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Ed
 Post subject: Assassin 280
PostPosted: Thu Dec 19, 2013 10:05 pm 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1043
Location: Sydney, Australia
Just the one hard-to-spot move. Nothing heavy. Can still make lots of progress even without spotting the key straight away - a feature I always try and build into my Assassins. It gets a SudokuSolver score of 1.60 and JSudoku gets it out easily.

Assassin 280

Image
code: paste into solver:
3x3::k:2816:1793:5378:5378:5378:3075:4100:4100:3333:2816:1793:7686:7686:5378:3075:3075:2311:3333:8712:3337:3337:7686:4362:5378:5643:2311:3333:8712:8712:6156:7686:4362:4362:5643:5643:5643:8712:6156:6156:6156:4362:1293:1293:2574:2574:8712:8712:6156:4623:4112:4112:5393:5393:5393:8712:2322:2322:4623:4112:7699:5393:2836:4117:1302:1302:4623:4623:7699:4375:4375:2836:4117:2328:2328:7699:7699:7699:4375:2329:2329:4117:
solution: 
+-------+-------+-------+
| 3 2 6 | 1 4 5 | 9 7 8 |
| 8 5 7 | 9 3 6 | 1 4 2 |
| 1 9 4 | 8 2 7 | 6 5 3 |
+-------+-------+-------+
| 2 4 3 | 6 5 9 | 8 1 7 |
| 9 8 5 | 7 1 3 | 2 6 4 |
| 7 6 1 | 4 8 2 | 3 9 5 |
+-------+-------+-------+
| 5 7 2 | 3 6 1 | 4 8 9 |
| 4 1 9 | 2 7 8 | 5 3 6 |
| 6 3 8 | 5 9 4 | 7 2 1 |
+-------+-------+-------+
Merry Christmas!
Ed
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wellbeback
 Post subject: Re: Assassin 280
PostPosted: Fri Dec 20, 2013 10:29 pm 
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 282
Location: California, out of London
Thanks again Ed. I'm guessing my step 1 is the hard-to-spot move - but maybe it's my step 4. or maybe something else entirely. This is the way I did it (after a little clean-up) but the more I looked at it afterwards the more I saw other paths. It will be interesting to see how others solved it.

Merry Christmas to you as well and to those (very few) others who still read this board :)

(Typo fixed courtesy of Ed. He beat you to it Andrew :) ).
2nd edit - more typos fixed thanks to Andrew! Ed & I need to do better :oops:
Hidden Text:
1. 16/2@r1c7 = {79}
Innies n3 -> r23c7 = +7
Innies n9 -> r78c7 = +9
Innies c789 -> r258c7 = +8
-> Must include a 1 plus 2 other numbers which total +7 (25) or (34)
Since r2c7 is already +7 with r3c7 -> it cannot be +7 with any other number in c7
-> it cannot be one of the two other numbers (in the innies c789 = +8) which total +7
-> r2c7 = 1
-> r3c7 = 6
-> 9/2@r2c8 = {45}
-> 13/3@r1c9 = {238}

2. r58c7 = +7
r5c7 from (234)
-> r8c7 from (543)
-> r7c7 from (456)
But 6 already in c7
-> r578c7 = [245] or [354]
-> 5/2@r5c6 = {23}
Also (45) in n6 in r456c9
Innies c9 -> r456c9 = +16 must be {457}
-> 16/3@r7c9 = {169}
and 11/2@r7c8 = {38}
and 9/2@r9c7 = {27}
Also -> 10/2@r5c8 = [64]

3. Innies c12 -> r357c2 = +24 - must be {789}
-> 13/2@r3c2 = [94] or [85]
Also 9/2@r7c2 = [72] or [81]
Since 1 or 2 also in 5/2@r8c1
-> 9/2@r9c1 from {36} or {45}

4. Between them the 7/2@r1c2 and the 13/2@r3c2 contain two of the numbers (456)
-> 11/2@r1c1 cannot be {56}
-> 11/2@r1c1 contains one of the numbers (789)

The 34/7@r3c1 is missing two numbers which total +11.
-> the two numbers in 11/2@r1c1 are either both in the 34/7@r3c1 or both not in it.
However, since whichever of (789) is in 11/2@r1c1 cannot go in r46c2 means that number cannot be in the 34/7@r3c1
-> The two values in 11/2@r1c1 are the missing values from 34/7@r3c1
-> the two values from r89c1 must be in the 34/7@r3c1 and can only go in r46c2
-> the value in r7c3 (1 or 2) must go in r456c1 in n4 and in the 7/2@r1c2 in n1
-> 7/2@r1c2 = [16] or {25} depending on whether 9/2@r7c2 is [81] or [72]

