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 Post subject: Assassin 310X
PostPosted: Thu Feb 26, 2015 12:49 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 310X

Got quite a hard one when starting JS, SS gave it 1.2 so I merged a couple of cages and SS gave this 1.6, which I think is overstated.

Funnily enough the JS solution appears to be harder for the unmerged one.



Image

JSCODE:
3x3:d:k:1025:1025:6914:6914:6914:2563:2563:5124:5124:3333:3333:6914:6150:6914:2055:2055:5124:5124:3336:3336:6150:6150:6150:4105:4105:2826:2827:2060:3336:4365:4365:6926:4105:4105:2826:2827:2060:1551:4365:6926:6926:6926:5648:5393:5393:9234:1551:5139:5139:6926:4372:5648:5648:5393:9234:5397:5397:5139:4372:4372:5648:5393:5393:9234:5397:5397:2070:2070:1815:1815:6427:6427:9234:9234:9234:9234:3354:3354:1815:6427:6427:

Solution
135276498
769843512
482951736
317492685
526138974
948567321
254719863
893624157
671385249


Afmob, Andrew I've got a few assassins on standby and I messed up the posting - corrected above.


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PostPosted: Fri Feb 27, 2015 7:35 pm 
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Joined: Mon Apr 21, 2008 9:44 am
Posts: 310
Location: MV, Germany
Thanks for the new Killer, HATMAN! It took me some time to find the critical moves but once you get them, the puzzle crumbles quickly. I'm actually glad that there is only one Assassin every week or two since solving the Killers and optimizing my walkthrough takes quite some time.

A310X Walkthrough:
1. R6789
a) Innie C89 = R6C8 = 2
b) Outies C12 = 11(3+1) <> 8,9; R9C4 <> 6,7
c) Outies N7 = 12(2) = [93/84/75]
d) Outies C12 = 11(3+1): R789C3 <= 8 since R9C4 >= 3 -> 1 locked in R789C3 for C3+N7; R789C3 <> 6,7
e) Innies+Outies C12: -10 = R9C34 - R78C2: R78C2 <> 2,3,4 since R9C4 >= 3

2. N4578
a) Innies N4578 = 7(2+1) <> 6,7,8,9; R4C2 <> 5
b) Innies C6789 = 29(4) = {5789} locked for C6
c) 17(3) @ N8 must have one of (1234) -> R7C5 = (1234)
d) Innies N4578 = 7(2+1): R4C2 <> 3 since R8C6 <> 3

3. C1234 !
a) Innie R12 = R2C4 = 8
b) 4(2) = {13} locked for R1+N1
c) ! 13(3): R3C1 <> 6,8 since 6{25}/8{14} blocked by Killer pairs (14,25) of 6(2) @ N4
d) 8 locked in 36(7) @ C1 for 36(7) = 189{2367/2457/3456} -> R9C3 = 1

4. R789 !
a) 7(3) = {124} -> 1 locked for R8; CPE: R8C89 <> 2,4
b) 8(2) <> 7
c) ! Hidden Killer triple (789) in R7C46 for N8 since 13(2) can only have one of them
d) Hidden Killer quad (1234) in R7C789 @ N9 since 25(4) must have exactly one of them -> R7C789 must have exactly one of (1234)
e) ! Hidden Killer quad (1234) in R7C13 for R7 since R7C789 can only have one of (1234)
f) Hidden Killer pair (56) in R7C789 @ N9 since 25(4) must have exactly one of (56) -> R7C789 must have exactly one of (56)
g) Hidden Killer pair (56) in R7C2 for R7 since R7C789 can only have one of them
h) 36(7) must have two of (789) @ N7 -> 36(7) = 12789{36/45} -> 2 locked for N7

5. C123
a) Outies C12 = 10(2+1) = 3+{34} -> R9C4 = 3; 3,4 locked for C3+N7
b) 3 locked in 8(2) @ N4 = {35} locked for C1+N4
c) 6(2) = [24] -> R5C2 = 2, R6C2 = 4
d) R1C1 = 1
e) Outie N7 = R6C1 = 9
f) 20(3) = 5{69/78} -> R6C4 = 5

