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Ed
 Post subject: Assassin 265
PostPosted: Fri Aug 02, 2013 7:09 am 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Found a really neat step pretty quickly on the first solve but then took much longer finding the more routine stuff to get it out. A challenging puzzle. SudokuSolver gives it 1.65, JSudoku has to use "1 recursive solve" (not sure what that means but it sounds hard!).

Assassin 265
Image

code: paste into solver:
3x3::k:3328:3328:3328:6145:6145:6145:2050:2050:7939:3328:3844:3844:6145:6149:6145:7939:7939:7939:5894:3844:6149:6149:3335:6149:6149:7939:5640:5894:2825:2825:5642:3335:7179:1804:1804:5640:5894:5642:5642:5642:3335:7179:7179:7179:5640:5894:1805:5902:5902:3335:5902:5902:3343:5640:3600:1805:5649:5649:5902:5650:5650:3343:2067:3600:5649:5649:2068:3861:3350:5650:5650:2067:3600:2583:2583:2068:3861:3350:2328:2328:2067:
solution: 
+-------+-------+-------+
| 6 4 1 | 9 8 2 | 3 5 7 |
| 2 3 7 | 4 5 1 | 9 6 8 |
| 9 5 8 | 3 7 6 | 2 1 4 |
+-------+-------+-------+
| 8 7 4 | 6 1 9 | 5 2 3 |
| 1 2 9 | 5 3 4 | 7 8 6 |
| 5 6 3 | 8 2 7 | 1 4 9 |
+-------+-------+-------+
| 7 1 6 | 2 4 3 | 8 9 5 |
| 3 9 5 | 1 6 8 | 4 7 2 |
| 4 8 2 | 7 9 5 | 6 3 1 |
+-------+-------+-------+
Cheerio
Ed
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Frank
 Post subject: Re: Assassin 265
PostPosted: Sun Aug 04, 2013 5:03 pm 
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Joined: Mon Apr 28, 2008 10:58 pm
Posts: 47
Location: Victoria, B.C., Canada
Too easy Ed! :).

Here is how I did it:
r3c9 is part of a hidden 20(3) cage, so r3c79 = [15] or [24].
If r3c79=[15] then the placement of the 1 in n6 makes the 7(2) cage = [61].
But the 8(2) cage in n3 is [26] which leaves no place for a 6 in n9.

So r3c79 = [24] and the rest is easy.

Many thanx - Cheers - Frank
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Andrew
 Post subject: Re: Assassin 265
PostPosted: Mon Aug 05, 2013 12:55 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Thanks Ed for your latest Assassin.

Easier than many Assassins and other recent puzzles on this forum but I won't go as far as Frank's comment. It took me a long time to reach the key first placements; the same ones as Frank found.

Here is my walkthrough for Assassin 265:
Prelims

a) R1C78 = {17/26/35}, no 4,8,9
b) R4C23 = {29/38/47/56}, no 1
c) R4C78 = {16/25/34}, no 7,8,9
d) R67C2 = {16/25/34}, no 7,8,9
e) R67C8 = {49/58/67}, no 1,2,3
f) R89C4 = {17/26/35}, no 4,8,9
g) R89C5 = {69/78}
h) R89C6 = {49/58/67}, no 1,2,3
i) R9C23 = {19/28/37/46}, no 5
j) R9C78 = {18/27/36/45}, no 9
k) 8(3) cage at R7C9 = {125/134}
l) 13(4) cage at R1C1 = {1237/1246/1345}, no 8,9
m) 13(4) cage at R3C5 = {1237/1246/1345}, no 8,9
n) 28(4) cage at R4C6 = {4789/5689}, no 1,2,3

