YAK 96 UA 96

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Jean-Christophe
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YAK 96 UA 96

Post by Jean-Christophe »

Here is Yet Another Killer #96

YAK 96 V1, rating "much too easy for regular members"

Image

3x3::k:7683:7683:7683:2833:2833:2833:8708:8708:8708:7683:7683:3353:3353:3338:2066:2066:8708:8708:7683:2581:4615:4615:3338:4872:4872:1046:8708:3343:2581:4615:4615:7937:4872:4872:1046:4112:3343:1804:1804:7937:7937:7937:3597:3597:4112:3343:2583:4617:4617:7937:4870:4870:3352:4112:8453:2583:4617:4617:3595:4870:4870:3352:6146:8453:8453:2835:2835:3595:3092:3092:6146:6146:8453:8453:8453:2574:2574:2574:6146:6146:6146:

YAK 96 V2: rating "not sure it's hard enough for regular members"

Image

3x3::k:8707:8707:8707:3345:3345:3345:7684:7684:7684:8707:8707:2329:2329:2826:3602:3602:7684:7684:8707:1813:4359:4359:2826:3848:3848:2070:7684:4111:1813:4359:4359:7681:3848:3848:2070:2320:4111:3340:3340:7681:7681:7681:3853:3853:2320:4111:1559:5129:5129:7681:6662:6662:2840:2320:7429:1559:5129:5129:2571:6662:6662:2840:7170:7429:7429:1811:1811:2571:2836:2836:7170:7170:7429:7429:7429:4110:4110:4110:7170:7170:7170:

SudokuSolver v3.0 scores:
V1: 0.90
V2: 1.79
JSudoku cannot solve V2
Last edited by Jean-Christophe on Sun Mar 30, 2008 10:49 am, edited 3 times in total.
Afmob
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Post by Afmob »

I think V1 is neither "reasonably hard" nor tricky. It can be solved using a (very) basic Killer technique.

JC, wouldn't it be better if you called YAK 96 UA 96 since no one else posted an UA so far and it would be a bit weird if there is a UA 95, but no UA 96 (the same goes for the YAK). Maybe we should sort out before every friday who is going to post the UA for the following week. Of course this doesn't include the Maverick Series since it doesn't appear weekly.

By the way, I broke my personal record for shortest wt again (sorry UA 93 :cry:).

UA 96 Walkthrough:

1. R5+C5
a) 13(2) <> {58} since it's a Killer pair of 14(2) @ C5
b) Killer pair (69) locked in 13(2) + 14(2) for C5
c) 31(5) = 9{1678/2578/3478/3568/4567} -> 9 locked for R5+N5
d) 14(2) @ R5 = {68} locked for R5+N6
e) 31(5) = 789{25/34} -> 7 locked for N5, 8 locked for C5+N5
f) 14(2) @ C5 = {59} locked for C5+N8
g) 13(2) = {67} locked for C5+N2
h) 7(2) <> 1
i) 16(3) = {259/349/457} <> 1
j) Hidden Single: R5C1 = 1 @ R5
j) 10(3) = 1{27/36} -> 1 locked for R9+N8
k) 11(3) = 2{18/45} -> 2 locked for R1+N2

2. C789
a) 8(2) = [17/35/53]
b) Innies N3 = 11(3) = {137} because R2C7+R3C8 = (1357)
-> {137} locked for N3, 7 locked for C7, 1 locked for R3
c) 13(2): R7C8 <> 5,7
d) Innies N9 = 21(3) = {489} locked for N9
e) 12(2), 13(2) <> 7
f) 7 locked in 16(3) = {457} locked for C9+N6
g) 13(2) = [94] -> R6C8 = 9, R7C8 = 4

