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Assassin 75
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sublue
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PostPosted: Fri Nov 02, 2007 2:15 am    Post subject: Assassin 75 Reply with quote

I didn't write down a WT for this one - it was extremely straight forward. The innies and outies were the basic, obvious ones. I took me only 45 minutes, and I'm a newbie working by hand. If you want to check against the answer, here's mine:

728194635
415763982
639528417
271649358
563281794
984357261
342816579
857932146
196475823

The real puzzle is this: What does "4" have to do with this?
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Para
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PostPosted: Fri Nov 02, 2007 5:24 am    Post subject: Re: Assassin 75 Reply with quote

sublue wrote:

The real puzzle is this: What does "4" have to do with this?


Maybe Ruud has 3 more puzzles for us Very Happy
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CathyW
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PostPosted: Fri Nov 02, 2007 1:39 pm    Post subject: Reply with quote

Well done Susan!

I'm struggling with it Confused Sad - hoping it will be one of those that falls fairly quickly once you've got the first few placements. Lunch break over so will have to persevere later - I feel a bout of combo crunching coming on.

I suppose the 4 could be to do with the quads resulting from the outies of r1, r9, c1, c9?
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PostPosted: Fri Nov 02, 2007 1:50 pm    Post subject: Reply with quote

Surely 4 relates to Vnumber of this weeks bonus puzzle. Brick wall Brick wall
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PostPosted: Fri Nov 02, 2007 7:20 pm    Post subject: Re: Assassin 75 Reply with quote

sublue wrote:
it was extremely straight forward...basic, obvious ...took me only 45 minutes...

Go on then, Susan, give us a hint! Smile
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sublue
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PostPosted: Fri Nov 02, 2007 8:27 pm    Post subject: Hint Reply with quote

I too, thought that the 4 had to do with quadruples innie-outies. If it does, I missed it.

I'm starting a WT, since I can't remember how I got the first placement to fall. If you saw the hint I'm editting out, ignore it - I apparently had a brain freeze.
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PostPosted: Fri Nov 02, 2007 9:16 pm    Post subject: Reply with quote

Well done Susan!

I started last night and after about 30-45 minutes had done the Prelims and 11 steps; no placements so far. Then I decided I'd better get to bed.

Haven't looked at it yet today. I've still got to do the Extreme Jigsaw on my other sudoku website. It's got to be submitted by tomorrow evening and they are usually stinkers.
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PostPosted: Fri Nov 02, 2007 9:17 pm    Post subject: Reply with quote

Finally got there. Smile Total time about 4 hours and I haven't done a WT! Fortunately it did fall fairly quickly after the first placement.

The outies - innies of N2, N4, N6 and N8 helped a bit, N2 and N4 more so after analysing the combo options.

First placement was 2 at r1c2 which followed a complex naked triple in N4.
Shocked

Maybe the 4 relates to the rotational symmetry i.e. many steps can be repeated in different units 4 times (e.g. r1,r9,c1,c9)!
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sublue
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PostPosted: Fri Nov 02, 2007 9:59 pm    Post subject: Now I can't do it... Reply with quote

My first placement was 9 at R3C3. But I'll be darned if I can figure out how I did it! Note to self: write this stuff down the *first* time!
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PostPosted: Fri Nov 02, 2007 11:22 pm    Post subject: Reply with quote

CathyW wrote:
Total time about 4 hours and I haven't done a WT!

And there's more.

Ruud
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gary w
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PostPosted: Sat Nov 03, 2007 1:13 am    Post subject: assassin 75 Reply with quote

After the wall this one wasn't too bad.Couldn't match Sue's time..took me about 2.5 hours.As per the others couldn't see where the 4s particularly entered into the solution..briefly outlined.



1. r4679c2=28=89 (4/7 or 5/6)
2. r158c2=13=2(4/7 or 5/6)
3. Thus r8c2=4/5/7.... combos in N7 and cells at r3c13(=15) -> 15(3) cage N7 does not contain a 9.
4. So r9c12=19 thus 8 @ r67c2
5. Further combo work utilising r3c1+r7c3=r4c3+7 shows r1c2=2 and r4c1-2
6.So r79c3={26}..using 5. above and combos for 15(4) cage N47 -> R79C3=26
7.R7C349=19..Innies of R789 -> r7c49={89} -> HS r6 r6c1=9 and r6c2=8.r5c1=5 r4c2=7
8.In N9 9 not in r8-> in r7 so r7c9=9 r7c4=8
9.Now deduce r4c2=5/7 thus in r456c3 1 cannot be in the 15(4) cage..would not be able to complete it.Thus r4c3=1
10.With r4c3=1 and r7c3=8 -> r3c6+r6c7=10

Mop up now



Look out for the variants!