Steps resulting from Prelims
1a. 8(3) cage at R7C9 = {125/134}, 1 locked for C9 and N9, clean-up: no 8 in R9C78
1b. R9C78 = {27/36} (cannot be {45} which clashes with 8(3) cage), no 4,5 in R9C78
1c. Killer pair 2,3 in 8(3) cage and R9C78, locked for N9
1d. R9C23 = {19/28/46} (cannot be {37} which clashes with R9C78), no 3,7 in R9C23
1e. R89C6 = {49/58} (cannot be {67} which clashes with R89C5), no 6,7 in R89C6
1f. Killer pair 8,9 in R89C5 and R89C6, locked for N8
1g. 13(4) cage at R1C1 = {1237/1246/1345}, 1 locked for N1
1h. 13(4) cage at R3C5 = {1237/1246/1345}, 1 locked for C5
1i. 28(4) cage at R4C6 = {4789/5689}, CPE no 8,9 in R5C4

2. Variable hidden killer pair 8,9 in 22(4) cage at R7C6 and R7C8 for N9, R7C8 cannot contain more than one of 8,9 -> 22(4) cage must contain at least one of 8,9) = {1489/…/3568} (cannot be {4567} which doesn’t contain 8 or 9)
2a. 1,2,3 only in R7C6 -> R7C6 = {123}

3. 45 rule on C1 2 innies R12C1 = 8 = {17/26/35}, no 4
3a. 45 rule on C1 2 outies R1C23 = 5 = {14/23}
3b. R1C23 + R1C78 = 13 = {1237/1246/1345}, 1 locked for R1, clean-up: no 7 in R2C1

4. 45 rule on N1 2 innies R3C13 = 17 = {89}, locked for R3 and N1

5. 45 rule on N3 2 innies R3C79 = 6 = [15/24/42]

6. 45 rule on C9 2 innies R12C9 = 15 = {69/78}
6a. 45 rule on C9 3 outies R2C78 + R3C8 = 16 = {169/178/349/358} (cannot be {268/367} which clash with R12C9, cannot be {259/457} which clash with R3C79), no 2

7. 45 rule on N2 2 outies R3C37 = 1 innie R3C5 + 3
7a. Min R3C37 = 9 -> min R3C5 = 6
7b. Max R3C37 = 10 -> R3C37 = [81/82/91], no 4 in R3C7, clean-up: no 2 in R3C9 (step 5)
[With hindsight I could have reduced R3C9 to {45} more directly using 45 rule on R123 3 innies R3C159 but I didn’t spot that 45 until later.]
7c. Killer pair 6,7 in R3C5 and R89C5, locked for C5
7d. Killer pair 4,5 in R3C9 and 8(3) cage at R7C9, locked for C9

8. R3C5 = {67} -> 13(4) cage at R3C5 = {1237/1246} (cannot be {1345} which doesn’t contain 6 or 7), no 5, 1,2 locked for C5 and N5

9. Hidden killer pair 2,3 in R456C9 and 8(3) cage at R7C9 for C9, 8(3) cage contains one of 2,3 -> R456C9 must contain one of 2,3
9a. Hidden killer triple 1,2,3 in R4C78, R456C9 and R6C7 for N6, R4C78 contains one of 1,2,3, R456C9 contains one of 2,3 -> R6C7 = {123}

10. 45 rule on N8 3 innies R7C456 = 9 = {135/234} (cannot be {126} because R7C5 only contains 3,4,5), no 6,7, 3 locked for R7 and N8, clean-up: no 4 in R6C2, no 5 in R89C4
10a. 45 rule on N7 1 outie R7C4 = 1 innie R7C2 + 1, no 5,6 in R7C2, no 1,4 in R7C4, clean-up: no 1,2 in R6C2
10b. R7C456 = {135/234}
10c. 2 of {234} must be in R7C4 (R7C456 cannot be [342] which clashes with R7C24 = [23], IOD blocker), no 2 in R7C6
10d. 2 in N8 only in R789C4, locked for C4

11. Naked quint {12345} in R7C24569, locked for R7, clean-up: no 8,9 in R6C8

12. 45 rule on N9 1 innie R7C8 = 1 outie R7C6 + 6, R7C6 = {13} -> R7C8 = {79}, R6C8 = {46}
12a. 5 in N6 only in R45C78, CPE no 5 in R4C6

