,12
81b. only combos with 9 are 4{15}9/5{35}9
81c. no 9 at r23c1

82. 45 on n7 - r7c2+r9c3 = r6c1+7
82a. r7c2+r9c3 = 8,10,11,12,13,15,16,17
82b. r6c1=1,3,6,8,9 - no 2

83. 45 on n9 - r7c8+r9c7 = r6c9
83a. r7c8+r9c7 = 5,7,9 - no 8 at r6c9

84. 45 on r4 - r4c14569 = r3c28+20
84a. innies r4c14569 total 29,30,32,35
84b. outies r3c28 need to total 9,10,12,15
84c. outies = 9=[36]10=[37]12=[not poss]15=[87]/[96] - no 6 r3c2
84d. innie total 32 not poss - no 4 at r4c1

85. revisit step 81. no 8 now at r23c1
85a. 13(3)={157}/{247} - no 3 - 7 locked for c1

86. 13(3)c1 & 16(3)c1 now lock 4 - nowhere else in c1

87. 45 on r9 - innies = 22={12469}/{12568}
87a. 31(5)n9 = {16789}/{45679}
87b. for {12568} - 2 locked at c12, c89={16}/{18}/{68} - {56} blocked by c5
87c. no 5 in c89

88. 45 on n2356 - r1c3 r5c3 r7c8 r7c9 r8c9 r7c5 equal 25
88a. c3 total 7,8,9,12,13
88b. r7c5=3,5,8
88c. 31(5) n9 blocks {14}/{15}/{58}in n9 cells
7+3->n9=15(3)=[483] - blocked by 14(3)c9
7+5->n9=13(3)=[283] - blocked by 14(3)c9
7+8->n9=10(3)=2{35} - blocked by 14(3)c9
8+3->n9=14(3)=2{48} - blocked by 14(3)c9
8+5->n9=12(3)=[435] - blocked by 14(3)c9
8+8->n9=9(3)={234}
9+3->n9=13(3)=[283] - blocked by 14(3)c9
9+5->n9=11(3)=2{18}/2{45}
9+8->n9=8(3) - not possible
12+3->n9=10(3)=[253] - blocked by 14(3)c9
12+5->n9=8(3) - not possible
12+8->n9=5(3) - not possible
13+3->n9=9(3)={234}
13+5->n9=7(3)-not possible
13+8->n9=4(3) - not possible
88d. no 5 r7c9 (whew!!)

89. 31(5)n9={16789}/{45679} - no 4 at r8c8

90. killer pair {45} in n1 - r1c3 and 13(3)atr23c1 - nowhere else in n1

That step 88 took it out of me! time for a rest!!

[edit] - ok I'll add a couple more

91. 45 on n3 n9 innies total 35
91a. r19c7={35}=8
91b.r37c8=8=[62]/9=[72]/10=[64]/11=[74]
91c.r23c9=10 or 11
91d.r78c9=9,7 or 5
r37=8, r23=11, r78=8 - not poss
r37=9, r23=10, r78=8 - not poss
r37=10, r23=11, r78=6 - not poss
r37=11, r23=10, r78=6 - not poss
r37=8, r23=10={82}/{73}, r78=9={18}/[45]
r37=9, r23=11={92}/{83}/{65}, r78=7=[25]
r37=10, r23=10={82}/{73}, r78=7=[25]
r37=11, r23=11={92}/{83}/{65}, r78=5={23}
91e. no 4 r23c9

92. 19(3) c9 blocks {257} in 14(3) c9 - no 5 at r8c9

93. 14(3) and 19(3) now lock 3, 8 for c9

94. 24(5)n3 - {12489}/{14568}/{12579} - no 7 at r2c8

95. innies in n3 - 21(4)={2379}/{3567} -24(5)n3 blocks combos with 2&6 - no 8,
95a. so 8 locked in 24(5)
95b. 24(5)={12489}/{14568} - no 7 in 24(5)

Regards
Richard

PsyMar · Post by **PsyMar** » Mon Apr 30, 2007 10:52 pm

Here's my solution for V1 -- all fairly standard killer stuff until step 49, which involves Medusa Coloring of a sort, with a bit of a twist at the end. It might be easier to write it as a double implication chain in Eureka notation, but I didn't.

I am, as always, quite prone to errors and quite slow to fix them even after they're pointed out. I don't mean to seem ungrateful, I'm just lazy.

Maybe I'll find a few days this summer to fix them all.

