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 Post subject: Assassin 249
PostPosted: Fri Nov 09, 2012 5:31 pm 
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Grand Master
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 694
Location: Saudi Arabia
Assassin 249
Djape did a very pleasant zero IQ weekend before last (code below) I could not see how to get into it and had to get a hint from JSudoku.

I swapped values to get this one at 1.25


Image

JS Code:
3x3::k:2828:2828:21:22:23:24:25:2561:2561:2828:5133:5133:26:2571:2571:27:5378:5378:2574:5133:28:29:1801:1290:1290:5378:5378:2574:3343:3343:30:1801:1544:1544:2819:2819:1552:1552:2321:2321:31:32:33:34:3844:1552:6162:6162:35:36:37:3335:5381:3844:38:39:6162:6162:40:41:3335:5381:3844:42:2068:2068:2068:43:4614:4614:5381:5381:44:45:4371:4371:46:4614:47:48:49:

Solution:
34 swap


Assassin 249 A - this one is at 1.95 but I'm not sure it is that much harder.

Image

JSCODE:
3x3::k:2572:2572:21:22:23:24:25:2305:2305:2572:5133:5133:26:2315:2315:27:5890:5890:2318:5133:28:29:1801:1034:1034:5890:5890:2318:3599:3599:30:1801:1544:1544:2819:2819:1808:1808:2065:2065:31:32:33:34:3588:1808:5906:5906:35:36:37:3591:5381:3588:38:39:5906:5906:40:41:3591:5381:3588:42:2068:2068:2068:43:4870:4870:5381:5381:44:45:4115:4115:46:4870:47:48:49:

Solution
78 swap


Pedja's original one SS gives 0.85

Image

(Moderator) Ed's image added.

JS Code:
3x3::k:2572:2572:21:22:23:24:25:2561:2561:2572:5133:5133:26:2571:2571:27:5634:5634:2574:5133:28:29:1801:1034:1034:5634:5634:2574:3343:3343:30:1801:1544:1544:2819:2819:1808:1808:2065:2065:31:32:33:34:3844:1808:6162:6162:35:36:37:3335:5381:3844:38:39:6162:6162:40:41:3335:5381:3844:42:2068:2068:2068:43:4870:4870:5381:5381:44:45:4371:4371:46:4870:47:48:49:

Solution
614795382
357182946
289643157
876431529
123569478
495278613
532816794
941327865
768954231


Last edited by HATMAN on Sat Nov 10, 2012 9:06 am, edited 1 time in total.

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 Post subject: Re: Assassin 249
PostPosted: Sat Nov 10, 2012 5:03 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
So this time we've got "3 for the price of 1"! I ought to be doing something else first but I couldn't resist the appearance of a new puzzle so did Djape's original version.

I used elimination solving but it's probably solvable using insertion solving.

Puzzle Diagram:
Image

Here is my walkthrough for Djape's puzzle:
Prelims

a) R1C89 = {19/28/37/46}, no 5
b) R2C56 = {19/28/37/46}, no 5
c) R34C1 = {19/28/37/46}, no 5
d) R34C5 = {16/25/34}, no 7,8,9
e) R3C67 = {13}
f) R4C23 = {49/58/67}, no 1,2,3
g) R4C67 = {15/24}
h) R4C89 = {29/38/47/56}, no 1
i) R5C34 = {17/26/35}, no 4,8,9
j) R67C7 = {49/58/67}, no 1,2,3
k) R9C34 = {89}
l) 10(3) cage at R1C1 = {127/136/145/235}, no 8,9
m) 20(3) cage at R2C2 = {389/479/569/578}, no 1,2,3
n) 7(3) cage at R5C1 = {124}
o) 8(3) cage at R8C2 = {125/134}
p) 19(3) cage at R8C6 = {289/379/469/478/568}, no 1

Steps resulting from Prelims
1a. Naked pair {13} in R3C67, locked for R3, clean-up: no 7,9 in R4C1, no 4,6 in R4C5
1b. Naked pair {89} in R9C34, locked for R9
1c. Naked triple {124} in 7(3) cage at R5C1, locked for N4, clean-up: no 6,8,9 in R3C1, no 9 in R4C23, no 6,7 in R5C4
1d. 8(3) cage at R8C2 = {125/134}, 1 locked for R8
1e. R4C89 = {29/38/47} (cannot be {56} which clashes with R4C23), no 5,6
1f. 9 in N4 only in R6C23, locked for R6 and 24(4) cage at R6C2, no 9 in R7C34, clean-up: no 4 in R7C7
1g. 2 of 19(3) cage at R8C6 must be in R9C6, no 2 in R8C67