6. N568
a) 8(2) = {26} locked for R8+N8
b) 17(3) @ N8 = {179} -> R6C6 = 7, R7C5 = 1, R7C6 = 9
c) R7C4 = 7 -> R6C3 = 8, R8C6 = 4, R9C7 = 2, R8C3 = 3, R7C3 = 4
d) 17(3) @ N4 = {467} since R45C3 = (67) -> R4C4 = 4; 6,7 locked for C3
e) 22(4) = 29{38/56} since R6C7 = (136) -> R5C7 = 9; R7C7 = (58)
f) Hidden Single: R5C5 = 3 @ D\, R8C2 = 9 @ N7
g) R6C5 = 6, R6C7 = 3 -> R7C7 = 8, R8C8 = 5, R2C2 = 6 -> R2C1 = 7, R2C8 = 1, R2C7 = 5 -> R2C6 = 3
h) 11(2) @ C8 = [38/47]

7. Rest is singles without considering diagonals.

Rating:
1.25 - Hard 1.25. I used some Hidden Killer subsets.


Last edited by Afmob on Fri Mar 06, 2015 9:28 pm, edited 1 time in total.

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 Post subject: Re: Assassin 310X
PostPosted: Sun Mar 01, 2015 9:59 am 
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Grand Master
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 694
Location: Saudi Arabia
What I am trying to do is that if there is a week or more gap then I post one.

My assassins come from when I start up JSudoku do create another variant and it generates a hard puzzle which as I did this week I increase in difficulty until it is at assassin level. As you know my "normal" puzzles are below assassin level so should not take up too much time.


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 Post subject: Re: Assassin 310X
PostPosted: Sun Mar 01, 2015 11:32 pm 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Thanks HATMAN for your latest Assassin.

A challenging one. I found it harder than Afmob did, partly because there was one 45 which I didn't spot.

Here is my walkthrough for Assassin 310X:
Thanks Afmob for showing me that I'd reached naked singles earlier than I'd realised, and for pointing out a typo.
Prelims

a) R1C12 = {13}
b) R1C67 = {19/28/37/46}, no 5
c) R2C12 = {49/58/67}, no 1,2,3
d) R2C67 = {17/26/35}, no 4,8,9
e) R34C8 = {29/38/47/56}, no 1
f) R34C9 = {29/38/47/56}, no 1
g) R45C1 = {17/26/35}, no 4,8,9
h) R56C2 = {15/24}
i) R8C45 = {17/26/35}, no 4,8,9
j) R9C56 = {49/58/67}, no 1,2,3
k) 20(3) cage at R6C3 = {389/479/569/578}, no 1,2
l) 7(3) cage at R8C6 = {124}

Steps resulting from Prelims
1a. Naked pair {13} in R1C12, locked for R1 and N1, clean-up: no 7,9 in R1C67
1b. Naked triple {124} in 7(3) cage at R8C6, CPE no 1,2,4 in R8C89

2. 45 rule on R12 1 innie R2C4 = 8, clean-up: no 2 in R1C7, no 5 in R2C12
2a. Max R67C4 = 16 -> min R6C3 = 4

3. 45 rule on C89 1 innie R6C8 = 2, clean-up: no 9 in R34C8, no 9 in R3C9, no 4 in R5C2
3a. 45 rule on C89 3 outies R567C7 = 20 = {389/479/569/578}, no 1

4. 45 rule on N7 2(1+1) outies R6C1 + R9C4 = 12 = {39/57}/[66/84], no 1,4 in R6C1, no 1,2 in R9C4

5. 45 rule on C12 4(3+1) outies R789C3 + R9C4 = 11
5a. Min R789C3 = 6 -> max R9C4 = 5, clean-up: min R6C1 = 7 (step 4)
5b. Min R9C4 = 3 -> max R789C3 = 8 -> R789C3 = {123/124/125/134}, no 6,7,8,9, 1 locked for C3 and N7

6. 13(3) cage at R3C1 = {148/157/238/247/256} (cannot be {139} which clashes with R1C2, cannot be {346} which clashes with R2C12), no 9
6a. Min R3C12 = 6 -> max R4C2 = 7

7. 45 rule on C6789 4 innies R5679C6 = 29 = {5789}, locked for C6, clean-up: no 1,3 in R2C7, no 7,9 in R9C5
7a. 17(3) cage at R6C6 must contain one of 1,2,3,4 -> R7C5 = {1234}

8. 45 rule on C12 2 innies R78C2 = 2 outies R9C34 + 10
8a. Min R9C34 = 4 -> min R78C2 = 14, no 2,3,4 in R78C2