3. R123
a) Hidden Single: R1C7 = 4 @ N3
b) 11(2) = {128} locked for R1+N2
c) 8(2) = {35} -> R2C6 = 5, R2C7 = 3
d) 3 locked in R3C46 for R3
e) R3C8 = 1, R3C7 = 7
f) 13(2) = {49} locked for R2
g) 10(2) = {28/46}
h) 19(4) = 27{19/46} -> 2 locked for R4
i) R3C5 = 6
j) Innies N1 = 15(3) = {249} because R2C3 = (49)
-> {249} locked for N1, 2 locked R3, 9 locked for C3

4. N47
a) R4C1 = 9 @ R4 -> 13(3) = [913] -> R6C1 = 3
b) 7(2) = {25} locked for R5+N4
c) 2 locked in R789C1 for N7
d) 10(2) @ N7 = [46/73]

5. R789
a) 4 locked in R8C46 for R8
b) 2,4,9 locked in 33(6) = 2479{38/56} -> 7 locked for N7
c) 11(2) <> {38} since it's a Killer pair of 12(2)
d) 11(2) = {56} -> R8C3 = 5, R8C4 = 6
e) 10(3) = {127} locked for R9+N8
f) 19(4) = 6{139/148/238} -> R6C6 = 6
g) Hidden Single: R7C3 = 1 @ C3

6. Rest is singles.

Rating: 0.75-1.0. I used a Killer pair to crack it.

I am not sure about the rating since I've never rated such an easy puzzle. If using a Killer pair automatically means a rating of at least 1.0 then it's an easy 1.0 Killer otherwise a 0.75.
Last edited by Afmob on Sun Mar 30, 2008 1:30 pm, edited 1 time in total.
Jean-Christophe
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Post by Jean-Christophe »

Hum, I didn't expected a WT nor a rating for this V1. :?

I know the regular members here usually eat "monsters" and this V1 could look very easy in comparison. But on other forums, most will find it "just the right difficulty".

I personally would not qualify "(very) basic" such techniques as killer naked pair or combination conflicts like you used. Also consider the number of In/outies with three cells in this puzzle, most useless at first.

Anyhow, thanks for the comments
sudokuEd
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Post by sudokuEd »

Jean-Christophe wrote:Hum, I didn't expected a WT nor a rating for this V1. :?
Why not? Every puzzle posted here gets a walk-through and a rating. I am very surprised and a bit shocked you would not expect this.
Jean-Christophe wrote:YAK 96 V1, rating "reasonably hard" yet tricky
Is this how you found it when you solved it? From your reaction to Afmob calling it "neither" I get the feeling you didn't find it "hard" nor "tricky". So why rate it like that? It seems you are rating this puzzle based on how some other forum might find it. I don't like that.
Afmob wrote:JC, wouldn't it be better if you called YAK 96 UA 96
I think this is a really sensible question given the way this forum is based around the Assassin killers. I am offended that you completely ignored this suggestion. [edit: Thanks to J-C for changing the thread title. Appreciated.]

Ed
Last edited by sudokuEd on Sat Mar 29, 2008 8:55 pm, edited 1 time in total.
Andrew
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Post by Andrew »

Jean-Christophe wrote:I know the regular members here usually eat "monsters" and this V1 could look very easy in comparison. But on other forums, most will find it "just the right difficulty".
You are right that on other forums this would be "just the right difficulty". I started doing killers on www.sudoku.org.uk, where I still do the killers without using elimination solving, and I know that a lot of forum members there find Assassins too hard.

Using elimination solving, YAK96 was an easy puzzle. I expect it could still be solved fairly easily without using elimination solving but I won't bother to do that.
Afmob wrote:I am not sure about the rating since I've never rated such an easy puzzle. If using a Killer pair automatically means a rating of at least 1.0 then it's an easy 1.0 Killer otherwise a 0.75.
I suppose I could say that it depends which type of killer pair. Afmob uses the term killer pair in three different ways in other walkthroughs. In this case it is the "classic" killer pair which can be spotted even without using elimination solving. The other two types, hidden killer pairs and clashes with killer pairs in other cages, are more difficult moves.

This puzzle only requires one killer pair and a few simple 45s so I'll rate it at 0.75.