13. Hidden killer pair 8,9 in R12C9 and R456C9 for C9, R12C9 contains one of 8,9 -> R456C9 must contain one of 8,9
13a. Hidden killer pair 8,9 in R456C9 and R5C78 for N6, R456C9 contains one of 8,9 -> R5C78 must contain one of 8,9
13b. Hidden killer pair 8,9 in R45C6 and R5C78 for 28(4) cage at R4C6, R5C78 contains one of 8,9 -> R45C6 must contain one of 8,9
13c. Killer pair 8,9 in R45C6 and R89C6, locked for C6

[Just spotted that I hadn’t looked deeply enough in step 3.]
14. R1C23 + R1C78 (step 3b) = {1237/1246/1345} = {1237}, locked for R1 or R1C23 = {14}, locked for N1 -> R12C1 (step 3) = {26/35} (cannot be [71], killer locking-out cages)
14a. 1 in N1 only in R1C23 (step 3a) = {14}, locked for R1 and N1, clean-up: no 7 in R1C78
14b. Killer pair 2,5 in R1C78 and R3C79, locked for N3

15. 14(3) cage at R7C1 = {149/158/167/347} (cannot be {239/257/356} which clash with R12C1, cannot be {248} which clashes with R7C2 + R9C23, killer ALS block), no 2
15a. 8,9 of {149/158} must be in R7C1 -> no 8,9 in R89C1
15b. 14(3) cage = {149/158/167}, 1 locked for N7 or 14(3) cage = {347} => killer pair 1,2 in R7C2 + R9C23, locked for N7 -> 1 in 14(3) cage + R7C2 + R9C23, locked for N7

16. 8 in N9 only in 22(4) cage at R7C6
16a. 22(4) cage = {1489/3478/3568} (cannot be {1678} which clashes with R9C78)
16b. Consider combinations for R7C456 (step 10) = {135/234}
16c. R7C456 = {135} = {35}1 => 22(4) cage = {1489}
or R7C456 = {234} = [243] => 2 in N7 only in R9C23 => R9C78 = {36}, locked for N9 => 22(4) cage = {3478}
-> 22(4) cage = {1489/3478}, no 5,6, 4 locked for R8 and N9, clean-up: no 9 in R9C6
16d. 6 in N9 only in R9C78 = {36}, locked for R9 and N9, clean-up: no 4 in R9C23, no 2 in R8C4, no 9 in R8C5
[Cracked. The rest is straightforward.]

17. Naked triple {125} in 8(3) cage at R7C9, locked for C9 -> R3C9 = 4, R3C7 = 2 (step 5), clean-up: no 6 in R1C78, no 5 in R4C8
17a. 3 in C9 only in R456C9, locked for N5 -> R6C7 = 1, R4C8 = 2 (hidden single in N6), R4C7 = 5, R1C78 = [35], R9C78 = [63], clean-up: no 3,5 in R2C1, no 6,9 in R4C23

18. 45 rule on R123 2 remaining innies R3C15 = 16 -> R3C1 = 9, R3C5 = 7, R3C3 = 8, clean-up: no 3 in R4C2, no 8 in R89C5, no 2 in R9C2
18a. R89C5 = [69], R1C5 = 8, clean-up: no 7 in R2C9 (step 6), no 1 in R9C23, no 4 in R9C6
18b. R9C23 = [82], R89C6 = [85], R9C9 = 1, R9C4 = 7, R8C4 = 1, R7C456 = [243], R7C2 = 1, R6C2 = 6, R6C8 = 4, R7C8 = 9, R7C7 = 8 (hidden single in N9), R8C78 = [47]

19. Naked pair {26} in R12C1, locked for C1 and N1 -> R7C1 = 7, R9C1 = 4, R8C1 = 3 (cage sum)
19a. R1C23 = [41], R4C2 = 7, RR4C3 = 4
19b. Naked pair {35} in R23C2, locked for C2 and N1 -> R2C3 = 7
19c. Naked pair {26} in R1C16, locked for R1 -> R1C4 = 9, R1C9 = 7, R2C9 = 8 (step 6)

20. R5C78 = [78] (hidden pair in N6) = 15 -> R45C6 = 13 = [94]

21. R2C5 = 5 (hidden single in C5), R23C4 = [43] (hidden pair in N2), R3C6 = 6 (cage sum)

22. R6C67 + R7C5 = [714] = 12 -> R6C34 = 11 = [38]

and the rest is naked singles.

