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HATMAN
PostPosted: Thu Sep 06, 2012 4:16 pm 
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Kenyan Killer 5 Assassin 245

The letters are the cage totals. Each letter represents a different cage total.
The Kenyan is on the easy side - provided you get the tricky bit.

This is the original puzzle intended at Paper Solvable level: it scores 0.95.
JS code:
3x3::k:24:5139:5139:5139:1552:1552:25:3078:26:4107:4107:4623:4623:4623:3078:3078:3078:7681:4107:4107:4616:4616:6151:6151:6146:7681:7681:6154:6154:4616:6151:6151:6146:6146:7681:1559:6154:3077:3077:4616:4611:6146:4108:4108:1559:6154:3077:3076:4611:4611:4617:4108:4108:27:3077:3076:3076:4611:4617:4617:4617:4116:4116:28:3076:3854:3854:2317:2317:4117:4117:3094:29:30:3089:3089:1554:1554:4117:3094:3094:

I found that removing the B(2) red cage this gives an SS score of 1.4 - this is Assassin 245.
JS code:
3x3::k:23:5139:5139:5139:1552:1552:24:3078:25:4107:4107:4623:4623:4623:3078:3078:3078:7681:4107:4107:4616:4616:6151:6151:6146:7681:7681:6154:6154:4616:6151:6151:6146:6146:7681:26:6154:3077:3077:4616:4611:6146:4108:4108:27:6154:3077:3076:4611:4611:4617:4108:4108:28:3077:3076:3076:4611:4617:4617:4617:4116:4116:29:3076:3854:3854:2317:2317:4117:4117:3094:30:31:3089:3089:1554:1554:4117:3094:3094:


If you also remove the green C(3) the score is 1.8: you may wish to try it.
JS Code:
3x3::k:22:5139:5139:5139:1552:1552:23:3078:24:4107:4107:4623:4623:4623:3078:3078:3078:7681:4107:4107:4616:4616:6151:6151:6146:7681:7681:6154:6154:4616:6151:6151:6146:6146:7681:25:6154:3077:3077:4616:4611:6146:4108:4108:26:6154:3077:3076:4611:4611:4617:4108:4108:27:3077:3076:3076:4611:4617:4617:4617:4116:4116:28:3076:3854:3854:2317:2317:4117:4117:29:30:31:3089:3089:1554:1554:4117:32:33:

Image
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Solution:
685742913
197835246
243196587
538274691
721689435
964513728
352468179
816927354
479351862
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Andrew
PostPosted: Fri Sep 07, 2012 1:50 am 
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Thanks HATMAN for another Kenyan Killer! :D

HATMAN wrote:
The Kenyan is on the easy side - provided you get the tricky bit.
I just solved the full version of the Kenyan part of the puzzle. I've left it to wellbeback to solve it without using the red cage and its accompanying ratio.
Using the full version:
there wasn't any tricky bit and, in fact, one ratio was redundant; I didn't use E:G, alternatively I could have omitted E:J. Also, unlike at least one of the earlier Kenyan Killers, it wasn't necessary to use the condition that each letter represents a separate total.

Here is my walkthrough for the Kenyan part of the puzzle:
1. C:F = 2:5 with 4-cell cages can only be 10:25 or 12:30. C is also the total for 2-cell and 3-cell cages so must be the smaller total.
[At this stage I incorrectly wrote C:H = 4:5 fixes C = 12, H = 15, F = 30. HATMAN asked “Why can’t C:H = 4:5 be C = 10, H = 8? Very careless of me to overlook that option! :oops: ]
Looking ahead to a later step the answer to this is simple. H:J = 3:5 but H = 8 isn’t a multiple of 3 or 5 -> H = 15, C = 12, F = 30
[Now I can return to my original steps]

2. B:F = 1:5, F = 30 -> B = 6

3. C:G = 1:2, C = 12, G is 4-cell cages -> G = 24

4. H:J = 3:5, H = 15, J is a 2-cell cage -> J = 9

5. D:G = 2:3, G = 24, D is total for 2-, 3- and 4-cell cages -> D = 16

6. A:D = 4:5, D = 16 -> A = 20

7. E:J = 1:2, J = 9 -> E = 18
Thanks HATMAN for pointing out my error. I hope the steps are correct now.

As with the earlier KK#3 and KK#4, which both led to Assassins, I'll solve the Assassin and one/both of the variants separately.


Last edited by Andrew on Sat Sep 08, 2012 1:20 am, edited 1 time in total.

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HATMAN
PostPosted: Fri Sep 07, 2012 7:21 am 
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Andrew

I'll have to work harder on these in future.

but why not C=10 H=8 on your first line?
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Andrew
PostPosted: Sat Sep 08, 2012 1:26 am 
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Thanks HATMAN for pointing out my error/oversight. Thanks for doing so politely and in tiny text. However I prefer feedback pointing out errors/omissions to be sent to me by PM. Then, whenever there's something which needs to be shown to a wider audience I add appropriate notes, as I've done in my walkthrough above for the Kenyan part of the puzzle. I also added an appropriate note to my Assassin 243 walkthrough after Ed had sent me a PM pointing out a flawed step.

Andrew
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HATMAN
PostPosted: Sat Sep 08, 2012 7:21 pm 
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will do Andrew
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Ed
PostPosted: Sun Sep 09, 2012 9:43 am 
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Another ripper Assassin. Thankyou HATMAN. I found it a lot of fun because it kept flowing using normal Assassin moves, yet, was very resistant to finally getting cracked. Step 17 (in this optimised path) was the step that was alluding me. Found it a lot later in the solution first time through.

Assassin 245
First, a pic:
Image


Walkthrough
28 steps:
Please let me know of any errors or anything that could be clearer.