Here is my walkthrough. It would have been shorter but I missed one simple 45 which Afmob used. Even so it's still one of my shortest walkthroughs.

Prelims

a) R2C34 = {49/58/67}, no 1,2,3
b) R23C5 = {49/58/67}, no 1,2,3
c) R2C67 = {17/26/35}, no 4,8,9
d) R34C2 = {19/28/37/46}, no 5
e) R34C8 = {13}, locked for C8
f) R5C23 = {16/25/34}, no 7,8,9
g) R5C78 = {59/68}
h) R67C2 = {19/28/37/46}, no 5
i) R67C8 = {49/58/67}, no 1,2,3
j) R78C5 = {59/68}
k) R8C34 = {29/38/47/56}, no 1
l) R8C67 = {39/48/57}, no 1,2,6
m) R1C456 = {128/137/146/236/245}, no 9
n) R9C456 = {127/136/145/235}, no 8,9
o) 34(6) cage in N3 = {136789/145789/235789/245689/345679}, 9 locked for N3
p) 31(5) cage in N5 = {16789/25789/34789/35689/45679}, 9 locked for N5
[I was careless with the Prelims. I should also have spotted 24(6) cage in N9 = {123459/123468/123567}, 1,2,3 locked for N9 when I would have also spotted 45 rule on N9 3 innies R7C78 + R8C7 = 21 = {489/579/678}.]

1. 45 rule on N3 3 innies R2C7 + R3C78 = {128/137/146/236} (cannot be {245} because R3C8 only contains 1,3), no 5, clean-up: no 3 in R2C6

2. R23C5 = {49/67} (cannot be {58} which clashes with R78C5), no 5,8

3. Killer pair 6,9 in R23C5 and R78C5, locked for C5

4. 9 in N5 locked in R5C46, locked for R5, clean-up: no 5 in R5C78
4a. Naked pair {68} in R5C78, locked for R5 and N6, clean-up: no 1 in R5C23, no 5,7 in R7C8

5. R456C9 = {259/349/457}, no 1

6. 45 rule on N6 4 innies R46C78 = {1239/1257/1347}
6a. 9 of {1239} must be in R6C8 = 9 -> no 9 in R46C7

7. 31(5) cage in N5 = {25789/34789} (only remaining combinations), no 1, 7,8 locked for N5, 8 locked for C5, clean-up: no 6 in R78C5

8. R5C1 = 1 (hidden single in R5), clean-up: no 9 in R3C2, no 9 in R7C2
8a. R46C1 = 12 = {39/48/57}, no 2,6

9. Naked pair {59} in R78C5, locked for C5 and N8, clean-up: no 4 in R23C5, no 2,6 in R8C3, no 3,7 in R8C7

10. Naked pair {67} in R23C5, locked for C5 and N2, clean-up: no 6,7 in R2C3, no 1,2 in R2C7
10a. 7 in N5 locked in R5C46, locked for R5

11. R5C46 = {79} (hidden pair in N5), locked for R5
11a. 31(5) cage in N5 = {34789} (only remaining combination), no 2, 3,4 locked for C5 and N5

12. R5C23 = {25} (cannot be {34} which clashes with R5C5), locked for R5 and N4, clean-up: no 8 in R3C2, no 7 in R46C1 (step 8a), no 8 in R7C2

13. R456C9 (step 5) = {349/457} (cannot be {259} because R5C9 only contains 3,4), no 2, 4 locked for C9 and N6, clean-up: no 9 in R7C8
13a. 2 in N6 locked in R46C7, locked for C7

14. R1C456 = {128/245}, no 3, 2 locked for R1 and N2, clean-up: no 6 in R2C7

15. 3 in N2 locked in R3C46, locked for R3 -> R34C8 = [13], R5C9 = 4, R5C5 = 3, clean-up: no 7 in R3C2, no 7,9 in R4C2, no 9 in R46C9 (step 13), no 9 in R6C1 (step 8a)