1. sole combination for 3/2 in n2 = {12} pair -> elim {12} from rest of n2/c5
2. sole combination for 8/2 in n8 = {35} pair -> elim {35} from rest of n8/c5
3. sole combination for 7/2 in n2 = {34} pair -> elim {34} from rest of n2/c6
4. sole combination for 13/2 in n8 = {67} pair -> elim {67} from rest of n8/c6
5. innies of c789 = r159c7 = 7/3 = {124} triple -> elim {124} from rest of c7
6. sole combination for 3/2 in r9 = {12} pair -> elim {12} from rest of r9
7. sole permutation for 10/2 in r9 = [64] -> make eliminations
8. permutations for 9/2 in r1 = [54|81]
9. 9 of r6 locked in n5 -> elim from rest of n5
10a. combinations for 14/2 in c4 = {59|68} -> no 7
10b. combinations for 10/2 in c4 = {19|28}
10c. 14/2 and 10/2 in c4 form killer pair {89} -> elim {89} from r1456c4
11. sole combination for 12/2 in r1 = {57} pair -> elim {57} from rest of r1 -> 9/2 in r1 = [81] -> 3/2 in r9 = [12] -> 3/2 in n2 = [21] && r5c7 = 4
12. sole combination for 10/2 in n8 = {28} -> elim {28} from rest of c4/n8 -> r7c5 = 9
13. sole combination for 14/2 in n2 = {59} -> elim {59} from rest of c4/n2 -> 12/2 in r1 = [57] -> r3c5 = 6
14. outies of n5 = r5c3 = 9
15. innies of r5 = r5c456 = 14/3; only permutation is [185]
16. innies of r1234 = 19/3 = {379|469} -> r4c6 = 9 -> r6c6 = 2
17. sole combination for 35/5 in n9 = {56789} -> elim from rest of n9
18. outies-innies of n9 = r6c9-r7c8 = 4 -> r6c9 = {578}
19. sole combination for 7/3 in c1 = {124} -> elim from rest of c1
20. outies-innies of n1 = r4c1-r3c2 = -5 -> r3c2 = {79}, r4c1 = {24}
21. r3c1 = 1 {hidden single in c1}
22. outies-innies of n7 = r6c1-r7c2 = 0 -> r6c1 = r7c2 = {3578}
23. r79c12+r8c1 = naked quintuple {35789} -> elim from rest of n7
24. r78c3+r8c2 = {124} = 7; rest of 19/5 in n7 = r9c12 = 12/2 = {39|57}
25. split cage 12/2 in r9c12 forms killer pair {35} with r9c5; elim 5 from r9c89
26. 8 of r9 locked in n7; elim from rest of n7
27. outies-innies of n3 = r4c9-r3c8 = -1 -> r3c8 != 5, r4c9 != 5
28a. combinations for 12/2 in c9 = {138|147|345}
28b. innies of c9 = r159c9 = 18/3 without 1 or 5 = {279|369} (378|468 conflict with 28a) -> no 4|8 in r159c9, no 9 in rest of c9
29. r9c8 = 8 (hidden single in r9)
30. combinations for 15/3 in c9 = {168|258|267|456} (348|357 conflict with 28a) -> no 3 in 15/3 in r9
31. outies-innies of n3 = r4c9-r3c8 = -1 -> r3c8 != 4, r4c9 != 7
32. outies of n3 = r4c789 = 15/3 = {168|258} (does not have 4 or 9; 267|357 conflicts with 9/2 in n6) -> no 3|7 in r4c789; elim 8 from rest of r4c789
33. outies of n1 = r4c123 = 11/3 = {146|245} (must have 2 or 4; 236 conflicts with r4c4) -> no 3|7 in r4c123 && no 4 in rest of r4 -> r4c5 = 7 -> r6c5 = 4
34. r4c4 = 3 (hidden single in r4) -> r6c4 = 6
35. 8 of r4 locked in n6 -> elim from rest of n6
36. combinations for 12/3 in c9 = {147|345} -> must have 4 in r78c9 -> elim from rest of c9/n9
37. combinations for 15/3 in c9 = {168|258} -> no 7, and r4c9 cannot be 6 as only permutation for {168} is [681]
38. outies-innies of n3 = r4c9-r3c8 = -1 -> r3c8 != 7
39. combinations for 28/5 in n3 = {24679|34579|34678} (does not have 1; must have 4 and 7 (4 and 7 of n3 locked in cage)
40. outies of n1 = r4c123 = 11/3 = {146|245} (by step 33) -> either 5 or 6 is in r4c2 (only spot for 5 or 6, and one of the two must be there)
41. r478c3 = naked triple {124} -> elim from rest of c3
42. r3c6 = 4 (hidden single) -> r2c6 = 3
43. 2 of n1 locked in r2 -> elim from rest of r2
44. combinations for 28/5 in n1 with no 1 or 2 = {34579|34678} -> no 3 in rest of n3
45. outies-innies of n3 = r4c9-r3c8 = -1 -> r4c9 != 2
46. permutations for 15/3 in c9 = [528|681] = [{56}{28}{18}]
47. innies of c9 = r159c9 = 18/3 = [369|639|927] = [{369}{236}{79}]
48. innies of c1 = r159c1 = 18/3 = [369|639|675|963] = [{369}{367}{359}]
49a. Medusa: (9)r3c8 blue <-> (2)r3c8 red <-> (2)r3c9 blue <-> (8) r3c9 red <-> (8) r4c9 blue <-> (1)r4c9 red
49b. Medusa continued: by cage sums in 15/3 in c9, (6)r2c9 red and (5)r2c9 blue <-> (5)r6c9 red <-> (7)r6c9 blue <-> (7)r9c9 red <-> (7)r9c2 blue
49c. (9)r3c8 and (7)r9c2 both blue -> blue elims all digits from r3c2 -> elim all blue digits mentioned, place all red digits mentioned
49d. naked singles: r4c8 = 6 -> r4c7 = 8 -> r4c2 = 5
50. r1c9 = hidden single (9) in c9
51. r5c9 = hidden single (2) in c9 -> r5c8 = 7
52. r7c8 = hidden single (1) in n9
53. 9 of n1 locked in c2 -> elim from rest of c2 -> naked singles and last-digit-in-cage moves solve it

