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 Post subject: Assassin 346
PostPosted: Thu Feb 01, 2018 9:03 am 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 754
Location: Sydney, Australia
Assassin 346
Attachment:
a346.JPG
a346.JPG [ 67.07 KiB | Viewed 1976 times ]

I'll try and post one of these at the start of each month (so still plenty of time to do ixsetf's new one :) ). Others are welcome to fill up the middles!

This puzzle started out as a version with a very elegant solution, but typing up my walkthrough realized I'd made a logical flaw. Then couldn't solve it! Had to make some adjustments. It's now quite fun since can keep making regular progress and doesn't finally crack completely until a long way in. As so often happens, this cage pattern has some interesting interactions that are not needed for an optimised solution. It gets an SSscore of 1.55 SS doesn't use my key step.

Please post if someone would like to try the version I couldn't solve. Would love to know how to do it. That version gets a 1.50

code: paste into solver:
3x3::k:9472:9472:9472:9472:769:769:7170:7170:7170:9472:6659:6659:6659:9472:7170:2820:4869:4869:9472:5382:5382:6659:2580:2580:2820:2312:4869:5382:5382:6153:6153:3861:3861:2820:2312:4869:6154:6153:6153:2059:2059:2059:4364:4364:4869:6154:2317:4878:6671:6671:4364:4364:5136:5136:6154:2317:4878:6671:6671:5137:5136:5136:9490:6154:6154:4878:3603:9490:5137:5137:5137:9490:3603:3603:3603:2055:2055:9490:9490:9490:9490:
solution:
Code:
+-------+-------+-------+
| 4 6 3 | 5 1 2 | 8 7 9 |
| 2 7 5 | 8 9 4 | 6 1 3 |
| 8 1 9 | 6 3 7 | 4 5 2 |
+-------+-------+-------+
| 3 8 2 | 7 6 9 | 1 4 5 |
| 5 9 6 | 1 4 3 | 7 2 8 |
| 1 4 7 | 2 8 5 | 3 9 6 |
+-------+-------+-------+
| 6 5 4 | 9 7 8 | 2 3 1 |
| 9 3 8 | 4 2 1 | 5 6 7 |
| 7 2 1 | 3 5 6 | 9 8 4 |
+-------+-------+-------+
Cheers
Ed


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 Post subject: Re: Assassin 346
PostPosted: Mon Feb 05, 2018 10:36 pm 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1594
Location: Lethbridge, Alberta, Canada
Thanks Ed for posting this Assassin! Great news that you plan to post a new one each month! :D

As you said, this is a puzzle where one continues to make progress.

Here is my walkthrough for Assassin 346:
Prelims

a) R1C56 = {12}
b) R3C56 = {19/28/37/46}, no 5
c) R34C8 = {18/27/36/45}, no 9
d) R4C56 = {69/78}
e) R67C2 = {18/27/36/45}, no 9
f) R9C45 = {17/26/35}, no 4,8,9
g) 11(3) cage at R2C7 = {128/137/146/236/245}, no 9
h) 8(3) cage at R5C4 = {125/134}
i) 19(3) cage at R6C3 = {289/379/469/478/568}, no 1
j) 28(4) cage at R1C7 = {4789/5689}, no 1,2,3
k) 26(4) cage at R2C2 = {2789/3689/4589/4679/5678}, no 1
l) 26(4) cage at R6C4 = {2789/3689/4589/4679/5678}, no 1
m) 14(4) cage at R8C4 = {1238/1247/1256/1346/2345}, no 9
n) 37(7) cage at R1C1 = {1246789/1345789/2345689}
o) 37(7) cage at R7C9 = {1246789/1345789/2345689}

Steps resulting from Prelims
1a. Naked pair {12} in R1C56, locked for R1 and N2, clean-up: no 8,9 in R3C56
1b. 8(3) cage at R5C4 = {125/134}, 1 locked for R5 and N5
1c. 28(4) cage at R1C7 = {4789/5689}, CPE no 8,9 in R1C4 + R2C789
1d. 3 in R1 only in R1C1234, locked for 37(7) cage at R1C1, no 3 in R23C1 + R2C4
1e. 37(7) cage must contain 8,9, CPE no 8,9 in R2C23
1f. Max R2C23 = 13 -> min R23C4 = 13, no 3 in R23C4

2. 45 rule on R1234 3 outies R5C239 = 23 = {689}, locked for R5
2a. 19(5) cage at R2C8 = {12349/12358/12367/13456} (cannot be {12457} because R5C9 only contains 6,8,9)
2b. R5C9 = {689} -> no 6,8,9 in R2C89 + R34C9
2c. 9 in N3 only in R1C789, locked for R1 and 28(4) cage at R1C7, no 9 in R2C6
2d. 24(4) cage at R4C3 = {1689/2589/2679/3489/3678/4569} (cannot be {3579/4578} because R5C23 contain two of 6,8,9)
2e. 1 of {1689} only in R4C3, other combinations only contain two of 6,8,9 -> no 6,8,9 in R4C3