9. Double hidden killer triple 1,2,4 in R1C67, 7(3) cage at R8C6 and rest of C67 for C67, R1C67 contains one of 2,4, 7(3) cage = {124} -> rest of C67 contains two of 1,2,4 -> 16(4) cage at R3C6 = {1258/1267/1348/1357/1456/2347/2356} (cannot be {1249} which contains all of 1,2,4) -> no 9 in R34C7
9a. 3 in C6 only in R2C67 = [35] or 16(4) cage -> 16(4) cage = {1267/1348/1357/2347/2356} (cannot be {1258/1456} which clashes with R2C67 = [35], blocking cages)

10. 9 in R3 only in 24(4) cage at R2C4 = {1689/2589/3489}, no 7

11. 13(3) cage at R3C1 cannot be 8[23] which clashes with R1C2 + R56C2 (killer ALS block), cannot be 8[41] which clashes with R56C2 -> no 8 in R3C1

12. 8 in C1 only in R6789C1, locked for 36(7) cage at R6C1, no 8 in R9C2
12a. 36(7) cage containing 8 must also contain 1 -> R9C3 = 1
12b. 7(3) cage at R8C6 = {124}, 1 locked for R8, clean-up: no 7 in R8C45

13. R45C1 = {17/26/35}, R56C2 = {15}/[24] -> combined cage R45C1 + R56C2 = {17}[24]/{26}{15}/{35}[24], 2 locked for N4

14. R78C3 (step 5b) = {23/24/25/34} -> 21(4) cage at R7C2 = {2379/2469/2478/2568/3459/3468} (cannot be {3567} because R78C3 cannot contain both of 3,5)
14a. 36(7) cage at R6C1 must contain 9
14b. R6C1 + R9C4 (step 4) = [75/84/93]
14c. 21(4) cage = {2469/2478/2568/3459/3468} (cannot be {2379} with R6C1 + R9C4 = [93] because R789C3 + R9C4 must total 11, step 5)
14d. R78C3 = {24/25/34} = 6,7 -> R9C34 = 4,5 (step 5) = [13/14], no 5 in R9C4, clean-up: no 7 in R6C1

15. Killer quad 1,2,3,4 in R7C5, R8C45, R8C6 and R9C4, locked for N8, clean-up: no 9 in R8C6
15a. Killer pair 5,6 in R8C45 and R9C56, locked for N8
15b. 9 in N8 only in R7C46, locked for R7

16. 45 rule on C6789 2 innies R59C6 = 1 outie R7C5 + 12
16a. R7C5 + R59C6 = 1{58}/3{78}/4[97] (cannot be 2[95] which clashes with R8C45, killer IOD clash) -> no 2 in R7C5
16b. 2 in N8 only in R8C456, locked for R8

17. Consider placements for R9C4 = {34}
R9C4 = 3 => R8C45 = {26}
or R9C4 = 4 => R9C7 = 2, R8C6 = 1, R7C5 = 3 => R8C45 = {26}
-> R8C45 = {26}, locked for R8 and N8, clean-up: no 7 in R9C6
17a. Naked pair {14} in R8C67, locked for R8 and 7(3) cage at R8C6 -> R9C7 = 2, clean-up: no 6 in R2C6
17b. Naked pair {58} in R9C56, locked for R9 and N8
17c. Naked pair {79} in R7C46, locked for R7

18. R78C3 (step 14d) = {25/34} (cannot be {24} because 2,4 only in R7C3)
18a. 2,4 only in R7C3 -> R7C3 = {24}
18b. R78C3 = {25/34} = 7 -> R9C34 = 4 (step 5) = [13] -> R9C4 = 3, R6C1 = 9 (step 4), clean-up: no 4 in R2C2

19. 8 in C1 only in R78C1, locked for N7
19a. R9C34 = [13] = 4 -> R78C2 = 14 (step 8) = [59], 9 placed for D/, R8C3 = 3 -> R7C3 = 4 (cage sum), placed for D/, clean-up: no 4 in R2C1, no 1 in R56C2
19b. Naked pair {67} in R9C12, locked for R9 and N7 -> R78C1 = [28], clean-up: no 6 in R45C1
19c. R7C5 = 1 -> R67C6 = 16 = [79], 7 placed for D\
19d. R2C12 = [76], 6 placed for D\, R2C7 = 5 -> R2C6 = 3, R2C8 = 1, placed for D/, clean-up: no 1 in R45C1, no 6 in R4C8, no 6 in R4C9
19e. R8C89 = [57], 5 placed for D\, clean-up: no 6 in R3C8, no 4 in R34C9
19f. R9C1 = 6, placed for D/
[Thanks Afmob for pointing out that it's naked singles from here, without using the diagonals. Afmob uses SimpleSudoku to check for singles. I only check manually for naked single when I think I may have reached that stage.]