Now I'll work on the hard version...

Andrew · Post by **Andrew** » Tue May 01, 2007 4:27 am

We now have 3 walkthroughs for the original Assassin 48 and all had different breakthroughs from the way that I finished it so here is my walkthrough. Looks like my way took Ruud's introductory comment more literally than the others.

All the other 3 have neat breakthroughs. Mine is a bit more routine but I hope still worth looking at.

Thanks for the corrections Ed. I've also added an extra comment to step 25.

1. R1C34 = {39/48/57}, no 1,2,6

2. R12C5 = {12}, locked for C5 and N2

3. R1C67 = [36/45/54/63/72/81], no 9, no 7,8 in R1C7

4. R23C4 = {59/68}

5. R23C6 = {34}, locked for C6 and N2, clean-up: no 8,9 in R1C3, no 5,6 in R1C7

6. R5C12 = {18/27/36/45}, no 9

7. R5C89 = {18/27/36/45}, no 9

8. R78C4 = {19/28/37/46}, no 5

9. R78C6 = {58/67}, no 1,2,9

10. R89C5 = {35}, locked for C5 and N8, clean-up: no 7 in R78C4, no 8 in R78C6
10a. Naked pair {67}in R78C6, locked for C6 and N8, clean-up: no 2,3 in R1C7, no 4 in R78C4

11. R9C67 = {12}, locked for R9

12. R9C45 = [64]

13. R234C1 = {124}, locked for C1, clean-up: no 5,7,8 in R5C2

14. 19(5) cage in N7 = 1{2349/2358/2457}, 1,2 locked for N7 [Edit. Unnecessary combinations removed.]

15. 35(5) cage in N9 = {56789}, locked for N9

16. Killer pair 8/9 in R23C4 and R78C4 for C4, clean-up: no 3,4 in R1C3

17. Naked pair {57} in R1C34, locked for R1 -> R1C67 = [81], R12C5 = [21], R9C67 = [12], clean-up: no 6 in R23C4, no 9 in R78C4
17a. Naked pair {59} in R23C4, locked for C4 and N2 -> R1C34 = [57]
17b. Naked pair {28} in R78C4, locked for C4 and N8

18. R3C5 = 6, R7C5 = 9 (naked singles)

19. R456C4 = {136} -> R5C3 = 9

20. R456C6 = {259} -> R5C7 = 4, clean-up: no 5 in R5C1, no 5 in R5C89

21. R5C6 = 5 (hidden single in R5)

22. 45 rule on R5 2 remaining innies R5C45 = 9 = [18]
[If this hadn’t fixed these two cells, there was killer triple 6/7/8 in R5C12, R5C5 and R5C89 for R5 which eliminates 6 from R5C4]

23. 45 rule on N9 1 outie R6C9 – 4 = 1 innie R7C8 -> R6C9 = {578}

24. 45 rule on N7 1 innie R7C2 = 1 outie R6C1 -> no 6 in R6C1, no 4 in R7C2 [Edited because of correction to step 14.]

25. Hidden triple {124} in R7C3 +R8C23
[Alternatively naked quint {35789} in R7C12 + R8C1 + R9C12, locked for N7. I think I saw the hidden triple first.]
25a. 19(3) cage (step 14) = 124{39/57}, no 8 -> R9C12 = {39/57}

26. Killer pair 3/5 in R9C12 and R9C5 for R9

27. 8 in R9 locked in R9C89, locked for N9

28. 8 in C1 locked in R678C1
28a. R678C1 = 8{39/57}, 9 only in R8C1 -> no 3 in R8C1

29. 45 rule on N1 1 innie R3C2 – 5 = 1 outie R4C1 -> R3C2 = {79}, R4C1 = {24} -> R3C1 = 1 (hidden single in C1)

30. 45 rule on N3 1 innie R3C8 – 1 = 1 outie R4C9 -> no 5 in R3C8, no 5,9 in R4C9