3. 14(4) cage at R8C4 = {1247/2345} (cannot be {1238/1256/1346) which clash with R9C45), no 6,8
3a. 14(4) cage at R8C4 = {1247/2345}, CPE no 2,4 in R8C123 + R9C456, clean-up: no 6 in R9C45
3b. Combined 14(4) cage + R9C45 contain 1,3,5,7, CPE no 1,3,5,7 in R9C6
3c. 6,8,9 in R9 only in R9C6789, locked for 37(7) cage at R7C9, no 6,8,9 in R78C9 + R8C5
3d. 37(7) cage must contain 4, CPE no 4 in R8C78
3e. 37(7) cage contains 6 so must contain 2, CPE no 2 in R8C78

4. 45 rule on N36 4(2+2) outies R26C6 + R7C78 = 14
4a. Min R26C6 = 6 -> max R7C78 = 8, no 8,9 in R7C78
4b. Max R7C78 = 8 -> min R6C89 = 12, no 1,2 in R6C89
4c. Min R2C6 + R7C78 = 7 -> max R6C4 = 7
4d. Min R5C78 + R6C6 = {234} = 9 -> max R6C7 = 8

5. 3 in R1 only in 37(7) cage at R1C1 = {1345789/2345689}, 28(4) cage at R1C7 = {4789/5689}
5a. Hidden killer pair 6,7 in 37(7) cage and 28(4) cage for R1, each contains one of 6,7 -> they must be in R1 -> no 6,7 in R23C1 + R2C56
5b. 45 rule on R1 4(2+2 or 3+1) outies R23C1 + R2C56 = 23
5c. R23C1 must contain at least one of 1,2 for 37(7) cage -> max R23C1 = 11 -> min R2C56 = 12 -> R2C56 = {48/49/58/59} (cannot be {89} which clashes with 26(4) cage at R2C2)
5d. R2C56 = 12,13,14 -> R23C1 = 9,10,11 = {18/19/28/29}, no 4,5
5e. 26(4) cage at R2C2 = {4679/5678} (cannot be {2789/3689/4589} which clash with R2C56), no 2,3
5f. 26(4) cage and R2C56 ‘see’ each other so form combined cage {456789}, 26(4) cage = {4679/5678} -> R2C56 {58}/[94] = 13 -> R23C1 = 10 = {19/28}
5g. 37(7) cage must contain 4, locked for R1
5h. 37(7) cage and combined cage {456789} must both contain 4, locked for N12, no 4 in R3C2356, clean-up: no 6 in R3C56
5i. Naked pair {37} in R3C56, locked for R3 and N2, clean-up: no 2,6 in R4C8
5j. 6 in N2 only in R123C4, locked for C4

6. 26(4) cage at R2C2 (step 5e) = {4679/5678}, 7 locked for R2 and N1
6a. 7 in R1 only in 28(4) cage at R1C7 = {4789} -> R1C789 = {789}, locked for R1 and N1, R2C6 = 4 -> R2C5 = 9 (step 5f), clean-up: no 6 in R4C6, no 1 in R4C8
6b. 37(7) cage at R1C1 = {2345689} -> R23C1 = {28}, locked for C1 and N1
6c. Naked triple {568} in R123C4, locked for C4, clean-up: no 3 in R9C5
6d. Combined 14(4) cage + R9C45 contain 1,3,5,7 (step 3b), 5 locked for R9

7. R3C23 = {19} (hidden pair in N1), locked for R3 and 21(4) cage at R3C2, clean-up: no 8 in R4C8
7a. R3C23 = {19} =10 -> R4C12 = 11 = [38]/{47/56}
7b. Combined cage R4C12 + R4C56 = [38][69]/{47}[69]/{56}{78}
7c. 24(4) cage at R4C3 cannot be [19]{68} which clashes with R4C12 + R4C56, no 1 in R4C3, no 9 in R4C4
7d. R4C6 = 9 (hidden single in R4) -> R4C5 = 6, clean-up: no 5 in R4C12
7e. 9 in R9 only in R9C789, locked for N9
7f. 1 in N4 only in R6C12, locked for R6
7g. 45 rule on N47 4(2+2) outies R3C23 + R48C4 = 21, R3C23 = {19} = 10 -> R48C4 = 11 = {47}, locked for C4, clean-up: no 1 in R9C5
7hi. R4C12 = [38] (cannot be {47} which clashes with R4C4), clean-up: no 6 in R3C8, no 1 in R6C2, no 1,6 in R7C2
7i. Naked pair {69} in R5C23, locked for R5 and N4 -> R5C9 = 8, clean-up: no 3 in R7C2
7j. 17(4) cage at R5C7 = {2357/2456}, CPE no 5 in R6C89
[The order of steps 7g to 7j has been changed.]

