SudokuSolver Forum

Sorry to do that (actually, no, I'm not ), but I have a SOK - a Special Occasion Killer. I know, nobody has yet volunteered for A110 so there should be no other killer - but, just as the title says, it's a special occasion.

Yesterday evening there was the first semi-final in the European Soccer Championship, Germany vs. Turkey. I wonder if you can deduce the final score from this killer.

BTW - how often can you find the score?

Rating by SSolver(3.2.0): 0.81 - sorry, not hard enough for a Messy One.

SOK 1



Code:

Solution:

Have fun!

Cheers,
Nasenbaer

There is another version available: click here. I think it's easier, but DON'T try to rate it with SSolver(3.2.0)! (Possible bug, see the post)

Cheers,
Nasenbaer

Even if we play bad, we still win (apart from the game vs. Croatia).

To celebrate our victory, I'll post my shortest wt so far (again and again ...). By the way, it was ok to post a non-Assassin Killer or a variant since it's not Friday. If it had been Friday and A110 hadn't been claimed at that point, then you wouldn't have been allowed to post a variant.

BTW, I counted the score 9 times. I counted the cage sums and cells that were next to each other (also diagonally).

SOK 1 Walkthrough:

1. R123
a) Outies R12 = 15(2) = [69] -> R3C4 = 6, R3C8 = 9
b) 9(3) = {126} -> 1,2 locked for R2
c) 18(3) = 9{36/45}
d) 15(3) = {168/258/267} <> 3,4 because other combos blocked by Killer pairs (34,35,56) of 18(3)
e) 32(5) = 89{267/357/456} -> 8,9 locked for R1

2. C789
a) Outies C9 = 6(3) = {123} locked for C8, 3 locked for N9
b) 3 locked in 16(4) @ N9 = 3{148/157/247/256} <> 9
c) 9 locked in 20(5) @ C9 = {12359} locked for C9
d) 15(3) = 6{18/27} -> 6 locked for C9+N1
e) 18(3) = {459} -> 4,5 locked for R2+N3

3. C123
a) 12(2) = {39} locked for R2+N2
b) Hidden Single: R1C3 = 9 @ N1
c) 32(5) = {35789} -> R1C7 = 3; {578} locked for R1+N2
d) R4C3 = 7, R4C2 = 9

4. N456
a) 9 locked in 45(9) @ N1 for C4+N5
b) 14(2) = {68} locked for R5+N5
c) 9(2) = {45} -> R3C5 = 4, R4C5 = 5
d) 6 locked in 45(9) @ N1 for R7+N7
e) Hidden Single: R4C8 = 6 @ 45(9) @ N2
f) 13(2) = {58} -> R4C1 = 8, R3C1 = 5
g) 15(2) = {78} locked for R6+N6

5. N78
a) 17(3) = {179} locked for C1+N7
b) 10(2) = {28} locked for C2+N7
c) 12(4) = {1236} because 4,5 only possible @ R7C4 -> R8C5 = 6

6. N14
a) 19(4) = {2467} -> R2C1 = 6, R2C2 = 7, R1C2 = 4, R1C1 = 2
b) R2C3 = 1, R3C2 = 3
c) 3,6 locked in 14(4) @ N4 = {1346} -> R6C2 = 1, R6C3 = 6; 4 locked for N4

7. Rest is singles.

Rating: (Easy) 1.0. I used Killer pairs once to eliminate some combos and thereby candidates.

Congratulations to all of you for Germany winning the semi-final yesterday.
Let's hope Germany will play a killer match on Sunday, so that we will not be but have to

On that basis SOK1 couldn't be a Messy One because Nasenbaer has already posted one.

I don't know if Ed is still planning to have a poll. It's a good idea but one problem is that it's been quite a long time since the first Messy One was posted. Since it doesn't look as if there are many people creating them (sorry Ed! ) maybe it's time for the three people who have already posted them to post their second/third/... ones?

It was a good final. I was supporting Germany but on the day Spain were the better team.

Now I've done 3 Nasenbaer puzzles in a row!

A fun puzzle and an easy one. I'll also rate it Easy 1.0.

I think my walkthrough is sufficiently different from Afmob's to post. It's always good to have at least two posted walkthroughs unless they would be too similar.

Here is my walkthrough for SOK 1.