1. "45" on n9: 1 innie r7c7 = 1

2. Two 12(4) cages at r5c2 & r6c3 = {1236/1245}(no 7,8,9)
2a. both must use 1&2 -> 1&2 locked for n4 & n7

3. Hidden single 1 in n7 -> r8c2 = 1
3a. Hidden single 1 in n4 -> r5c3 = 1

4. 12(2)r9c3: no 1,2,6
4a. 1 in r9 only in 6(2)n8 = {15} only: both locked for n8 and 5 for r9
4b. no 7 in r9c34

5. 16(2)n9 = {79} only: both locked for r7 and n9

6. 16(3)n9 = {268/358}(no 4)
6a. 9(2)n8 = {27/36}(no 4,8,9) = [6/7..]
6b. 15(2)r8c3 = {69/78}(no 1,2,3,4,5) = [6/7..]
6c. Killer pair 6,7 in 9(2) and 15(2): both locked for r8

7. "45" on r8: 1 outie r9c7 + 4 = 2 innies r8c19
7a. -> no 4 in r8c1 since it would force r9c7 to equal r8c9 (IOU)
7b. hidden single 4 in r8 -> r8c9 = 4
7c. r9c89 = 8 = {26} only: both locked for r9 and n9

8. 5 in n9 only in r8: locked for r8

[note: Andrew's step 7 is a much easier way to do this next step]
9. "45" on n4 & n7: 6 innies r4c3 + r8c13 + r9c123 = 42
9a. Three of the innies in n7 must have 7,8 & 9 for n7 -> r4c3 + the other 2 cells in n7 = 18 = [9]{{36}/[8]{46}
9b. -> r4c3 = (89)
9c. must have 6 -> r8c3 = 6
9d. r8c4 = 9
9e. 9(2)n8 = {27} only: 2 locked for n8
9f. no 3 in r9c3

10. 12(4)r6c3 = {1245} only
10b. r9c3 sees all cells of 12(4)r6c3 -> no 4 in r9c3 (CPE)
10c. no 8 in r9c3

11. 12(4)r5c2: {1245} blocked by r7c2 = (245)
11a. = {1236} only
11b. -> r79c2 sees all of r56c2+r7c1 -> no 2 or 3 in r79c2 (CPE)
11c. r456c1 sees all of 12(4)r5c3 -> no 3 in r456c1 (CPE)
11d. 6 in r56c2 locked for n4 & c2
11e. 2 must be in 12(4)r6c3 and is only in c3: 2 locked for c3

12. naked pair {89} in r49c3: both locked for c3

13. 3 & 7 in c3 only in n1: both locked for n1

14. 16(4)n1: {1456} must have {45} in c2 but this is blocked by r7c2 = (45)
14a. = {1249/1258}(no 6)
14b. must have 2: 2 locked for n1

15. Hidden single 6 in n1 -> r1c1 = 6

16. 30(4)r2c9 = {6789} only
16a. r12c8 sees all of 30(4) -> no 6,7,8,9 in r12c8
16b. caged X-wing 6,7 & 9 with 16(2)n9 and r9c89: all locked for c89

17. 1,3 & 5 for c9 only in r1456c9 -> r456c9 must have at least two of 1,3,5
17a. ->16(4)n6: {1258/1348/1357/1456/2356} all blocked
17b. = {1249/1267/2347}(no 5,8) =[1/3..]
17c. must have 2: 2 locked for n6

18. Hidden killer pair 1,3 in n6: 16(4) must have one of 1 or 3 (step 17b)-> r456c9 can only have one of 1 or 3 -> r1c9 must have one of 1 or 3 for c9 -> r1c9 = (13)
18a. Killer pair 1,3 in n6 between 16(4) and r456c9: 3 locked for n6

19. r456c9 can only have one of 1/3 -> other two cells must have 5 & 8: both locked for n6 and 8 for c9
19a. -> 8 in 30(4)r2c9 must be in r3c8 only

20. 6(2)n2 = {15/24}(no 3,6,7,8,9)
20a. 8 in r1 only in 20(3)r1c2 = {389/578}(no 1,2,4) = [7/9..]
20b. 3 in {389} must be in r1c3 -> no 3 in r1c4

21. Hidden killer pair 7,9 in r1: 20(3)r1c2 has only one of 7 or 9 -> r1c7 must have 7 or 9 = (79)

22. 2 & 4 in r1 only in 6(2)n4 or r1c8: can't both be in r1c8 -> 6(2) = {24} only; both locked for r2 and n2
[note, a shortcut to this step would be can't be {15} because it would lead to multiple solutions with r9c56 (Killer UR)]

23. Naked triple 6,7,9 in n3 in r1c7 + r23c9: all locked for n3
23a. 6 locked for c9 and 30(4)r2c9 -> no 6 in r4c8

[Andrew really likes this next step but it's nothing really. Locking for one of the two areas leaves a hidden single: which is the case in Andrew's WT after his step 24a. He doesn't miss those often these days. I have a nice cheaters button that highlights all the 1s for me whenever I ask it so I have no excuse!]
24. 2 in r4 only in n5 and 2 in c4 only in n5 -> r4c4 = 2
24a. 24(4)r3c5 must have 2 = {2589/2679}(no 1,3,4)

25. 18(4)r3c3 must have 8 or 9 for r4c3 = {1359/1368/1458}(no 7)
25a. must have 1 -> r3c4 = 1

26. 3 in n2 only in r2: 3 locked for r2

27. Hidden single 1 in n1 -> r2c1 = 1

[Andrew uses a good way to do the next (his step 25) which would have eliminated my doing step 26 first]
28. 12(4)r1c8: must have two of 2,4,5 for r2c78 -> {1236} blocked
28a. = {1245} only
28b. r1c8 = 1, r2c678 = [5]{24}: 2 & 4 locked for n3 and r2

Cracked
Cheerio
Ed
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Andrew
PostPosted: Mon Sep 10, 2012 5:32 am 
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Thanks HATMAN for another enjoyable and challenging Assassin, after we'd solved the Kenyan part of the puzzle!