Rating Comment:
I'll rate my walkthrough for A265 at Easy 1.5. I used locking-out and similar steps.
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Ed
 Post subject: Re: Assassin 265
PostPosted: Thu Aug 08, 2013 10:16 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Nice one Frank! Really well spotted. My way is much closer to Andrew's though we got to the final breakthrough quite differently and in a completely different area. I've put my key step right near the beginning even though it doesn't become really useful till step 15.

Assassin 265
20 steps:
Preliminaries courtesy of SudokuSolver (Solver menu>Options>scoring>OK>F10)
Cage 15(2) n8 - cells only uses 6789
Cage 7(2) n47 - cells do not use 789
Cage 7(2) n6 - cells do not use 789
Cage 8(2) n3 - cells do not use 489
Cage 8(2) n8 - cells do not use 489
Cage 13(2) n8 - cells do not use 123
Cage 13(2) n69 - cells do not use 123
Cage 9(2) n9 - cells do not use 9
Cage 10(2) n7 - cells do not use 5
Cage 11(2) n4 - cells do not use 1
Cage 8(3) n9 - cells do not use 6789
Cage 28(4) n56 - cells do not use 123
Cage 13(4) n25 - cells do not use 89
Cage 13(4) n1 - cells do not use 89

My key step is available from near the start.
1. "45" on c1: 2 innies r12c1 = 8 = {17/26/35}(no 4)

2. "45" on c1: 2 outies r1c23 = 5 = {14/23}(no 5,6,7) = [1/2..]

3. The two 8(2)r1c6 and h8(2)r12c1 must have different combos since they have the same cage sum but r1c1 sees both of r1c78
3a. but can't both be {17/26} because of the h5(2)r1c23 = [1/2..]
3b. -> at least one of the 8(2) must have {35}: r1c23 sees both of those -> no 3 in r1c23
3c. r1c23 = {14} only: both locked for r1 and n1
3d. no 7 in 8(2)n3
3e. h8(2)r12c1 = {26/35}(no 7)

Rest from here is pretty bog standard Assassin stuff.
4. "45" on n1: 2 innies r3c13 = 17 = {89} only: both locked for n1 and r3

5. "45" on n2: 1 innie r3c5 + 3 = 2 outies r3c37
5a. min. 2 outies = [81] = 9 -> min. r3c5 = 6 (no 1,2,3,4,5)

6. 13(4)r3c5 must have 6/7 for r3c5 = [7]{123}/[6]{124}(no 5, no 6 or 7 in r456c5)
6a. must have both 1 & 2: both locked for c5 and n5

7. 15(2)n8 = {69/78} = [6/7..]
7a. Killer pair 6,7 in c5 in r3c5 + 15(2): both locked for c5

8. "45" on n8: 3 innies r7c456 = 9
8a. but {126} blocked since no 1,2,6 in r7c5
8b. = {135/234}(no 6,7,8,9)
8c. must have 3: 3 locked for r7 and n8
8d. no 5 in 8(2)n8

9. 8(3)n9 = {125/134} = [4/5..]
9a. must have 1: 1 locked for c9 and n9
9b. no 8 in 9(2)n9

10. 9(2)n9: {45} blocked by 8(3)n9 = [4/5]
10a. 9(2) = {27/36}(no 4,5) = [3/7..]