-> r127c2 = [168] or [{25}7] and both of those options prevent 13/2@r3c2 from being [85]

-> 13/2@r3c2 = [94]
-> 9/2@r2c8 = [45]

5. Innies - Outies n12 -> r4c4 = r3c15 + 3
-> Max r3c15 = +6.
Since (45) already in r3 -> r3c15 cannot be +6
-> r3c15 = max +5 and not include a 4
-> r3c1 and r3c5 both from (123) and r4c4 from (678)

-> HS 9 in 30/4@r2c3 -> r2c4 = 9
Also HS 7 in n1 -> r2c3 = 7
-> r34c4 = [86]
Also -> HS 7 in n2 -> r3c6 = 7
-> r12c6 = {56}

Also Innies n1 -> r1c3+r3c1 = +7
-> 11/2@r1c1 = {38} and those are the two values not in the 34/7@r3c1
-> 3 in n4 in r46c3
-> 3 in n7 in r89c2

Also Innies n12 -> r3c15 = {12}
-> (HS 3 in r3) r3c9 = 3

6. Whichever of (12) is in r7c3 goes in n4 in r456c1 and therefore the other of (12) goes in r3c1
-> (since 3 in r12c1) -> (123) locked in r123456c1 in c1
-> 5/2@r8c1 = [41]
-> 9/2@r7c2 = [72] and r5c2 = 8
-> 2 in n4 in r46c1
-> r3c1 = 1
-> r3c5 = 2
Also 7/2@r1c2 = {25}
-> r1c3 = 6
-> r12c6 = [56]
-> 7/2@r1c2 = [25]
-> r12c9 = [82]
-> 11/2@r1c1 = [38]
-> r2c5 = 3 and r1c45 = {14}

7. Since 4 now placed in r8 -> r78c7 = [45]
-> 5/2@r5c6 = [32]
-> r89c6 = [84]
Also 9/2@r9c7 = [72]
-> 16/2@r1c7 = [97]
-> r46c7 = {38} and r46c8 = {19}
-> Only solution for 22/4@r3c7 and 21/4@r6c7 are [6817] and [3954] respectively

8. 9 in n5 locked in r4c56 or r5c5 (in fact r4c6 is NS 9)
-> 17/4@r3c5 = [2591]

9. HS 8 in n5 -> r6c5 = 8
-> 16/2@r6c5 = [826]
-> HS 4 in n5 -> r6c4 = 4
-> NS 7 in n5 -> r5c4 = 7

10. Outies n4 -> r7c1 = 5
-> r89c3 = [98]
-> 9/2@r9c1 = [63]
Also -> 11/2@r7c8 = [83]
-> HS 3 in n8 -> r7c4 = 3
-> r8c34 = [92]
etc.


Last edited by wellbeback on Mon Dec 23, 2013 7:20 am, edited 1 time in total.

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Andrew
 Post subject: Re: Assassin 280
PostPosted: Mon Dec 23, 2013 4:44 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1894
Location: Lethbridge, Alberta, Canada
Thanks Ed for an interesting Assassin!

I found the same two early key steps which wellbeback used. He got a bit more at the end of step 4, giving a more direct solving path.

wellbeback wrote:
(Typo fixed courtesy of Ed. He beat you to it Andrew :) ).
That's because I didn't start on this puzzle until I'd finished A279 and posted my WT. ;)

Thanks Ed for your comments and corrections to typos.
Here is my walkthrough for Assassin 280:
Prelims

a) R12C1 = {29/38/47/56}, no 1
b) R12C2 = {16/25/34}, no 7,8,9
c) R1C78 = {79}
d) R23C8 = {18/27/36/45}, no 9
e) R3C23 = {49/58/67}, no 1,2,3
f) R5C67 = {14/23}
g) R5C89 = {19/28/37/46}, no 5
h) R7C23 = {18/27/36/45}, no 9
i) R78C8 = {29/38/47/56}, no 1
j) R8C12 = {14/23}
k) R9C12 = {18/27/36/45}, no 9
l) R9C78 = {18/27/36/45}, no 9
m) 30(4) cage at R2C3 = {6789}

