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 Post subject: Assassin 219
PostPosted: Fri Aug 19, 2011 1:34 am 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Time for a break from lots of big cages. Still get a purist's nightmare! :) Fortunately, there is a neat (though quite long) way through that doesn't use any sneaky contradictions (chains). Took some good digging to find! Satisfying.

Assassin 219
Image
code: copy & paste into solver:
3x3::k:3328:3073:3073:3073:2818:3587:3587:3587:4356:3328:4101:4101:1798:2818:2567:3336:3336:4356:6153:4101:2314:1798:5131:2567:5388:3336:5645:6153:2314:2314:5131:5131:5131:5388:5388:5645:6153:6153:6153:4878:4878:2319:5645:5645:5645:1808:4881:4881:4881:4878:2319:5394:5394:5394:1808:6163:6163:2580:2837:3862:5394:1303:5394:2328:2328:6163:2580:2837:3862:3862:1303:2585:2328:3866:3866:3866:5659:5659:5659:5659:2585:
solution:
+-------+-------+-------+
| 5 1 9 | 2 6 7 | 4 3 8 |
| 8 7 3 | 4 5 1 | 2 6 9 |
| 2 6 4 | 3 8 9 | 7 5 1 |
+-------+-------+-------+
| 9 3 2 | 7 1 4 | 6 8 5 |
| 7 5 1 | 8 2 6 | 3 9 4 |
| 4 8 6 | 5 9 3 | 1 7 2 |
+-------+-------+-------+
| 3 9 8 | 1 7 2 | 5 4 6 |
| 6 2 7 | 9 4 5 | 8 1 3 |
| 1 4 5 | 6 3 8 | 9 2 7 |
+-------+-------+-------+

Cheers
Ed


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 Post subject: Re: Assassin 219
PostPosted: Mon Aug 22, 2011 9:02 pm 
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Joined: Mon Apr 21, 2008 11:35 pm
Posts: 49
Thanks, Ed! I really enjoyed trying to march through this this weekend. You're right - it did take awhile - and I had notes all over the page - but, I finally cracked it!


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 Post subject: Re: Assassin 219
PostPosted: Tue Aug 23, 2011 1:44 am 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Thanks Ed for another tough Assassin. As you said it takes a long time to get out.

Rating Comment:
I'll agree with the SS score of 1.5. My technically hardest step was the one where I did permutation analysis for the 19(3) cage at R5C4, overlapping with hidden cage R45C4 which between them had to contain both of 8,9 for N5. I hope that analysis, within a nonet, isn't considered to be a chain. Even if that step proved unnecessary, I'd still rate my walkthrough at 1.5 because it's long and some steps were hard to find.

Here is my walkthrough for A219:
Prelims

a) R12C1 = {49/58/67}, no 1,2,3
b) R12C5 = {29/38/47/56}, no 1
c) R12C9 = {89}
d) R23C4 = {16/25/34}, no 7,8,9
e) R23C6 = {19/28/37/46}, no 5
f) R56C6 = {18/27/36/45}, no 9
g) R67C1 = {16/25/34}, no 7,8,9
h) R78C4 = {19/28/37/46}, no 5
i) R78C5 = {29/38/47/56}, no 1
j) R78C8 = {14/23}
k) R89C9 = {19/28/37/46}, no 5
l) 9(3) cage at R3C3 = {126/135/234}, no 7,8,9
m) 21(3) cage at R3C7 = {489/579/678}, no 1,2,3
n) 19(3) cage at R5C4 = {289/379/469/478/568}, no 1
o) 19(3) cage at R6C2 = {289/379/469/478/568}, no 1
p) 24(3) cage at R7C2 = {789}
q) 9(3) cage at R8C1 = {126/135/234}, no 7,8,9

Steps resulting from Prelims
1a. Naked pair {89} in R12C9, locked for C9 and N3, clean-up: no 1,2 in R89C9
1b. Naked triple {789} in 24(3) cage at R7C2, locked for N7