Regards

Gary
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sublue
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PostPosted: Sat Nov 03, 2007 6:41 am    Post subject: Mea Culpa Reply with quote

After spending several hours trying to recreate the incredibly fast way I solved this puzzle, I am now convinced that I made an intuitive leap and correctly guessed my first placement at R3C3. Everything fell out from there. I have re-solved it, and can't figure out where I got my original placement from, so I went with what I can defend.

Editted once for corrections and clarifications. Thanks Andrew!
1) Prelims:
a) N1 11(2) no 1.
b) N1 4(2) = {13} locked.
c) N25 21(3) no 1,2,3.
d) N3 8(2) no 4,8,9.
e) N3 17(2) = {89} locked.
f) N3 10(3) no 8,9.
g) N4 22(3) no 1,2,3,4. 9 locked to N4 22(3).
h) N56 10(3) no 8,9.
i) N7 12(2) no 1,2,6
j) N7 10(2) no 5.
k) N78 19(3) no 1.
l) N9 11(3) no 8
m) N9 11(2) no 1
n) N9 9(2) no 9.

Solving:
2) Cleanup from 1b), 1e) and 1g).
a) N1 11(2) no 8.
b) N7 12(3) no 9 R8C3.
c) N7 10(2) no 7,9 R9C1.
d) N1 11(2) no 2 R1C1.

3) N1 2 innies R3C13 = 15 -> no 2,4,5.

4) N3 2 innies R13C7 = 10 -> no 1,2,5.

5) N7 2 innies R79C3 = 8 -> no 4,8,9. No 7 at R7C3.

6) N9 2 innies R7C79 = 14 -> no 1,2,3,4,7.

7) C1: 4 outies: R4679C2 = 28 -> no 2 in R479C2.
a) Cleanup: R9C1 no 8
b) 8 and 9 are locked into R4679C2
c) R8C3 no 3,4.

8) R12 3 innies R2C259 = 9 -> no 7 at R2C59.

9) C89 3 innies R259C8 = 19 -> no 1 at R59C8.

10) R789 3 innies R7C349 = 19 -> no 1 R7C34. No 7 at R9C3.

11) N89 4 innies R7C49 and R89C4 = 30 -> R789C4 no 2,3.

12) C3 has locked 9 in R13C3 since n/e.
a) R2C1 no 2, R3C3 no 6.

13) N14 2 innies and 1 outie: R34C3 = R7C3 +8 -> R3C3 <> 8 and R4C3 <>7 or 8.
a) Cleanup: R3C1 <> 7.
b) N14 15(3) R3C1 = {68} -> R4C1 no 6,7,8.

14) N69 1 outie and 2 innies: R3C7 + 3 = R67C7 ->R6C7 no 3,5,6,7.

15) N1: 8 must be in either R1C3 or R3C1. 9 locked in R13C3 (step 12) -> R1C3 = {89}

16) N1 15(3) has locked 2 since n/e.

16b) N1 15(3) R1C2 and R2C3 = {245}

17) Deleted. Bwahahaha!

18) N1: 6 locked in R123C1 since n/e. 6 locked for C1.
a) Cleanup: R9C2 no 4.
b) R6C2 no 7.

19) N7: 1 locked to R789C1 since n/e. Locked for C1.

20) deleted

21) N14 15(3) no 6, 8 at R4C2.

22) N89 15(4): R7C7 = {5689} -> R7C56 and R8C5 min 6, max 10 -> no 8,9.

23) C12 3 innies R158C2 = 13 -> R15C2 no 5.

24) N8: 1 innie and 2 outies: R7C4 + 3 = R7C7 + R9C3 -> R7C4 no 6 or 9, R9C3 no 3.

25) N78 19(3): no 7 at R89C4.

26) N7 2 innies: R79C3 = 8 -> R7C3 <> 5.

27) R789 3 innies: R7C349 =19 -> R7C9 no 6.

28) N9 2innies: R7C79 = 14 -> R7C7 no 8.

29) R123: 3 innies R349C3 = 16 -> R4C3 no 6.

30) C789: 3 innies R167C7 = 13 -> R1C7 no 4.

31) N3: 2 innies R13C7 =10 -> R3C7 no 6.

32) N23 13(3): R12C6 no 7.

33) R123: 3 innies R349C3 = 16 -> R3C6 no 4.

34) N8: 1 innie and 2 outies: R7C4 + 3 = R7C7 + R9C3
a) N78 19(3) if R9C3 = 2, then R89C4 = {89} which means R7C4 <> 8.
b) Therefore, no combo can have 9 at R7C7.