11. 10(2)n7: {37} blocked by 9(2)n9 = [3/7]
11a. 10(2) = {19/28/46}(no 3,7)

12. "45" on n7: 1 innie r7c2 + 1 = 1 outie r7c4
12a. r7c2 = (124), r7c4 = (235)
12b. r6c2 = (356)

13. "45" on n9: 1 innie r7c8 - 6 = 1 outie r7c6
13a. r7c6 = (123), r7c8 = (789)
13b. r6c8 = (456)

14. 23(4)r3c1: max. any two cells = 17 -> min. any two cells = 6: ie, cannot have both {14} -> 14(3)r7c1 must have at least one of 1,4 for c1 (Hidden killer pair)
14a. also, 10(2)n7 has one of 1,2,4; r7c2 has one of 1,2,4 -> 14(3)cannot have more than one of 1,2,4
14b.-> 14(3)n7 has exactly one of 1,2,4 = {158/167/239/257/347}
14c.-> all three of 1,2,4 taken for n7: all locked for n7 (Killer triple)

15. 14(3)n7: {239/257} blocked by r12c1 = {26/35} = [2/3,2/5..]
15a. 14(3) = {158/167/347}(no 2,9) = [3/6/8..]

16. 22(4)r7c3 = {2389/2569/2578/3568}
16a. all combos must have 2 in r7c4 except {3568}
16b. but {368}[5] blocked by 14(3)n7 = [3/6/8]
16c. -> no 5 in r7c4
16d. no 4 in r7c2 (IODn7 = -1)
16e. no 3 in r6c2

17. Naked triple {123} in r7c246: all locked for r7 [note: Andrew's WT spotted that there is a naked quint in r7 at this point - nice find!]
17a. 8(3)n9 must have 4 or 5 which must be in r7c9
17b. no 4,5 in r89c9

18. 4 in n7 in 10(2) = {46} or in 14(3) -> 6 in 14(3) must also have 4 or there would be no 4 for n7 (Locking-out cages)
18a. -> {167} blocked from 14(3)
18b. 14(3) = {158/347}(no 6) = [3/8..]

19. 22(4)r7c3 = {2389/2569/2578/3568}
19a. but [2]{389} blocked by 14(3) = [3/8]
19b. = {2569/2578/3568}
19c. must have 5: locked for n7

20. Hidden single 5 in r9: r9c6 = 5

cracked.
Cheerio
Ed
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Pinata
 Post subject: Re: Assassin 265
PostPosted: Thu Aug 08, 2013 10:49 am 
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Posts: 92
I enjoyed the assassin, it was nice and flowing
My Walkthrough:
Prelims
a) 13(4) in R1C1 must be [1237], [1246], [1345] no 8, 9
b) 13(4) in R3C5 must be [1237], [1246], [1345] no 8, 9
c) 8(2) in R1C7 must be [17], [26], [35] no 4, 8, 9
d) 31(5) in R1C9 must be [16789], [25789], [34789], [35689], [45679]
e) 11(2) in R4C2 must be [29], [38], [47], [56] no 1
f) 7(2) in R6C2 must be [16], [25], [34] no 7, 8, 9
g) 28(4) in R4C6 must be [4789], [5689] no 1, 2, 3
h) 7(2) in R4C7 must be [16], [25], [34] no 7, 8, 9
i) 13(2) in R6C8 must be [49], [58], [67] no 1, 2, 3
j) 10(2) in R9C2 must be [19], [28], [37], [46] no 5
k) 8(2) in R8C4 must be [17], [26], [35] no 4, 8, 9
l) 15(2) in R8C5 must be [69], [78] no 1, 2, 3, 4, 5
m) 13(2) in R8C6 must be [49], [58] no 1, 2, 3, 6, 7 (cannot be [67] because the 15(2) in R8C5 won’t have a valid combination
n) 9(2) in R9C7 must be [18], [27], [36], [45] no 9
o) 8(3) in R7C9 must be [125], [134] no 6, 7, 8, 9

Walkthrough
1) Locked Candidates- 1 Locked to R3456C5
1a) CPE- No 1 in R12789C5

2) Locked Candidates- 1 Locked to R789C9
2a) CPE- No 1 in R123456C9, R789C78
2b) Cage Blocker- 9(2) in R9C7 can’t be [18]- no 8 in R9C78

