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Assassin 87

 
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mhparker
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Joined: 20 Jan 2007
Posts: 345
Location: Germany

PostPosted: Fri Jan 25, 2008 12:41 pm    Post subject: Assassin 87 Reply with quote

Hi folks,

Afmob (on A73 V1.5 thread) wrote:
...a lot of little moves...

This puzzle proved to be the opposite for me, with progression possible in big leaps and bounds. Hopefully, I haven't made one of those logic errors that result in a lucky fake breakthrough Exclamation I've checked the WT through already, so it should be OK (unless I was blind in the same place twice...).

Rating: 1.0 ? Will be interested to hear what the rest of you think.

Edit: Just seen that SS rates this at 1.88! Shocked


Assassin 87 Walkthrough

Prelims

a) 10(2) at R1C3 = {19/28/37/46} (no 5)
b) 15(2) at R12C5 = {69/78}
c) 5(2) at R1C6 = {14/23}
d) 11(2) at R23C4, R9C34 and R9C67 = {29/38/47/56} (no 1)
e) 9(2) at R23C6 = {18/27/36/45} (no 9)
f) 12(2) at R34C5 = {{39/48/57} (no 1,2,6)
g) 14(4) at R3C8 = {1238/1247/1256/1346/2345} (no 9)
h) 24(3) at R6C6 = {789}
i) 7(2) at R78C1 = {16/25/34} (no 7..9)
j) 14(2) at R7C9 = {59/68}
k) 13(4) at R8C7 = {1237/1246/1345} (no 8,9); 1 locked for N9

1. Innies N2: R1C46+R3C5 = 10(3) = {127/136/145/235} (no 8,9)
1a. R1C46 cannot sum to 5 due to R1C7
1b. -> no 5 in R3C5
1c. possible permutations are: [127/217]
(Note: [613] blocked because it would force a 4 into both of R1C37)
1d. -> R3C5 = 7
1e. -> R4C5 = 5
1f. R1C46 = {12}, locked for R1 and N2
1g. cleanup: no 3,4,6,7 in R1C3; no 8 in R12C5; no 4,9 in R23C4; no 8 in R23C6

2. Naked pair (NP) at R12C5 = {69}, locked for C5 and N2
2a. cleanup: no 5 in R23C4; no 3 in R23C6
2b. NP at R23C4 = {38}, locked for C4
2c. NP at R23C6 = {45}, locked for C6
2d. cleanup: no 3,8 in R9C3; no 6,7 in R9C7

3. Innies C6789: R58C6 = 10(2) = {19/28/37} (no 6)
3a. -> R58C6 and R67C6 form killer triple (KT) on {789} in C6
3b. -> no 7,8,9 elsewhere in C6
3c. cleanup: no 2,3,4 in R9C7

4. Innies C1234: R58C4 = 11(2) = {29}/{47}/[65] (no 1)
4a. no 6 in R8C4

5. 13(4) at R8C7 = {1237/1246/1345}
5a. -> must have exactly 2 of {234}
5b. only other place for {234} in N9 is R7C8
5c. -> R7C8 = {234} (no 5..9)

6. Outies N36: R14C6+R7C8 = 9(2+1)
6a. R1C6 = {12} -> R4C6+R7C8 must sum to 7 or 8 = [34/62] (only possible permutations)
6b. -> R4C6 = {36}, R7C8 = {24}

7. Innies N9: R7C78+R9C7 = 18(3) = [729/945] (no 8) (only possible permutations)
7a. 9 locked in R79C7 for C7 and N9
7b. cleanup: no 3 in R9C6; no 5 in R78C9

8. Naked pair (NP) at R78C9 = {68}, locked for C9 and N9

9! Innie/Outie (I/O) diff. C9: R67C8 = R19C9 + 8
9a. Max. R67C8 = 13 -> max. R19C9 = 5
9b. -> R19C9 = [31/32/41]
9c. -> R1C9 = {34}; R9C9 = {12}
9d. Min. R19C9 = 4 -> min. R67C8 = 12
9e. -> R67C8 = [84/94]
9f. -> R7C8 = 4; R6C8 = {89}
9g. -> R79C7 = [95] (step 7)
9h. -> R9C6 = 6 (cage sum)
9i. -> R4C6 = 3
9j. -> R1C6 = 2 (step 6)
9k. -> R1C7 = 3 (cage sum); R1C34 = [91]
9l. -> R19C9 = [41] (step 9b); R12C5 = [69]
9m. -> R6C8 = 9 (step 9.)
9n. cleanup: no 7,8 in R58C6 (step 3); no 2 in R9C4; no 3 in R8C1