2. 45 rule on R4 3 innies R4C145 = 15 = {168/258/267/348/357/456} (cannot be {159/249} because R4C1 only contains 3,6,8), no 9

3. 9 in R4 only in R4C89 = {29}, locked for R4 and N6, clean-up: no 4 in R4C67
3a. Naked pair {15} in R4C67, locked for R4 -> R4C5 = 3, R3C5 = 4

4. R4C23 = {67} (only remaining combination), locked for R4 and N4 -> R4C1 = 8, R3C1 = 2, R4C4 = 4, clean-up: no 6 in R2C5, no 6,7 in R2C6

5. R5C2 = 2 (hidden single in N4)

6. R5C34 = {35} (only remaining combination) -> R5C3 = 3, R5C4 = 5, R4C67 = [15], R3C67 = [31], clean-up: no 7,9 in R2C5, no 9 in R1C89, no 8 in R67C7

7. R6C23 = {59} = 14 -> R7C34 = 10 = [28/46/73/82]

8. 10(3) cage at R1C1 = {136/145}, no 7, 1 locked for N3

9. 45 rule on N1 2 remaining innies R13C3 = 13 = [49/58/85] (cannot be {67} which clashes with R4C3)
9a. Naked quad {4589} in R1369C3, locked for C3, clean-up: no 2,6 in R7C4 (step 7)

10. R8C3 = 1, R7C3 = 2 (hidden singles in C3), R7C4 = 8 (step 7), R9C89 = [89], clean-up: no 5 in R13C3 (step 9)
10a. 8(3) cage at R8C2 = {125/134} -> R8C2 = {45}

11. R13C3 = [49], R6C23 = [95], clean-up: no 5 in 10(3) cage at R1C1 (step 8), no 6 in R1C89
11a. Naked triple {136} in 10(3) cage, locked for N1 -> R2C3 = 7, R4C23 = [76]
11b. Naked pair {58} in R23C2, locked for C2 -> R8C2 = 4, R8C4 = 3 (step 10a)
11c. Naked pair {36} in R79C2, locked for C2 and N7 -> R1C2 = 1

12. R2C4 = 1 (hidden single in C4), clean-up: no 9 in R2C6
12a. Naked pair {28} in R2C56, locked for R2 and N2 -> R23C2 = [58]

13. 45 rule on N3 2 remaining innies R12C7 = 12 = [39/84/93]
13a. R67C7 = {67} (cannot be [49] which clashes with R12C7), locked for C7
13b. R12C7 = {39} (cannot be [84] which clashes with R5C7), locked for C7 and N3 -> R8C7 = 8, R5C7 = 4, R9C7 = 2, R56C1 = [14], clean-up: no 7 in R1C89

14. Naked pair {46} in R2C89, locked for R2 and N3 -> R12C1 = [63], R13C4 = [76], R12C7 = [39], R6C4 = 2

15. R8C5 = 2 (hidden single in R8), R2C56 = [82]

16. Naked pair {67} in R6C57, locked for R6 -> R6C6 = 8

17. Max R56C9 = 11 -> min R7C9 = 4
17a. R69C9 = {13} (hidden pair in C9)

18. 21(4) cage at R6C8 cannot contain both of 1,3 -> no 1,3 in R7C8
18a. R9C89 = {13} (hidden pair in N9), locked for R9 -> R79C2 = [36]

19. R9C6 = 4 (hidden single in R9), R8C6 = 7 (cage sum), R9C5 = 5, R7C4 = 6, R67C7 = [67]

20. 15(3) cage at R5C9 = {347/358} (cannot be {159} because 5,9 only in R7C9), no 9 -> R6C9 = 3, R6C8 = 1

21. 21(4) cage at R6C8 = 1{569} (only remaining combination), 5 locked for N9 -> R7C9 = 4, R5C9 = 8 (step 20)

and the rest is naked singles.

Rating Comment:
I'll rate my walkthrough for Djape's puzzle at Easy 1.0. I can't go any lower because I used elimination solving; I may well have given the same rating if I'd tried insertion solving.