3) Outies of C1 = 5
3a) CPE- No 5, 6, 7, 8, 9 in R1C23
3b) Hidden Killer Pair R1C23 must be [14], if it is [23] then R12C1 would have to be [17] but then the 8(2) in R1C7 wouldn’t have a valid combination
3c) Locked Pair- [14] Locked to R1C23
3d) CPE no [14] in R1C1456789, R23C123
3e) Cage Blocker- 8(2) in R1C7 can’t be [17] no 7 in R1C78, R12C1
3f) Innies of C1 = 8- R12C1 must contain 2, 3, 5, 6

4) Innies of N1 = 17- R3C13 must contain [89]
4a) Locked pair- [89] Locked to R3C13
4b) no 8, 9 in R12C123, R3C2456789

5) Innies of C9 = 15- R12C9 must contain 6, 7, 8, 9

6) Innies of N3 = 6- R3C7 must contain 1, 2, 4 and R3C9 must contain 2, 4, 5
6a) Combination blocker- 31(5) in R1C9 can’t be [25789] as neither the 8(2) in R1C7 or R3C79 would have a valid combination, no 2 in R12C9, R2C78, R3C8
6b) Combination blocker 31(5) in R1C9 can’t contain a 5 as neither the 8(2) in R1C7 or R3C79 would have a valid combination, no 5 in R12C9, R2C78, R3C8
The 31(5) in R1C9 must be either 16789 or 34789

7) Outies of N9 = 7
7a) R6C8 must contain 4, 5, 6 and R7C6 must contain 1, 2, 3
7b) as R6C8 must contain 4, 5,6 that means R7C8 must contain 7, 8, 9

8) IOD- R3C37- R3C5 = 3
8a) as R3C3 must contain [89], which is locked to [89] R3C13, 4 can be removed from R3C7 and 2 can be removed from R3C9, 1, 2, 3, 4, 5 can be removed from R3C5 which must contain [67]
8b) combination locker- 13(4) in R3C5 must be [1237], [1246], R456C5 must contain 1, 2, 3, 4
8c) Locked Pair- [12] Locked to R456C5, no [12] in R123789C5, R456C46

9) Combination blocker- 22(4) in R3C9 can’t be [4567] as the innies of C9 or the 8(3) in R7C9 won’t have a combination no [45] in R456C9

10) Combination Blocker- 9(2) in R9C7 can’t be [45] as 8(3) in R7C9 wouldn’t have a valid combination R9C78 must contain 2, 3, 6, 7

11) Innies of N8 = 9 no 7, 8, 9 in R7C456

12) Cell Blocker- as R3C5 must contain [67] and 15(2) in R8C5 must be [69], [78] then [67] can be deleted from R124567C5

13) Cell blocker- R7C4 can’t contain a 6 as N8 innies can’t be [126]

14) Locked Candidates- 3 Locked to R7C456
14a) CPE- No 3 in R7C123789
14b) cage blocker- 8(2) in R8C4 can’t be [35]
14c) CPE- no 4 in R6C2

15) Outies of N7 = 8
15a) R6C2 must contain 3, 5, 6 and R7C4 must contain 2, 3, 5
15b) R7C2 must contain 1, 2, 4

16) Hidden Quadruple- 6, 7, 8, 9 only in R7C1378

17) Combination Blocker- 10(2) in R9C2 can’t be [37] as 9(2) in R9C7 would have no combination

18) Cell Blocker- R1C9 can’t be 6 as 24(5) in R1C4 wouldn’t have a valid combination

19) CPE- No 9 in R2C9

20) Cell Blocker R1C1 can’t contain a 2 as 24(5) in R1C4 wouldn’t have a valid combination as there would be no solution for R3C5.
20a) CPE no 6 in R2C1

21) Cage blocker- no 8, 9 in R89C1

22) Combination Blocker- 14(3) in R7C1 can’t be [167] no [67] in R89C1 because 14(3) is [167], 10(2) in R9C2 is [28], R7C2 is 4 leaves no valid combination for 22(4) in R7C3.