10. Hidden single (HS) at R3C9 = 9
10a. -> R24C9 = [52] (only possible permutation)
10b. -> R23C6 = [45]

11. Innie N3: R3C8 = 1
11a. -> R4C78 = [46] (only possible permutation)

12. HS in C8/N6 at R5C8 = 5
12a. -> split 9(2) at R56C7 = {18}, locked for C7

13. 18(4) at R3C2 cannot have both of {79} due to cage sum
13a. -> no 7,9 in R4C23
13b. -> R4C23 = {18}, locked for N4
13c. -> split 9(2) at R3C2+R4C4 = [27] (only possible permutation)
13d. cleanup: no 4 in R9C3; no 4 in R58C4 (step 4)

14. R67C6 = [87]
14a. -> R56C7 = [81]

15. R3C7 = 6; R4C1 = 9 (both naked singles)
15a. split 9(2) at R23C1 = [18/63] (only possible permutations)
15b. -> R3C3 = 4 (HS@R3/N1)

16. Innies N7: R7C23+R9C3 = 16(3)
16a. -> R9C3 can't be 2, because {68} in R7C23 blocked by R7C9
16b. -> R9C34 = [74]
16c. -> split 9(2) at R7C23 = {18/36} (no 2,5)
16d. -> R7C23 and R7C9 form KP on {68}, locked for R7
16e. cleanup: no 1 in R8C1

17. 12(3) at R6C4+R7C34 = [615] (only possible permutation)
17a. -> R7C2 = 8 (step 16c)
17b. -> R4C23 = [18]; R78C9 = [68]
17c. cleanup: no 2,6 in R8C1

18. NP at R79C1 = {23}, locked for C1 and N7
18a. cleanup: no 6 in R2C1 (step 15a)

19. HS in C1 at R5C1 = 6
19a. -> split 7(2) at R6C12 = [43] (only possible permutation)
19b. cleanup: no 3 in R7C1

Now only naked singles left.

_________________
Cheers,
Mike
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Afmob
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Joined: 22 Sep 2007
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PostPosted: Fri Jan 25, 2008 6:34 pm    Post subject: Reply with quote

That was a fun assassin! Like Mike I'm surprised that SS rated this one so high. My key moves are similar to Mike's one though I used a different reasoning to apply them.

A87 Walkthrough:

1. R89
a) Outies = 11(3) <> 9
b) Outies = 11(3) must have two of (1234) and they are only possible @ R7C15 -> R7C15 <> 5,6,7,8
c) 7(2): R8C1 <> 1,2
d) 14(2): R8C9 <> 5
e) 13(4) = 1{237/246/345} -> 1 locked for N9

2. N2 !
a) Innies = 10(3) <> 8,9
b) Innies = 10(3): R1C4 <> 7 because R3C5 <> 1,2
c) 10(2): R1C3 <> 1,2,3
d) 12(2): R4C5 <> 3,4
e) Outies = 17(2+1): R1C3 <> 4 because R1C7 <= 4
f) ! Outies = 17(2+1): R4C5 <> 7 because R1C37 would be 10(2) which conflicts with 10(2) @ R1C34
g) R1C4 <> 6
h) 12(2): R3C5 <> 5
i) Innies = 10(3) must have one of (567) and they are only possible @ R3C5 -> R3C5 = 7

3. C456
a) Innies N2 = 10(3) = {127} -> {12} locked for R1+N2
b) 12(2) = [75] -> R4C5 = 5
c) 10(2) = [82/91]
d) 15(2) = {69} locked for C5+N2
e) 11(2) @ N2 = {38} locked for C4+N2
f) 9(2) = {45} locked for C6
g) 11(2) @ N7 <> 3,8
h) 11(2) @ N9: R9C7 <> 6,7
i) Innies C1234 = 11(2) <> 1; R8C4 <> 6
j) Innies C6789 = 10(2) <> 6
k) Killer triple (789) locked in Innies C6789 + R67C6 for C6 -> R49C6 <> 7,8,9
l) 11(2) @ N9: R9C7 <> 2,3,4