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 Post subject: Re: Assassin 249
PostPosted: Sun Nov 11, 2012 10:44 am 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Another classy Assassin. Thanks HATMAN! It came out quite quickly for me after seeing some heavy moves early on (2,3,4,6 & 7). Haven't had time to check if my way also does the A version. The SudokuSolver score should always be a bit low for zero killers since there are not as many unimportant steps since there are fewer cages. SS just does a weighted step count. Fewer steps, lower score (unless it uses more harder steps!!).

A249
23 steps:
Did prelims this time because all the cages have to get used by the end anyway [edit: in order to get a unique solution. See Andrew's question next post].
Prelims
i. 11(3)n1: no 9
ii. 20(3)n1: no 1,2
iii. 10(2) cages at r1c8, r2c5 and r3c1: no 5
iv. 7(2)r3c5: no 7,8,9
v. 5(2)r3c6 = {14/23}: no 5..9
vi. 13(2) cages at r4c2 & r6c7: no 1,2,3
vii. 6(2)r4c6 = {15/24}: no 3,6,7,8,9
viii. 11(2)r4c8: no 1
ix. 6(3)n4 = {123}
x. 9(2)r5c3: no 9
xi. 8(3)r8c2: no 6,7,8,9
xii. 17(2)r9c3 = {89}

1. 6(3)n4 = {123} only: all locked for n4
1a. no 7,8,9 in r3c1
1b. no 6,7,8 in r5c4

2. "45" in c1: 2 outies r15c2 + 18 = 3 innies r789c9
2a. max. r789c9 = 6+7+9 = 22 (can't have both 8&9 because of r9c3 = (89))
2b. -> r15c2 = 3 or 4 = {12/13}: 1 locked for c2
2c. r789c1 = 21 or 22 = {579/679/678}(no 1,2,3,4)
2d. must have 7: 7 locked for n7 & c1
2e. no 3 in r3c1

3. "45" on r4: 3 innies r4c145 = 15
3a. but {258/456} blocked by 6(2) = [2/5, 4/5..]
3b. = {159/168/249/267/348/357} = [2/4/5/8..]
3c. -> {58}{24} blocked from combined cage 13(2)r4c2+6(2)r4c6
3d. 13(2)r4c2 = {49/67}(no 5,8)

4. 4 in c1 in 10(2)r3c1 = {46} or in 11(3)r1c1 -> 6 in 11(3) must also have 4 or there would be no 4 for c1 -> {236} blocked from 11(3) (Locking-out cages)
4a. 11(3) = {128/146/245}(no 3)=[4/8,4/5/8..]

5. 20(3)n1 = {389/479/569/578}=[4/8/9,4/5/8..] (no eliminations yet)

6. combining steps 4a & 5, if r13c3 = 13:
6a. {49/58} blocked by 11(3)+20(3)
6b. can only be {67} (no eliminations yet)

7. "45" on n1: 1 outie r4c1+4 = 2 innies r13c3
7a. from step 6b. the only way for r13c3 = 13 is {67}
7b. but r4c1 = 9 and r13c3 = {67} clashes with 13(2)r4c2 = [9]/{67}
7c. no 9 in r4c1
7d. no 1 in r3c1

8. 3 in c1 only in r56c1-> no 3 in r5c2

9. naked pair {12} in r15c2: 2 locked for c2

10. 1 in n7 only in c3: locked for c3
[Andrew noticed that there is actually a hidden pair {12} here in r78c3. Missed that.]

11. 1 in n1 only in 11(3) = {128/146}(no 5)

12. 5 in c1 only r789 = {579} only (step 2c.): all locked for n7

13. r9c34 = [89]

14. 8(3)r8c2: must have 3 or 4 for r8c2 = {134} only: all locked for r8

15. naked triple {346} in r789c2: locked for c2 and n7
15a. no 7,9 in r4c3

16. r78c3 = [21]

17. 24(4)r6c2 must have 2 = {2589/2679}(no 1,3,4)
17a. must have 9 which is only in r6c23: 9 locked for r6 and n4

18. r4c23 = [76]
18a. no 4 in r3c1

19. 24(4)r6c2 (step 17): {2679} blocked by 6&7 only in r7c4
19a. = {2589} only

20. 9(2)r5c3 = {45} only: both locked for r5

21. 9 in r4 only in 11(2)r4c8 = {29} only: both locked for n6 and 2 for r4
21a. 6(2)r4c6 = {15} only: both locked for r4
21b. 7(2)r3c5 = {34} only: both locked for c5
21c. no 6,7 in r2c6

22. 20(3)r2c2: {479} blocked by 4&7 only in r2c3
22a. = {389/578}(no 4)
22b. must have 8 -> 8 locked for c2 & n1

23. naked pair {59} in r5c23: 5 locked for n4 and r6 and no 5 in r7c4
23a. r7c4 = 8

on from there.
Cheerio
Ed


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 Post subject: Re: Assassin 249
PostPosted: Sun Nov 11, 2012 10:53 pm 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Thanks HATMAN for another interesting Assassin. Challenging the way I solved it; I missed Ed’s step 2 so had to work harder. If I’d found that step, maybe Ed and I wouldn’t both have posted walkthroughs.