2. Killer pair 3,4 in R78C8 and R89C9, locked for N9

3. 45 rule on R9 2 innies R9C19 = 8 = [17/26/53], clean-up: no 6 in R8C9

4. 45 rule on N5 1 outie R3C5 = 1 innie R6C4 + 3, R3C5 = {56789}, R6C4 = {23456}

5. 19(3) cage at R6C2 = {289/379/469/478/568}
5a. 2,3 of {289/379} must be in R6C4 -> no 2,3 in R6C23

6. 45 rule on R1 3 outies R2C159 = 22 = {589/679}, 9 locked for R2, clean-up: no 9 in R1C1, no 7,8,9 in R1C5, no 1 in R3C6

7. Hidden killer triple 7,8,9 in R12C1 and R345C1 for C1, R12C1 contains one of 7,8,9 -> R345C1 must contain two of 7,8,9
7a. 24(5) cage at R3C1 cannot contain more than two of 7,8,9 -> no 7,8,9 in R5C23

8. 45 rule on R6789 2 innies R6C56 = 12 = {48/57}/[93], no 1,2,6, no 3 in R6C5, clean-up: no 3,7,8 in R5C6

9. 21(3) cage at R3C7 = {489/579/678}
9a. 4 of {489} must be in R3C7 -> no 4 in R4C78

10. 45 rule on R9 3 outies R8C129 = 11 = {137/146/236} (cannot be {245} because 9(3) cage at R8C1 cannot be {25}2), no 5
10a. 4 of {146} must be in R8C9 -> no 4 in R8C12

11. 9(3) cage at R8C1 = {126/135}, 1 locked for N7, clean-up: no 6 in R6C1

12. Max R9C23 = 11 -> min R9C4 = 4

13. 45 rule on C123 3 outies R169C4 = 13
13a. Min R69C4 = 6 -> max R1C4 = 7

14. 45 rule on C1234 2 innies R45C4 = 15 = {69/78}

15. R169C4 = 13 (step 13) = {139/148/157/238/247/256/346}
15a. Killer quad 6,7,8,9 in R169C4, R45C4, R78C4, locked for C4, clean-up: no 1 in R23C4

16. R5C45 cannot be 15 (it would clash with R45C4, CCC) -> no 4 in R6C5, clean-up: no 8 in R6C6 (step 7), no 1 in R5C6
16a. Similarly, and possibly harder to spot, R5C4 + R6C5 cannot be 15 (because of the same CCC) -> no 4 in R5C5

17. 1 in N5 only in R4C56, locked for R4
17a. Min R4C23 = 5 -> max R3C3 = 4

18. 45 rule on C1 3(2+1) outies R58C2 + R5C3 = 8
18a. Min R58C2 = 3 -> max R5C3 = 5

19. 45 rule on N1245 2(1+1) innies R1C6 + R6C1 = 11, no 1 in R1C6 + R6C1, clean-up: no 6 in R7C1
19a. Max R6C1 = 5 -> min R1C6 = 6

20. 1 in N4 only in R5C123, locked for R5 and 24(5) cage at R3C1, no 1 in R3C1
20a. 1 in N6 only in R6C789, locked for 21(5) cage at R6C7, no 1 in R7C79

21. 45 rule on N7 1 outie R9C4 = 1 innie R7C1 + 3, no 4,9 in R9C4
21a. R169C4 = 13 (step 13), min R69C4 = 7 -> max R1C4 = 6

22. 45 rule on N2 3 innies R1C46 + R3C5 = 17 = {179/269/278/368/467} (cannot be {359/458} which clash with R23C4), no 5, clean-up: no 2 in R6C4 (step 4)
22a. 1,2,3,4 only in R1C4 -> R1C4 = {1234}

23. 20(4) cage at R3C5 must contain 1 = {1289/1379/1469/1478/1568} -> R4C56 = {12345}

24. 19(3) cage at R5C4 = {289/379/568}
24a. 2,3 of {289/379} must be in R5C5 -> no 7,9 in R5C5

25. Killer quad 6,7,8,9 in R12C5, R3C5, R56C5 and R78C5, locked for C5

26. 8,9 in N5 only in R45C4 (step 14) = {69/78} and 19(3) cage at R5C4 (step 24) = {289/379/568} -> 19(3) cage can only be [829/928/739/658/685] (using interactions between these overlapping cages which place both of 8,9), no 6 in R5C5, no 7 in R6C5, clean-up: no 5 in R6C6 (step 8), no 4 in R5C6

