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 Post subject: Assassin 215
PostPosted: Fri Jun 24, 2011 1:34 am 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1043
Location: Sydney, Australia
Not technical like the last couple. All common moves. Lots of spots to explore and makes you think. One to enjoy!

Assassin 215

NOTE: 1-9 cannot repeat on the diagonals
Image


Code: paste into solver:
3x3:d:k:3585:3585:9730:9730:9730:9730:2307:2307:6660:3585:3589:3589:3846:9730:9730:6660:6660:5895:1544:1544:6153:3846:3846:9730:6660:5895:5895:6666:6153:6153:6153:3846:6660:2571:5895:5644:6666:6666:6153:2573:2573:2571:2571:5644:5644:6666:1294:1294:4879:6160:2065:5650:4883:5644:2068:4879:4879:6160:6160:2065:5650:4883:4883:2068:4879:6421:6421:3094:3094:5650:3607:3607:4879:6421:6421:1816:1816:1305:1305:3607:3607:
solution:
+-------+-------+-------+
| 7 3 2 | 8 9 1 | 4 5 6 |
| 4 8 6 | 3 5 7 | 2 1 9 |
| 1 5 9 | 4 2 6 | 8 3 7 |
+-------+-------+-------+
| 3 2 7 | 1 6 9 | 5 4 8 |
| 8 6 5 | 7 3 4 | 1 9 2 |
| 9 4 1 | 2 8 5 | 7 6 3 |
+-------+-------+-------+
| 2 1 4 | 9 7 3 | 6 8 5 |
| 6 7 3 | 5 4 8 | 9 2 1 |
| 5 9 8 | 6 1 2 | 3 7 4 |
+-------+-------+-------+
Cheers
Ed


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 Post subject: Re: Assassin 215
PostPosted: Sun Jun 26, 2011 2:51 am 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1894
Location: Lethbridge, Alberta, Canada
Ed wrote:
Not technical like the last couple. All common moves. Lots of spots to explore and makes you think. One to enjoy!
Many thanks Ed for your latest Assassin. A very good description for it. This was a fun puzzle! :D

No difficult steps although I found it hard to spot a couple of my steps.

Rating Comment:
I'll rate my walkthrough for A215 at Easy 1.5, because of step 14 and maybe step 19 is also in this rating. Some people might rate step 14 lower but since it's a bit of a "trick" I think it deserves to be in the 1.5 range.

Here is my walkthrough for A215:
As usual I've stated placements on the diagonals for anyone doing manual eliminations or solving on paper.

Prelims

a) R1C78 = {18/27/36/45}, no 9
b) R2C23 = {59/68}
c) R3C12 = {15/24}
d) R5C45 = {19/28/37/46}, no 5
e) R6C23 = {14/23}
f) R67C6 = {17/26/35}, no 4,8,9
g) R78C1 = {17/26/35}, no 4,8,9
h) R8C56 = {39/48/57}, no 1,2,6
i) R9C45 = {16/25/34}, no 7,8,9
j) R9C67 = {14/23}
k) 10(3) cage at R4C7 = {127/136/145/235}, no 8,9
l) 24(3) cage at R6C5 = {789}
m) 22(3) cage at R6C7 = {589/679}
n) 19(3) cage at R6C8 = {289/379/469/478/568}, no 1
o) 26(4) cage at R4C1 = {2789/3689/4589/4679/5678}, no 1
p) 14(4) cage at R8C8 = {1238/1247/1256/1346/2345}, no 9

Steps resulting from Prelims
1a. 22(3) cage at R6C7 = {589/679}, 9 locked for C7
1b. 9 in R9 only in R9C123, locked for N7
1c. 24(3) cage at R6C5 = {789}, CPE no 7,8,9 in R8C5, clean-up: no 3,4,5 in R8C6
1d. Naked triple {789} in R7C45 + R8C6, locked for N8, clean-up: no 1 in R6C6

2. R9C45 = {16/25} (cannot be {34} which clashes with R9C67), no 3,4
2a. Combined cage R9C4567 = 12 = {1236/1245}, 1,2 locked for R9
2b. Min R9C89 = 9 (R9C89 cannot be {34/35} which clash with R9C4567) -> max R8C89 = 5, no 5,6,7,8 in R8C89
[I originally used killer pair 1,2 in step 2a but the combined cage is the simplest way to lead into step 2b.]