As Ed said it was resistant in the later stages, although I think we experienced that in different ways. My step 18 was somewhat similar to Ed's step 17 in that we worked in the same area but saw it differently. Then Ed had a really neat step 24 which cracked the puzzle; I don't think I've ever spotted a step like that, my final breakthrough was my much heavier step 29.

Here is my walkthrough for hidden Assassin 245:
Thanks Ed for pointing out a couple of typos and that I'd missed a hidden single after step 24a; I realised that after thinking about your breakthrough step really being two steps combined.

Prelims

a) R1C56 = {15/24}
b) R7C89 = {79}
c) R8C34 = {69/78}
d) R8C56 = {18/27/36/45}, no 9
e) R9C34 = {39/48/57}, no 1,2,6
f) R9C56 = {15/24}
g) 20(3) cage at R1C2 = {389/479/569/578}, no 1,2
h) 12(4) cage at R1C8 = {1236/1245}, no 7,8,9
i) 30(4) cage at R2C9 = {6789}
j) 12(4) cage at R5C2 = {1236/1245}, no 7,8,9
k) 12(4) cage at R6C3 = {1236/1245}, no 7,8,9

Steps resulting from Prelims
1a. Naked pair {79} in R7C89, locked for R7 and N9
1b. R8C56 = {18/27/36} (cannot be {45} which clashes with R9C56), no 4,5
1c. Naked quad {6789} in 30(4) cage at R2C9, CPE no 6 in R12C8
1d. Caged X-Wing pair for 7,9 in 30(4) cage at R2C9 and R7C89, no other 7,9 in C89
1e. 9 in N8 only in R89C4, locked for C4
[I originally also included 12(4) cage at R5C2 and 12(4) cage at R6C3 contain both 1,2 for N47, locked for N47, but I’ll leave that for now in case there’s a simpler way.]

2. 45 rule on N9 1 innie R7C7 = 1
2a. 12(4) cage at R5C2 = {1236/1245}, 1 locked for N4
2b. 12(4) cage at R6C3 = {1236/1245} -> R8C2 = 1, clean-up: no 8 in R8C56
2c. R5C3 = 1 (hidden single in N4)
2d. 12(4) cage at R5C2 = {1236/1245}, CPE no 2 in R456C1
2e. Killer pair 6,7 in R8C34 and R8C56, locked for R8
2f. 8 in R7 only in R7C456, locked for N8, clean-up: no 7 in R8C3, no 4 in R9C3

3. 1 in N8 only in R9C56 = {15}, locked for R9 and N8, clean-up: no 7 in R9C34

4. 12(3) cage at R8C9 = {246} (only remaining combination, cannot be {345} which clashes with R9C34), locked for N9, 6 also locked for R9
4a. Caged X-Wing for 6 in 30(4) cage at R2C9 and 12(3) cage, no other 6 in C89
4b. 5 in N9 only in R8C78, locked for R8
4c. Killer pair 3,8 in R9C34 and R9C7, locked for R9

5. 7 in R9 only in R9C12
5a. 45 rule on R89 3 remaining innies R8C1 + R9C12 = 19 = {379/478}, no 2
5b. 2 in R9 only in R9C89, locked for N9 -> R8C9 = 4

6. 2 in R8 only in R8C56 = {27}, locked for R8 and N8, clean-up: no 8 in R8C3
6a. Naked pair {69} in R8C34, locked for R8

7. 45 rule on N47 (using R8C1 + R9C12 = 19, step 5a) 3 remaining innies R489C3 = 23 = {689}, locked for C3, clean-up: no 9 in R9C4
7a. R8C4 = 9 (hidden single in N8), R8C3 = 6
7b. 7 in C3 only in R123C3, locked for N1

8. 12(4) cage at R6C3 = {1245} (only remaining combination), no 3
8a. 3 in N7 only in R78C1, locked for C1
8b. 3 in C3 only in R123C3, locked for N1

9. 12(4) cage at R5C2 = {1236} (only remaining combination, cannot be {1245} which clashes with R7C2), no 4,5, 6 locked for C2 and N4
9a. 12(4) cage at R5C2 = {1236}, CPE no 2 in R7C2
9b. 12(4) cage at R6C3 (step 7) = {1245}, 2 locked for C3
9c. 5 in N7 only in R7C23, locked for 12(4) cage at R6C3, no 5 in R6C3

10. 18(4) cage at R6C6 contains 1 = {1368/1458/1467} (cannot be {1269/1278/1359} because 2,5,7,9 only in R6C6), no 2,9
10a. 5,7 of {1458/1467} must be in R6C6 -> no 4 in R6C6

11. 5,7 in N4 only in 24(4) cage at R4C1 = {3579/4578}, no 2
11a. 3 of {3579} must be in R4C2 -> no 9 in R4C2

12. 20(3) cage at R1C2 = {389/479/569/578}
12a. 3 of {389} must be in R1C3 -> no 3 in R1C4
12b. 9 of {479} must be in R1C2 -> no 4 in R1C2

13. 18(3) cage at R2C3 = {279/369/378/459/468/567} (cannot be {189} because no 1,8,9 in R2C3), no 1
13a. 9 of {279} must be in R2C5 -> no 2 in R2C5
13b. 9 of {459} must be in R2C5, 4 of {468} must be in R2C3 -> no 4 in R2C5