16. Naked pair {57} in R46C9, locked for C9 and N6, R6C8 = 9, R7C8 = 4, clean-up: no 6 in R6C2, no 1 in R7C2, no 8 in R8C6
16a. 1 in N6 locked in R46C7, locked for C7

17. R9C456 = {127/136}, no 4, 1 locked for R9 and N8

18. 4 in N8 locked in R8C46, locked for R8, clean-up: no 7 in R8C4

19. 45 rule on R12 3 outies R3C159 = 19 = {469/478/568} (cannot be {289} because R3C5 only contains 6,7), no 2
19a. 4,5 only in R3C1 -> R3C1 = {45}
19b. 8,9 only in R3C9 -> R3C9 = {89}
19c. 2 in R3 locked in R3C23, locked for N1

20. 45 rule on R1234 3 outies R4C159 = 20 = {479} (cannot be {578} because 5,7 only in R4C9) -> R4C159 = [947], R6C159 = [385], clean-up: no 6 in R4C2, no 7 in R7C2

21. Naked pair {68} in R4C23, locked for N4, clean-up: no 2 in R7C2

22. R2C7 + R3C78 (step 1) = {137} -> R23C7 = [37], R2C6 = 5, R23C5 = [76], clean-up: no 4 in R1C456 (step 14), no 8 in R2C34

Now for those who find it acceptable, there’s R6C6 = 6 to prevent UR but since I prefer to demonstrate that a puzzle has a unique solution I won’t use it. I think it’s a UR in this case because the remaining cells in the 19(4) cages at R3C6 and R6C6 can’t be used to determine the placing of 1,2 in R4C67 and R6C67.

23. R4C4 = 5 (hidden single in R4)

24. R3C6 = 9 (cage sum), R2C34 = [94], R5C46 = [97], clean-up: no 4 in R3C1 (step 19), no 7 in R8C3, no 2 in R8C4, no 5 in R8C7
24a. R3C19 = [58], R3C4 = 3, clean-up: no 8 in R8C3

25. Naked pair {26} in R2C89, locked for R2 and N3 -> R1C789 = [459], R2C12 = [81]

26. R8C6 = 4 (hidden single in C6), R8C7 = 8, R5C78 = [68], R8C4 = 6, R8C3 = 5, R78C5 = [59], R79C7 = [95], R5C23 = [52], R3C23 = [24], R4C2 = 8, R4C3 = 6, R6C3 = 7, R6C2 = 4, R7C2 = 6, R1C123 = [673], R89C2 = [39], R79C3 = [18], R67C4 = [28], R6C67 = [61], R4C67 = [12], R1C456 = [128], R9C45 = [71], R9C6 = 2 (step 17)

and the rest is naked singles

6 7 3 1 2 8 4 5 9
8 1 9 4 7 5 3 2 6
5 2 4 3 6 9 7 1 8
9 8 6 5 4 1 2 3 7
1 5 2 9 3 7 6 8 4
3 4 7 2 8 6 1 9 5
7 6 1 8 5 3 9 4 2
2 3 5 6 9 4 8 7 1
4 9 8 7 1 2 5 6 3
Last edited by Andrew on Thu Apr 03, 2008 10:42 pm, edited 1 time in total.
mhparker
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Post by mhparker »

Andrew wrote:Using elimination solving, YAK96 was an easy puzzle. I expect it could still be solved fairly easily without using elimination solving but I won't bother to do that.
I just did it on paper, and found the puzzle quite enjoyable, so thanks Jean-Christophe for posting it!

I can understand Jean-Christophe being unfamiliar with our rating system. However, one acceptable alternative for this would be to simply publish the SudokuSolver score instead.
Cheers,
Mike
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Post by Afmob »

UA 96 V2 was quite challenging and interesting at the same time. It had a regular X-Wing which you don't find that often in Killers. Thanks JC!

Compared to ND 9 my moves wheren't much more complicated but it took me some time to find them where as ND 9 flowed quite well. That's why I rate UA 96 V2 higher.