1. Naked pair {79} in R1C78, locked for R1 and N3, clean-up: no 2,4 in R2C1, no 2 in R23C8

2. 45 rule on N3 2 innies R23C7 = 7, no 8 in R23C7

3. 45 rule on N9 2 innies R78C7 = 9, no 9 in R78C7

4. 45 rule on C789 3 innies R258C7 = 8 = {125/134}, 1 locked for C7, clean-up: no 8 in R9C8
4a. R2C7 = 1 (R25C7 and R28C7 cannot total 7, which would clash with R23C7 = 7, CCC), R3C7 = 6 (step 2), clean-up: no 6 in R1C2, no 3,8 in R23C8, no 7 in R3C23, no 4 in R5C6, no 2,3,8 in R7C7, no 3 in R8C7 (both step 3), no 3 in R9C8
4b. Naked pair {45} in R23C8, locked for C8 and N3, clean-up: no 6 in R5C9, no 6,7 in R78C8, no 4,5 in R9C7
4c. Naked triple {238} in 13(3) cage at R1C9, locked for C9, clean-up: no 2,7,8 in R5C8
4d. Killer pair {45} in R3C23 and R3C8, locked for R3
4e. R2C7 = 1 -> R12C6 = 11 = [29/47]/{38/56}, no 2,4 in R2C6
4f. Max R8C7 = 5 -> min R78C6 = 12, no 1,2 in R89C6
4g. 4 in N2 only in R1C456 + R2C5, CPE no 4 in R1C3

[Note. 45 rule on C1 2 outies R46C2 = 2 innies R89C1.
If these don’t total 11, then they must be the same pair of numbers. However if they total 11 then R46C2 must contain either the pair of numbers in R12C1 or the pair of numbers in R89C1. At this stage I can’t see how to use this, so it’s only a note.]

5. 45 rule on C12 3 innies R357C2 = 24 = {789}, locked for C2, clean-up: no 8,9 in R3C3, R7C3 = {12}, no 1,2 in R9C1
5a. Killer pair 1,2 in R7C3 and R8C12, locked for N7, clean-up: no 7,8 in R9C1
5b. Killer pair 3,4 in R8C12 and R9C12, locked for N7

[Now to follow up on the earlier note …]
6. 45 rule on C1 2 outies R46C2 = 2 innies R89C1
6a. 34(7) cage is missing two numbers totalling 11, these must be in R12C1 (since R89C1 cannot total 11), no other numbers are missing from the 34(7) cage -> R46C2 and R89C1 must contain the same pair of numbers
6b. 3,4 in N7 only in R89C12 -> 3,4 must be in R4689C2, locked for C2
6c. R12C1 = [29/38/47/83] (cannot be {56} which clashes with R12C2), no 5,6 in R12C1
[Note. An alternative way to eliminate 5,6 from R12C1, without using step 6a, is
R12C1 cannot be {56} which clashes with R12C2 + R3C3, killer ALS block,
but this doesn’t make any eliminations from R12C2.]

7. 45 rule on N1 3 innies R12C3 + R3C1 = 14
7a. Min R2C3 = 6 -> max R1C3 + R3C1 = 8, no 8,9 in R1C3 + R3C1
7b. R12C3 + R3C1 = 14 = {149/167/239/257/347} (cannot be {158/356} which clash with R12C2, cannot be {248} which clashes with R3C23), no 8 in R2C3
7c. Naked quad {6789} in 30(4) cage at R2C3, 8 locked for C4

8. 45 rule on C9 2 innies R46C9 = 1 outie R5C8, IOU no 6 in R6C9

9. 6 in C9 only in 16(3) cage at R7C9 = {169} (only remaining combination), locked for C9 and N9, clean-up: no 1,9 in R5C8, no 2 in R78C8, no 3,8 in R9C7
9a. Naked pair {38} in R78C8, locked for C8 -> R5C8 = 6, R5C9 = 4, clean-up: no 1 in R5C6
9b. Naked pair {23} in R5C67, locked for R5
9c. Naked pair {27} in R9C78, locked for R9 and N9
9d. Naked pair {57} in R46C9, locked for N6
9e. 8 in R9 only in R9C356, CPE no 8 in R7C6 + R8C5

10. R3C7 = 6 -> R4C789 = 16 = {178/259} (cannot be {358} because 3,8 only in R4C7), no 3 in R4C7
10a. 21(4) cage at R6C7 = {1578/3459} (cannot be {1389/2379} because R7C7 only contains 4,5, cannot be {1479/2478} which clash with R4C789), no 2
10b. 3,8 only in R6C7 -> R6C7 = {38}