In his walkthrough Ed wrote:
Did prelims this time because all the cages have to get used by the end anyway.
Isn't that a good reason to do the Prelims every time?

Here is my walkthrough for Assassin 249:
I’ve added an explanation for step 6f and corrected a typo in step 1b
Prelims

a) R1C89 = {19/28/37/46}, no 5
b) R2C56 = {19/28/37/46}, no 5
c) R34C1 = {19/28/37/46}, no 5
d) R34C5 = {16/25/34}, no 7,8,9
e) R3C67 = {14/23}
f) R4C23 = {49/58/67}, no 1,2,3
g) R4C67 = {15/24}
h) R4C89 = {29/38/47/56}, no 1
i) R5C34 = {18/27/36/45}, no 9
j) R67C7 = {49/58/67}, no 1,2,3
k) R9C34 = {89}
l) 11(3) cage at R1C1 = {128/137/146/236/245}, no 9
m) 20(3) cage at R2C2 = {389/479/569/578}, no 1,2,3
n) 6(3) cage at R5C1 = {123}
o) 8(3) cage at R8C2 = {125/134}

Steps resulting from Prelims
1a. Naked pair {89} in R9C34, locked for R9
1b. Naked triple {123} in 6(3) cage at R5C1, locked for N4, clean-up: no 7,8,9 in R3C1, no 6,7,8 in R5C4
1c. 8(3) cage at R8C2 = {125/134}, 1 locked for R8

2. R4C23 = {49/58/67}, R4C67 = {15/24} -> combined cage R4C2367 = {49}{15}/{58/24}/{67}{15}{67}{24}
2a. R4C89 = {29/38} (cannot be {47/56} which clash with combined cage)

3. Hidden killer pair 6,7 in R4C145 and R4C23 for R4, R4C23 contains both or neither of 6,7 -> R4C145 must contain both or neither of 6,7
3a. 45 rule on R4 3 innies R4C145 = 15 = {159/249/267/348} (cannot be {258/456} which clash with R4C67, cannot be {168/357} which only contain one of 6,7)
3b. R4C145 = {159/249/267/348}, R4C89 (step 2a) = {29/38} -> combined cage R4C14589 = {159}{38}/{249}{38}/{267}{38}/{348}{29}
3c. R4C23 = {49/67} (cannot be {58} which clashes with combined cage)
3d. 5 in N4 only in R5C3 + R6C23, CPE no 5 in R7C3

4. 45 rule on R34 6(5+1) innies R3C23489 + R4C4 = 38, max R3C23489 = 35 -> min R4C4 = 3
4a. R4C145 (step 3a) = {159/249/267/348}
4b. 1,2 of {159/267} must be in R4C5 -> no 5,6 in R4C5, clean-up: no 1,2 in R3C5

5. 11(3) cage at R1C1 = {128/137/146/236/245}
5a. 7,8 of {128/137} must be in R12C1 (R12C1 cannot be {12/13} which clash with 6(3) cage at R5C1, ALS block), no 7,8 in R1C2

6. Variable hidden pair 8,9 in R78C1 and R9C3 for N7, R9C3 = {89} -> R78C1 cannot contain both of 8,9
6a. 11(3) cage at R1C1 and R34C1 cannot contain both of 8,9 in C1 (11(3) cage = {128} blocks [19/28] from R34C1)
6b. R124C1 cannot contain both of 8,9, R78C1 cannot contain both of 8,9 -> each group must contain one of 8,9
6c. Killer pair 8,9 in R78C1 and R9C3, locked for N7
6d. R124C1 contains one of 8,9 -> R123C1 must contain at least one of 1,2
6e. Killer triple 1,2,3 in R123C1 and R56C1, locked for C1
6f. R123C1 contains one of 1,2, R56C1 = {123} -> 3 in R56C1, locked for C1 and N4, clean-up: no 7 in R4C1
[Note. For step 6f 11(3) cage at R1C1 cannot contain 3 in R12C1 and R34C1 cannot be [37] because R124C1 must contain one of 8,9, step 6b, so either 11(3) cage = {128} or R34C1 = [19/28].]