UA 96 V2 Walkthrough:

1. R456
a) 13(2) <> {67} since it's a Killer pair of 15(2)
b) Killer pair (89) locked in 13(2) + 15(2) for R5
c) 7(2) <> {25} because it's a Killer pair of 6(2)
d) Killer pair (14) locked in 7(2) + 6(2) for C2
e) 13(2): R5C3 <> 9
f) Innies R1234 = 22(3) = 9{58/67} -> 9 locked for R4; R4C15 <> 5,6
g) 9(3) = 1{26/35} because R4C9 = (56) -> 1 locked for C9+N6; R56C9 <> 5,6
h) 8(2): R3C8 <> 7
i) Innies R6789 = 16(3): R6C15 <> 1,2,3 because R6C9 <= 3

2. R123
a) Outies R12 = 21(3) <> 1,2,3
b) 11(2): R2C5 <> 8,9
c) Innies N1 = 11(3) <> 9
d) Innies N3 = 15(3): R3C7 <> 3 because 4,7 only possible there

3. C4789
a) Innies C1234 = 22(3) = 9{58/67} -> 9 locked for C4
b) Innies C1234 = 22(3): R19C4 <> 5 because R5C4 <> 8,9
c) 13(3): R1C56 <> 6,7,8,9 because R1C4 >= 6
d) Outies C89 = 11(3) <> 9; R19C8 <> 5,6,7,8
e) 15(2): R5C8 <> 6

5. R5+C5
a) Hidden Killer triple (123) in R5C19 + 30(5) for R5 -> R5C1 = (123); 30(5) <> {45678}
b) 30(5) = 9{1578/2478/2568/3468} -> 9 locked for C5+N5
c) 11(2) <> 2
d) 10(2) <> 1

6. R789
a) Innies N7 = 16(3): Hidden Killer triple (789) in R7C3 -> R7C3 = (789)
b) Innies N7 = 16(3): R8C3 <> 1 because R7C23 <= 14
c) 7(2): R8C4 <> 6
d) Outies R89 = 11(3) <> 9
e) Outies R89 = 11(3): R7C1 <> 7,8 because R7C59 >= 5

7. C5+R5 !
a) Innies+Outies R5: 13 = R46C5 - R5C19 -> R6C5 <> 4,5,6
b) Killer quad (6789) locked in 11(2) + 10(2) + R46C5 for C5
c) 13(2) + 15(2) = h28(4) = 89{47/56}
d) ! 30(5) <> 1 since {15789} blocked by Killer pair (57) of h28(4)
e) 6 in C5 must be in 11(2) xor 10(2) -> 11(2) + 10(2) = h21(4) = 6{258/348/357}
-> 11(2) <> {47}
f) 30(5): R5C6 <> 7 because R5C4 <> 2,4

8. N123 !
a) 9(2) <> {36} since R2C5 + 14(2) must have two of (356)
b) ! Innies N2 = 21(4): R3C6 <> 7 because:
- <> 7{158/239/248/356} because if R3C6 = 7 -> Outies R12 = 21(3) = {489} -> 11(2) = [38]
- <> {1479} since only placement is [1947] (since 14(2) = [95] -> 9(2) <> 4)
-> blocked by Outies R12 = 21(3) = {489}
c) 7 locked in R123C4 for C4

9. R456
a) 7 locked in 15(2) = {78} locked for R5+N6
b) 13(2) = {49} -> R5C2 = 9, R5C3 = 4
c) 30(5) = 569{28/37} -> 5,6 locked for N5
d) 8(2) <> 1
e) 7(2): R3C2 <> 3
f) 6(2): R7C2 <> 2

10. C456
a) Innies C1234 = 22(3): R19C4 <> 6 because R5C4 = (56)
b) 9(2): R2C4 <> 5
c) 7(2): R8C4 <> 3

11. C1289+C57 !
a) Outies C89 = 11(3) = 1{28/37} -> 1 locked for C7
b) Outies C12 = 15(3) = 4{29/38/56} <> 1,7
c) ! X-Wing (1) in R19C57 -> 1 locked for R19