11. 45 rule on N4 3(2+1) outies R37C1 + R5C4 = 13
11a. Min R37C1 = 6 -> max R5C4 = 7
[Alternatively 45 rule on C123 4 innies R1289C3 = 1 outie R5C4 + 23, max R1289C3 = 30 -> max R5C7 = 7.]
11b. R3C1 + R5C4 cannot total 5 -> no 8 in R7C1

12. 30(5) cage at R7C6 = {15789/24789/25689/34689/35679/45678}
12a. 17(3) cage at R8C6 = {359/458} (cannot be {368} because R8C7 only contains 4,5, cannot be {467} which clashes with 30(5) cage), no 6,7
12b. 17(3) cage = {359/458}, CPE no 5 in R8C45
12c. R12C6 (step 4e) = [29/47/56/65] (cannot be {38} which clashes with 17(3) cage), no 3,8 in R12C6

13. Killer quad 6,7,8,9 in R12C6, R23C4 and 21(5) cage at R1C3 (must contain at least one of 7,8,9 in N2 and may contain 6 in R1C3), locked for N2
13a. 21(5) cage at R1C3 = {12459/12468/13458/13467/23457} (cannot be {12369} = 6{1239} which clashes with R3C5, cannot be {12378} which contains both of 7,8 in N2, cannot be {12567} = 6{1257} which clashes with R12C6)
13b. 6 in R1 only in R1C3 or in R12C6 (step 12c) = [65] (locking-out cages) -> 5 of 21(5) cage can only be in R1C3, no 5 in R1C45 + R2C5
13c. 5,6 of 21(5) cage only in R1C3 -> R1C3 = {56}

14. 5 in N2 only in R12C6 (step 12c) = {56}, locked for C6 and N2
14a. 17(3) cage at R8C6 (step 12a) = {359/458} -> R8C7 = 5, R7C7 = 4
14b. 21(4) cage at R6C7 (step 10a) = {3459} (only remaining combination) -> R6C789 = [395], R5C67 = [32], R4C789 = [817]
14b. 17(3) cage = {458} (only remaining combination), 4,8 locked for C6 and N8
14c. Naked pair {56} in R1C36, locked for R1, clean-up: no 2 in R2C2

15. Naked pair {56} in R1C3 + R2C2, locked for N1 -> R3C3 = 4, R3C2 = 9, R23C8 = [45], R2C3 = 7, R234C4 = [986], clean-up: no 2 in R1C1
15a. Naked pair {38} in R12C1, locked for C1, clean-up: no 2 in R8C2, no 6 in R9C2
15b. R3C6 = 7 (hidden single in R3) -> 21(5) cage at R1C3 (step 13a) = {13467/23457}, 3 locked for N2
15c. R3C9 = 3 (hidden single in R3)
15d. R4C3 = 3 (hidden single in C3)

16. R37C1 + R5C4 = 13 (step 11)
16a. Max R37C1 = 11 -> no 1 in R5C4
16b. R5C4 = {57} -> R37C1 = 6,8 = [15/17/26]
16c. 7 in N7 only in R7C12, locked for R7
16d. 9 in N7 only in R89C3, locked for C3

17. R6C6 = {12} -> 16(3) cage at R6C5 = {268} (only remaining combination, cannot be {169/259} because 6,9 only in R7C5, cannot be {178} because 7,8 only in R6C5) -> R6C6 = 2, R67C5 = [86], R4C6 = 9, R7C6 = 1, R7C3 = 2, R7C2 = 7, clean-up: no 3 in R8C2
17a. Naked pair {14} in R8C12, locked for R8 and N7 -> R89C6 = [84], clean-up: no 5 in R9C12
17b. R9C12 = [63], R7C1 = 5, R89C3 = [98]

18. R46C2 and R89C1 must contain the same pair of numbers (step 6a), R9C1 = 6, R8C1 = {14} -> R6C2 = 6, R4C2 = 4

and the rest is naked singles.

Rating Comment:
I'll rate my walkthrough for A280 at 1.5. Step 6a is effectively a "clone" step and therefore in the 1.5 range; I also used locking-out cages in step 13b.

Merry Christmas and a Happy New Year to all on these forums. :santa:


Last edited by Andrew on Tue Jan 21, 2014 4:48 am, edited 1 time in total.

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Ed
 Post subject: Re: Assassin 280
PostPosted: Tue Dec 24, 2013 3:51 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1043
Location: Sydney, Australia
Really interesting WTs from wellbeback and Andrew! Still struggling to really understand how they are using the missing values in large cages but am enjoying learning from them. I used a completely different move to crack it which I think is much simpler, see step 8. Very routine after that if you don't have time to do the whole WT. [Many thanks for Andrew for some clarifications to improve this.]