7. R4C145 (step 3a) = {159/249/267/348}
7a. 7 of {267} must be in R4C4 -> no 6 in R4C4
7b. 6 in R4 only in R4C123, locked for N4, clean-up: no 3 in R5C4

8. R124C1 contains one of 8,9 (step 6b)
8a. 11(3) cage at R1C1 = {128/146/245} (cannot be {137} because R12C1 = {17} clashes with R34C1 = [28] + R56C1, killer ALS block}, cannot be {236} because R12C1 = {26} clashes with R34C1 = [19] + R56C1, killer ALS block), no 3,7
8b. 1,2 of {146/245} must be in R1C2 (R12C1 cannot be {14/16} which clash with R34C1 = [28] + R56C1, killer ALS block, R12C1 cannot be {24/25} which clash with R34C1 = [19] + R56C1, killer ALS block), 8 of {128} only in R12C1 -> R1C2 = {12}
8c. Naked pair {12} in R15C2, locked for C2
8d. 7 in C1 only in R789C1, locked for N7

9. R78C3 = {12} (hidden pair in N7), locked for C3
9a. 8(3) cage at R8C2 = {125/134}
9b. 5 of {125} must be in R8C2 -> no 5 in R8C4
9c. R7C3 = {12} -> 24(4) cage at R6C2 = {1689/2589} (cannot be {2679} which clashes with R4C23), no 1,2,3,4,7 in R6C23 + R7C4
9d. 24(4) cage R6C2 = {1689/2589}, CPE no 8,9 in R6C4

10. R124C1 contains one of 8,9 (step 6b)
10a. 11(3) cage at R1C1 = {128/146/245} (step 8a) -> R12C1 = {18/28/45/46}, R34C1 = [19/28/46/64] -> combined cage R1234C1 = {18}{46}/{28}{46}/{45}[19/28]/{46}[19/28] (cannot be {28}[19] because R124C1 only contains one of 8,9), 4 locked for C1
10b. 3,4 in N7 only in R789C2, locked for C2, clean-up: no 9 in R4C3

11. 20(3) cage at R2C2 = {389/479/569/578}
11a. 3,4 of {389/479} must be in R2C3, 9 of {569} must be in R23C2 (R23C2 cannot be {56} which clashes with R789C2, ALS block), no 9 in R2C3

12. 45 rule on N4 4 innies R4C1 + R5C2 + R6C23 = 26 contains 5 = {4589} (only remaining combination, cannot be {5678} = [67]{58} because R4C159 = [672] clashes with R5C3 = [72]), locked for N4, clean-up: no 4 in R3C1, no 2 in R5C4
12a. Naked pair {67} in R4C23, locked for R4

13. 45 rule on N1 3 innies R1C3 + R3C13 = 14 = {239/257/356} (cannot be {149/158/248} which clash with 11(3) cage at R1C1, cannot be {167} which clashes with R4C3, cannot be {347} because no 3,4,7 in R3C1), no 1,4,8 clean-up: no 9 in R4C1
13a. R3C1 = {26} -> no 6 in R13C3

14. 9 in C1 only in R78C1, locked for N1 -> R9C34 = [89], clean-up: no 1 in R5C4
14a. Naked pair {45} in R5C34, locked for R5
14b. 9 in N4 only in R6C23, locked for R6, clean-up: no 4 in R7C7

15. 9 in R4 only in R4C89 = {29}, locked for R4 and N6, clean-up: no 5 in R3C5, no 4 in R4C67
15a. Naked pair {15} in R4C67, locked for R4, clean-up: no 6 in R3C5
15b. Naked pair {34} in R34C5, locked for C5, clean-up: no 6,7 in R2C6
15c. Killer pair 3,4 in R3C5 and R3C67, locked for R3
15d. 3 in R4 only in R4C45, locked for N5

16. 1 in N1 only in 11(3) cage at R1C1 = {128/146} (step 8a), no 5
16a. 5 in C1 only in R789C1, locked for N7
16b. Naked triple {346} in R789C2, locked for C2 and N7 -> R4C23 = [76]

17. 20(3) cage at R2C2 = {389/578} (cannot be {479 because 4,7 only in R2C3), no 4, 8 locked for N1