4. C6789
a) Innies N9 = 18(3): R7C8 = (234) because R79C7 >= 12 and {567} blocked by Killer pair (56) of 14(2)
b) Outies N36 = 9(2+1): R7C8 <> 3 because R14C6 <> 4,5
c) 3 locked in 13(4) = 13{27/45} <> 6
d) 6 locked in 14(2) = {68} locked for C9+N9
e) 9 locked in R79C7 for C7
f) 24(3) = {789}; 8 locked for C6
g) Innies C1234 = 10(2) = {19/37}
h) 11(2) = [29/65]
i) 16(3) = {259/349/457}
j) Outies N36 = 9(2+1) = 2+[16] / 4+[23] -> R4C6 = (36)

5. R789
a) Killer pair (24) locked in Outies R89 + R7C8 for R7
b) 23(4) <> 1 because R7C8 = (24)
c) Hidden Single: R9C9 = 1 @ C9

6. N1
a) Innies+Outies: -2 = R4C1 - R1C3+R3C2
-> R3C2 = (123) because R1C3 = (89)
-> R3C2 = (789) because R1C3+R3C2 >= 9

7. C789 !
a) ! Innies+Outies C9: 9 = R67C8 - R1C9
-> R1C9 <> 5,7,9 because R67C8 <= 13
-> R67C8 = 12/13(2) = [9/8]4 -> R7C8 = 4
b) 13(4) = {1237} locked for N9
c) 24(3) = {789} -> R7C7 = 9; 7 locked for C6
d) Outies N36 = 9(2+1) = 4+[23] -> R1C6 = 2, R4C6 = 3
e) 5(2) = {23} -> R1C7 = 3
f) R1C9 = 4
g) 16(3) = {259} locked for C9, 5 locked for N3
h) 23(4) = {3479} -> R6C8 = 9, {37} locked for N6
i) R4C9 = 2
j) 14(3) = {158} locked for N6

8. R123
a) 10(2) = {19} -> R1C3 = 9, R1C4 = 1
b) 5 locked in R1C12 for N1
c) 14(4) = {1346} -> R3C8 = 1, R4C7 = 4, R4C8 = 6
d) 18(4) = {1278} -> R3C2 = 2, R4C4 = 7, {18} locked for N4
e) 18(3) = 9{18/36} -> R4C1 = 9; R2C1 <> 8

9. N8
a) R9C6 = 6, R7C4 = 5
b) 23(5) = {12389} locked
c) 11(2) @ R9C3 = {47} -> R9C3 = 7, R9C4 = 4
d) 12(3) = {156} -> R6C4 = 6, R7C3 = 1

10. N17
a) Hidden Single: R4C2 = 1 @ N4, R2C1 = 1 @ N1 -> R3C1 = 8
b) 7(2) = [25/34]
c) 22(4) = 69{25/34} because 38{29/56} blocked by Killer pairs (23,35) of 7(2)
d) Hidden Single: R7C2 = 8 @ N7
e) R1C5 = 6
f) Hidden Single: R5C1 = 6 @ C1
g) 21(4) = 68{25/34}; R6C1 <> 5

11. Rest is singles.

Rating: Hard 1.0. I thought about rating it 1.25 but we had enough of them already Wink.


Last edited by Afmob on Mon Jan 28, 2008 5:38 am; edited 2 times in total
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Andrew
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Joined: 11 Aug 2006
Posts: 300
Location: Lethbridge, Alberta

PostPosted: Sun Jan 27, 2008 7:27 am    Post subject: Reply with quote

Yes, a fun puzzle. Comparing Mike's and Afmob's walkthroughs with mine, there seems to be a general direction that has to be followed although it's definitely not a narrow solving path and not hard to find.

They both got one key move, Mike's step 9 and Afmob's step 7a, which made their solutions more direct than mine. We all got the other key moves although not necessarily using the same logic.

Afmob also had interesting steps 1a, 1b and 5a. I donít think they were used by Mike and I didnít spot them. Remainder of this comment deleted, I'd been looking at an incorrect grid state.