31. 45 rule on R1234 3 innies R4C456 = 19 = [379/649] -> R4C6 = 9, R6C6 = 2

32. 9 in N6 locked in R6C78, 13(3) cage in N69 = {139}, no 1,3 in R45C8, R6C7 = {39}, R6C8 = {139}, R7C8 = {13}, clean-up: no 6 in R5C9, no 8 in R6C9 (step 23}

33. 4 in N9 locked in R78C9, locked for C9

34. 8 in R6 locked in R6C123, locked for N4
34a. 45 rule on N7 3 outies R6C123 = 16 = 8{17/35}, no 4,6

35. R6C5 = 4 (hidden single in R6) -> R4C5 = 7, clean-up: no 8 in R3C8 (step 30)

36. R6C4 = 6 (hidden single in R6) -> R4C4 = 3, clean-up: no 4 in R3C8 (step 30)

37. Naked triple {124} in R478C4, locked for C3 [Edit. My usual error of calling a naked ... a killer ...]

38. R3C6 = 4 (hidden single in R3) -> R2C6 = 3

39. 2 in R3 locked in R3C89, locked for N3

40. 4 in N4 locked in R4C123
40a. 45 rule on N1 3 outies R4C123 = 11 = [254/452/461] -> R4C2 = {56}

41. 8 in N6 locked in R4C789
41a. 45 rule on N3 3 outies R4C789 = 15 = 8{16/25}
41b. 1 only in R4C9 -> no 6 in R4C9, clean-up: no 7 in R3C8 (step 30)

42. 45 rule on C1 3 innies R159C1 = 18 = 6{39/57}
42a. 5 only in R9C1 -> no 7 in R9C1, clean-up: no 5 in R9C2 (step 25a)

43. 45 rule on N3 3 innies R2C9 + R3C89 = 16 = 2{59/68}, no 3,7, clean-up: no 2 in R4C9 (step 30)
43a. 6 only in R2C9 -> no 8 in R2C9

44. R678C9 = 4{17/35}
44a. R234C9 = {168/258} (cannot be {159} which clashes with R678C9) = 8{16/25}, no 9, 8 locked for C9
44b. 2 only in R3C9 -> no 5 in R3C9

45. 9 in C9 locked in R19C9
45a.45 rule on C9 3 innies R159C9 = 18 = 9{27/36}
45b. 2 only in R5C9 -> no 7 in R5C9, clean-up: no 2 in R5C8
45c. 6 only in R1C9 -> no 3 in R1C9 [Edit. Typo corrected]

46. R9C8 = 8 (hidden single in R9)

47. 6 in C9 locked in R12C9, locked for N3

48. R4C789 (step 41a) = 8{16/25}, 2 only in R4C8 -> no 5 in R4C8
48a. 1 only in R4C9 and no 8 in R4C8 -> no 6 in R4C7
[This also comes from the 16(3) cage in N36 with 2 locked in R34C8]

49. 6 in C7 locked in R78C7, locked for N9

50. 16(3) cage in N47 = 8{17/35} (step 34a with R7C2 = R6C1 from step 24), 1 only in R6C2 -> no 7 in R6C2

[While looking for how to make further progress, I spotted that R6C1 = R7C2 (step 24) produces a situation similar to step 34 in Mike’s Assassin 46 walkthrough; Ed told me that he did a similar move in Ruud’s second Special Killer-X. However it didn’t help here because at this stage, R6C1/R7C2 contain the same numbers as R123C3. I haven’t put this comment after step 24 because this process doesn’t work the other way round; it can’t eliminate candidates from R123C3.]

Ruud wrote "You really need wings to complete this journey."

These are presumably the four 5 cell cages, the wings of a butterfly. Until now I've made limited use of the two lower wings in N7 and N9. Now the upper left wing in N1 provides the final breakthrough, in the spirit of Ruud's introductory message.

51. 28(5) cage in N1 = 368{29/47}
51a. 28(5) cage cannot be {23689} because R2C2 = 2 => R1C12 = {69} clashes with R1C9 [Edit. Typo corrected.]
51b. 28(5) cage = {34678}, no 2,9, locked for N1
51c. 4 in N1 locked in R12C2, locked for C2
51d. 4 in N7 locked in R78C3, locked for C3
[Step 51a could have been used immediately after step 45c, although I didn't see it at that stage. By leaving it in the order that I made these steps, step 51 proves to much more powerful as can be seen below.]

and the rest is naked singles although it’s quicker if naked pairs and hidden singles are also used

rcbroughton · Post by **rcbroughton** » Tue May 01, 2007 8:11 am

Hmm - this one is refusing to give in

91. 45 rule on n1 n3 innies total 40 - eliminate 7 at r3c1
96a. innies of n1 total 19
96b. innies of n3 total 21
96c. r17c1=[45]/[53] ([43] blocked by innies of r1
96d. so when r17c1=[53] rest of innies in n1 total 14, rest of innies in n3 total 18
innies n1={14}9/{24}8/{34} - 7 not used
innies n3=7{29}/{567}
96e. and when r17c1=[45] rest of innies in n1 total 15, rest of innies in n3 total 16
innies n1={25}8/[753]
innies n3=[367] - no 7 at r3c1