14. Min R34C3 = 11 -> max R35C4 = 7, no 7,8 in R35C4, no 6 in R3C4

15. 16(4) cage at R2C1 = {1249/1258} (cannot be {1456} which clashes with R123C3, ALS block), no 6
15a. R1C1 = 6 (hidden single in N1)
15b. Killer pair 4,5 in 16(4) cage and R123C3, locked for N1

16. 45 rule on R1 3 remaining innies R1C789 = 13 = {139/157/238/247} (cannot be {148} which clashes with R1C56)
16a. 7 of {157/247} must be in R1C7 -> no 4,5 in R1C7
16b. Killer quad 6,7,8,9 in R1C789 and 30(4) cage at R2C9, locked for N3
16c. 6 in N3 only in R2C9 + R3C89, locked for 30(4) cage, no 6 in R4C8

17. 4,6 in N6 only in R4C7 and 16(4) cage at R5C7 -> 16(4) cage must contain at least one of 4,6 = {1249/1267/1456/2347} (cannot be {1348/2356} which clash with R456C9, ALS block), no 8
17a. 1 of {1456} must be in R6C8 -> no 5 in R6C8

18. 16(4) cage at R5C7 (step 17) = {1249/1267/1456/2347}
18a. Killer quad 1,2,3,5 in 16(4) cage and R456C9, locked for N6
18b. Hidden killer quad 1,2,3,5 in 16(4) cage and R456C9 for N6, 16(4) cage contains two of 1,2,3,5 -> R456C9 must contain two of 1,2,3,5 -> R456C9 = {158/238/358}, 8 locked for C9 and N6

19. 30(4) cage at R2C9 = {6789} -> R3C8 = 8
19a. 8 in R1 only in 20(3) cage at R1C2 (step 12) = {389/578}, no 4

20. 30(4) cage at R2C9 = {6789}, 6 locked for C9 -> R9C89 = [62]

21. 2 in N6 only in 16(4) cage at R5C7 (step 17) = {1249/1267/2347}, no 5
21a. 5 in N6 only in R456C9, locked for C9

22. Hidden killer pair 4,6 in R4C7 and 16(4) cage at R5C7 for N6, 16(4) cage contains one of 4,6 -> R4C7 = {46}
[Killer pair 7,9 in R4C8 and 16(4) cage at R5C7 is technically simpler, but my mind was focussed on 4,6 in N6.]
22a. Max R34C7 = 11 -> min R45C6 = 13, no 1,2,3 in R45C6

23. R1C789 (step 16) = {139/157} (cannot be {247} because R1C9 only contains 1,3), no 2,4, 1 locked for R1 and N3, clean-up: no 5 in R1C56
23a. 7,9 only in R1C7 -> R1C7 = {79}

24. Naked pair {24} in R1C56, locked for N2
24a. 2 in C4 only in R456C4, locked for N5
[Here I missed R4C4 = 2 (hidden single in R4), which would have simplified step 28. I haven’t re-worked from here, even though I consider that I missed an “obvious” step, because that would remove the difficult but interesting step 29, which was my final breakthrough step. Even though Ed commented that I don’t miss many hidden singles these days, they are still hard to spot.]

25. 12(4) cage at R1C8 = {1245} (only remaining combination, cannot be {1236} which clashes with R1C9), no 3,6
25a. 2,4 only in R2C78 -> R2C78 = {24}, locked for R2 and N3
25b. Naked triple {358} in R389C7, locked for C7
25c. Naked pair {15} in R29C6, locked for C6

26. 3 in R2 only in 18(3) cage at R2C3 = {369/378}, no 5
26a. 9 of {369} must be in R2C5 -> no 6 in R2C5
26b. Hidden killer pair 6,7 in 18(3) cage and R2C9 for R2, 18(3) cage contains one of 6,7 -> R2C9 = {67}

27. 16(4) cage at R2C1 (step 15) = {1249/1258}
27a. 9 of {1249} must be in R2C2 -> no 9 in R2C1 + R3C12

28. 18(4) cage at R3C3 = {1278/1359/1368/1458/2349} (cannot be {1269} because 2,6 only in R5C4, cannot be {1467/2367/2457/3456} because R4C3 only contains 8,9, cannot be {2358} which clashes with R3C7)
28a. 1 of {1278/1359/1368/1458} must be in R3C4, 3 of {2349} must be in R3C4 -> R3C4 = {13}

29. 45 rule on N12 5(1+4) innies R2C6 + R3C3456 = 24 = 5 + 19(4) (cannot be R2C6 = 1, R3C4 = 3, R3C356 = 20 because R2C6 + R3C3456 cannot be 143{79} which clashes with 18(3) cage at R2C3, ALS block, and cannot be 153{69} which clashes with R3C7) -> R2C6 = 5, R1C89 = [13], R1C7 = 9 (step 23), R1C2 = 8, R2C4 = 7, R1C3 = 5, R2C12 = [19], R3C7 = 5, R9C56 = [51]
29a. R8C8 = 8 (hidden single in R8)

30. Naked pair {24} in R67C3, locked for C3 and 12(4) cage at R6C3 -> R7C2 = 5

31. 18(3) cage at R2C3 (step 26) = {378} (only remaining combination) -> R2C3 = 7, R2C45 = {38}, locked for R2 and N2 -> R23C9 = [67], R4C8 = 9, R7C89 = [79]

32. R3C34 = [31], R4C3 = 8, -> R5C4 = 6 (step 28), R9C3 = 9, R9C4 = 3

33. R3C7 = 5, R4C5 = 6 (hidden single in R4) -> R45C6 = 13 = [49], R1C56 = [42], R8C56 = [27], R3C56 = [96], R7C56 = [68], R6C6 = 3

34. R3C56 = [96] = 15 -> R4C45 = 9 = [27]

and the rest is naked singles.