Afmob wrote:
Rating: Hard 1.0. I thought about rating it 1.25 but we had enough of them already Wink.

I immediately spotted the Wink but I'll still make the general comment that ratings shouldn't be influenced by which ones are more common. Mike's rating definitions suggest that typical Assassins, apart from the early ones, will normally be rated 1.25 so there will be more of that rating than anything else.

I'm not sure between hard 1.0 and easier 1.25 but I think I'll rate A87 an easier 1.25 for the way I solved it. However I'm expecting Ed to rate it 1.0 because of the key step that I missed.

Here is my walkthrough for A87 (original step 35 deleted and step 36 renumbered).

Prelims

a) R1C34 = {19/28/37/46}, no 5
b) R12C5 = {69/78}
c) R1C67 = {14/23}
d) R23C4 = {29/38/47/56}, no 1
e) R23C6 = {18/27/36/45}, no 9
f) R34C5 = {39/48/57}, no 1,2,6
g) R78C1 = {16/25/34}, no 7,8,9
h) R78C9 = {59/68}
i) R9C34 = {29/38/47/56}, no 1
j) R9C67 = {29/38/47/56}, no 1
k) 24(3) cage at R6C6 = {789}
l) 14(4) cage at R3C8 = {1238/1247/1256/1346/2345}, no 9
m) 13(4) cage in N9 = {1237/1246/1345}, no 8,9, 1 locked for N9

1. 45 rule on N2 3 innies R1C46 + R3C5 = 10 = {127/136/145/235}, no 8,9, clean-up: no 1,2 in R1C3, no 3,4 in R4C5
1a. Cannot be {136} because [613] would make R1C37 = [44]
1b. Cannot be {145} because R1C46 = {14} clashes with R1C67 = {14}
1c. Cannot be {235} because R1C46 = {23} clashes with R1C67 = {23}
1d. -> R1C46 + R3C5 = {127} -> R1C46 = {12}, locked for R1 and N2, R3C5 = 7, R4C5 = 5, clean-up: R1C3 = {89}, no 8 in R12C5, no 4,9 in R23C4, no 8 in R23C6

2. Naked pair {69} in R12C5, locked for C5 and N2, clean-up: no 5 in R23C4, no 3 in R23C6
2a. Naked pair {38} in R23C4, locked for C4, clean-up: no 3,8 in R9C3
2b. Naked pair {45} in R23C6, locked for C6, clean-up: no 6,7 in R9C7

3. 45 rule on C1234 2 innies R58C4 = 11 = {29/47}/[65], no 1, no 6 in R8C4

4. 45 rule on C6789 2 innies R58C6 = 10 = {19/28/37}, no 6

5. Killer triple 7,8,9 in R58C6 and R67C6, locked for C6, clean-up: no 2,3,4 in R9C7

6. 45 rule on N9 3 innies R7C78 + R9C7 = 18 = {279/378/459} (cannot be {369) which clashes with 13(4) cage, cannot be {468} because 4,6 only in R7C8, cannot be {567} which clashes with R78C9), no 6
6a. 2,3,4 only in R7C8 -> R7C8 = {234}
6b. 8 of {378} must be in R9C7 -> no 8 in R7C7

7. 8 in 24(3) cage locked in R67C6, locked for C6, clean-up: no 2 in R58C6 (step 4)

8. 45 rule on N36 2 outies R4C6 + R7C8 = 1 innie R1C7 + 4
8a. R1C7 = {34} -> R4C6 + R7C8 = 7,8 = [34/62], R4C6 = {36}, R7C8 = {24}, clean-up: no 8 in R9C7 (step 6), no 3 in R9C6

9. R7C78 + R9C7 (step 6) = {279/459}, 9 locked in R79C7 for C7 and N9, clean-up: no 5 in R78C9

10. Naked pair {68} in R78C9, locked for C9 and N9

11. R234C9 = {259/349/457}, no 1

12. 45 rule on N14 2 outies R4C4 + R7C2 = 1 innie R1C3 + 6
12a. Min R1C3 = 8 -> min R4C4 + R7C2 = 14, no 1,2,3,4

13. 14(4) cage at R3C8 = {1238/1256/1346/2345} (cannot be {1247} because R4C6 only contains 3,6), no 7