20. Naked pair {35} in R45C1, locked for C1 and N4 -> R1C1 = 1, placed for D\
20a. R56C2 = [24], R3C12 = [48] -> R4C2 = 1 (cage sum), clean-up: no 3,7 in R4C8, no 3 in R4C9

21. R3C6 = 1 (hidden single in C6), R4C6 = 2, placed for D/
21a. R34C6 = [12] = 3 -> R34C7 = 13 = [76], R3C8 = 3 -> R4C8 = 8

22. R1C9 = 8, placed for D/, R5C5 = 3, placed for D\
22a. R1C67 = [64], R1C8 = 9, R9C8 = 4, R9C9 = 9, placed for D\

and the rest is naked singles, without using the diagonals.

Rating Comment:
I'll rate my walkthrough at 1.5.

Thanks HATMAN for clarifying your plans for posting your stock of future Assassins. May I add that if the previous Assassin has been a particularly difficult one, and posting of walkthroughs has been slow, then a two week gap will be welcome.


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 Post subject: Re: Assassin 310X
PostPosted: Thu Mar 05, 2015 9:45 am 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1043
Location: Sydney, Australia
HATMAN wrote:
What I am trying to do is that if there is a week or more gap then I post one.
Great! I only post mine every 2nd week. I can't do more than this since it is so time-consuming.

I found this a really tough puzzle first time through. I had to try again since I knew I must have missed something. I had (my step 4: Afmob doesn't even give it a ! - his 2a]. Much easier in this optimized solution. Used some same steps from either of the other WTs (especially Andrew's 13) - to make a very different solution. Quite happy with it.

A310 start
15 steps:
Preliminaries
Cage 4(2) n1 - cells ={13}
Cage 6(2) n4 - cells only uses 1245
Cage 8(2) n23 - cells do not use 489
Cage 8(2) n4 - cells do not use 489
Cage 8(2) n8 - cells do not use 489
Cage 13(2) n1 - cells do not use 123
Cage 13(2) n8 - cells do not use 123
Cage 10(2) n23 - cells do not use 5
Cage 11(2) n36 - cells do not use 1
Cage 11(2) n36 - cells do not use 1
Cage 7(3) n89 - cells ={124}
Cage 20(3) n458 - cells do not use 12

1. 4(2)n1 = {13}: both locked for n1 and r1
1a. no 7,9 in 10(2)r1c6
1b. "45" on r12: 1 innie r2c4 = 8
1c. no 5 in 13(2)n1

2. "45" on c6789: 4 innies r5679c6 = 29 = {5789} only: all locked for c6
2a. no 7,9 in r9c5
2b. no 2 in r1c7
2c. no 1,3 in r2c7

3. Combined cages 8(2)n4+6(2)n4 = {17+24}{26+15}{35+24}
1a. must have 2, locked for n4

4. "45" on n4578: 3 innies r4c26+r8c6 = 7
4a. Max. any one of those = 4
4b. =[124/142/412/421](no 3 in r4c26)
4c. must have 2 in r48c6: locked for c6
4d. -> 10(2)r1c6 = {46} only: both locked for r1
4e. no 6 in r2c7

5. Caged x-wing on 4 in 10(2)r1c6 and 7(3)r8c6: 4 locked for c67

6. 3 & 6 in c6 only in n2: locked for n2

7. 27(5)r1c3: {14679} blocked by 1,4,6 only in r2c35
7a. {24678} must have 4 & 6 in r2 but this clashes with 13(2)n1 = {49/67} = 4 or 6
7b. = {12789/14589/15678/24579} Note: must have 8 in r1c3 or 4 in r2

8. 3 in r2 in 8(2)r2c6 = [35] or 3 is in 20(4)n3 -> 5 in 20(4) must also have 3 or there would be no 3 for r2 -> {1568/2459/2567} all blocked (Locking out cages)
8a. the only combo left with 5 is {3458} with 4 in r2 and 8 in r1: but this is blocked by 27(5)r1c3 (step 7b)
8b. no 5 in 20(4)n3