Amended Rating Comment:
When I first posted my walkthrough for hidden A245, I rated it at Easy 1.5 because step 29 feels like it's more than in the 1.25 range. However after seeing Ed’s breakthrough step and realising that I’d missed an “obvious” hidden single after step 24a, I think the rating for this Assassin should be Hard 1.25, based on the analysis in N6, using either my way or Ed’s way to do it.
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Andrew
PostPosted: Tue Sep 11, 2012 3:15 am 
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Then I had a go at the Paper Solvable version and was happy to find that I was able to get it out using only Paper Solvable steps. I'm not particularly good at this; it needs a different approach from elimination or insertion solving.

Here is my modified walkthrough for the Paper Solvable version:
I’ll try to solve this version using Paper Solvable steps, including repeating steps used previously rather than referring back to those previous steps. However I will use some knowledge which I got from solving A245 first and from going through Ed’s walkthrough for that puzzle.

1. 45 rule on N9 1 innie R7C7 = 1

2. 12(4) cage at R5C2 must contain 1, locked for N4 -> 12(4) cage at R6C3 must contain 1, only possible in R8C2 -> R8C2 = 1

3. 12(4) cage at R5C2 must contain 1, only possible in R5C3 -> R5C3= 1

4. Step 4 deleted. It was unnecessarily difficult at this stage, although I’ve used it at step 10. Steps 5 and 10 have been slightly modified.

5. 16(2) cage at R7C8 = {79} -> 16(3) cage at R8C7 cannot contain 4 (because {349/457} would clash with 16(2) cage
15(2) cage at R8C3 = {69/78}
1 in N8 can only be in 6(2) cage at R9C5 = {15} -> 9(2) cage at R8C5 cannot be {45}
45 rule on R8 2 innies R8C19 = 1 outie R9C7 -> no 4 in R8C1 (IOU)
-> R8C9 = 4 (only possible place for 4 in R8)

6. 6(2) cage at R4C9 = {15} (only remaining combination) -> R4C9 = 1, R5C9 = 5

7. 1 in N8 can only be in 6(2) cage at R9C5 = {15} -> 12(3) cage at R8C9 = 4{26} (cannot be 4{35} which would clash with 6(2) cage at R9C5)
30(4) cage at R2C9 and 12(3) cage at R8C9 both contain 6 for C89, caged X-Wing
45 rule on N6 3 innies R4C78 + R6C9 = 23 = {689} -> R4C7 = 6

8. 30(4) cage at R2C9 = {6789}, 45 rule on N6 2 remaining innies R4C8 + R6C9 = 17 = {89} -> R34C8 must be {89} (otherwise R23C9 would clash with R6C9)
16(2) cage at R7C8 = {79} -> R7C8 = 7, R7C9 = 9, R6C9 = 8, R4C8 = 9, R3C8 = 8

9. 1 in N8 can only be in 6(2) cage at R9C5 = {15} -> 12(3) cage at R8C9 = 4{26} (cannot be 4{35} which would clash with 6(2) cage at R9C5)
30(4) cage at R2C9 = {6789} -> R23C9 = {67}, 12(3) cage at R8C9 = 4{26} -> R9C9 = 2, R9C8 = 6
R1C9 = 3 (only possible place for 3 in C9)

10. 45 rule on R89 3 innies R8C1 + R9C12 = 19
45 rule on N47 (using R8C1 + R9C12 = 19), 3 remaining innies R489C3 = 23 = {689} -> R8C3 = 6 (cannot place 6 in R49C3), R8C4 = 9, R4C3 = 8, R9C3 = 9, R9C4 = 3

11. 1 in N8 can only be in 6(2) cage at R9C5 = {15}
45 rule on R8 1 remaining innie R8C1 = 1 outie R9C7 can only be 8 because no 2,3,5,7 in R9C7 -> R9C7 = 8, R8C1 = 8

12. 1 in N8 can only be in 6(2) cage at R9C5 = {15} -> R9C12 = {47} (only remaining places in R9)
12(4) cage at R6C3 = {1245} (only remaining combination because R8C3 = 6) -> R6C3 = 4 (only possible place for 4 in 12(4) cage), R7C23 = {25} -> R7C1 = 3 (only remaining place for 3 in N7)
12(4) cage at R5C2 = {1236} (only remaining combination because R7C1 = 3) -> R56C2 = {26} -> R7C2 = 5, R7C3 = 2, R4C2 = 3 (only remaining place for 3 in N4)

13. 12(4) cage at R5C2 = {1236} (only remaining combination because R7C1 = 3) -> R56C2 = {26} -> 2,6 in N1 can only be in R1C1 or in 16(4) cage at R2C1, 16(4) cage cannot contain both of {26} in R23C1 because R23C2 = 8 clashes with 1,3,5 in C2 -> R1C1 must contain one of 2,6
3,5,7 in C3 only in R123C3 = 15, 45 rule on N1 2 remaining innies R1C12 = 14 -> R1C1 = 6, R1C2 = 8
45 rule on R1 2 remaining innies R1C78 = 10 = {19} (only remaining combination) -> R1C8 = 1, R1C7 = 9

14. 30(4) cage at R2C9 = 30 = {6789} -> R23C9 = {67} = 13
45 rule on N3 1 remaining innie R3C7 = 1 remaining outie R2C6 -> no 3 in R2C6 -> 12(4) cage at R1C8 = {1245} (only remaining combination) -> R2C678 = {245}
6(2) cage at R1C5 = {24} (only remaining combination) -> R2C6 = 5 -> R3C7 = 5
R8C8 = 5 (only remaining place for 5 in N9), R8C7 = 3 (remaining place in N9)
R9C5 = 5 (only remaining place for 5 in N8), R9C6 = 1 (only remaining place for 1 in N8)