18. R4C1 = 8 (hidden single in C1), R3C1 = 2, R1C2 = 1, R5C2 = 2, clean-up: no 9 in R1C89, no 3 in R3C67
18a. Naked pair {14} in R3C67, locked for R3 -> R34C5 = [34], R4C4 = 3, R5C34 = [45], R4C67 = [15], R3C67 = [41], clean-up: no 6,7,9 in R2C5, no 8 in R67C7
18b. R1C89 = {28/37} (cannot be {46} which clashes with R1C1), no 4,6

19. 8(3) cage at R8C2 = {134} (only remaining combination) = [314], R7C3 = 2

20. 24(4) cage at R6C2 (step 9c) = {2589} (only remaining combination) -> R7C4 = 8

21. 45 rule on N3 2 remaining innies R12C7 = 13 = {49/67}
21a. Naked quad {4679} in R1267C7, locked for C7

22. 18(3) cage at R8C6 = {378} (only remaining combination, cannot be {567} because 5,6,7 only in R89C6) -> R8C7 = 8, R89C6 = [73], R59C7 = [32], R56C1 = [13]
22a. 1,2 in N8 only in R789C8, locked for C8 -> R2C5 = 8, R2C6 = 2
22b. Naked pair {67} in R13C4, locked for C4 and N2 -> R26C4 = [12]
22c. Naked pair {59} in R1C56, locked for R1, clean-up: no 4 in R2C7 (step 21)
22d. R8C5 = 2 (hidden single in R8)

23. 6 in R8 only in R8C89, locked for N9 and 21(4) cage at R6C8, no 6 in R6C8, clean-up: no 7 in R7C7
23a. 21(4) cage at R6C8 contains 6 = {1569/3567} (cannot be {3468} because 3,4,8 only in R67C8), no 4,8
23b. R6C8 = {17} -> no 1,7 in R7C8

[I saw this 45 a long time ago but it couldn’t be used until low value candidates in R5C8 had been eliminated.]
24. 45 rule on C89 3(1+2) innies R5C8 + R9C89 = 12
24a. Min R5C8 = 6 -> max R9C89 = 6, no 7
24b. Min R9C89 = 5 -> max R5C8 = 7
24c. R9C89 = {14/15}, 1 locked for R9 and N9
24d. R7C5 = 1 (hidden single in R7), R9C1 = 7 (hidden single in R9)

25. 8 in N6 only in R56C9, locked for C9, clean-up: no 2 in R1C8
25a. 15(3) cage at R5C9 contains 8 = {348} (only remaining combination, cannot be {168} because no 1,6,8 in R7C9) = [843], R6C7 = 6, R7C7 = 7

and the rest is naked singles.

Solution:
+-------+-------+-------+
| 6 1 3 | 7 9 5 | 4 8 2 |
| 4 5 7 | 1 8 2 | 9 3 6 |
| 2 8 9 | 6 3 4 | 1 5 7 |
+-------+-------+-------+
| 8 7 6 | 3 4 1 | 5 2 9 |
| 1 2 4 | 5 6 9 | 3 7 8 |
| 3 9 5 | 2 7 8 | 6 1 4 |
+-------+-------+-------+
| 5 4 2 | 8 1 6 | 7 9 3 |
| 9 3 1 | 4 2 7 | 8 6 5 |
| 7 6 8 | 9 5 3 | 2 4 1 |
+-------+-------+-------+

Rating Comment:
I’ll rate my walkthrough for A249 at least 1.5.


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 Post subject: Re: Assassin 249
PostPosted: Thu Nov 15, 2012 4:40 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Thanks HATMAN for a challenging variant.

HATMAN wrote:
Assassin 249 A - this one is at 1.95 but I'm not sure it is that much harder.
It’s harder but definitely not by the difference between the SS scores. I think it starts more easily than A249.
I used:
a couple of forcing chains. The difficult part was finding the simplest and most effective forcing chains, particularly the second one. There may be some equivalent “fishes”, but I know very little about “fishes”.

This time I used Ed’s early step, so definitely no heavy analysis (which I’d used for A249 because I’d missed that step).