Prelims

a) R2C56 = {39/48/57}, no 1,2,6
b) R34C1 = {49/58/67}, no 1,2,3
c) R34C5 = {18/27/36/45}, no 9
d) R4C23 = {79}, locked for R4 and N4, clean-up: no 4,6 in R3C1, no 2 in R3C5
e) R4C67 = {15/24}
f) R5C56 = {59/68}
g) R6C78 = {69/78}
h) R89C2 = {19/28/37/46}, no 5
i) R8C34 = {17/26/35}, no 4,8,9
j) R8C67 = {17/26/35}, no 4,8,9
k) 9(3) cage at R2C3 = {126/135/234}, no 7,8,9
l) 12(4) cage at R6C5 = {1236/1245}, no 7,8,9, CPE no 1,2 in R9C5
m) R1C34567 = {26789/35789/45689}, no 1, 8,9 locked for R1
n) R9C34567 = {26789/35789/45689}, no 1, 8,9 locked for R9, clean-up: no 1,2 in R8C2

1. 45 rule on R1 3 outies R2C129 = 21 = {489/579/678}, no 1,2,3
1a. Killer triple 7,8,9 in R2C129 and R2C56, locked for R2

2. 45 rule on R2 2 outies R3C48 = 15 = [69], clean-up: no 4 in R4C1, no 3 in R4C5, no 6 in R6C7, no 2 in R8C3
2a. R3C4 = 6 -> R2C34 = 3 = {12}, locked for R2
2b. R4C8 = 9 -> R2C78 = 9 = {36/45}

3. 45 rule on R89 1 outie R7C1 = 1 innie R8C5 + 1, no 1,8,9 in R7C1

4. 45 rule on C9 3 outies R189C8 = 6 = {123}, locked for C8, clean-up: no 6 in R2C7 (step 2b)
4a. Killer triple 1,2,3 in R8C34, R8C67 and R8C8, locked for R8, clean-up: no 7 in R9C2

5. 15(3) cage at R1C8 = {168/258/267/348/357} (cannot be {456} which clash with R2C8)
[Oops! I missed that {348/357} also clash with R2C78. This would have simplified the next few steps. However I’ve left the walkthrough as I did it because I got there fairly quickly with step 7a.]
5a. R1C8 = {123} -> no 1,2,3 in R1C9
5b. 8 of {348} must be in R2C9 -> no 4 in R2C9

6. 16(4) cage in N9 = {1249/1258/1267/1348/1357/2347/2356} (cannot be {1456} because R89C8 must contain two of 1,2,3)
6a. R89C8 = {123} -> no 1,2,3 in R9C9
6b. 45 rule on C9 4 innies R1289C9 = 25 = {4579/4678}, 4,7 locked for C9

7. 45 rule on N3 3 innies R1C7 + R3C79 = 12 = {138/147/237/246} (cannot be {156/345} which clash with R2C78), no 5
7a. Hidden killer pair 1,2 in R1C7 + R3C79 and R1C8 for N3 -> R1C8 = {12}
7b. 15(3) cage at R1C8 (step 5) = {168/258/267}, no 4
7c. R1C7 + R3C79 (step 7) = {138/147/237} (cannot be {246} which clashes with 15(3) cage), no 6

8. 3 in C8 locked in R89C8, locked for N9, clean-up: no 5 in R8C6
8a. 16(4) cage in N9 (step 6) = {1348/1357/2347/2356}, no 9

9. 1,2,3,9 in C9 locked in R34567C9 = {12359} (only remaining combination), locked for C9

10. 15(3) cage at R1C8 (step 7b) = {168/267}, 6 locked for C9 and N3, clean-up: no 3 in R2C7 (step 2b)
10a. 16(4) cage in N9 (step 8a) = {1348/2347}, 4 locked for N9

11. Naked pair {45} in R2C78, locked for R2 and N3, clean-up: no 7,8 in R2C56
11a. Naked pair {39} in R2C56, locked for R2 and N2, clean-up: no 6 in R4C5
11b. R1C3 = 9 (hidden single in R1), R4C23 = [97], clean-up: no 1 in R8C4, no 1 in R9C2
11c. R8C1 = 9 (hidden single in C1), R79C1 = 8 = {26/35}/[71], no 4, no 7 in R9C1

12. 9 in 45(9) cage at R3C2 locked in R56C4, locked for C4 and N5, clean-up: no 5 in R5C56
12a. Naked pair {68} in R5C56, locked for R5 and N5, clean-up: no 1 in R3C5
12b. Naked quint {12345} in R4C45679, locked for R4, clean-up: no 8 in R3C1