23) Cell Blocker- no 6 in R1C456 as 8(2) in R1C7 wouldn’t have a valid combination

24) Cell Blocker- No 6 in R9C5 as 9(2) in R9C7 would have to be [27] and R9C4 would have to be 1, leaving 10(2) in R9C2 with no valid combination
24a) CPE- No 9 in R8C5

25) Consider Permutations for 22(4) in R7C3- This cage can’t contain a 7
25a) If R7C3, R8C23 is [179] and the 10(2) in R9C2 is [28] then the 14(3) in R7C1 would have to be [356], which would leave no solution for R12C1.
25b) If R7C3, R8C23 is [179] and the 10(2) in R9C2 is [46], then the 14(3) in R7C1 wouldn’t have a valid combination
25c) If R7C3, R8C23 is [478] and the 10(2) in R9C2 is [19] then the 14(3) would have to be [356] which would leave no solution for R12C1
25d) If R7C3, R8C23 is [479] and 10(2) in R9C2 is [28] then 14(3) in R7C1 would have to be [356], which would leave no solution for R12C1.
25e) If R7C3, R8C23 is [578] and 10(2) in R9C2 is [19], then 14(3) in R7C1 would have no combination
25f) If R7C3, R8C23 is [578] and 10(2) in R9C2 is [46], then 14(3) is [239] then that would leave R12C1 without a valid combination
25g) CPE- no 1 in R7C3, R8C23
25h) if R7C3, R8C23 is [278] and 10(2) in R9C2 is [19] then 14(2) in R7C1 would have to be [356] which would leave no combination for R12C1
25i) If R7C3, R8C23 is [278] and 10(2) in R9C2 is [46] then 14(3) in R7C1 would have no valid combination
25j) if R7C3, R8C23 is [467] and 10(2) in R9C2 is [28] then 14(2) in R7C1 would have no valid combination
25k) If 25h) if R7C3, R8C23 is [467] and 10(2) in R9C2 is [19] then 14(2) would have no valid combination

26) Hidden Single- R7C1 = 7
26a) CPE- No 7 in R7C23456789, R89C123, R12345689C1
26b) CPE- no 6 in R6C8

27) Killer Triplet- 14(3) must be [347], can’t be [257] as that would leave R12C1 without a valid combination
27a) Naked Killer Triplet- [347] locked to R789C1
27b) no [347] in R123456C1, R789C23
27c) CPE- no 6 in R9C23 which must contain 1, 2, 8, 9
27d) CPE- no 5 in R7C4
27e) CPE no 3 in R6C2
27f) R7C4 must contain 2, 3
27g) R7C2 must contain 1, 2
27h) R6C2 must contain 5, 6
27i). CPE no 5 in R12C1

28) Naked Single- R1C1 must contain a 6
28a) CPE- No 6 in R1C234569
28b) No 6 in R23C123
28c) No 6 in R23456789C1

29) Naked Single- R2C1 must be 2
29a) no 2 in R2C23, R3C2, 15(3) must be [357]
29b) no 2 in R13456789C1, R123C23, R2C23456789
29c) Combination Locker 23(4) in R3C1 must be [1589]

30) CCC- as R12C1 must be [26], 8(2) in R1C7 must be [35]
30a) Naked Killer Pair- [35] locked to R1C78
30b) no [35] in R1C1234569, R23C789
30c) Combination Locker- 31(5) in R1C9 must be [16789]
30e) CPE- No 4 in R1C9, R2C789, R3C8

31) Naked Single- R3C9 must be 4
31a) CPE no 4 in R12C789, R3C12345678, R456789C9
31b) Hidden Killer Triplet, 8(3) in R7C9 must be [125]
31c) CPE no 3 in R789C9
31d) R456C9 must be [369]/[378]
31e) Combination Blocker- 7(2) in R4C7 can’t be [34]
31f) CPE- R6C7 must contain 1, 2
31g) Combination Blocker- 11(2) in R4C2 can’t be [56]

32) Locked Pair- [15] locked to R456C1
32a) no [15] in R456C23

Puzzle is Cracked Now, Only Requiring Singles

rating:
I found this a nice puzzle, easy 1.50


Thank you to Ed and Andrew for spotting any errors or clarifications in my walkthrough, which has been corrected.
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