8. R6C5 = 8 (hidden single in N5), R7C4 = 9 (hidden single in N8) -> R6C4 + R7C5 = 9 = [27], R3C56 = [37], R8C4 = 4, R9C5 = 5 -> R9C4 = 3, R5C45 = [14] -> R5C6 = 3 (cage sum)
8a. R4C4 = 7, R5C34 = {69} -> R4C3 = 2 (cage sum), clean-up: no 2 in R3C8
8b. R6C6 = 5, R5C78 = {27}, locked for R5 and N6 -> R6C7 = 3 (cage sum), clean-up: no 4 in R7C2
8c. Naked pair {17} in R9C13, locked for R9 and N7, R9C2 = 2, R7C2 = 5 -> R6C2 = 4
8d. R6C3 = 7, 8 in C3 only in 19(3) cage at R6C3 = [748]
8e. R6C89 = {69} = 15 -> R7C78 = 5 = [23]
8f. 37(7) cage at R7C9 contains 6 so must contain 2 -> R8C5 = 2
8g. R7C19 = [61], R7C6 = 8 -> R89C6 = [16]
8h. R78C6 = [81] = 9 -> R8C78 = 11 = {56}, locked for R8

9. R5C9 = 8 -> R2C89 + R34C9 = 11 = [1325]

and the rest is naked singles.
Thanks Ed for pointing a couple of minor errors in step 7.

Rating Comment:
I'll rate my walkthrough for A346 at Hard 1.25 for my use of combined cages.
I'd originally used a short forcing chain for step 5h, which I'd have rated at Easy 1.5, but the double 4 locked for N12 is more direct.

Ed. Please post your original version. I'd like to have a try at it before I have a go at ixsetf's Human Solvable (but I've since changed my mind and done that one); there's still one of HATMAN's Human Solvables which I haven't yet tried.


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 Post subject: Re: Assassin 346
PostPosted: Wed Feb 07, 2018 8:14 pm 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 754
Location: Sydney, Australia
Thanks for your WT Andrew! Really enjoyed it with easy-enough to follow combination work. Very different WT to mine. I started with "45"s.

I will get us up to 350 but no promises after that. A busy year coming up.

Will post the V1.5 shortly.

A346
start:
Cage clean-up not done unless specified
Preliminaries courtesy of SudokuSolver
Cage 3(2) n2 - cells ={12}
Cage 15(2) n5 - cells only uses 6789
Cage 8(2) n8 - cells do not use 489
Cage 9(2) n47 - cells do not use 9
Cage 9(2) n36 - cells do not use 9
Cage 10(2) n2 - cells do not use 5
Cage 8(3) n5 - cells do not use 6789
Cage 11(3) n36 - cells do not use 9
Cage 19(3) n47 - cells do not use 1
Cage 28(4) n23 - cells do not use 123
Cage 14(4) n78 - cells do not use 9
Cage 26(4) n12 - cells do not use 1
Cage 26(4) n58 - cells do not use 1

1. 14(4)r8c4 = {1238/1247/1256/1346/2345} = [2/6..]
1a. -> {26} blocked from 8(2)r9c4
1b. = {17/35}(no 2,6) = [1/3,1/5..]
1c. -> {1238/1256/1346} all blocked from 14(4)
1d. = {1247/2345}(no 6,8)
1e. must use 2 and 4 -> no 2,4 in r9c6 nor r8c123

SS doesn't use this step - my key one.
2. "45" on c1234: 1 innie r5c4 + 28 = four outies r2679c5
2a. Max. 4 outies = 30 -> r5c4 = (12)
2b. r2679c5 = 29/30 = {5789/6789}(no 1,2,3,4): 7,8,9, locked for c5
2c. r4c56 = [69]
2d. -> r2679c5 = {5789} = 29 -> r5c4 = 1
2e. -> r9c45 = [35](only permutation)
2f. r3c6 = (678), no 1 in r3c5

3. 14(4)r8c4 = {1247} only
3a. no 7 in r9c6 nor r8c123
3a. 1 only in r9 in n7: locked for r9 and n7

4. hidden single 3 in n2 -> r3c5 = 3, r3c6 = 7

5. "45" on r56789: 3 innies r5c239 = 23 = {689}: all locked for r5

6. 3 in n5 only in r56c6 -> no 3 in r5c78 (CPE)

7. r5c4 = 1 -> r5c56 = 7 = [25/43] = [4/5..]