97. revisiting step 88 - we can also remove {29} from r23c9 using the same logic (should have seen it at the time)

98. innies of n3 now={3567} - locked for n3

99. 45 rule on c1&c9 (yes!!) innies r159c19=28 - removes 5 from r9c1 - here's how
99a. innies c1 r159c1=16={169}/{268}/{358} -others blocked by 13(3) & 16(3)c1
99b. innies c9 r159c9=12={129}/{147}/{246} -others blocked by 19(3) & 14(3)c9
99c. r1c19 can't be [34] - blocked by 12(2) & 11(2) r1
99d. r9c19 can't be [89]/[54] - blocked by 11(2) & 12(2) r9
99e. r5c19 can't be [91]/[97]/[82]/[51]/[57] - blocked by h(23)5 r5
99f. working the combinations (I'll include them later for anyone interested!) - no 5 r9c1

Rgds
Richard

PsyMar · Post by **PsyMar** » Tue May 01, 2007 2:37 pm

I just thought I'd take a crack at the hevvie myself, and found it isn't as hard as it was cracked up to be. I got stuck at the *very end*, thinking there were two solutions -- I honestly had 70 of 81 digits placed, and the rest were all pairs, and I thought there might be a dual solution, but sumocue said otherwise. I've left out the one step I didn't get, just so you guys can get it.

Edit: rcbroughton has pointed out a glaring error I made in step 3 which invalidates the entire thing, even though I got the right solution by luck. Argh. Sorry people.

I'll leave this up here as a lesson for others to avoid.

1. sole combination for 3/2 in n8 = {12} pair -> elim from rest of n8/c4
2. combinations for 9/2 in c5 = {36|45} -> no 7..9
3. combinations for 15 in n8 = {59|68} -> no 1..4|7; forms killer pair {56} with 9/2 in n8 -> elim from rest of n8
4. innies of n8 = r7c5+r9c46 = 18/3 with no 1|2|5|6 = {378} triple (sole combination) -> elim from rest of n8
5. sole combination for 9/2 in c5 = {45} pair -> elim from rest of n8/c5
6. r78c6 = naked pair {69} -> elim from rest of c6
7. combinations for 12/4 in c67 = {1236|1245} -> no 7..9; elim 1&2 from r5c45
8. combinations for 7/2 in n2 = {25|34} -> no 1|7|8
9. r23456c6 = naked quad {12345} in c6 -> elim from rest of c6
10. 1 of c6 locked in 12/4 and in n5 -> elim from rest of n5 and from r5c7
11. permutations for 12/2 in r9 = [75|84] = [{78}{45}]
12. permutations for 11/2 in r1 = [74|83] = [{78}{34}]
13. outies of c6 = r159c7 = 12/3; r19c7 = {34|35|45} = 7..9 -> r5c7 = 3..5
14. sole combination for 12/4 in c67 = {1245} -> no 3; elim 4&5 from r5c45
15. r59c7 = naked pair {45} -> elim {45} from rest of c7 -> 11/2 in r1 = [83] -> 12/2 in r9 = [75] -> r5c7 = 4 && 9/2 in n8 = [54]
16. r456c6 = naked triple {125} -> elim from rest of n5/c6
17. 7/2 in n2 = {34} naked pair -> elim from rest of n2
18. sole combination for 14/2 in n2 = {59} -> no 1|2|6|7; elim from rest of n2/c4
19. sole permutation for 12/2 in r1 = [57]
20. innies of n2 = r3c5 = 1
21. r12c5 = naked pair {26} -> elim from rest of c5
22. outies of n5 = r5c3+r7c5 = 11/2 = {38} pair (sole combination) -> no other digits; elim 3&8 from r5c5+r7c33
23. sole combination for 14/2 in r5 = {59} pair (68 would conflict with r5c34) -> elim from rest of r5/n4 -> r5c5 = 7
24. sole combination for 8/2 in r5 = {26} pair -> elim from rest of r5/n6 -> r5c6 = 1
25. r59c4 = naked pair {38} -> elim from rest of c4
26. sole combination for 11/2 in r9 = {38} pair -> elim from rest of r9
27. r59c3 = naked pair {38} -> elim from rest of c3
28. outies-innies of n9 = r6c9-r7c8 = 5 -> r6c9 = {789} && r7c8 = {234}
29. 5 of c9 locked in 19/3 -> sole combination for 19/3 is {568} triple -> elim from rest of c9 -> 8/2 in r5 = [62]
30. 6 of r9 locked in n7 -> not in rest of n7
31. sole combination for 31/5 in n9 = {16789} (no 5, and must have two of {129} due to r9c89) -> no 2..4; elim {16789} from rest of n9
32. r7c8 = 2 (hidden single) -> 3/2 in n8 = [12]
33. sole combination for 14/3 in c9 = {347} -> 7 in r6c9
34. r9c89 = naked pair {19} -> elim from rest of r9/n9
35. 6 of n9 locked in c7 -> elim from rest of c7
36. r78c9 = naked pair {34} -> elim 4 from r1c9
37. unique corner: r19c89 cannot all be {19} so r1c8 = 4
38. sole combination for 24/5 in n3 with 4 and without 3, 5 or 6 = {12489} -> no 7, elim from rest of n3
39. sole permutation for 11/3 in c78 = [713] -> r8c8 = 8
40. sole permutation for 16/3 in c78 = [952] -> 12/4 in c67 = [5142] -> r4c9 = 8 -> r4c5 = 9
41. outies of n1 = r4c123 = 13/3 = {247} triple (sole combination) -> elim {247} from rest of r4/n4 -> r4c4 = 6 -> r6c4 = 4
42. outies-innies of n1 = r4c1-r3c2 = -1 -> r3c2 = {38} && r4c1 = {27}
43. innies of n1 = r159c1 = [952|196] (only permutations) -> r1c1 = {19}
44. r1c19 = naked pair {19} -> r1c2 != 1|9 -> r1c2 = {26}
45. r19c2 = naked pair {26} -> elim from rest of c2
46. combinations for 13/3 in c1 (must have 2 or 7) = {238|247} -> no 1|5|6|9 && elim 2 from rest of c1 -> r9c1 = 6 -> r9c2 = 2 -> r1c2 = 6 -> 8/2 in c5 = [26] -> 19/3 in c9 = [568] -> 14/2 in c4 = [95] -> r2c8 = 1 -> r1c9 = 9 -> r1c1 = 1 && r9c8 = 9 && r9c9 = 1
47. r3c3 = 9 (hidden single in n1)
48. r6c3 = 6 (hidden single in n4)
49. r6c2 = 1 (hidden single in n4) -> r7c2 = 7 (last digit in cage) -> 22/5 in n7 = [49162] (left-right, top-bottom) -> 16/3 in c1 = [853] -> lots of naked singles and last-digit-in-cage moves -- just *one more* tricky move solves it, but I admit asking sudocue as I didn't see it and thought there might be multiple solutions (there aren't.)