27. 1,2 in R6 only in R6C1789
27a. 45 rule on R789 4 outies R6C1789 = 14 = {1238/1247/1256}, no 9

28. 45 rule on N5 4 innies R4C456 + R6C4 = 17 and must contain 1 = {1268/1349/1367/1457} (cannot be {1259/1358} which clash with 19(3) cage at R5C4)
28a. 7,8 of {1268/1367} must be in R4C4 -> no 6 in R4C4, clean-up: no 9 in R5C4 (step 14)

29. 45 rule on R6789 3 outies R5C456 = 16 = [682/736/826] (cannot be [835] because 19(3) cage at R5C4 cannot be [838]), no 5, clean-up: no 4 in R6C6, no 8 in R6C5 (step 8)

30. 19(3) cage at R5C4 (step 24) = {289/568} (cannot be {379} which clashes with R6C6), no 3, 7

31. Naked triple {268} in R5C456, locked for R5 and N5, clean-up: no 9 in R3C5 (step 4)

32. R169C4 (step 15) = {148/157/256/346} (cannot be {238/247} which clash with R23C4)
32a. 6 of {256} must be in R9C4 -> no 5 in R9C4, clean-up: no 2 in R7C1 (step 21), no 5 in R6C1, no 6 in R1C6 (step 19)

33. 19(3) cage at R6C2 = {469/478/568} (cannot be {379} which clashes with R6C6), no 3, clean-up: no 6 in R3C5 (step 4)
33a. R6C4 = {45} -> no 4,5 in R6C23

34. Killer pair 4,5 in R23C4 and R6C4, locked for C4, clean-up: no 6 in R78C4

35. 20(4) cage at R3C5 (step 23) = {1379/1478}, no 5

36. 19(3) cage at R6C2 (step 33) = {469/478/568}, R6C56 (step 8) = 12 = [57/93]
36a. 5 in N5 only in R6C45, locked for R6
36b. 5 in N5 only in R6C45 -> 19(3) cage at R6C2 = {568} or R6C56 = [57] -> 19(3) cage at R6C2 = {469/568} (cannot be {478}, locking-out cages), no 7, 6 locked for R6 and N4

37. 7 in N4 only in R45C1, locked for C1, clean-up: no 6 in R12C1

38. R2C159 (step 6) = {589} (only remaining combination, cannot be {679} because 6,7 only in R2C5), locked for R2, clean-up: no 4,5 in R1C5, no 2 in R3C4, no 2 in R3C6

39. 45 rule on R1 3 innies R1C159 = 19 = {289/469/568}, no 3, clean-up: no 8 in R2C5

40. R1C46 + R3C5 (step 22) = {179/278}, no 3, 7 locked for N2, clean-up: no 3 in R23C6
[Ed pointed out that an alternative way to eliminate 3 from R1C4 is
45 rule on N14 2 outies R16C4 = 1 innie R6C1 + 3, no 3 in R1C4 (IOU)]


[Cracked at last!]

41. 3 in N2 only in R23C4 = {34}, locked for C4 and N2 -> R6C4 = 5, R6C5 = 9, R6C6 = 3 (step 8), R5C6 = 6, R5C45 = [82], R4C4 = 7, R123C5 = [658], R9C4 = 6, R3C6 = 9, R2C6 = 1, R1C46 = [27], R4C56 = [14], R7C1 = 3 (step 21), R6C1 = 4, clean-up: no 8 in R1C1, no 9 in R2C1, no 3 in R8C5, no 2 in R8C8, no 4 in R8C9, no 2 in R9C1 (step 3)

42. R12C1 = [58], R12C9 = [89]

43. R9C5 = 3 (hidden single in C5), R9C1 = 1. R89C9 = [37], clean-up: no 2 in R7C8
43a. Naked pair {68} in R6C23, locked for R6 and N4
43b. Naked pair {14} in R78C8, locked for C8 and N9 -> R1C8 = 3, R1C7 = 4 (cage sum)