35) N9:2 innies R7C79 = 14 -> R7C9 no 5.

36) N9: R7C7 = {56} -> N9 11(2) R78C8 cannot be 5,6.

37) N69 16(3) R6C89 no 8,9.

38) C1: 4 outies: R4679C2 = 28 which must include 8 and 9. R67C2 no 5, since the 6 the 5 would have to have would eliminate either an 8 or 9, and we need both.

39) R6789: 4 outies R5C1236 = 15 -> R5C36 no 6 or8.

40) C6789: 4 outies R4789C5 = 15 -> R9C5 no 9.

41) C12: 3 innies R158C2 = 13, and 2 is locked in this split, hidden 13(3) since n/e.
a) Determine whether R15C2 contains the 2. Assume R5C5 = 2. Cleanup as you go hoping to force a contradiction, or work through to end of puzzle.
b) Then R1C2 = 4
c) N1 11(2) = {56}
d) R3C1 = 8.
e) R1C3 = 9
f) R2C3 = 2
g) R3C3 = 7
h) N14 15(3) has R3C1 = 8, and R4C12 = {34/57}. I can’t make a 15 out of this, there R5C5 cannot be 2.

42) R1C2 = 2.
a) Cleanup R1C89 no 6.

43) N3 10(3) locked 2s since n/e. -> no 4 in R2C9 and R3C89.

44) HS R3C7 = 4.

45) N46 23(4) : R45C7 no 1.

46) R123 2 innies: R3C16 = 14 -> {68} locked pair for R3.

47) N3: 1 innies R1C7 = 6.
a) Cleanup R2C1 <> 5.

48) NS R7C7 = 5.
a) Cleanup R89C9 no 4.

49) N9: 1 innie R7C9 = 9.
a) Cleanup R6C89 no 7.
b) R78C8 no 2.

50) N9 4 locked in R789C8 for N9 and C8.
a) Cleanup R6C9 no 3.

51) N6 1 innie R6C7 = 2.
a) Cleanup: R6C89 no 5.

52) N9 11(3) R9C8 <> 6 since R89C7 cannot make a 5. -> 6 is locked in N9 9(2)

53) N9 9(2) = {3,6} locked pair for N9 and C9.
a) Cleanup: R78C8 no 8. -> R78C8 = {4,7} NP locked for N9 and C8.
b) Cleanup: R6C8 no 1.
c) R1C8 no 5
d) R1C9 no 1.

54) HS R9C8 = 2.

55) N7 2 innies R79C3 = 8 -> R7C3 no 6.

56) N3 has locked 3 in R13C8 locked for N3 and C8.
a) Cleanup: R6C8 = 6 -> R7C9 = 1

57) N36 23(4) Has locked 3 in R45C7. Remaining 2 cells sum to 16 -> must be 7 and 9. No 5,8.
a) NS R5C8 = 9.
b) NS R2C8 = 8
c) NS R2C7 = 9.
d) NS R4C8 = 5

58) NP R45C7 = {3.7}

59) HS R8C2 = 5.

60) And the rest are naked and hidden singles.
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Last edited by sublue on Fri Nov 09, 2007 1:28 am; edited 1 time in total
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Afmob
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PostPosted: Sat Nov 03, 2007 6:59 pm    Post subject: Reply with quote

That one was tougher than I thought. I had to use some forcing chains (Considering placement with 2 possibilites) to solve this one so rating should be about 1.25.

A75 Walkthrough:

1. C123
a) 4(2) = {13} locked for C2+N1
b) Innies C12 = 13(3) = 2{47/56} -> 2 locked for C2
c) Innies N1 = 15(2) = {69/78}
d) 15(3) @ R1C2 <> 7 because {267} is blocked by Killer pair (67) of Innies of N1
e) 11(2) <> 8
f) 12(2) = [48/57/75]
g) 10(2): R9C1 <> 7,8,9
h) Innies N7 = 8(2) = {17/26/35}, R7C3 <> 7

2. N14
a) 22(3) = 9{58/67} -> 9 locked for N4
b) 9 locked in R123C3 for N1
c) 11(2) = {47/56}
d) Killer pair (67) locked in 11(2) + Innies of N1 for N1
e) Innies N1 = 15(2): R3C3 <> 6