15. R34C7 = [56] = 11 -> R45C6 = 13 = {49} (only remaining combination) -> R4C6 = 4, R5C6 = 9
6(2) cage at R1C5 = {24} (only remaining combination) -> R1C5 = 4, R1C6 = 2
R1C3 = 5 (only remaining place for 5 in R1), R1C4 = 7 (remaining place in R1)
R8C5 = 2 (only remaining place for 2 in R8), R8C6 = 7 (remaining place in R8)

16. R7C4 = 4 (only remaining place for 4 in R7)
R7C56 = {68} (only remaining places for 6,8 in R7) = 14, R7C7 = 1 -> R6C6 = 3 (cage total)
R5C8 = 3 (only remaining place for 3 in N6)

17. R2C8 = 4 (only remaining place for 4 in C8), R2C7 = 2 (cage total), R6C8 = 2 (remaining place in C8)
R5C7 = 4 (only remaining place for 4 in C7), R6C7 = 7 (remaining place in C7)

18. R7C1 = 3 -> 12(4) cage at R5C2 = {1236} -> R5C2 = 2, R6C2 = 6

19. 5,7,9 in N4 only possible in R456C1 -> R6C1 = 9, R5C1 = 7, R4C1 = 5
R9C2 = 7 (only remaining place for 7 in N7), R9C1 = 4 (remaining place in N7)
R3C12 = [24] (only remaining places for 2,4 in N1), R2C12 = [19] (only remaining places for 1,9 in N1)

20. 45 rule on R2 1 remaining innie R2C9 = 6, R3C9 = 7 (only remaining place in C9)

21. 3,7,8 in R2 only possible in 18(3) cage at R2C3 -> R2C3 = 7, R2C4 = 8, R2C5 = 3

22. R3C3 = 3 (only remaining place in N1)
R34C3 = [38] = 11 -> R35C4 = 7 = {16} (cannot be {25} because 2,5 blocked from R3C4) -> R5C4 = 6, R3C4 = 1

23. R5C5 = 8 (only remaining place in R5), R7C6 = 8 (only remaining place for 8 in R7), R7C5 = 6 (only remaining place in R7)

24. R6C5 = 1 (only remaining place for 1 in R6), R6C4 = 5 (only remaining place in R6)

25. R4C4 = 2 (only remaining place in C4), R4C5 = 7 (only remaining place in R4), R3C5 = 9 (only remaining place in C5), R3C6 = 6 (last cell)
I’ve slightly simplified my solving path, deleting my original step 4 which was unnecessarily difficult at that stage, although I’ve used it at step 10. If anyone who is better at Paper Solving can find a simpler way to continue after my step 9, please feel free to post it in this thread.

Rating Comment:
I won't rate my Paper Solvable walkthrough. However the SS score of 0.95 seems much too low. I found it necessary to do my hardest step (step 10, previously used at step 4) fairly early. I wonder how Sudoku Solver managed to obtain R489C3 = 23 using steps which only give a score of 0.95.
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Andrew
PostPosted: Fri Sep 14, 2012 1:25 am 
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Joined: Wed Apr 23, 2008 6:04 pm
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Location: Lethbridge, Alberta, Canada
Thanks HATMAN for another interesting and challenging variant. The further removal of cage C definitely made it into a V2 but I also found it to be the most enjoyable of the three versions. :D

Here is my walkthrough for hidden Assassin 245 V2:
Prelims

a) R1C56 = {15/24}
b) R7C89 = {79}
c) R8C34 = {69/78}
d) R8C56 = {18/27/36/45}, no 9
e) R9C34 = {39/48/57}, no 1,2,6
f) R9C56 = {15/24}
g) 20(3) cage at R1C2 = {389/479/569/578}, no 1,2
h) 12(4) cage at R1C8 = {1236/1245}, no 7,8,9
i) 30(4) cage at R2C9 = {6789}
j) 12(4) cage at R5C2 = {1236/1245}, no 7,8,9
k) 12(4) cage at R6C3 = {1236/1245}, no 7,8,9

Steps resulting from Prelims
1a. Naked pair {79} in R7C89, locked for R7 and N9
1b. R8C56 = {18/27/36} (cannot be {45} which clashes with R9C56), no 4,5
1c. Naked quad {6789} in 30(4) cage at R2C9, CPE no 6 in R12C8
1d. 12(4) cage at R5C2 and 12(4) cage at R6C3 contain both 1,2 for N47, locked for N47
1e. Caged X-Wing pair for 7,9 in 30(4) cage at R2C9 and R7C89, no other 7,9 in C89
1f. 9 in N8 only in R89C4, locked for C4
[Because the easy start in A245 is no longer available, I’ve kept step 1d here rather than hoping for technically simpler steps to become available.]

2. 16(3) cage at R8C7 = {268/358}, no 1,4 8 locked for N9
2a. 8 in R7 only in R7C456, locked for N8, clean-up: no 7 in R8C3, no 1 in R8C56, no 4 in R9C3
2b. Killer pair 6,7 in R8C34 and R8C56, locked for R8
2c. 6 of {268} must be in R9C7 -> no 2 in R9C7
2d. 7 in R8 only in R8C456, locked for N8, clean-up: no 5 in R9C3

3. 45 rule on N6 4 outies R7C123 + R8C2 = 1 innie R4C3 + 3
3a. Min R7C123 + R8C2 = 10 -> min R4C3 = 7

4. 45 rule on R1 4 innies R1C1789 = 19 = {1369/1378/1567/2368/2467} (cannot be {1279/1459/2359/2458/3457} which clash with R1C56, cannot be {1468} which clashes with 20(3) cage at R1C2)
4a. Killer quad 6,7,8,9 in R1C789 and 30(4) cage at R2C9, locked for N3
4b. Hidden killer quad 6,7,8,9 in R1C789 and 30(4) cage at R2C9 for N3, 30(4) cage contains three of 6,7,8,9 in N3 -> R1C789 must contain one of 6,7,8,9 -> R1C1 = {6789} (because R1C1789 contains two of 6,7,8,9)