8. 17(4)r5c7 must have 2 off {2457} for r5c78 = {1259/1457/2348/2357/2456}
8a. but {1259} blocked by 1,9 only in r6c7
8b. {457}[1] blocked by r5c56 (step 7)
8c. {24}+{38}/{245}[6] blocked by r5c5
8b. = {2357} only (no 1,4,6,8,9)

9. "45" on r6789: 3 outies r5c178 = 14 = {257}: only valid combination: all locked for r5

10. r5c56 = [43], r6c7 = 3 (hidden single in 17(4))
10a. 7 must be in 17(4) only in r5: locked for r5 and n6

11. 3(2)n2 = {12} both locked for r1 and 2 for n2

12. 3 in r1 only in n1: locked for n1

13. 26(4)r2c2: can't have both 8,9 because of r2c5 = {4679/5678}(no 2)
13a. must have 7: locked for n1 and r2
13b. killer pair 8,9 in 26(4) and r2c5 -> no 8 in r2c6
(edit: Andrew noticed I could have locked 789 in r1c789 -> r2c6 = 4 at this point rather than the round-a-bout way I did the following few steps. Thanks for telling me!
13c. 28(4)r1c7 must have 8 and 9 which are only in r1c789: locked for r1 and n3

14. 37(7)r1c1 must have 3 for r1 = {2345689}(no 1)
14a. must have 2 which is only in c1: locked for c1 and n1
14b. r5c1 = 5
14c. r5c78 = {27}: 2 locked for n6 and no 2 in r6c6
14d. r6c6 = 5

15. 24(5)r5c1 = [5]{1369/1378/1468}
15a. must have 1 -> r6c1 = 1

16. 2 in n5 only in c4: locked for c4
16a. -> 1,2 which must be in 14(4)r8c4 is only in c23 -> r9c23 = {12}: 2 locked for r9 and n7

17. 26(4)r2c2: {4679} blocked by r2c6
17a. = {5678} only -> no 8,6 in r2c56
17b. r2c56 = [94]

18. hidden pair 1,9 in n1 -> r3c23 = {19}: both locked for r3

19. "45" on n47: 4 outies r3c23 + r48c4 = 21: r3c23 = 10 -> r48c4 = 11 = [74] only
19a. r3c23 = 10 -> r4c12 = 11 = {38} only: both locked for n4 and 8 for r4

20. 2 and 8 must be in 37(7)r1c1 -> r23c1 = {28}: 8 locked for c1 and n1

much easier now
Cheers
Ed


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 Post subject: Re: Assassin 346
PostPosted: Thu Feb 08, 2018 8:12 am 
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Posts: 754
Location: Sydney, Australia
A346 V1.5
Attachment:
a346v1.5.JPG
a346v1.5.JPG [ 64.92 KiB | Viewed 1930 times ]

code: paste into solver:
3x3::k:9472:9472:9472:9472:769:769:7170:7170:7170:9472:6659:6659:6659:9472:7170:2820:4869:4869:9472:5382:5382:6659:6407:6407:2820:2312:4869:5382:5382:6153:6153:6407:6407:2820:2312:4869:6154:6153:6153:2059:2059:2059:4364:4364:4869:6154:2317:4878:6671:6671:4364:4364:5136:5136:6154:2317:4878:6671:6671:5137:5136:5136:9490:6154:6154:4878:3603:9490:5137:5137:5137:9490:3603:3603:3603:2068:2068:9490:9490:9490:9490:
Same solution as the V1. Andrew's V1 WT works till step 5h. Mine to 2b. It gets an SSscore of 1.50
[edit: as Andrew remembered in his following post - this was the original puzzle I found but couldn't solve it]

Cheers
Ed


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 Post subject: Re: Assassin 346
PostPosted: Fri Feb 09, 2018 10:11 pm 
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Posts: 159
Location: California, out of London
Hi Ed - Welcome back :)

Tough puzzle. I loved the implications of the interactions for the shapes and numbers in n123!
Here is my WT for V1.5. I'm sure there's a better way to do my step 4!

Assassin 346 V1.5 WT:
1. Outies r1234 = r5c239 = +23(3) = {689}
Since min r5c9 = 6 -> 9 in n3 only in r1c789
-> 9 in 37(7)r1c1 in r23c1 or r2c5.

2. 3(2)r1 = {12}
28(4)r1c7 = {4789} or {5689}
-> 3 in r1 in r1c1234
37(7)r1 is missing two numbers +8(2) = {17} or {26} and they must be in n1 in r23c23
The missing numbers cannot both be in r2c23 (would put r23c4 = +18(2))
and since whatever is in r2c5 must go in n1 in r3c23 -> they cannot both go in r3c23.
This also means whatever goes in r1c4 goes in n1 in r2c23.