44. Naked pair {26} in R28C1, locked for C1 -> R45C1 = [97]

45. R3C9 = 1 (hidden single in N3), R6C9 = 2, R6C8 = 7, R6C7 = 1

46. R6C789 = 172] = 10 -> R7C79 = 11 = {56}, locked for R7 and N9

47. 13(3) cage at R2C7 = {256} (only remaining combination) -> R3C8 = 5, R2C78 = {26}, locked for R2 and N3, R3C7 = 7, R5C8 = 9

48. 21(3) cage at R3C7 = {678} (only remaining combination) -> R4C78 = {68}, locked for R4 -> R4C9 = 5

49. 9(3) cage at R3C3 = {234} (only remaining combination) -> R3C3 = 4, R4C23 = {23}, locked for N4

50. 16(3) cage at R2C2 = {367} (only remaining combination) -> R3C2 = 6

and the rest is naked singles.

I was fairly slow to spot some important 45s, particularly the ones in steps 19 and 29; if I’d spotted them earlier my solving path might have been a bit shorter. Also, after working through wellbeback's neat breakthrough and Ed's walkthrough, I realise that I missed the 45 on C6789 which would have given an early placement.

An interesting thing I noticed, after going through my walkthroughs for A218 and A219, is that they both used killer quad 6,7,8,9 to eliminate the {16} combination from a 7(2) cage, in C3 for A218 (step 9) and in C4 for A219 (step 15a).


Last edited by Andrew on Sun Aug 28, 2011 4:59 pm, edited 2 times in total.

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 Post subject: Re: Assassin 219
PostPosted: Wed Aug 24, 2011 12:26 am 
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Grand Master
Grand Master

Joined: Tue Jun 16, 2009 9:31 pm
Posts: 282
Location: California, out of London
I found a neat move which opens it up quite quickly. I'm not sure I've ever used anything exactly like it before.
I've tried to write it avoiding chains, and hope I haven't made a mistake!

Hidden Text:
Some setup steps...

1. Innies r6789 -> r6c56 = +12 -> No (126) in r6c56 -> (378) not in r5c6.
Also r5c456 = +16.

2. Innies c1234 -> r45c4 = +15 -> No (12345) in r45c4.
Also -> r5c45 cannot be = +15 -> 1 not in r5c6.

3. Outies - Innies c5 (and using r45c4 = +15) -> r4c6 = r9c5 + 1 -> 1 not in r4c6.

4. 1 cannot go in 19/3s at r5c4 or r6c2
-> HS 1 in n5 -> r4c5 = 1.

5. Remaining cells in n5: r4c6 + r5c56 + r6c4 = +17.
[r5c6,r6c4] cannot be [2,3] (would put two 7s in r6)
-> r5c6 + r6c4 = +6 or more (Since r5c6 cannot be 3 and 1 already in n5)
-> r5c5 + r4c6 = +11 or less.

Here it is...

6. Because of the 11/2s and 1 already in c5: (Edited to make clearer hopefully).
-> r9c5 is in range 2-8. (1 less than r4c6).
Whatever adds to it to make +11 (i.e., 9-3) in c5 cannot go in the existing 11/2 cages or in r4c5 where the 1 is.
-> Whatever that is must be in r3, 5, or 6 in c5.
I.e., r9c5 + one of r356c5 = +11.

-> (by 3 above) r4c6 + one of r356c5 = +12.
But (by 1 above) r6c5 + r6c6 already = +12 so r6c5 + r4c6 cannot = +12
and (by 5 above) r5c5 + r4c6 is max +11

-> r3c5 + r4c6 = +12
-> r4c4 = 7, etc., etc.


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 Post subject: Re: Assassin 219
PostPosted: Sat Aug 27, 2011 1:17 am 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Wow wellbeback! That's one of the best shortcuts ever! :applause: Took me a couple of days of thinking to see how you did that.
Another way to explain the start of wellbeback's step 6:
6. "45" on c5: 4 remaining innies r3569c5 = 22
6a. Those 4 cells must have the four digits missing from the two 11(2) cages in c5
6b. to avoid clashing with the two 11(2) cages, the h22(4) can also be broken into two 11(2) cages
6c. -> r9c5 + one of r356c5 make up an implied 11(2)
Andrew and I worked a lot, lot harder. In truth, I found making my walk-through tedious after realising that the move I really liked (step 18) could be done an easier way (from reading Andrew's WT). My beginning uses some nice Assassin moves before the tedious analysis sets in.

