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 Post subject: Assassin 330 X
PostPosted: Fri Feb 12, 2016 11:16 am 
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Grand Master
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 693
Location: Saudi Arabia
Assassin 330 X
Quite a pretty one.

SS gives it 1.25, JS uses 4 smaller fishes, so probably about right.
I tried number swapping to make it harder but it goes off scale, even swapping 8 and 9 which I expected to give a small increase gives 4.5 on SS and JS fails.


Image

JS Code:
3x3:d:k:4353:21:4353:3088:22:3604:4355:23:4355:24:4353:3088:3088:25:3604:3604:4355:26:4353:5906:4353:27:28:29:4355:1807:4355:5906:5906:30:31:32:33:34:1807:1807:35:36:37:38:39:40:41:42:43:2317:2317:44:45:46:47:48:4369:4369:4612:2317:4612:49:50:51:4354:4369:4354:52:4612:4883:4883:53:2062:2062:4354:54:4612:55:4612:4883:56:2062:4354:57:4354:

Solution:
792415386
845362719
163789245
984637521
526194837
317528964
258946173
639271458
471853692


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 Post subject: Re: Assassin 330 X
PostPosted: Sun Feb 14, 2016 12:42 am 
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Grand Master
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 280
Location: California, out of London
Very pretty! Thanks HATMAN. Not too tough... Only one less than straightforward move for me. Don't know if it's a fish or not.

Assassin 330 X WT:
1. 9 on D\ in n5
9 on D/ in n5
-> r5c5 = 9

2. 8 on D\ in n5
-> 8 on D/ in n7
-> 18(5)n7 = {12348}

3. Innies n3 -> Min r3c8 = 4
-> 7(3)n36 = [4{12}]
-> 17(5)n3 = {12356}
-> 7 in D/ in n5 in r4c6 or r6c4.

4. Min r7c2 = 5 -> 1 in r6c12
Innies n9 -> Min r8c7 = 4 -> 1 in r89c6
-> HS 1 in n5 -> r5c4 = 1

5. Min r1289c6 is +10 -> Max r2c7 + r8c7 = +12.
Since r8c7 is Min 4 -> r2c7 is Max 8.
Min r2c7 = 7 -> Max r12c6 = +7(2)
Innies n1 -> Min r2c3 = 4 -> Max r12c4 = +8(2)
-> 9 in n2 in r3c46
-> 9 in n1 in r1c2 or r2c1
-> 9 in 23(3) in r4c12 in n4
-> HS 9 in n7 -> r8c3 = 9
-> 9 in n8 in r7c46
-> HS 9 in n9 -> r9c8 = 9
-> HS 9 in n3 -> r2c9 = 9
-> HS 9 in n6 -> r6c7 = 9
Also HS 9 in n1 -> r1c2 = 9
-> r4c1 = 9

6. r34c2 = {68}
-> 9(3)r6c1 = [{13}5]
Also -> r8c1 = 6
-> r9c2 = 7

7! Since 1 already in c4 and c6 -> Min r12c4 + r12c6 = {2345} = +14(4)
-> Max r2c3 + r2c7 = +12
Since r2c7 from (78) and r2c3 is Min 4 -> r2c3 from (45)
r8c7 is also from (45)

17(5) in n1 and n9 are both from {12347} or {12356}.
They cannot be the same since that would put only five different numbers into 6 cells on D\.
Since one of the 17(5)s in n19 is {12347} -> r2c3 and r8c7 cannot both be 4.
Since one of the 17(5)s in n19 is {12356} -> r2c3 and r8c7 cannot both be 5.
-> one of r2c3 and r8c7 is 5 and the other is 4.
Since both (r2c3 + r2c7) and (r8c7 + r2c7) are Max +12 -> r2c7 cannot be 8.
-> r2c7 = 7
-> r1c8 = 8
-> r8c9 = 8
-> 8 in n6 in r45c7

8. Finishing
7 in n1 only in corners.
-> 17(5)n1 = {12347}
-> r2c3 = 5
-> r12c4 = {34}
-> r12c6 = [52]
-> r89c6 = {13}
-> r8c7 = 4
-> 17(5)n9 = {12356}
-> r7c8 = 7
-> r6c89 = [64]
Also r89c4 = [28]
Also HS 8 in D\ -> r6c6 = 8
Also HS 6 in n1 -> r3c2 = 6
-> r2c1 = r4c2 = 8
-> r7c3 = 8
etc.


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 Post subject: Re: Assassin 330 X
PostPosted: Wed Feb 17, 2016 5:02 am 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
Thanks HATMAN for your latest Assassin. Nice cage pattern!

You wrote:
JS uses 4 smaller fishes
Maybe only one is needed, if the right one is found?

Some interesting analysis in wellbeback's step 7!

I was a bit slow on this puzzle because I spent quite a lot of time looking for a routine continuation after my step 9, before I decided that there probably wasn't one.