3. Outies r1 = r23c1 + r2c56 = +23(4)
Since whatever is in r2c6 must go in r1 in n1 (r1c123) -> the h+23(4) must be four different numbers.
Since one of (12) must be in r23c1 -> max r23c1 = {29} = +11(2)
-> min r2c56 = +12(2) (contains at least one of (789))
-> 2 cannot go in 26(4)r2c2

4. Outies r1 = r23c1 + r2c56 = +23(4)
Either:
a) Missing numbers from 37(7) are (17)
-> 7 in r2c23, 1 in r3c23, 2 in r23c1 -> Outies r1 = {2489}
-> Since 26(4) can have at most one of (489) to go with the 7 -> 26(4) = {5678}
-> 3 in n2 in r3c56
Also 7 in r1 in n3 r1c789 -> 28(4) = {4789}
Also 7 in r3 in r3c56
-> 25(4)r3c5 = [{37}{69}]
Also since whatever is in r1c4 is in r2c23 -> r123c4 = {568}
-> r23c1 = {28}
-> r2c56 = [94]
-> r3c23 = {19}

Or:
b) Missing numbers from 37(7) are (26)
-> 6 in r2c23, 2 in r3c23, 1 in r23c1 -> Outies r1 = {1589}
-> Since 26(4) can have at most one of (589) to go with the 6 -> 26(4) = {4679}
-> 3 in n2 in r3c56
Also 6 in r1 in n3 r1c789 -> 28(4) = {5689}
Also 6 in r3 in r3c56
-> 25(4)r3c5 = [{36}{79}]
Also since whatever is in r1c4 is in r2c23 -> r123c4 = {479} with 9 in r23c4.
-> r23c1 = {19}
-> r2c56 = {58}
-> r3c23 = {2(5|8))
But since r5c23 contains at least one of (68) -> r3c23 cannot be {25}
-> r3c23 = {28}
-> r2c56 = [85]

5. The cells in 14(4)r8c4 and 8(2)r9c4 are all buddies - must be {123457}
-> 8(2) from {17} or {35}
-> 14(4) from {2345} or {1247}
Also {689) in r9 all in r9c6789
Since whatever is in r8c4 must be in r9 in r9c789
-> r9c6 from (689}

6. Consider option 4.b) above.
It has r3c23 = {28}, r4c56 = {79}, r123c4 = {479}
Puts r4c12 = {56}
Puts r5c23 = {89}
Puts r4c34 = [43]
But since remaining outies of n47 = r48c4 = +11(2)
this would put r4c8 = 8 which contradicts max r8c4 = 7.
-> Option 4.a) is the correct one.

i.e.,
r1c789 = {789}
r123c4 = {568} with 8 in r23c4.
r2c56 = [94]
26(4)= {5678} with 7 in r2c23 and 8 in r23c4
r1c1234 = {3456} with r1c4 from (56)
r23c1 = {28}
r3c23 = {19}
25(4)r3c5 = [{37}69]

7. r4c12 from [38] or {47}
-> r5c23 cannot be {68} (Would put r4c34 = {37})
-> 9 in r5c23
8(3) n5 contains a 1 -> r6c7 cannot be a 9
-> 9 in r6c89

8. Remaining outies n47 = r48c4 = +11(2)
Since neither call can be 9 and r123c4 = {568}
-> r48c4 = {47}
-> r4c12 = [38]
-> r5c23 = {69} and r5c9 = 8
Since remaining outies n36 = r6c6 + r7c78 = +10(3)
-> HS 8 in r6/n5 -> r6c5 = 8
Since Max r6c89 = +16(2) -> Min r7c78 = +4(2) -> Max r6c6 = 6
-> HS 7 in n5 -> r4c4 = 7
-> r8c4 = 4
Also r4c3 = 2

9. Since 6 already in c5 -> r7c5 not 6 -> r6c4 not 3
-> Remaining innies n5 r6c46 = +7(2) = [25]
-> 8(3) = {134} with 4 in r5c5

Also r7c78 = +5(2) -> r6c89 = {69}
Also r4c789 = {145}
-> 17(4)n6 = [{27}3]
Also r5c1 = 5

Also r7c5 = 7
-> 8(2)n5 = [35]
-> 8(3)n5 = [143]
Also r3c56 = [37]

10. Also r9c123 = {127}
-> 19(3)c3 = [748]
-> r6c12 = [14]
-> r7c2 = 5
-> r8c2 = 3, r7c1 = 6
Also r9c123 = [721]
-> r3c23 = [19]
-> r5c23 = [96]
-> r12c3 = [35]
-> r1c4 = 5
-> r1c12 = [46]

11. r7c78 = +5(2) must be [23]
-> r5c78 = [72]
-> r23c9 = [32]
-> r23c1 = [28]
-> r23c4 = [86]
-> r2c78 = [61]
-> r34c7 = [41]
-> r34c8 = [54]
-> r4c9 = 5
-> r8c7 = 5

etc.


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 Post subject: Re: Assassin 346
PostPosted: Sat Feb 10, 2018 10:06 pm 
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Location: Lethbridge, Alberta, Canada
Thanks Ed for posting your original, harder version as a V1.5

Here is my walkthrough for A346 V1.5, with a little help from Ed's step 2 for A346:
My start for Assassin 346 also applies for this variant apart from a couple of prelims and a couple of clean-ups, which have been omitted.