14. 12(3) at R6C4 = {129/147/156/237/246/345} (cannot be {138} because 3,8 only in R7C3), no 8
14a. R67C4 = {12} would clash with R1C4 -> no 9 in R7C3

15. 4 remaining innies in N5 R46C46 = 24 = {1689/2679/3489/3678}
15a. 1,2,4 of {1689/2679/3489} must be in R6C4 -> no 9 in R6C4

16. 45 rule on N1 2 innies R1C3 + R3C2 = 1 outie R4C1 + 2
16a. Min R1C3 + R3C2 = 9 -> min R4C1 = 7
16b. Max R4C1 = 9 -> max R1C3 + R3C2 = 11, max R3C2 = 3

17. 45 rule on N3 2 innies R1C7 + R3C8 = 1 outie R4C9 + 2
17a. R4C9 = {23479} -> R1C7 + R3C8 = 4,5,6,9,11 = [31/32/41/42/36/45/38], no 3,4 in R3C8

18. 45 rule on N3 4 innies R1C7 + R2C9 + R3C89 = 18 = {1359/2349/2358/2457/3456} (cannot be {1269/1278} because R1C7 only contains 3,4, cannot be {1368/1458/1467} because 1,6,8 only in R3C8, cannot be {2367} because {27} cannot be placed in R23C9)
18a. 1,6,8 of {1359/2358/3456} must be in R3C8, 5 of {2457} cannot be in R3C8 because {27} cannot be placed in R23C9 -> no 5 in R3C8

19. 14(4) cage at R3C8 (step 13) = {1238/1346}, CPE no 1 in R56C8
19a. 3 locked in R4C678, locked for R4

20. 45 rule on N6 3 innies R4C789 = 1 outie R7C8 + 8
20a. R7C8 = {24} -> R4C789 = 10,12 = {127/145/235/129/147/237/246/345} (cannot be {136/138/156} because R4C9 only contains 2,4,7,9), no 8

21. 14(4) cage at R3C8 (step 19) = {1238/1346}
21a. 8 must be in R3C8 -> no 2 in R3C8

22. R1C7 + R3C8 = R4C9 + 2 (step 17)
22a. R4C9 = {2479} -> R1C7 + R3C8 = 4,6,9,11 = [31/36/38] (no valid permutations for R1C7 + R3C8 = 6) -> R1C7 = 3, R4C9 = {279}, R1C6 = 2, R1C4 = 1, R1C3 = 9, R12C5 = [69], R9C6 = 6, R9C7 = 5, R4C6 = 3, clean-up: no 7 in R58C6 (step 4), no 2 in R9C4
22b. R3C9 = 9 (hidden single in R3), R24C9 = 7 = [52], R23C6 = [45]
22c. R7C7 = 9 (hidden single in C7)

23. R4C4 + R7C2 = R1C3 + 6 (step 12), R1C3 = 9 -> R4C4 + R7C2 = 15 = [78/96]
23a. R1C3 + R3C2 = R4C1 + 2 (step 16), R1C3 = 9 -> R4C1 = R3C2 + 7, R4C1 = {89}, R3C2 = {12}

24. Naked pair {68} in R7C29, locked for R7 -> R7C6 = 7, R6C6 = 8, clean-up: no 1 in R8C1, no 4 in R9C3

25. 45 rule on N3 1 remaining innie R3C8 = 1, R3C2 = 2, R4C1 = 9 (step 23a), R4C4 = 7

26. Naked pair {46} in R4C78, locked for R4 and N6
26a. Naked pair {18} in R4C23, locked for N4

27. 45 rule on N8 2 remaining innies R79C4 = 9 = [54], R9C3 = 7, clean-up: no 2 in R8C1
27a. R1C9 = 4 (hidden single in C9)

28. 12(3) cage at R6C4 = {156} (only remaining combination, cannot be {345} because 3,4 only in R7C3) -> R6C4 = 6, R7C3 = 1, R4C23 = [18], clean-up: no 6 in R8C1
[Alternatively R6C4 = 6 comes from remaining innie in N5]

29. R6C8 = 9 (hidden single in R6)

30. 45 rule on N7 1 remaining innie R7C2 = 8, R78C9 = [68]
30a. R9C5 = 8, R9C2 = 9, R9C9 = 1 (hidden singles in R9)