Here is my walkthrough for Assassin 330 X:
Prelims

a) 23(3) cage at R3C2 = {689}
b) 7(3) cage at R3C8 = {124}
c) 9(3) cage at R6C1 = {126/135/234}, no 7,8,9
d) 19(3) cage at R8C3 = {289/379/469/478/568}, no 1
e) 8(3) cage at R8C6 = {125/134}
f) 17(5) cage at R1C1 = {12347/12356}, no 8,9
g) 17(5) cage at R1C7 = {12347/12356}, no 8,9
h) 18(5) cage at R7C1 = {12348/12357/12456}, no 9
i) 17(5) cage at R7C7 = {12347/12356}, no 8,9

1. 17(5) cage at R1C7 contains 1,2,3, locked for N3
1a. R3C8 = 4
1b. Naked pair {12} in R4C89, locked for R4 and N4
1c. 17(5) cage at R1C7 = {12356} (only remaining combination), locked for N3

Other steps resulting from Prelims
2a. 17(5) cage at R1C1 = {12347/12356}, 1,2,3 locked for N1
2b. 18(5) cage at R7C1 = {12348/12357/12456}, 1,2 locked for N7
2c. 17(5) cage at R7C7 = {12347/12356}, 1,2,3 locked for N9
2d. Naked triple {689} in 23(3) cage at R3C2, CPE no 6,8,9 in R56C2

3. 8(3) cage at R8C6 = {125/134} -> R89C6 = {12/13}, 1 locked for C6 and N8
3a. Killer pair 4,5 in R8C7 and 17(5) cage at R7C7, locked for N9

4. 8,9 on D\ only in R4C4 + R5C5 + R6C6, locked for N5
4a. R5C5 = 9 (hidden single on D/)
4b. 8 on D/ only in R7C3 + R8C2 + R9C1 -> 18(5) cage at R7C1 = {12348}, locked for N7, no 8 in R7C1 + R9C3
4c. 7 on D/ only in R4C6 + R6C4, locked for N5

5. R7C2 = {56} -> 9(3) cage at R6C1 = {126/135}, R6C12 = {12/13}, 1 locked for R6 and N7
5a. R5C4 = 1 (hidden single in N5)

6. 14(3) cage at R1C6 = {248/257/347} (cannot be {239} which clashes with 8(3) cage at R8C7, cannot be {356} because R2C7 only contains 7,8,9), no 6,9
6a. R2C7 = {78} -> no 7,8 in R12C6
6b. Killer pair 2,3 in 14(3) cage and 8(3) cage, locked for C6
6c. 2 in N5 only in R6C45, locked for R6

7. 12(3) cage at R1C4 = {246/345} (cannot be {237} which clashes with 14(3) cage at R1C6), no 7,8,9
7a. 12(3) cage = {246/345}, CPE no 4 in R2C56
7b. Killer pair 2,3 in 12(3) cage and 14(3) cage, locked for N2
7c. 9 in N2 only in R3C46, locked for R3
7d. 23(3) cage at R3C2 = {689}, 9 locked for R4 and N4

8. R8C3 = 9 (hidden single in C3) -> R89C4 = 10 = {28/37/46}, no 5
8a. R9C8 = 9 (hidden single in R9)
8b. R1C2 = 9 (hidden single in R1)
8c. R2C9 = 9 (hidden single in R2)
8d. Naked pair {68} in R34C2, locked for C2 and 23(3) cage -> R4C1 = 9
8e. R7C2 = 5 -> R6C12 = 4 = {13}, locked for R6 and N4
8f. R9C2 = 7, R8C1 = 6, clean-up: no 3 in R8C4, no 4 in R9C4
8g. 5 in N8 only in R89C5, locked for C5
8h. 3 in N5 only in R4C45, locked for R4

9. 17(3) cage at R6C8 = {458/467} -> R6C9 = 4, R67C8 = [58/67/76]
9a. Killer pair 7,8 in R1C8 and R67C8, locked for C8
9b. R6C7 = 9 (hidden single in N6)

10. Consider placements for 8 in C8
R1C8 = 8 => R2C7 = 7
or R7C8 = 8 => R8C9 = 7
-> no 7 in R7C7
[Cracked. The rest is fairly straightforward.]

11. 7 on D\ only in R1C1 + R3C1 -> 17(5) cage at R1C1 = {12347}, 4,7 locked for N1, no 7 in R1C3 + R3C1
11a. 12(3) cage at R1C4 (step 7a) = {246/345}
11b. R2C3 = {56} -> no 5,6 in R12C4

12. Naked quad {2345} in R12C46, locked for N2
12a. 5 in N1 only in R2C13, locked for R2
12b. R1C6 = 5 (hidden single in N2) -> R2C67 = 9 = [27], R1C8 = 8, R8C9 = 8 (hidden single in N9)
12c. Naked pair {34} in R12C4, locked for C4 -> R89C4 (step 8) = [28]
12d. R12C4 = {34} -> R2C3 = 5 (cage sum), R2C1 = 8, R34C2 = [68]
12e. R89C6 = {13} -> R8C7 = 4 (cage sum)
12f. 17(5) cage at R7C7 = {12356}, locked for N9
12g. R7C8 = 7 -> R6C8 = 6 (cage sum), R4C7 = 5, R5C8 = 3, R2C8 = 1, placed for D/, R8C2 = 3, placed for D/

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

Rating Comment:
I'll rate my walkthrough for A330 X at Easy 1.5 because I used a short forcing chain.
However my step 10 may well be a recognised type of fish, with body at R17C8 and fins at R2C7 and R8C9, but I'll stick to my rating since I saw it as a forcing chain.
SudokuSolver scoring summary said that it used XY-Chain Simple 3. Maybe that was my step 10.


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