Prelims

a) R1C56 = {12}
b) R34C8 = {18/27/36/45}, no 9
c) R67C2 = {18/27/36/45}, no 9
d) R9C45 = {17/26/35}, no 4,8,9
e) 11(3) cage at R2C7 = {128/137/146/236/245}, no 9
f) 8(3) cage at R5C4 = {125/134}
g) 19(3) cage at R6C3 = {289/379/469/478/568}, no 1
h) 28(4) cage at R1C7 = {4789/5689}, no 1,2,3
i) 26(4) cage at R2C2 = {2789/3689/4589/4679/5678}, no 1
j) 26(4) cage at R6C4 = {2789/3689/4589/4679/5678}, no 1
k) 14(4) cage at R8C4 = {1238/1247/1256/1346/2345}, no 9
l) 37(7) cage at R1C1 = {1246789/1345789/2345689}
m) 37(7) cage at R7C9 = {1246789/1345789/2345689}

Steps resulting from Prelims
1a. Naked pair {12} in R1C56, locked for R1 and N2
1b. 8(3) cage at R5C4 = {125/134}, 1 locked for R5 and N5
1c. 28(4) cage at R1C7 = {4789/5689}, CPE no 8,9 in R1C4 + R2C789
1d. 3 in R1 only in R1C1234, locked for 37(7) cage at R1C1, no 3 in R23C1 + R2C4
1e. 37(7) cage must contain 8,9, CPE no 8,9 in R2C23
1f. Max R2C23 = 13 -> min R23C4 = 13, no 3 in R23C4

2. 45 rule on R1234 3 outies R5C239 = 23 = {689}, locked for R5
2a. 19(5) cage at R2C8 = {12349/12358/12367/13456} (cannot be {12457} because R5C9 only contains 6,8,9)
2b. R5C9 = {689} -> no 6,8,9 in R2C89 + R34C9
2c. 9 in N3 only in R1C789, locked for R1 and 28(4) cage at R1C7, no 9 in R2C6
2d. 24(4) cage at R4C3 = {1689/2589/2679/3489/3678/4569} (cannot be {3579/4578} because R5C23 contain two of 6,8,9)
2e. 1 of {1689} only in R4C3, other combinations only contain two of 6,8,9 -> no 6,8,9 in R4C3

3. 14(4) cage at R8C4 = {1247/2345} (cannot be {1238/1256/1346) which clash with R9C45), no 6,8
3a. 14(4) cage at R8C4 = {1247/2345}, CPE no 2,4 in R8C123 + R9C456, clean-up: no 6 in R9C45
3b. Combined 14(4) cage + R9C45 contain 1,3,5,7, CPE no 1,3,5,7 in R9C6
3c. 6,8,9 in R9 only in R9C6789, locked for 37(7) cage at R7C9, no 6,8,9 in R78C9 + R8C5
3d. 37(7) cage must contain 4, CPE no 4 in R8C78
3e. 37(7) cage contains 6 so must contain 2, CPE no 2 in R8C78

4. 45 rule on N36 4(2+2) outies R26C6 + R7C78 = 14
4a. Min R26C6 = 6 -> max R7C78 = 8, no 8,9 in R7C78
4b. Max R7C78 = 8 -> min R6C89 = 12, no 1,2 in R6C89
4c. Min R2C6 + R7C78 = 7 -> max R6C4 = 7
4d. Min R5C78 + R6C6 = {234} = 9 -> max R6C7 = 8

5. 3 in R1 only in 37(7) cage at R1C1 = {1345789/2345689}, 28(4) cage at R1C7 = {4789/5689}
5a. Hidden killer pair 6,7 in 37(7) cage and 28(4) cage for R1, each contains one of 6,7 -> they must be in R1 -> no 6,7 in R23C1 + R2C56
5b. 45 rule on R1 4(2+2 or 3+1) outies R23C1 + R2C56 = 23
5c. R23C1 must contain at least one of 1,2 for 37(7) cage -> max R23C1 = 11 -> min R2C56 = 12 -> R2C56 = {48/49/58/59} (cannot be {89} which clashes with 26(4) cage at R2C2)
5d. R2C56 = 12,13,14 -> R23C1 = 9,10,11 = {18/19/28/29}, no 4,5
5e. 26(4) cage at R2C2 = {4679/5678} (cannot be {2789/3689/4589} which clash with R2C56), no 2,3
5f. 26(4) cage and R2C56 ‘see’ each other so form combined cage {456789}, 26(4) cage = {4679/5678} -> R2C56 {58}/[94] = 13 -> R23C1 = 10 = {19/28}
5g. 37(7) cage must contain 4, locked for R1
5h. 37(7) cage and combined cage {456789} must both contain 4, locked for N12, no 4 in R3C2356
5i. Combined cage = {456789}, 8,9 locked for N2
5j. Max R3C56 = 13 -> min R4C56 = 12, no 2 in R4C56
5k. Hidden killer pair 1,2 in R23C1 and R3C23 for N1, R23C1 contains one of 1,2 -> R3C23 must contain one of 1,2
5l. 21(4) cage at R3C2 can only contain one of 1,2 -> no 1,2 in R4C12