Here's as far as I got:
.-----------.-----------.-----.-----------.-----------.
|(26) |(12) |(8) |(11) |(24) |
| 1 6 | 5 7 | 2 | 8 3 | 4 9 |
:-----. '-----.-----: :-----.-----' .-----:
|(13) | |(14) | |(7) | |(19) |
| 247 | 38 27 | 9 | 6 | 34 | 28 1 | 5 |
| :-----. | :-----: | .-----: |
| |(14) | | |(28) | | |(11) | |
| 24 | 38 | 9 | 5 | 1 | 34 | 28 | 7 | 6 |
| | '-----+-----: :-----+-----' | |
| | |(21) | |(12) | | |
| 27 | 4 27 | 6 | 9 | 5 | 1 3 | 8 |
:-----'-----.-----' | | '-----.-----'-----:
|(14) | | | |(8) |
| 9 5 | 3 8 | 7 | 1 4 | 6 2 |
:-----.-----'-----. | | .-----'-----.-----:
|(16) |(14) | | | |(16) |(14) |
| 8 | 1 6 | 4 | 3 | 2 | 9 5 | 7 |
| | .-----+-----: :-----+-----. | |
| | |(22) |(3) | |(15) |(31) | | |
| 5 | 7 | 4 | 1 | 8 | 9 | 6 | 2 | 3 |
| :-----' | :-----: | '-----: |
| | | |(9) | | | |
| 3 | 9 1 | 2 | 5 | 6 | 7 8 | 4 |
:-----' .-----'-----: :-----'-----. '-----:
| |(11) | |(12) | |
| 6 2 | 8 3 | 4 | 7 5 | 9 1 |
'-----------'-----------'-----'-----------'-----------'

rcbroughton · Post by **rcbroughton** » Tue May 01, 2007 3:48 pm

PsyMar wrote:I just thought I'd take a crack at the hevvie myself, and found it isn't as hard as it was cracked up to be.

Hmm - we're still struggling with it

PsyMar wrote:3. combinations for 15 in n8 = {59|68} -> no 1..4|7; forms killer pair {56} with 9/2 in n8 -> elim from rest of n8

Why can't 15(2) in n8 be {78}? - {59}=14 and {68}=14.
I'm still seeing a 5 elsewhere in n8 ...