Here is my walkthrough for Assassin 249A:
Prelims

a) R1C89 = {18/27/36/45}, no 9
b) R2C56 = {18/27/36/45}, no 9
c) R34C1 = {18/27/36/45}, no 9
d) R34C5 = {16/25/34}, no 7,8,9
e) R3C67 = {13}
f) R4C23 = {59/68}
g) R4C67 = {15/24}
h) R4C89 = {29/38/47/56}, no 1
i) R5C34 = {17/26/35}, no 4,8,9
j) R67C7 = {59/68}
k) R9C34 = {79}
l) 10(3) cage at R1C1 = {127/136/145/235}, no 8,9
m) 20(3) cage at R2C2 = {389/479/569/578}, no 1,2,3
n) 6(3) cage at R5C1 = {124}
o) 8(3) cage at R8C2 = {125/134}
p) 19(3) cage at R8C6 = {289/379/469/478/568}, no 1

Steps resulting from Prelims
1a. Naked pair {13} in R3C67, locked for R3, clean-up: no 6,8 in R4C1, no 4,6 in R4C5
1b. Naked triple {124} in 7(3) cage at R5C1, locked for N4, clean-up: no 5,7,8 in R3C1, no 6,7 in R5C4
1c. 8(3) cage at R8C2 = {125/134}, 1 locked for R8
1d. Naked pair {79} in R9C34, locked for R9
1e. R4C89 = {29/38/47} (cannot be {56} which clashes with R4C23), no 5,6

2. 45 rule on R4 3 innies R4C145 = 14 = {167/239/347} (cannot be {149/257} which clash with R4C67, cannot be {158/356} which clash with R4C23, cannot be {248} because R4C1 only contains 3,5,7) -> R4C1 = {37}, R4C4 = {469}, R4C5 = {123}, clean-up: no 4 in R3C1, no 2 in R3C5

[Having seen Ed’s walkthrough for A149, I’ll use innies-outies for C1 even though I didn’t spot it when I solved A149.]
3. 45 rule on C1 3 innies R789C1 = 2 outies R15C2 + 19
3a. Max R789C1 = 23 (cannot be {789} which clashes with R9C3) -> max R15C2 = 4 -> R15C2 = {12/13}, 1 locked for C2
3b. R15C2 = 3,4 -> R789C1 = 22,23 = {589/689} (cannot be {679} which clashes with R9C3), 8,9 locked for N7 -> R9C34 = [79], clean-up: no 1 in R5C4
3c. 9 in N2 only in R1C56, locked for R1

4. R4C145 (step 2) = {167/347}, no 2 -> R4C1 = 7, R3C1 = 2, clean-up: no 5 in R3C5, no 4 in R4C89
4a. Naked pair {14} in R56C1, locked for C1 and N4 -> R5C2 = 2, clean-up: no 6 in R5C3
4b. R1C2 = 1 (hidden single in C2), R12C1 = 9 = {36}, locked for C1 and N1, clean-up: no 8 in R1C89
4c. Naked triple {589} in R789C1, locked for N7
4d. Naked triple {346} in R789C2, locked for C2 and N7, clean-up: no 8 in R4C3
4e. R1C89 = {27/45} (cannot be {36} which clashes with R1C1), no 3,6
4f. 45 rule on N1 2 remaining innies R13C3 = 13 = [49/58/85], no 4 in R3C3

5. 8(3) cage at R8C2 = {134} (only remaining combination, cannot be {125} because R8C2 only contains 3,4) -> R8C3 = 1, R8C24 = {34}, locked for R8

6. R7C3 = 2 -> 23(4) cage at R6C2 = {2579/2678} (cannot be {2489} which clashes with R4C23) -> R7C4 = 7, R6C23 = [59/86/95]
6a. 8 in N4 only in R46C2, locked for C2
6b. 20(3) cage at R2C2 = {479/578}
6c. 4,8 only in R2C3 -> R2C3 = {48}

7. R5C3 = 3 (hidden single in N4), R5C4 = 5, clean-up: no 1 in R4C7
7a. 1 in R4 only in R4C56, locked for N5

8. R2C4 = 1 (hidden single in C4), R3C67 = [31], clean-up: no 6,8 in R2C56
8a. 20(3) cage at R2C2 (step 6b) = {479/578}
8b. 9 of {479} must be in R2C2 (R2C23 cannot be [74] which clashes with R2C56), no 9 in R3C2

9. 45 rule on N3 2 remaining innies R12C7 = 12 = [39/48/84] (cannot be {57} which clashes with R1C89)
9a. Killer pair 8,9 in R12C7 and R67C7, locked for C7

10. 19(3) cage at R8C6 = {478/568}, no 2, 8 locked for C6 and N8
10a. 2 in N8 only in R89C5, locked for C5, clean-up: no 7 in R2C6