5. Consider placements for 1,2 in N7
1,2 in R7C1 => R6C3 = 1,2 (hidden single in N6)
or 1,2 in R78C2 => R5C3 = 1,2 (hidden single in N6)
or 1,2 in R7C3
-> 1,2 must be in R567C3, locked for C3
5a. 1,2 in N1 only in 16(4) cage at R2C1 = {1249/1258/1267}, no 3

6. 4 in R8 only in R8C129
6a. 45 rule on R8 3 innies R8C129 = 1 outie R9C7 + 5
6b. Max R9C7 = 8 -> max R8C129 = 13 containing 4 = {148} -> no 9 in R8C1

7. 9 in R8 only in R8C34 = {69}, locked for R8, clean-up: no 3 in R8C56
7a. Naked pair {27} in R8C56, locked for R8 and N8, clean-up: no 4 in R9C56
7b. Naked pair {15} in R9C56, locked for R9 and N8, clean-up: no 7 in R9C3
7c. 2 in R9 only in R9C89, locked for N9
[Ed mentioned in his A245 walkthrough that there’s a short-cut. R1C56 must be {24} because {15} would cause a killer UR with R9C56. However I’m one of the solvers who doesn’t use UR, because it doesn’t solve the complete puzzle.]

8. 16(3) cage at R8C7 (step 2) = {358} (only remaining combination), locked for N9, 5 also locked for R8
8a. Killer pair 3,8 in R9C34 and R9C7, locked for R9

[Step 7 of my A245 walkthrough no longer works, because the total of the “zero cells” in N7 isn’t yet known. However Ed’s 45 on N47 still works.]
9. 45 rule on N47 6(1+5) innies R4C3 + R8C13 + R9C123 = 42
9a. All of 7,8,9 in N7 are in R8C13 + R9C123 -> R4C3 + two remaining values in R8C13 + R9C123 = 18 = 8{46}/9{36} -> R4C3 = {89}
9b. R4C3 + two remaining values in R8C13 + R9C123 = 8{46}/9{36}, 6 locked for N9

10. 7 in C3 only in R123C3, locked for N1
10a. 16(4) cage at R2C1 (step 5a) = {1249/1258}, no 6

11. Min R34C3 = 11 -> max R35C4 = 7, no 7,8 in R35C4
11a. 16(4) cage can only contain one of 8,9, R4C3 = {89} -> no 8,9 in R3C3

[At this stage I saw the following; however after spotting the next stage I’ll leave this for now in case it’s not necessary …
12(4) cage at R5C2 and 12(4) cage at R6C3 cannot both be {1245} (I won’t list all the permutations but I think this can be demonstrated by looking at the placements for 2,5 in R7C123 and showing that there would be a clash in C2 or C3 if both 12(4) cages have the same combination) -> one of these 12(4) cages must be {1236} -> R56C23 must contain 6, locked for N4
Maybe somebody can show a more elegant way to demonstrate that these 12(4) cages cannot have the same combination?]

12. Consider permutations for R4C3 + two remaining values in R8C13 + R9C123 (step 9a) = 8{46}/9{36}
R4C3 = 8 => no 3 in R8C13 + R9C123 => R9C3 = 9 => R8C3 = 6
or R4C3 = 9 => R8C3 = 6
-> R8C3 = 6, R8C4 = 9, clean-up: no 3 in R9C3
12a. Naked pair {89} in R49C3, locked for C3
12b. 12(4) cage at R6C3 = {1245} (only remaining combination), no 3
12c. Naked pair {14} in R8C28, locked for R8
12d. 3 in N7 only in R78C1, locked for C1
12e. R9C89 = {26} (hidden pair in R9), locked for N9
12f. Grouped X-Wing for 6 in 30(4) cage at R2C9 and R9C89, no other 6 in C89

13. 20(3) cage at R1C2 = {389/479/569/578}
13a. 3 of {389} must be in R1C3 -> no 3 in R1C24
13b. 9 of {569} must be in R1C2 -> no 6 in R1C2
13c. 3 in N1 only in R123C3, locked for C3

14. R1C1 = 6 (hidden single in N1)
14a. R1C1789 (step 4) = {1369/1567/2368/2467} -> R1C789 = {139/157/238/247}
14b. 7,9 of {139/157/247} must be in R1C7 -> no 1,4,5 in R1C7

15. Hidden killer pair 8,9 in R1C2 and 16(4) cage at R2C1 for N1, 16(4) cage contains one of 8,9 -> R1C2 = {89}

16. 7 in N4 only in 24(4) cage at R4C1 = {3579/4578} (cannot be {3678} because 3,6 only in R4C2), no 6, 5 locked for N4
16a. 3 of {3579} must be in R4C2 -> no 9 in R4C2

17. 6 in N4 only in R56C2 -> 12(4) cage at R5C2 = {1236} (only remaining combination), no 4,5

18. 6 in N3 only in R2C9 + R3C89, locked for 30(4) cage at R2C9, no 6 in R4C8

19. 18(3) cage at R2C3 = {279/369/378/459/468/567} (cannot be {189} because no 1,8,9 in R2C3), no 1
19a. 9 of {279} must be in R2C5 -> no 2 in R2C5

20. 18(4) cage at R6C6 = {1368/1458/1467/3456} (cannot be {1269/1278/1359/2349/2358/2367/2457} because 2,5,7,9 only in R6C6), no 2,9
20a. 1 of {1368} must be in R7C7, 5,7 of {1458/1467/3456} must be in R6C6 -> no 1,4 in R6C6