[Since I’ve seen Ed’s WT for Assassin 346 and seen how powerful his step 2 was, it’s time to “borrow” that step for this puzzle. I ought to have spotted that 45, since I’d seen the one for R1234.]
6. 45 rule on C1234 4 outies R2679C5 = 1 innie R5C4 + 28
6a. Max R2679C5 = 30 -> max R5C4 = 2
6b. R2679C5 = 29,30 = {5789/6789}, 7,8,9 locked for C5, clean-up: no 5,7 in R9C4
6c. 25(4) cage at R3C5 = {3679/4579/4678} (cannot be {3589} because 8,9 only in R4C6) -> R3C6 = 7, R4C6 = {89}, clean-up: no 2 in R4C8
6d. Killer pair 5,6 in R2679C5 and R34C5, locked for C5
6e. 26(4) cage at R2C2 (step 5e) = {4679/5678}, 7 locked for R2 and N1
6f. 7 in R1 only in 28(4) cage at R1C7 = {4789) -> R2C6 = 4, R1C789 = {789}, 8 locked for R1 and N3, clean-up: no 1 in R4C8
6g. 26(4) cage = {5678}, 8 locked for C4, R2C5 = 9 (hidden single in N2), clean-up: no 1 in R23C1 (step 5f)
6h. Naked pair {28} in R23C1, locked for C1
6i. R3C23 = {19} (hidden pair in N1), locked for R3 and 21(4) cage at R3C2, clean-up: no 8 in R4C8
6j. R3C23 = {19} = 10 -> R4C12 = 11 = [38]/{47/56}, no 3 in R4C2
6k. 3 in R2 only in R2C789, locked for N3, clean-up: no 6 in R4C8
6l. R3C5 = 3 (hidden single in R3)
6m. R3C56 = [37] = 10 -> R4C56 = 15 = [69], clean-up: no 5 in R4C12

7. 45 rule on N47 4(2+2) outies R3C23 + R48C4 = 21, R3C23 = {19} = 10 -> R48C4 = 11 = {47} locked for C4
7a. R4C12 = [38] (cannot be {47} which clashes with R4C4), clean-up: no 6 in R3C8, no 1 in R6C2, no 1,6 in R7C2
7b. 9 in C1 only in 24(5) cage at R5C1 = {13479/13569} -> R8C2 = 3, 1 locked for C1, clean-up: no 6 in R6C2
7c. 14(4) cage at R8C4 = {1247} (only remaining combination), 1,2 locked for R9 and N7 -> R9C4 = 3, R9C5 = 5, R6C9 = 1 (hidden single in C1), clean-up: no 7 in R6C2
7d. R567C4 = [129] (hidden triple in C4), R8C1 = 9 (hidden single in C1), clean-up: no 7 in R7C2
7e. R5C45 = [14] -> R5C6 = 3 (cage sum)
7f. R8C4 = 4 (hidden single in N8), R9C1 = 7
7g. R4C4 = 7, R5C23 = {69} -> R4C3 = 2 (cage sum), clean-up: no 2 in R3C8
7h. R6C6 = 5, R5C78 = {27}, 7 locked for N6, R6C7 = 3 (cage sum)
7i. R6C3 = 7 (hidden single in N4), 8 in C3 only in 19(3) cage at R6C3 = [748]
7j. R6C89 = {69} (hidden pair in R6}, locked for N6
7k. R6C89 = {69} = 15 -> R7C78 = 5 = [23], R5C78 = [72]
7l. Naked pair {45} in R34C8, locked for C8

8. R2C8 = 1, R5C9 = 8 -> R234C9 = 10 = [325]
8a. 37(7) cage at R7C9 contains 6 so must contain 2 -> R8C5 = 2
8b. R67C5 = [87], R7C6 = 8 (hidden single in R7) -> R8C678 = 12 = [156]

and the rest is naked singles.

Rating Comment:
I'll also rate my WT for A346 V1.5 at Hard 1.25. While this puzzle is harder at a human level, it's technically at the same level as A346.


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 Post subject: Re: Assassin 346
PostPosted: Tue Feb 13, 2018 9:43 pm 
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Grand Master
Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 754
Location: Sydney, Australia
This is a hybrid optimised start for the V1.5 from Andrew's WT and SudokuSolver. It found one nice step which makes an easier solution overall I think. Not that hard a puzzle after all!