11. Consider placements for R2C3 = {48}
R2C3 = 4 => R23C2 = 16 = [97] => R2C7 = 8 => R1C7 = 4 (step 9)
or R2C3 = 8 => R1C3 = 4 (hidden single in C1)
-> R1C37 contains 4, locked for R1, clean-up: no 5 in R1C89
and R2C37 contains 8, locked for R2
11a. Naked pair {27} in R1C89, locked for R1 and N3
11b. R2C56 = [72] (hidden pair in N2), clean-up: no 4 in R4C7
11c. R3C2 = 7 (hidden single in N1)
11d. 4 in N2 only in R3C45, locked for R3
11e. 4 in R4 only in R4C46, locked for N5

12. R1C56 = {59} (hidden pair in N2), locked for R1, clean-up: no 8 in R3C3 (step 4f)
12a. Naked triple {468} in R134C4, locked for C4 -> R8C24 = [43], R6C4 = 2

13. Consider placement for R34C5 = [43/61]
13a. R34C5 = [43] => R4C89 = {29}, locked for N6 => R4C7 = 5, R67C7 = {68}, locked for C7 => R12C7 = [39] (step 9)
or R34C5 = [61] => 6 in N3 only in R2C89, locked for R2 => R12C1 = [63] => R1C7 = 3 (hidden single in R1C7)
-> R1C7 = 3, R2C7 = 9 (step 9), R12C1 = [63], R1C34 = [48], R2C23 = [58], R3C3 = 9, clean-up: no 5 in R67C7
13b. Naked pair {46} in R2C89, locked for N3
[The rest is fairly straightforward.]

14. Naked pair {68} in R67C7, locked for C7
14a. R6C23 (step 6) = [95] (cannot be [86] which clashes with R6C7), R4C23 = [86], R4C4 = 4, R3C45 = [64], R4C56 = [31], R4C7 = 5, R8C7 = 7, R59C7 = [42], R56C1 = [14]
14b. Naked pair {29} in R4C89, locked for N6
14c. Naked pair {68} in R6C57, locked for R6 -> R6C6 = 7

15. R8C7 = 7 -> R89C6 = 12 = [84]

[As with A249, I saw this 45 a long time ago but it’s only useful after low numbers have been eliminated from R5C8.]
16. 45 rule on C89 3(1+2) innies R5C8 + R9C89 = 12
16a. Min R6C8 = 6 -> max R9C89 = 6 -> R9C89 = {13/15}, 1 locked for R9 and N9
16b. R9C89 = {13/15} = 4,6 -> R5C8 = {68}

17. R5C9 = 7 (hidden single in N6), R1C89 = [72], R4C89 = [29]

18. R5C9 = 7 -> R67C9 = 7 = [16/34]
18a. Naked pair {46} in R27C9, locked for C9 -> R8C9 = 5, R8C1 = 9, R8C8 = 6
18b. R8C89 = [65] = 11 -> R67C8 = 10 = [19]

and the rest is naked singles.

Solution:
+-------+-------+-------+
| 6 1 4 | 8 9 5 | 3 7 2 |
| 3 5 8 | 1 7 2 | 9 4 6 |
| 2 7 9 | 6 4 3 | 1 5 8 |
+-------+-------+-------+
| 7 8 6 | 4 3 1 | 5 2 9 |
| 1 2 3 | 5 6 9 | 4 8 7 |
| 4 9 5 | 2 8 7 | 6 1 3 |
+-------+-------+-------+
| 5 3 2 | 7 1 6 | 8 9 4 |
| 9 4 1 | 3 2 8 | 7 6 5 |
| 8 6 7 | 9 5 4 | 2 3 1 |
+-------+-------+-------+

Rating Comment:
I’ll rate my walkthrough for A249A at Hard 1.5. I used two forcing chains; I think the second one is long enough to be rated at the top end of the 1.5 range.

It’s surprising how much difference interchanging a pair of numbers in each row makes. Djape’s original puzzle was easy by the standards of this forum, a typical “hard newspaper killer”, A249 is Assassin level while A249A is a genuine V1.5.


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 Post subject: Re: Assassin 249
PostPosted: Sat Nov 17, 2012 1:01 pm 
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 694
Location: Saudi Arabia
Andrew this is particularly the case with zero killers as they normally have a tight solution path that is number dependent. It took me a while to find swaps that gave solvable killers.


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