Para · Post by **Para** » Tue May 01, 2007 4:56 pm

I think Richard has a point. Seems like a logic flaw that didn't mess up the puzzle. I made one in my first ever walk-through. As your solution shows R78C6 = [96] and R89C5 = [54], so {56} are locked in these 2 cages in the end

PsyMar wrote: I've left out the one step I didn't get, just so you guys can get it.

R2C23 = 10 = [37]; [82] clashes with R2C7.
Not that hard.

greetings

Para

rcbroughton · Post by **rcbroughton** » Tue May 01, 2007 9:42 pm

Moving on . . . couple of ugly moves are all I could find, then a rather complex analysis of combinations that yields another placement:

100. can't have a 3 at r2c3 or r3c3 as they would eliminates all 3s in r9
-> r2c3=3 -> r2c6=4 -> r3c6=3 -> r1c7=3 -> r9c7=5 -> r1c3=5 -> r5c3=8 -> r5c4=3

101. Can't place a 3 at r2c9 as it would eliminate all places for 4 in r3 -> r2c9=3 -> r1c7=5 -> r1c3=4 -> r2c6=4
101a. 19(3)n39 no 7 at r3c9

102. 45 rule on rows 1 to 2. Outies r3c3 r3c7 r3c1 r4c1 r3c4 r3c6 r3c9 r4c9 equal 44
102a. r4c19 potentially [18]/[19]/[28]/[29]/[78]/[79] - only possible placements are:
479315
7____8

165493
7____9

425396
7____8

569413
7____9

469325
7____8

175483
7____9

495326
7____8

265483
7____9

all the others blocked by no valid combos for 13(3)c1 or 19(3)c9

102b. only 7 allowed at r4c1 and no 1 at r3c3

103. 14(3)n14 can't be {347} - no 3 at r3c2

[For anyone intereseted in the combinations not allowed at step 102, here they are:

Code: Select all

279485 179485 589426 269485 285496 295486_529486 589426
1____9 1____9 1____9 1____9 1____9 1____9_1____9 1____9

529486 179485 269485 285496 295486 589426
2____8 2____8 2____8 2____8 2____8 2____8

169485 185496 195486 519486 589416
2____9 2____9 2____9 2____9 2____9

569423 275483 175493 579413 265493
7____8 7____8 7____8 7____8 7____8

425386 485326 415396 469315 495316
7____9 7____9 7____9 7____9 7____9

[/color]

Rgds
Richard

PsyMar · Post by **PsyMar** » Wed May 02, 2007 1:45 pm

rcbroughton wrote:
PsyMar wrote:I just thought I'd take a crack at the hevvie myself, and found it isn't as hard as it was cracked up to be.
Hmm - we're still struggling with it

PsyMar wrote:3. combinations for 15 in n8 = {59|68} -> no 1..4|7; forms killer pair {56} with 9/2 in n8 -> elim from rest of n8
Why can't 15(2) in n8 be {78}? - {59}=14 and {68}=14.
I'm still seeing a 5 elsewhere in n8 ...

Ah crud. Looks like I gave combinations for the 15 as if it were a 14, and proceeded to screw up from there and got lucky and solved it. Can't believe I did that. Sorry.

mhparker · Post by **mhparker** » Wed May 02, 2007 4:01 pm

I gave combinations for the 15 as if it were a 14, and ... got lucky and solved it.

Can I take your result please?

I can't remember when I last saw a puzzle that was so gridlocked...

In the meantime, here's the next batch of minor steps, and - in case you want to ask - no, they don't achieve anything spectacular. It's a bit like trying to crack a safe armed with only a pile of sandpaper - not so much breaking in, more like just smoothing off a few rough edges.

104. i/o difference(n1): r4c23 = r1c3 + 1 -> min. of r4c23 = 5
-> no 1 in r4c2
(reason: r4c23 = {14} would imply r1c3 = 5, since r14c3 are peers)

105a. i/o difference(n6): r4c9 + r6c789 = r3c8 + 22
-> no 6 in r6c8
(reason: would force r3c8 to 7 -> r4c9 + r6c79 = 23 = {689} -> 2 6's in n6)
105b. No 8 in r4c7

106. i/o difference(n7): r6c123 = r9c3 + 7
-> no 8 in r6c3
(reason: would leave 3 or 7 in r9c3 -> r6c123 = 10/3 or 14/3. In the 10/3 case, r6c3=8 would blow the cage sum. In the latter case, it would require either of {45} in r6c12 - unavailable)

107a. Nishio: no 3 in r7c2
(reason: r7c2=3 -> r1c1=3 -> r5c3=3 ->r6c5=3 -> r9c4=3 -> nowhere to place 3 in c7)
107b. No 9 in either of r6c23

Maybe it's time to start on the hypotheticals again...

Edit: Here's the current grid state (just in case somebody else wants to join in this nightmare!).
(Richard - Please check the candidates here. It would be bordering on a miracle if I haven't made at least a minor mistake somewhere. Thanks.)