13. 45 rule on R1 4 innies R1C1289 = 13 = {1237/1246} (cannot be {1345} because R1C9 only contains 6,7), no 5
13a. R1C9 = {67} -> no 6,7 in R1C12
13b. R1C9 = 6 (hidden single in R1)
13c. 5 in R1 locked in R1C456, locked for N2, clean-up: no 4 in R4C5

14. 45 rule on R9 4 innies R9C1289 = 13 = {1237/1246/1345}
14a. R9C9 = {47} -> no 4 in R9C2, clean-up: no 6 in R8C2

15. 45 rule on N4 3 innies R4C1 + R5C23 = 15 = {258/348/456} (cannot be {168} because 6,8 only in R4C1), no 1 in R5C23

16. 6 in 45(9) cage at R3C2 locked in R7C23, locked for R7 and N7, clean-up: no 2 in R79C1 (step 11c) , no 4 in R8C2, no 2 in R8C4

17. R4C8 = 6 (hidden single in 45(9) cage at R3C6), R4C1 = 8, R3C1 = 5, clean-up: no 9 in R6C7, no 3 in R79C1 (step 11c)
17a. R79C1 = [71], R2C1 = 6, R8C2 = 8, R9C2 = 2, R2C2 = 7, R2C9 = 8, R1C8 = 1 (step 10), R89C8 = [23], R9C9 = 7 (step 14), R8C9 = 4, clean-up: no 7 in R8C4, no 1,6 in R8C6, no 6 in R8C7

18. Naked pair {35} in R8C34, locked for R8 -> R8C567 = [671], R5C56 = [86], clean-up: no 1 in R4C5, no 5 in R4C6

19. R2C12 = [67] = 13 -> R1C12 = 6 = [24]

and the rest is naked singles

It looks like a lot of the early steps can also be used in the variant which splits R8C67 and adds those cells to adjacent larger cages. If I do solve it, I'll tidy up step 5 and simplify the steps immediately following.

Many thanks for the variant! It was fun and I found it a lot easier that SOK1, which was also fairly easy. I've made some comments about the SS solution at the end.

This puzzle was very unusual because one would normally expect that splitting a 2-cell cage and adding the cells to larger cages would make it harder. In this case it made it easier!

I'll rate SOK1 Variant as 0.75. Maybe it ought to be even lower but I don't know whether step 3 would normally appear in a newspaper "deadly" so I think 0.5 would be too low; anyway I'm not sure if it would be so easy without elimination solving, which is how newspaper puzzles are normally solved.

Here is my walkthrough, probably the shortest I've posted since I started listing Preliminaries first and using clean-up routinely. I've started with the same first two steps as for SOK1 but then step 3, the key breakthrough, is more powerful than for SOK1 which is why the Variant is a lot easier. Step 1 may not be necessary but it does simplify R2.

Prelims

a) R2C56 = {39/48/57}, no 1,2,6
b) R34C1 = {49/58/67}, no 1,2,3
c) R34C5 = {18/27/36/45}, no 9
d) R4C23 = {79}, locked for R4 and N4, clean-up: no 4,6 in R3C1, no 2 in R3C5
e) R4C67 = {15/24}
f) R5C56 = {59/68}
g) R6C78 = {69/78}
h) R89C2 = {19/28/37/46}, no 5
i) R8C34 = {17/26/35}, no 4,8,9
j) 9(3) cage at R2C3 = {126/135/234}, no 7,8,9
k) 19(4) cage at R6C5 = {12349/12358/12368/12457/13456}, CPE no 1 in R9C5
l) R1C34567 = {26789/35789/45689}, no 1, 8,9 locked for R1
m) 17(5) cage in N9 = {12347/12356}, no 8,9, 1,2,3 locked for N9
n) R9C34567 = {26789/35789/45689}, no 1, 8,9 locked for R9, clean-up: no 1,2 in R8C2

1. 45 rule on R1 3 outies R2C129 = 21 = {489/579/678}, no 1,2,3
1a. Killer triple 7,8,9 in R2C129 and R2C56, locked for R2

2. 45 rule on R2 2 outies R3C48 = 15 = [69], clean-up: no 4 in R4C1, no 3 in R4C5, no 6 in R6C7, no 2 in R8C3
2a. R3C4 = 6 -> R2C34 = 3 = {12}, locked for R2
2b. R4C8 = 9 -> R2C78 = 9 = {36/45}