12. N13
a) Innies N1 = 11(3): R3C3 <> 7 because 3 only possible there, R3C3 <> 1 because R2C3 <> 4,6
b) Innies N3 = 15(3): R3C7 <> 6,9 because (147) only possible there

13. R789 !
a) ! Consider placement of 1 in R9 -> R7C2 <> 1:
- i) R9C5 = 1 -> 7(2): R8C3 <> 6 -> Innies N7 = 16(3) <> 1
- ii) R9C7 = 1 -> R2C8 = 1 (HS @ N3) -> R3C2 = 1 (HS @ N1) -> R7C2 <> 1
b) 1 locked in 29(6) for C1
c) 6(2): R6C2 <> 5

14. N12
a) 1 locked in Innies N1 = 11(3) = 1{28/37/46} <> 5
b) Innies N1 = 11(3): R3C2 <> 6 because 4 only possible there
c) 7(2): R4C2 <> 1
d) 16(3) = 8[26/35] because R4C2 = (36) blocks {367} -> R4C1 = 8

15. R456
a) Innies R1234 = 22(3) = {589} -> R4C5 = 9, R4C9 = 5
b) 9(3) = {135} locked for C9+N6
c) 8(2) = {26} locked for C8
d) 11(2) = {47} -> R6C8 = 4, R7C8 = 7

16. C789
a) Innies N9 = 17(3) = 7{28/46}
b) 11(2): R8C6 <> 2,4,6,8
c) 5 locked in Innies N3 = 15(3) = 5{28/46} for N3; R23C7 <> 2,6 because R3C8 = (26)
d) Hidden Single: R6C7 = 9 @ N6
e) Naked pair (26) locked in R4C78 for R4

17. N47
a) R4C2 = 3 -> R3C2 = 4, R5C1 = 2 -> R6C1 = 6, R6C2 = 1, R7C2 = 5
b) Hidden Single: R7C3 = 9 @ R7
c) Innies N7 = 16(3) = {259} -> R8C3 = 2
d) 7(2) @ N7 = {25} -> R8C4 = 5
e) R4C3 = 7, R6C3 = 5
f) 20(4) = {2459} -> R6C4 = 2, R7C4 = 4

18. Rest is singles.

Rating: 1.75. I used combo analysis and a forcing chain.
Last edited by Afmob on Wed Apr 02, 2008 3:23 pm, edited 2 times in total.
Jean-Christophe
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Post by Jean-Christophe »

Thanks for the WT, I'll study it.

Here are the "tricky" moves which could help simplifying the WT. Indeed I designed both versions around these:
Triple click to read what I wrote:For V2:
In/Outies @ r5 -> r46c5-r5c19 = 13
Max r46c5 = 17 -> Max r5c19 = 17-13 = 4
Min r5c19 = 3 -> Min r46c5 = 13+3 = 16
-> r46c5 = 16|17 = {9(7|8)} -> 9 locked for n5, c5
r5c19 = 3|4 = {1(2|3)} -> 1 locked for r5

Overlaps @ r5&c5 -> r159c5+r5c19 = 11 = {11225|11234|12233} = {12(125|134|233)}
-> r19c5 = {1234}, r5c5 = {2345}
Last edited by Jean-Christophe on Fri Apr 04, 2008 6:46 pm, edited 1 time in total.
gary w
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Post by gary w »

I certainly wouldn't say v1 was "much too easy" for regular assassinators.
It was relatively easy to show that the 13(2) cage N2 ={67} but then I didn't spot for quite a while that the 11(3) cage couldn't contain a 3.Given that the innies in N3 are{137} this meant that r2c7=3,r3c7=7 and r34c8=13.It's all over now..
Took me over an hour so I'ld rate it as harder than virtually all the "deadly" killers that appear in The Times..0.75??

Regards

Gary
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