3. C789
a) 17(2) = {89} locked for R2+N3
b) Innies N3 = 10(2) = {37/46}
c) 10(3) @ N3 <> 6 because {136} blocked by Killer pair (36) of Innies of N3
d) Innies N9 = 14(2) = {59/68}
e) 11(2) <> 5,6 since {56} blocked by Killer pair (56) of Innies of N9
f) Innies C89 = 19(3) -> no 1
g) Innies C89 = 19(3) <> 3 because {379} blocked by Killer triple (379) of 11(2)
h) Innies C789 = 13(3): R6C7 <> 5,6,7 since sum would be > 13
i) 11(3): R89C7 <> 7 since R9C8 <> 1,3

4. C123
a) 6 locked R123C1 for C1
b) 15(3) @ R1C2 must have 8 xor 9 -> only possible @ R1C3 -> R1C3 = (89)
c) 22(3): R6C2 <> 7 because R56C1 <> 6
d) 10(2): R9C2 <> 4
e) Innies C12 = 13(3): R5C2 <> 5 since R18C2 <> 6
f) 15(3) @ R3C1: R4C1 <> 7,8 because R3C1+R4C2 is at least 10

5. N789
a) Innies N89 = 30(3+1) -> R7C9 <> 5 since R789C4 would be 25
b) Innies N89 = 30(3+1) -> R789C4 <> 1,2,3 because it's at least 21
c) 15(4): R7C56+R8C5 <> 8,9 because R7C7 >= 5
d) Innies N9 = 14(2): R7C7 <> 9
e) Innies N789 = 19(3) -> no 1

6. N7
a) Innies N7 = 8(2) = {26/35}
b) 1 locked in R789C1 for C1
c) 7,8,9 only possible @ 15(3), 12(2), 10(2) but none of them can have both -> 10(2) must have 7,8 xor 9
-> 10(2) = [19/28/37]

7. C123
a) Innies C123 = 16(3): R4C3 <> 8 since R39C3 >= 9
b) 15(3) @ R3C1: R4C2 <> 8 because R34C1 >= 8
c) 15(3) @ N7 <> {456} since it's blocked by Killer pair (56) of Innies of N7
d) 2 possibilites of (123) locked in R789C1 for C1 and R4C1
is the only place @ C1 where one candidate of (123) is possible -> R4C1 = (23)
e) Innies C12 = 13(3): R1C2 <> 5 because R8C2 <> 2,6
f) Innies C123 = 16(3): R4C3 <> 7 since R39C3 would be >= 10

8. R6789
a) 14(3) <> 9 because 9{14/23} blocked by Killer pairs (14,23) of 15(4)
b) 9 locked in R789C4 for C4
c) Outies = 15(4) must have 1 because {2346} impossible since R5C1 <> 2,3,4,6
-> 1 locked for R5

9. R12
a) Innies = 9(3) -> no 7
b) 17(3): R1C5 <> 1 because R12C4 <> 9
c) 13(3): R2C6 <> 7 because R1C2 blocks {24} and R1C7 <> 1,5

10. N9
a) Consider placement of 9 in N9 -> 11(3) <> 5:
Either Innies N9 = [59] -> no {245} in 11(3) or 11(2) = {29} -> no {245} in 11(3)

11. N7+C2 !
a) Consider placement of 6 in N7 -> 10(2) <> 2,8:
- i) 6 in 15(3) = 6{18/27} -> Innies 8(2) = {35} -> 12(2) = {48} -> 15(3) = {267} -> 10(2) = {19}
- ii) 6 in Innies 8(2) = {26} -> 10(2) = {19/37} -> 15(3) = {159/348/357}
-> 10(2) <> 2,8 and 15(3) = {159/267/348/357}
b) ! Consider placement of 8 in C2 -> R2C3 <> 2:
- i) 8 in 22(3) = {589} -> R56C1 = {59} -> 11(2) = {47} -> R1C2 = 2 -> R2C3 <> 2
- ii) 8 in 15(3) @ N7 = {348} -> Innies N7 = {26} -> R2C3 <> 2
c) 15(3) @ R1C2 = 2{49/58} -> R1C2 = 2

12. N36
a) 8(2) <> 6
b) 2 locked in 10(3) @ N3 = 2{17/35}
c) 4 locked in Innies N3 = 10(2) = {46} -> locked for C7
d) 13(3) <> 9 since R1C7 = (46)
e) 7 locked in R45C7 for N6
f) 17(3) <> 1 since it has no 7
g) 7 locked in 23(4) -> 23(4) <> 5 because R3C7 = (46)
h) Hidden Single: R7C7 = 5 @ C7

13. N9
a) Innies = 14(2) = [59] -> R7C9 = 9
b) 11(2) <> 2
c) 9(2) <> 4
d) 11(3) <> 4 since R89C7 <> 4,6
e) 4 locked in 11(2) = {47} locked for C8+N9
f) 9(2) <> 2
g) 16(3) = 9{16/25/34}, R6C9 <> 3