31. Naked pair {37} in R56C9, locked for N6
31a. R7C8 = 4 (cage sum), R4C78 = [46], R6C7 = 1, R5C78 = [85]

32. Naked pair {23} in R79C1, locked for C1 and N7

33. R4C1 = 9 -> R23C1 = 9 = [18], R23C4 = [83], R3C7 = 6, R3C3 = 4

34. Naked pair {57} in R1C12, locked for R1 and N1 -> R1C8 = 8

35. R5C1 = 6 (hidden single in C1)
35a. R5C1 + R7C2 = 14 -> R6C12 = 7 = [43]

and the rest is naked singles


Last edited by Andrew on Fri Feb 01, 2008 6:15 am; edited 1 time in total
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Caida
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PostPosted: Mon Jan 28, 2008 10:50 pm    Post subject: walkthough Reply with quote

I finally finished Assassin 87.

When I first tried it a few days ago I kept winding up eliminating all the 7s in the bottom right corner. After many hours of frustration I gave up and then came back to it today. Realized that I had drawn the cages incorrectly in my excel spreadsheet Embarassed

I didn't find it too difficult (once I drew the cages correctly) - and my walkthrough doesn't seem to be significantly different from Mike and Andrew. I'd give it a rating of 1.25 - not bad - but not incredibly easy.

My walkthrough is below. (with edits - thanks Afmob!! and Andrew!!)

Cheers,

Caida

Assassin 87 walkthrough

Preliminaries:

a. 10(2)n12 = {19/28/37/46} (no 5)
b. 15(2)n2 = {69/78} (no 1..5)
c. 5(2)n23 = {14/23} (no 5..9)
d. 11(2)n2 and n78 and n89 = {29/38/47/56} (no 1)
e. 12(2)n25 = {39/48/57} (no 1,2,6)
f. 9(2)n2 = {18/27/36/45} (no 9)
g. 14(4)n356 = {1238/1247/1256/1346/2345} (no 9)
h. 24(3)n589 = {789} (no 1..6)
i. 7(2)n7 = {16/25/34} (no 7..9)
j. 14(2)n9 = {59/68} (no 1..4,7)
k. 13(4)n9 = {1237/1246/1345} (no 8,9) -> 1 locked for n9
l. 9 locked in c5 in 15(2)n2 and 12(2)n25 -> no 9 elsewhere in c5


1. Innies n2: r1c46+r3c5 = 10(3) = {127} (no 3,5,4,6,8,9)
Note: combo [14]{5} blocked by 5(2)n23
combo [613] blocked by 10(2)n12 and 5(2)n23 (would need two 4s in r1)
combo {23}[5] blocked by 5(2)n23
1a. r3c5 = 7; r4c5 = 5
1b. r1c3 = 8,9
1c. r1c7 = 3,4
1d. pair {12} locked in r1 and n2 in r1c46 -> no 1,2 elsewhere in r1 and n2
1e. 15(2)n2 = {69} -> locked for n2 and c5
1f. 9(2)n2 = {45} -> locked for n2 and c6
1g. 11(2)n2 = {38} -> locked for c4
1h. r9c3 no 3,8
1j. r9c7 no 6,7

2. Innies c1234: r58c4 = 11(2) = {29/47}/[65] (no 1)
2a. r8c4 no 6

3. Innies c6789: r58c6 = 10(2) = {19/28/37} (no 6)
3a. killer triple {789} in h10(2) (r58c6) and r67c6 -> no 7,8,9 elsewhere in c6
3b. -> r9c7 no 2,3,4

4. Outies n14: r14c4+r7c2 = 16(2+1) = [169/196/178/295/268/277]
4a. r4c4 no 1,2,4
4b. r7c2 no 1,2,3,4

5. Outies n36: r14c6+r7c8 = 9(2+1) = [126/162/216/135/234]
5a. -> r7c8 no 3,7,8,9
5b. 3 in n9 locked in 13(4)n9 = {1237/1345} (no 6) (contains either 2 or 4 but not both)
5c. -> only other place for 2 and 4 in n9 is r7c8
5d. -> r7c8 no 5,6
5e. 6 in n9 locked in 14(2)n9 = {68} -> locked for n9 and c9
5f. -> r9c6 no 3
5g. 8 in 24(3)n589 locked in c6 -> no 8 elsewhere in c6
5h. r58c6 no 2 (step 3)