3. 45 rule on N1 2 innies R13C3 = 11 = {29/38/47} (cannot be {56} which clashes with R2C23), no 1,5,6

4. 45 rule on N3 2 outies R4C68 = 13 = {49/58/67}, no 1,2,3

5. 45 rule on N7 2 outies R68C4 = 7 = {16/25/34}, no 7,8,9

6. 45 rule on N12 1 innie R3C3 = 1 outie R4C5 + 3, no 2,3 in R3C3, R4C5 = {1456}, clean-up: no 8,9 in R1C3 (step 3)

7. 45 rule on N124 2 outies R4C45 = 7 = [16/25/34/61], R4C4 = {1236}

8. 45 rule on R6789 2 innies R6C19 = 12 = {39/48/57}, no 1,2,6

9. 45 rule on D/ 2(1+1) outies R2C7 + R7C2 = 1 innie R5C5
9a. Min R2C7 + R7C2 = 2 -> min R5C5 = 2, clean-up: no 9 in R5C4

10. 45 rule on C789 2 outies R45C6 = 1 innie R9C7 + 10
10a. Min R45C6 = 11, no 1 in R5C6

11. 14(3) cage at R1C1 = {167/239/347} (cannot be {149/257} which clash with R3C12, cannot be {158/356} which clash with R2C23, cannot be {248} which clashes with R13C3), no 5,8

12. 38(7) cage at R1C3 = {1256789/1346789/2345789}, 8,9 locked for N2

13. 19(3) cage at R6C8 = {289/379/469/478/568}
13a. 2,3 of {289/379} must be in R7C89 (R7C89 cannot be {79/89} which clash with R7C45, ALS block), no 2,3 in R6C8

[I saw this 45 much earlier but decided to use the 45 on N12 (step 6) instead. It was only now that I saw the key point, triggered because I realised that it led to step 15; I’d been looking at the hidden killer pair, then spotted that step 14 made it powerful.]
14. 45 rule on N2 2(1+1) outies R1C3 + R4C5 = 8 = [26/35/71] (cannot be [44] because R1C3 + R4C5 “see” all cells in N2), no 4 in R1C3, no 4 in R4C5, clean-up: no 7 in R3C3 (step 3), no 3 in R4C4 (step 7)

15. Hidden killer pair 1,2 in R6C23 and R6C46 for R6, R6C23 contains one of 1,2 -> R6C46 must contain one of 1,2
15a. Killer pair 1,2 in R4C45 and R6C46, locked for N5, clean-up: no 8,9 in R5C45

16. R4C6 + R6C5 = {89} (hidden pair in N5), clean-up: R4C8 = {45} (step 4)
16a. 7 in 24(3) cage at R6C5 must be in R7C45, locked for R7 and N8, clean-up: no 1 in R8C1, no 5 in R8C5
16b. Naked pair {89} in R48C6, locked for C6

17. R17C4 = {89} (hidden pair in C4)
17a. R7C5 = 7 (hidden single in N8), clean-up: no 3 in R5C4

18. 45 rule on C789 3 outies R459C6 = 15, min R4C6 = 8 -> max R59C6 = 7, no 7 in R5C6

19. R4C45 = 7 (step 7), R4C6 + R6C5 = {89} (step 16) = 17
19a. 45 rule on N5 3 remaining innies R5C6 + R6C46 = 11 = {137/146/245} (cannot be {236} which clashes with R4C45)
19b. 3 of {137} must be in R5C6 -> no 3 in R6C46, clean-up: no 5 in R7C6, no 4 in R8C4 (step 5)
19c. 6 of {146} must be in R6C6 -> no 6 in R5C6 + R6C4, clean-up: no 1 in R8C4 (step 5)
19d. 5 of {245} must be in R6C46 (R6C46 cannot be [42] which clashes with R6C23), no 5 in R5C6
19e. 3 in N5 only in R5C56, locked for R5
19f. 5 in N8 only in R8C4 + R9C45, CPE no 5 in R9C23
[I ought to have spotted killer pair 3,4 in R5C45 and R5C6, locked for R5 and N5.]