14. N8
a) 4 locked in R9C456 for N8
b) 15(4) = 5{127/136} -> 1 locked for N8
c) 9 locked in 19(3) = 9{28/37/46}
d) 5 locked in 14(3) = 5{27/36}
e) Hidden Single: R9C4 = 4
f) 19(3) = {469} -> R9C3 = 6, R8C4 = 9
g) Hidden Single: R7C4 = 8, R6C2 = 8 @ C2

15. C123
a) 22(3) = {589} -> {59} locked for C1+N4
b) 11(2) = {47} locked for C1+N1
c) 15(3) @ R1C2 = {258} -> R1C3 = 8, R2C3 = 5
d) 15(3) @ R3C1 = {267} -> R3C1 = 6, R4C1 = 2, R4C2 = 7
e) 15(4) = {2346} -> R5C2 = 6, R7C3 = 2 -> {34} locked for C3+N4
f) R3C3 = 9, R3C7 = 4, R1C7 = 6, R4C3 = 1, R7C2 = 4, R7C8 = 7, R8C8 = 4
g) R8C2 = 5, R8C3 = 7, R9C2 = 9 -> R9C1 = 1, R7C1 = 3, R8C1 = 8

16. N2
a) 13(3) = 6{25/34}
b) 9 locked 17(3) = 9{17/26/35} -> R1C5 = 9
c) 17(3) <> 3,5 since (35) is a Killer pair of 13(3)
d) 17(3) = {179} -> {17} locked for C4+N2
e) 6 locked in 22(4) = 69{25/34} -> R2C5 = 6, {25} locked for R3+N2
f) 4 locked in 13(3) = {346} -> {34} locked for C6
g) 21(3) = {489} -> R3C6 = 8, R4C5 = 4, R4C6 = 9

17. N89
a) 15(4) = {1356} -> R7C5 = 1, R7C6 = 6, R8C5 = 3
b) 11(3) = {128} -> R8C7 = 1, {28} locked for R9+N9
c) R8C9 = 6, R9C9 = 3, R8C6 = 2

18. N36
a) Hidden Single: R5C7 = 7 @ N6
b) 23(4) = {3479} -> R4C7 = 3, R5C8 = 9
c) 10(3) @ N6 = {127} -> R6C7 = 2, R5C6 = 1, R6C6 = 7
d) 5 locked in 8(2) @ R1 -> 8(2) = [35] -> R1C8 = 3, R1C9 = 5

19. Rest is singles.

And I thought I get some relaxation after the pain that was the Brick Wall. Twisted Evil


Last edited by Afmob on Thu Nov 08, 2007 8:51 pm; edited 2 times in total
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Andrew
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PostPosted: Sat Nov 03, 2007 8:19 pm    Post subject: Reply with quote

Ruud wrote:
Number 4 plays an important role in your life this week.

I'm still working on this puzzle and have found that there are some interesting interactions between 2-cell split-cages and 3-cell split cages sharing a common cell. Just a thought, could those sets of 4 interacting cells make Ruud's number 4? Even if they aren't, those interactions seem to be the theme for A75.

Alternatively Cathy suggested the 4 might be for the rotational symmetry. That's as good a suggestion as my one.

I'm pleased to see that Afmob has given a rating with his walkthrough. On what I've done so far it feels like a 1.25.


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gary w
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PostPosted: Sat Nov 03, 2007 9:42 pm    Post subject: Ruud's "and there's more" Reply with quote

This variant turned out to be a jigsaw with a disjoint 9-cage.
It was somewhat easier than any of the assassins I have looked at to date..it took me about 1.5 hours to solve.



1. r5c5=4
2. r5c2+r8c5=17=[89] and r3c23={79}
3. Thus r7c2+r8c1= 3 or 5 if r3c2 is 9/7 respectively.
4. Now 9 at r3c2 -> r5c2=8 r6c2=7 (nowhere else in N at r4c3 for the 7)
and r8c1=1/2...thus cannot complete the 18(3) cage r8c1.Thus
r3c23=79.Thus 9 in 22(3) cage at r1c6 cannot be at r2c6 otherwise cannot place a 9 in r1.therefore in r2 HS r2c9=9.
5. r2c5+r5c8=9.
6. this really helped to crack it...with r3c2=7 r7c2+r8c1=5 so r78c3+r9c12=22 so r8c2(<>4) + r9c5=9 which must be 18

work to do but pretty much a mop up now.



Regards

Gary
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