6. h9(2+1) (step 5) = [162/234]
6a. -> r4c6 no 1,2

7. 23(4)n69 contains either a 2 or a 4 in n9 = {489/579}[2]/{289/379}[4] -> no 1,6
7a. -> r6c8 no 2,4 (if 23(4) contains both 2 and 4 then r6c8 = 8)

8. 9 in n9 locked in c7 -> no 9 elsewhere in c7

9. 16(3)n36 = {259/349/457} (no 1)
9a. -> r9c9 = 1 (single)

10. Innies r567: r7c159 = 11(3) = {12}[8]/{14}[6]/{23}[6]
10a. -> r7c15 = {1234} (no 5,6,8)
10b. -> r8c1 no 1,2
10c. killer pair {24} in r7 in c15 + c8 -> no 2,4 elsewhere in r7

11. 12(3)n578 = [219/291/417/471/156/165/615/651/237/435] (no 8)
11a. -> r6c4 no 7,9

12. Innies n5: r4c46+r6c46 = 24(4) = [9348/7368/9618/7629/9627]
12a. -> r4c4 no 6
12b. -> r7c2 no 9 (step 4)
12c. -> 18(4)n146 cannot have both 7 and 9
12d. -> r3c2+r4c23 no 7,9

13. Innies n8: r7c46+r9c46 = 22(4) = [6952/5962/5872/1876/5926/5746]
Note: combos [7942/9742] blocked by r7c7
combo [7852] blocked by 11(2)n89 and 24(3)n589
13a. -> r7c4 no 7,9
13b. -> r9c4 no 9
13c. -> r9c3 no 2

14. Outies and Innie c123: r149c4 Ė r7c3 = 11
14a. -> max r149c4 = 18 -> max r7c3 = 7 (no 9)
14b. 9 in r7 locked in 24(3)n589
14c. -> r6c6 no 9
14d. r6c4 no 2 (step 11)
14e. r6c6 = 8 (step 12)
14f. pair {79} locked for r7 in r7c67 -> no 7,9 elsewhere in r7
14g. r7c4 no 1 (step 13)
14h. r56c9 no 2,4 (step 7)
Andrew pointed out that another way to say this is: 23(4)n69 must now have three odd numbers - so no 2,4 I like this way of looking at the sums but don't often spot these until after the fact - perhaps with more practice I'll be able to use this.

15. 23(4)n69 = {379}[4]/{579}[2]
15a. -> 7,9 locked in n6 in 23(4)n69 -> no 7,9 elsewhere in n6
15b. 14(3)n6 = {158/248} (no 3,6) -> 8 locked in n6 and r5 in 14(3) -> no 8 elsewhere in n6 and r5
Note: combo {356} blocked by 23(4)n69
15c. 6 in n6 locked in 14(4)n356 in r4 -> no 6 elsewhere in 14(4)n356 and r4
15d. single: r4c6 = 3
15c. 14(4)n356 = {1346} (no 2,5,8)
15d. r5c8 no 1,4 (CPE with 14(4)n356)
15e. 14(3)n6 = {158} (combo {248} blocked by r4c9) 1,5,8 locked for n6
15f. r4c9 = 2 (single)
15g. r7c8 = 4 (step 15)

Singles and cage sums are left (finally!!!)


Last edited by Caida on Wed Jan 30, 2008 10:24 pm; edited 1 time in total
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gary w
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Joined: 07 Sep 2007
Posts: 84
Location: south wales

PostPosted: Wed Jan 30, 2008 12:45 am    Post subject: Reply with quote

Well I know everything seems to be rated 1.25 these days but I'ld have to rate this one about.....1.25.It was roughly as difficult as many recent ones,I certainly found it no easier.

I needed to use the O-I of c9 and the combos in N9 to show that there was a HS in c9 viz, r9c9=1.In c9 3 then had to be in r5/6 which meant the only option for r8c7=4.The O for N3/6 then -> r14c6 and I was well on my way to solving it..Took about 2 hours...long enough !!!

Regards

Gary
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