A280 start
14 steps:
Preliminaries courtesy of SudokuSolver
Cage 16(2) n3 - cells ={79}
Cage 5(2) n56 - cells only uses 1234
Cage 5(2) n7 - cells only uses 1234
Cage 7(2) n1 - cells do not use 789
Cage 13(2) n1 - cells do not use 123
Cage 9(2) n3 - cells do not use 9
Cage 9(2) n7 - cells do not use 9
Cage 9(2) n9 - cells do not use 9
Cage 9(2) n7 - cells do not use 9
Cage 10(2) n6 - cells do not use 5
Cage 11(2) n9 - cells do not use 1
Cage 11(2) n1 - cells do not use 1
Cage 30(4) n125 - cells ={6789}

1. "45" on n69: 2 innies r58c7 - 1 = 1 outie r3c3
1a. -> no 1 in r5c7 nor r8c7 since the other innie can't equal the one outie (IOU: Andrew and wellbeback saw this as CCC)
1b. no 4 in r5c6

2. "45" on n369: 3 innies r258c7 = 8 = {125/134}(no 6,7,8,9)
2a. -> r2c7 = 1
2b. r58c7 = 7 = [25]/{34}(no 2 in r8c7)
2c. no 6 in r1c2
2d. no 8 in 9(2)n3
2e. no 8 in r9c8

3. "45" on n3: 1 remaining innie r3c7 = 6
3a. no 3 in 9(2)n3
3b. no 3 in r9c8
3c. no 7 in 13(2)n1

4. 16(2)n3 = {79} only: both locked for r1 and n3
4a. -> 9(2)n3 = {45} only: both locked for c8 and n3
4b. -> 13(3)n3 = {238} only: all locked for c9
4c. no 4,5 in r9c7
4d. no 6,7 in 11(2)n9
4e. no 6 in r5c9
4f. no 8,7,2 in r5c8
4g. no 2,4 in r2c1

5. "45" on c12: 3 innies r357c2 = 24 = {789} only: all locked for c2
5a. no 1,2 in r9c1
5b. r7c3 from (12)
5c. r3c3 = (45)

6. "45" on n12: 1 outie r4c4 - 3 = 2 innies r3c15
6a. ->max. r3c15 = 6 (no 7,8,9 in r3c15)

7. "45" on n1: 3 innies r12c3+r3c1 = 14
7a. but {158} blocked by {49} in 13(2)n1 and 7(2)n1 = {16/25/34}=[1/4/5]
7b. but {248} blocked by 13(2)n1 = [85/49] = [4/8]
7c. but {356} blocked by 7(2)n1 = {16/25/34} = [3/5/6]
7d. h14(3) = {149/167/239/257/347}(no 8)
7e. must have 7 or 9 which are only in r2c3 -> r2c3 = (79)

Key step
8. 7 & 9 in r3 are only in r3c246 but both 7 & 9 can't be in r3c24 since they both see r2c3 = (79)
8a. -> r3c6 must have one of 7 or 9 only, ie r3c6 = (79) (Hidden killer pair)

Much easier now.
9. Naked pair {45} in r3c38: both locked for r3

10. "45" on n12: r4c4 - 3 = r3c15 which must have 1 for r3
10a. -> r3c15 = {12/13} = 3,4 -> r4c4 = (67)
10b. can only have one of 2 or 3 in r3c15 -> r3c9 must have one of 2 or 3 (no 8) (Hidden killer pair)
[edit: Andrew suggested the alternative " "naked triple {789} in r3c246". I can't find my SS walkthrough file to confirm that. Thanks Andrew!]

11. 30(4)r2c3 = {6789} only
11a. 6 & 8 locked for c4
11b. 8 locked for n2
11c. no 9 in r2c56 since they see all 9 in 30(4)

12. 7 in n1 only in r2: locked for r2

13. r2c7 = 1 -> r12c6 = 11 = {56} only: both locked for n2 and c6
13a. r4c4 = 6
13b. -> r3c15 = 3 = {12} only (IODn12=-3): both locked for r3
13c. r3c9 = 3

14. 3 & 4 in n2 only in 21(5)r1c3 = {13467/23457}(no 9)
14a. r3c6 = 7
14b. 21(5) must have 5 or 6 which are only in r1c3 -> r1c3 = (56)
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