9. 5 in r1 only in 27(5)r1c3: -> no 5 in r2c35

10. Hidden single 5 in r2 -> r2c7 = 5, r2c6 = 3
10a. no 6 in r4c89

11. 16(4)r3c6 = {1267} only (no 3,8,9)
11a. 7 only in c7: locked for c7

12. 3 in r3 only in one of 11(2)r3c89 -> one of r4c89 must have 8: 8 locked for r4 and n6

13. Hidden single 8 in c7 -> r7c7 = 8: placed for D\

14. "45" on c89: 1 innie r6c8 = 2
14a. r56c7 = 12 (cage sum) = {39} only: both locked for n6
14a. no 2,8,9 in r3c89,

15. 8 in r3 only in 13(3)r3c1 = {148} only -> r4c2 = 1, r3c12 = {48}: both locked for n1 and 4 for r3

on from there.
Cheers
Ed


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 Post subject: Re: Assassin 310X
PostPosted: Sun Mar 08, 2015 7:30 pm 
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Grand Master
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 282
Location: California, out of London
Some good steps from the other WTs I didn't spot! I didn't go through them in too much detail, but it looks like we all had some common elements and some differences.
Nice one HATMAN!

WT for 310X:
0. 4(2)n1 = {13}
7(3)r8c6 = {124}
Innies r12 -> r2c4 = 8
Innies c89 -> r6c8 = 2
Innies c6789 -> r5679c6 = +29 = {5789}

1. 8(2)n4 and 6(2)n4 between them contain a 2.
Innies n4578 = r4c26+r8c6 = +7.
Since r8c6 from (124), and r4c2 not 2, and 5 already in c6
-> only possibilities are for them to be {124} with 2 in r48c6.
-> r4c2 from (14)
1 or 4 also in 6(2)n4
-> (14) locked in c2n4
-> 4(2)n1 = [13]

2. Outies n7 -> r6c1+r9c4 = +12
-> min r9c4 = 3
Outies c12-> r789c3 + r9c4 = +11
-> max r9c4 = 5
-> r6c1 from (789)

3. r789c3 range from +6 to +8
-> Whichever of (987) is in r6c1 must go in n7 in r78c2
r78c3 are max +7
-> r78c2 min +14
-> whichever of (345) is in r9c4 cannot pair with (987) in r78c2 in n7/c2 -> must go in n7/c3 in r78c3
-> Since outies c12 = r789c3+r9c4 = +11 -> r9c4 cannot be 5.
-> Outies c12 = r789c3+r9c4 from {134}[3] or {124}[4]
-> (14) locked in c3/n7.
-> 4 in r23c1
Also r6c1 from (89)

4! Outies r89 -> r6c1+r7c123 = +20
-> Outies c12 + Outies r89 - Outies n7 = r7c12 + r89c3 + 2*r7c3 = +19
-> r89c12 = 26 + r7c3

Now, 4 somewhere in r789c3. (And r6c1 from (89))

a) 4 in r7c3 puts (6789) in 36(7) puts remaining cells in 36(7) = {123} -> 4 not in r9c4
b) 4 in r8c3 puts 4 in r9c7 -> 4 not in r9c4
c) 4 in r9c3 -> 4 not in r9c4

-> Outies n7 = [93]

5. The values (1234) in r8 - One goes in r8c3, one in r8c45 (not 3), and two in r8c67 (not 3)
-> r8c3 = 3
-> 8(2)n4 = {35}
-> 6(2)n4 = {24}
-> r4c2 = 1
-> r48c6 = {24}
-> 1 in r89c7

6. NT r456c3 = {678}
-> r3c12 = [48]
-> 13(2)n1 = {67}
-> r123c3 = {259}

7. NT r123c6 = {136}
-> NP 10(2)r1 = [64]
-> 7(3)r8c6 = [4{12}]
-> r4c6 = 2

8. 1 in n3 in r2c89
-> r23c6 = [31]
-> r2c7 = 5
-> r34c7 = {67}
-> r567c7 = {389}

9. 3 in r3/n3 in r3c89
-> 8 in r4c89
-> r567c7 = [938]
9 in n7 in r8c2 (since r7c123 = +11)
-> 21(4)n7 = [5493]
-> r9c3 = 1
-> r89c7 = [12]

10. 8 in r1 in r1c89
-> 20(4)n3 = {1289}
-> r3c789 = {367}
-> r3c345 = {259}
etc.

(Not used diagonals yet! Probably quicker way than above if I did!)


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