21. Consider combinations for 24(4) cage at R4C1 (step 16) = {3579/4578}
24(4) cage = {3579} => R6C3 = 4 (hidden single in N4) => R8C2 = 1
or 24(4) cage = {4578}, locked for N4 => naked pair {12} in R56C3, locked for N4
-> no 1 in R56C2
21a. 1 in N4 only in R56C3, locked for C3

[A slightly harder forcing chain, but still a fairly short one …]
22. 16(4) cage at R5C7 = {1249/1258/1267/1348/1357/1456/2347/2356}
Consider combinations for 16(4) cage
16(4) cage = {1267}, locked for N6 => R4C38 = {89}, locked for R4 => R4C8 = 9 (hidden single in N6) => 8 in N6 only in R56C9, locked for C9 and N6
or 16(4) cage contains one of 6,7,8,9, hidden killer quad 6,7,8,9 in R4C78 + R456C9 for N6 => R4C7 = {679}, R4C8 = {79}, R456C9 must contain 8, locked for C9 and N6
-> R456C9 must contain 8, locked for C9 and N6

23. 30(4) cage at R2C9 = {6789} -> R3C8 = 8
23a. R1C7 + R23C9 = {679} (hidden triple in N3) -> R1C7 = {79}
23b. 6 in N3 only in R23C9, locked for C9 -> R9C89 = [62]
23c. 20(3) cage at R1C2 = {389/578} (cannot be {479} which clashes with R1C7), no 4

24. 16(4) cage at R5C7 (step 22) = {1249/1267/2347/2356} (cannot be {1357/1456} which clash with R456C9, ALS block), 2 locked for N6
24a. 16(4) cage at R5C7 (step 22) = {1249/1267/2347} (cannot be {2356} because R456C9 = {148} (only remaining values for R456C9) would clash with R8C9), no 5
24b. Killer pair 7,9 in R4C8 and 16(4) cage, locked for N6

25. Max R34C7 = 11 -> min R45C6 = 13, no 1,2,3 in R45C6
25a. Max R45C6 = 17 -> min R34C7 = 7, no 1 in R4C7

[Another slightly harder forcing chain …]
26. 16(4) cage at R5C7 (step 24a) = {1249/1267/2347}
26a. Consider combinations for R1C789 (step 14a) = {139/157/247}
R1C789 = {139} = 9{13} => 12(4) cage at R1C8 = {1245} (cannot be {1236} which clashes with R1C9) => R1C8 = 1
or R1C789 = {157/247} => R1C7 = 7
-> 16(4) cage = {1249/2347} (because R1C8 = 1 or R1C7 = 7), no 6, 4 locked for N6

27. R4C7 = 6 (hidden single in N6)
27a. 5 in N6 only in R456C9, locked for C9

28. 2 in R4 only in R4C45, locked for N5 and 24(4) cage at R3C5, no 2 in R3C56
28a. 24(4) cage = {2589/2679}, no 1,3,4
28b. 2,8 of {2589} must be in R4C45 -> no 5 in R4C45

29. 12(4) cage at R1C8 = {1236/1245}
29a. 6 of {1236} must be in R2C6 -> no 3 in R2C6

30. 3 in C6 only in R67C6, locked for 18(4) cage at R6C6, no 3 in R7C5
30a. 18(4) cage at R6C6 (step 20) = = {1368/3456} (cannot be {1458/1467} which don’t contain 3), no 7
30b. R7C7 = {14} -> no 4 in R7C56
30c. 4 in N8 only in R79C4, locked for C4

31. 2 in R1 only in R1C56 and R1C8, CPE no 2 in R2C6
31a. 12(4) cage at R1C8 = {1236/1245}, 2 locked for N3

32. Consider combinations for 24(4) cage at R3C5 (step 28a) = {2589/2679}
32a. 24(4) cage = {2589}, 5 locked for R3 => 5 in N3 only in 12(4) cage at R1C8 = {1245}
or 24(4) cage = {2679}, 6 locked for N2 => 12(4) cage at R1C8 = {1245} (only remaining combination)
-> 12(4) cage at R1C8 = {1245}, no 3,6

33. R1C789 (step 14a) = {139/157/247}
33a. 4 of {247} must be in R1C9 -> no 4 in R1C8
33b. 12(4) cage at R1C8 = {1245}, 4 locked for R2
33c. 4 in N1 only in R3C123, locked for R3

[Just spotted …]
34. Naked quad {2789} in R4C3458, locked for R4
34a. Naked pair {45} in R4C16, locked for R4 -> R4C2 = 3, R4C9 = 1, R8C9 = 4, R7C7 = 1, R1C9 = 3, R8C2 = 1
34b. R5C3 = 1 (hidden single in N4)
[And at last I’ve reached the original A245 puzzle]

35. Naked pair {26} in R56C2, locked for C2, N4 and 12(4) cage at R5C2 -> R6C3 = 4, R7C1 = 3, R8C1 = 8, R9C3 = 9, R9C4 = 3, R4C1 = 5, R3C7 = 5, R4C6 = 4, R5C6 = 9 (cage sum), R56C1 = [79], R9C12 = [47]

36. R4C3 = 8, R4C45 = {27}, locked for R4, N5 and 24(4) cage at R3C5, no 7 in R3C56 -> R3C56 = [96]

37. 12(4) cage at R1C8 = {1245} -> R2C6 = 5, clean-up: no 1 in R1C56
37a. R1C56 = [42]

and the rest is naked singles.

Rating Comment:
I’ll rate my walkthrough for hidden A245 V2 at 1.75. I used several forcing chains. The two harder ones felt as if they should be in the 1.75 range, rather than the 1.5 range.
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