Wellbeback - wow! I couldn't follow all of step 4 in my head but I learned a lot from what I could follow. Thanks! I was doing something similar in n789 but was forgetting one part of the logic chain you used. That's why I got tricked.

A346V1.5
Hybrid Andrew and SudokuSolver optimized start:
No cage clean-up done unless stated.
Prelims courtesy of SudokuSolver

a) R1C56 = {12}
b) R34C8 = {18/27/36/45}, no 9
c) R67C2 = {18/27/36/45}, no 9
d) R9C45 = {17/26/35}, no 4,8,9
e) 11(3) cage at R2C7 = {128/137/146/236/245}, no 9
f) 8(3) cage at R5C4 = {125/134}
g) 19(3) cage at R6C3 = {289/379/469/478/568}, no 1
h) 28(4) cage at R1C7 = {4789/5689}, no 1,2,3
i) 26(4) cage at R2C2 = {2789/3689/4589/4679/5678}, no 1
j) 26(4) cage at R6C4 = {2789/3689/4589/4679/5678}, no 1
k) 14(4) cage at R8C4 = {1238/1247/1256/1346/2345}, no 9
l) 37(7) cage at R1C1 = {1246789/1345789/2345689}
m) 37(7) cage at R7C9 = {1246789/1345789/2345689}

1. 3(2)n2 = {12}: both locked for r1 and n2

2. 28(4)r1c7 = {4789/5689}
2a. must have 8,9 -> no 8,9 in r1c4 since it sees all 8,9 in 28(4) (CPE)
2b. note: 28(4) = one of 6,7

3. 3 in r1 must be in 37(7) = {1345789/2345689}
3a. must have 8,9 -> no 8,9 in r2c23 since it sees all 8,9 in 37(7) (CPE)
3b. note: 37(7) = one of 6,7

SS isn't coded to do this next step but is the key one from Andrew's WT
4. hidden killer pair 6,7 in r1 with each of 37(7) and 28(4) having one of and must be in r1 -> no 6,7 in r23c1 nor r2c56

5. 14(4)r8c4 = {1238/1247/1256/1346/2345} = 2 or 6..
5a. -> {26} blocked from 8(2)r9c4 = {17/35}(no 2,6)

6. "45" on c1234: 1 innie r5c4 + 28 = 4 outies r2679c5
6a. max. r2679c5 = 30 -> r5c4 = (12)
6b. r2679c5 = 29/30 = {5789/6789}(no 1,2,3,4): 7,8,9 all locked for c5
6c. no 5,7 in r9c4

Found this one on SudokuSolver
7. "45" on c1234: 2 innies r59c4 + 20 = 3 outies r267c5
7a. max. 3 outies = 24 -> max. r59c4 = 4 -> = [13/21]
7b. must have 1: locked for c4
7c. r267c5 = 23/24 = {689/789}(no 5)

8. 26(4)r2c2: {2789/3689/4589} all blocked by r2c5
8a. = {4679/5678}(no 2,3)
8b. = one of 8,9
8c. note: must have 7

9. Killer pair 8,9 in r23c4 and r2c5: locked for n2

10. 25(4)r3c5 can only have one of 8,9 in r4c6
10a. = {3679/4579/4678}
10b. r4c6 = (89)
10c. must have 7 -> r3c6 = 7
10d. -> 7 which must be in 26(4)r2c2 only in n1: 7 locked for n1 and r2

11. 37(7)r1c1 = {2345689}(no 1)
11a. must have 2 which is only in c1: 2 locked for c1 and n1

12. 28(4)r2c6 must have 4,5 for r2c6 -> r1c789 = {789/689}(no 4,5)
12a. must have 8,9: both locked for r1 and n3

13. naked quad {3456} in r1c1234: 6 locked for r1, no 3,4,5 in r23c1

14. naked quad {789} in r1c789 = 24 -> r2c6 = 4 (cage sum)
14a. deleted

15. 26(4)r2c2 = {5678} only (no 9)
15a. -> r2c5 = 9 (hidden single n2)

16. r3c23 = {19} only: hidden pair in n1: 1 locked for r3
16a. no 1,9 in r4c12

17. "45" on n12: 3 remaining innies r3c235 = 13
17a. r3c23 = 10 -> r3c5 = 3

18. r3c56 = 10 -> r4c56 = 15 = [69] only permutation

19. r267c5 = 24 -> r59c4 = 4 = [13](step 7)
19a. r9c5 = 5

20. naked triple {568} in r123c4: all locked for c4

21. r56c5 = {78} = 15 -> r67c4 = 11 = [29] only permutation

22. r3c23 = 10 -> r4c12 = 11: but {47} blocked by r4c4
22a. = {38} only: both locked for r4 and n4

etc. Andrew shows the key areas to look in if needed
Cheers
Ed


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