Code: Select all

.---------------------.---------------------.----------.---------------------.---------------------.
| 12369      123679   | 45         78       | 1267     | 68         35       | 1249       1249     |
&#58;----------.          '----------.----------&#58;          &#58;----------.----------'          .----------&#58;
| 1245     | 1236789    126789   | 59       | 1267     | 34       | 1289       12489    | 567      |
|          &#58;----------.          |          &#58;----------&#58;          |          .----------&#58;          |
| 1245     | 89       | 26789    | 59       | 12       | 34       | 1289     | 67       | 356      |
|          |          '----------+----------&#58;          &#58;----------+----------'          |          |
| 7        | 24         124      | 6        | 89       | 5        | 12         3        | 89       |
&#58;----------'----------.----------'          |          |          '----------.----------'----------&#58;
| 5689       5689     | 358        38       | 79       | 12         4        | 1267       1267     |
&#58;----------.----------'----------.          |          |          .----------'----------.----------&#58;
| 13689    | 12368      12368    | 47       | 348      | 12       | 679        58       | 579      |
|          |          .----------+----------&#58;          &#58;----------+----------.          |          |
| 1345689  | 579      | 1346789  | 12       | 358      | 6789     | 16789    | 24       | 12348    |
|          &#58;----------'          |          &#58;----------&#58;          |          '----------&#58;          |
| 1345689  | 13456789   1346789  | 12       | 345      | 6789     | 16789      156789   | 12348    |
&#58;----------'          .----------'----------&#58;          &#58;----------'----------.          '----------&#58;
| 12689      1245689  | 378        348      | 456      | 79         35       | 14689      1469     |
'---------------------'---------------------'----------'---------------------'---------------------'

mhparker · Post by **mhparker** » Wed May 02, 2007 5:52 pm

I wrote:Maybe it's time to start on the hypotheticals again...

How about this one?:

108a. (remaining)outies - innies, n3 = 6 -> [r4c9-r4c7-r1c7] = [813|925]
108b. Therefore r4c9 = 9 -> r4c7 = 2 -> 8/2(r5c89) = {17}
Similarly, r4c9 = 8 -> r4c7 = 1 -> 8/2(r5c89) = {26}
108c. The only possible candidate position for {89} in 19/3(c9) is r4c9
108d. Therefore, from 104b and 104c, if 19/3 = {9..}, r5c9 = {17}.
Similarly, if 19/3 = {8..}, r5c9 = {26}.
108e. Now examine all combinations of combinations for both 19/3(c9) and 14/3(c9):
{379}{149} - impossible, repeating digits
{379}{158} - contains 9 in 19/3 and both of {17}, hence clashes w/ r5c9 (step 104d)
{379}{239} - impossible, repeating digits
{379}{347} - impossible, repeating digits
{568}{149} - ok
{568}{158} - impossible, repeating digits
{568}{239} - ok
{568}{347} - ok
108f. Hence, we can reject combinations {379} for 19/3 and {158} for 14/3
-> 19/3(c9) = {568} (no 3,7,9), locked for c9, {56} locked for n3.

This step reduces the puzzle to only naked and hidden singles.

P.S. What's the situation with the Lite version? Is anybody planning on writing a walkthrough for it?

rcbroughton · Post by **rcbroughton** » Wed May 02, 2007 7:01 pm

or from your markup pic there is another immediate placement that breaks the puzzle . . .

108. 45 on n4 - innies r5c3+r6c1+7=r3c2+r7c2+3
108a. r37c2=[85]/[87]/[89]/[95]/[97] - 13,14,15,16,17
108b. so r5c3+r6c1 must total 9,10,11,12,13
108c. r5c3+r6c1 can't be [56] or [58] because of 14(2)n3
108d. possibilites [36]/[38]/[39]/[81]/[83] - 9,11,12
108e. r5c3={38} - remove 5
108f. r37c2 can't be 14, 17 - no 9 at r7c2
108g. cleanup 21(4)n45 = 6{38}4

and the rest falls very quickly

rcbroughton · Post by **rcbroughton** » Wed May 02, 2007 7:19 pm

PsyMar wrote:Ah crud. Looks like I gave combinations for the 15 as if it were a 14, and proceeded to screw up from there and got lucky and solved it. Can't believe I did that. Sorry.

No problems - we've all been there, done that at some stage . . .

mhparker · Post by **mhparker** » Wed May 02, 2007 9:16 pm

rcbroughton wrote:or from your markup pic there is another immediate placement that breaks the puzzle . . .

Nice one, Richard! Builds on the results of the earlier moves very neatly.

Glyn · Post by **Glyn** » Wed May 02, 2007 10:27 pm

Well done Mike and Richard, been following this one with interest, got lost at one stage so the grid state from Mike was very useful.

I had followed both options for the bivalue cells at the ends of the critical column 6 that broke the puzzle so when I saw that cell fall I knew it was over.

All the best,

Glyn