20. 10(3) cage at R4C7 = {136/145/235} (cannot be {127} because R5C6 only contains 3,4,5), no 7
20a. R5C6 = {34} -> no 3,4 in R45C7
20b. Killer pair 5,6 in R45C7 and 22(3) cage at R6C7, locked for C7, clean-up: no 3,4 in R1C8

21. 3 in N6 only in R46C9, locked for C9
21a. 22(4) cage at R4C9 contains 3 -> remaining three cells total 19, no 1 in 22(4) cage

22. 1 in N6 only in R45C7, locked for C7, clean-up: no 8 in R1C8, no 4 in R9C6
22a. 10(3) cage at R4C7 (step 20) contains 1 = {136/145}, no 2

23. R8C5 = 4 (hidden single in N8), R8C6 = 8, R4C6 = 9, placed for D/, R6C5 = 8, R7C4 = 9, R1C4 = 8, R4C8 = 4 (step 4), clean-up: no 1 in R1C8, no 6 in R5C4, no 4 in R6C1 (step 8)

24. 2,3 in N6 only in 22(4) cage at R4C9 = {2389} (only remaining combination), locked for N6, clean-up: no 5,7 in R6C1 (step 8)
24a. 7 in N6 only in R6C78, locked for R6, clean-up: no 1 in R7C6

25. Naked pair {39} in R6C19, locked for R6, clean-up: no 2 in R6C23
25a. Naked pair {14} in R6C23, locked for R6 and N4, clean-up: no 3,6 in R8C4 (step 5)
25b. Naked pair {25} in R68C4, locked for C4, clean-up: no 5 in R4C5 (step 7), no 2,5 in R9C5

26. Naked pair {16} in R4C45, locked for R4 and N5 -> R4C7 = 5, clean-up: no 4 in R5C4, no 2 in R7C6
26a. Naked pair {16} in R4C45, CPE no 1,6 in R23C4
26b. R5C4 = 7, R5C5 = 3, placed for both diagonals, R5C6 = 4, R5C7 = 1 (hidden single in R5)

27. 22(3) cage at R6C7 = {679} (only remaining combination) -> R7C7 = 6, placed for D\, R6C78 = [76], R8C7 = 9, R4C4 = 1, placed for D\, R4C5 = 6, R9C4 = 6, R9C5 = 1, R8C8 = 2, placed for D\, R8C9 = 1, R6C6 = 5, placed for D\, R7C6 = 3, R6C4 = 2, placed for D/, R9C6 = 2, R9C7 = 3, R8C4 = 5

28. R6C8 = 6 -> R7C89 = 13 = {58}, locked for R7 and N9 -> R9C8 = 7, R9C9 = 4, placed for D\, R1C8 = 5, R1C7 = 4, R3C7 = 8, R2C8 = 1, both placed for D/, R2C7 = 2, R7C23 = [14], R7C1 = 2, R8C1 = 6, R8C2 = 7, placed for D/, R1C9 = 6

and the rest is naked singles.


Last edited by Andrew on Sun Jul 03, 2011 5:07 am, edited 1 time in total.

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 Post subject: Re: Assassin 215
PostPosted: Thu Jun 30, 2011 9:50 pm 
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Grand Master
Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 1043
Location: Sydney, Australia
Thanks Andrew for your WT and enjoying this one. That is especially pleasing since your way is completely different to mine! I saw Andrew's step 14 the first time I worked on this one but couldn't make it effective like his step 15. Instead, I found what's now step 1 in this optimised walkthrough.