3. 45 rule on C9 4 (3+1) outies R189C8 + R8C7 = 7 = {123} + 1 -> R8C7 = 1, R1C8 = 1, R89C8 = {23}, locked for C8 and N9, clean-up: no 6 in R2C7 (step 2b), no 5 in R4C6, no 7 in R8C34
3a. Killer pair 2,3 in R8C34 and R8C8, locked for R8, clean-up: no 7 in R9C2

4. 15(3) cage at R1C8 = {168} (only remaining combination) -> R12C9 = [68], clean-up: no 4 in R2C56, no 3 in R2C7 (step 2b)

5. R8C78 + R9C8 = 6 -> R89C9 = 11 = {47}, only remaining combination, locked for C9 and N9

6. Naked pair {45} in R2C78, locked for R2 and N3, clean-up: no 7 in R2C56
6a. Naked pair {39} in R2C56, locked for R2 and N2, clean-up: no 6 in R4C5
6b. Naked pair {67} in R2C12, locked for N1, clean-up: no 6 in R4C1
6c. Naked pair {58} in R34C1, locked for C1
6d. 7 in N3 locked in R13C7, locked for C7, clean-up: no 8 in R6C8

7. R1C3 = 9 (hidden single in R1), R4C23 = [97], clean-up: no 1 in R9C2
7a. 9 in C1 locked in R78C1
7b. R789C1 = {179/269}, no 3,4

8. 9 in 45(9) cage at R3C2 locked in R56C4, locked for C4 and N5, clean-up: no 5 in R5C56
8a. Naked pair {68} in R5C56, locked for R5 and N5, clean-up: no 1 in R3C5
8b. Naked quint {12345} in R4C45679, locked for R4 -> R34C1 = [58], R4C8 = 6, R6C8 = 7, R6C7 = 8, clean-up: no 4 in R4C5

9. R2C12 = {67} = 13 -> R1C12 = 6 = {24}, locked for R1 and N1 -> R2C34 = [12], clean-up: no 6 in R8C3
9a. Naked pair {35} in R8C34, locked for R8 -> R89C8 = [23], clean-up: no 7 in R8C2

10. R1C7 = 3 (hidden single in R1), R3C79 = [72], clean-up: no 2 in R4C5
10a. R7C8 = 8 (hidden single in C8)

11. Naked pair {38} in R3C23, locked for R3 and 45(9) cage -> R3C56 = [41], R4C5 = 5
11a. Naked pair {24} in R4C67, locked for R4 -> R4C4 = 1, R4C9 = 3
11b. Naked pair {59} in R7C79, locked for R7 and N9 -> R9C7 = 6, clean-up: no 4 in R8C2

12. 1 in N7 locked in R79C1, locked for C1
12a. R789C1 (step 7b) = {179} (only remaining combination) -> R8C1 = 9, R79C1 = {17}, locked for C1 and N7 -> R2C12 = [67]
12b. R5C1 = 3, R5C4 = 7, R5C9 = 1 (hidden singles in R5), then R5C7 = 9 (hidden single in R5), R67C9 = [59], R7C7 = 5, R2C78 = [45], R4C67 = [42], R5C8 = 4, R6C4 = 9

13. Naked pair {25} in R5C23, locked for N4 and 45(9) cage -> R6C1 = 4

and the rest is naked singles


After solving this puzzle I had a look at the SS solution that Nasenbaer posted. There are two main things that I'll comment on.

First, SS didn't find my step 3 which I find rather surprising. It comes immediately from using mental arithmetic or Ruud's combination calculator for 7(3+1).

Second Nasenbaer comments on steps 21 and 26 of the SS solution. Richard has already said that he'll fix those steps in the next version of SS.

Superficially those steps appear correct if one takes LOL at face value. However on checking the 45 rule for these innies/outies the innies total 5 more than the outies so LOL cannot be used. The only time I've successfully used LOL in a killer was for one of Ruud's Assassins where there was a 45(9) cage overlapping with N5 and then it was clear that the innies had to equal the outies and to contain the same numbers. LOL is usually fairly easy to use for normal Jigsaws except when the same number can appear twice in the innies and outies. However one clearly has to be careful when using LOL for Killers and Jigsaw Killers.