A215 start
12 steps:
Note: this is an optimised solution so many obvious eliminations are not included. However, I do try and clean-up. Please let me know of any corrections or clarifications. Thanks.

Prelims
i. 9(2)n3: no 9
ii. 14(2)n1: no 1..4,7
iii. 6(2)n1: no 3,6..9
iv. 26(4)n4: no 1
v. 10(3)r4c7: no 8,9
vi. 10(2)n5: no 5
vii. 5(2)n4: no 5..9
viii. 24(3)r6c5 = {789}
ix. 8(2)r6c6: no 4,8,9
x. 22(3)r6c7: no 1..4
xi. 19(3)r6c8: no 1
xii. 8(2)n7: no 4,8,9
xiii. 12(2)n8: no 1,2,6
xiv. 14(4)n9: no 9
xv. 7(2)n8: no 7,8,9
xvi. 5(2)r9c6: no 5..9

1. "45" on r89: 1 innie r8c7 = 4 outies r6c4 + r7c123
1a. -> max. 4 outies = 9
1b. min r7c123 = 6 -> max r6c4 = 3
1c. min. r6c4 + r7c23 = 6 (same cage)-> max. r7c1 = 3
1d. min. r6c4 + r7c1 = 2 -> max r7c23 = 7 (no 6,7,8,9)(can't have 6 since r7c123 can't be [1]{16})
1e. min. r6c4 + r7c123 = [1]{[1]23} = 7 -> min. 1 innie r8c7 = 7
1f. r8c1 = (567)

2. "45" on n7: 2 outies r68c4 = 7
2a. = [16/25/34]: r8c4 = (456)

Andrew's steps 1c & d is an easier way to do the next step
3. "45" on n8: 1 outie r6c5 + 2 = 3 innies r7c6 + r8c4 + r9c6
3a. max. 3 innies = 11 -> max. r7c6 = 6 (no 7)
3b. no 1 in r6c6

The next step is pretty technical.....
4. Hidden killer triple 7,8,9 in r7...ie r7c789 must have exactly one of 7,8,9 for r7 since r7c45 has two of 7,8,9 (no eliminations yet)
4a. r8c7 from (789)
4b. ->Hidden killer triple 7,8,9 in n9 -> 14(4) must have 7,8
4c. 14(4)n9 = {1238/1247}(no 5,6)
4d. must have 1 & 2: both locked for n9
4e. r9c6 = (12)

This next step is just to help the final clean-out of the puzzle but is not needed to get the crack started.
5. "45" on c89: 1 outie r1c7 + 3 = 2 innies r1c9 + r2c8
5a. -> no 3 in r1c9 nor r2c8 (IOU)

Back on track
6. 7(2)n8: {34} blocked by r9c7
6a. = {16/25}(no 3,4) = [1/2..]

7. Killer pair 1,2 in 7(2)n8 (step 6a) & r9c6: both locked for r9 and n8
7a. no 6,7 in r6c6

8. 5 & 6 in n9 only in r7: both locked for r7

9. r7c6 = 3, r6c6 = 5 (Placed for D\)
9a. no 5 in r8c1
9b. no 9 in 12(2)n8
9c. no 7 in r8c5

10. naked triple {124} at r7c123: locked for r7 & n7
10a. sums to 7 -> max. r6c4 = 2 (step 1)
10b. and min. r6c4+r7c123= 8 -> min. r8c7 = 8 (step 1)
10c. no 4 in r8c4 (h7(2)r68c4)

11. 4 in n8 only in 12(2) = {48} only: both locked for r8 and 8 for n8

12. r8c7 = 9
12a. -> r6c4 + r7c123 = 9 (step 1)
12b. -> r6c4 = 2 (Placed for D/)

On from there. Mostly naked stuff and hidden singles. Try "45" on n12: 1 outie r4c5 + 3 = 1 innie r3c3, then innies n1 if you get stuck.
Cheers
Ed


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