A219
30 steps:
This is an optimised solution so some possible eliminations are left out. However, I do try and do clean-up as I go. Please let me know of any corrections of clarifications

1. "45" on c1234: 2 innies r45c4 = 15 = {69/78}

2. "45" on c1234: 2 outies r56c5 - 4 = 1 innie r4c4
2a. -> no 4 in r56c5 (IOU)

3. "45" on r6789: 2 innies r6c56 = 12 (no 1,2,6)
3a. & no 3 in r6c5; no 8 in r6c6
3b. no 1,3,7,8 in r5c6

4. "45" on r6789: 2 outies r5c45 - 7 = 1 innie r6c6
4a. -> no 7 in r5c45 (IOU)
4b. no 8 in r4c4 (h15(2)r45c4)

5. "45" on c6789: 1 innie r4c6 - 1 = 1 outie r9c5
5a. no 1 in r4c6; no 9 in r9c5

6. "45" on n5: 1 innie r6c4 + 3 = 1 outie r3c5
6a. r6c4 = 2..6; r3c5 = 5..9

7. r4c5 = 1 (hsingle n5)
7a. split 19(3)r3c5; min. r3c5+r4c4 = 11 -> max. r4c6 = 8
7b. no 2 in r4c6; no 8 in r9c5 (both from IODc6789 = -1)

8. 19(3)r5c4: min. r5c4+r6c5 = 11 -> max. r5c5 = 8

9. "45" on r6789: 1 innie r6c5 - 3 = 1 outie r5c6 = [9->6..] (no elims yet)
9a. 9 in n5 only in r6c5 or h15(2)r45c4 -> 6 locked for n5 (Locking Cages)
9b. no 9 in r3c5 (IODn5 = +3)
9c. no 5 in r9c5 (IODc6789 = -1)

10. split 19(3)r3c5 = {379/469/478/568}
10a. 3/4 in {379/478} must be in r4c6 -> no 7 in r4c6

11. "45" on n5: 3 remaining innies r4c46+r6c4 = 16
11a. {358} blocked by r4c4 from (679)
11b. {367} blocked by 6 & 7 only in r4c4
11b. = {259/268/349/457} = [7->5; at most one of 5/6..]

12. 7 in n5 only in h12(2)r6c56 or h16(3) (step 11b)
12a. -> 5 locked for n5 (Locking cages)
12b. no 4 in r6c6
12c. no 8 in r6c5 (h12(2)r6c56)

13. h12(2)r6c56 = [93]/{57} = [7/9..]
13a. -> in 19(3)r6c2: {379} blocked
13b. -> 19(3) = {289/469/478/568}(no 3)
13c. has only one of 2/4/5 -> no 2,4,5 in r6c23
13d. no 6 in r3c5 (IODn5 = +3)

14. deleted

15. from step 11b. h16(3) can't have both 5 & 6
15a. split cage 19(3)r3c5 = {379/469/478/568}
15b. but {568} cannot have both 5 & 6 in n5
15b. -> no 5 in r4c6
15c. no 4 in r9c5 (IODc6789 = -1)

16. 5 in n5 only in r6: locked for r6
16a. no 2 in r7c1

Finally move somewhere else!
17. "45" on n7: 2 outies r6c1+r9c4 = 10
17a. no 1,2,3,5 in r9c4

Ready for the crucial, hardest move for me to find. [but note that Andrew does this an easier way: his steps 15 & 22 both work at this point]
18. r3c5 sees all 5s in c4 directly through n2 or indirectly through IODn5 = +3
18a. -> no 5 in r3c5 (Anti-clone CPE)
18b. no 2 in r6c4 (IODn5 = +3)

19. "45" on r6789: 3 outies r5c456 = 16 and must have 2 for n5
19a. = {268} only: all locked for n5 and r5
19a. no 5 in r6c6
19b. no 7 in r6c5
19c. r9c5 from (23) (IODc6789 = -1)

20. 22(4)r9c5: r9c5 from (23) -> min. r9c678 = 19 (no 1)

21. "45" on c123: 3 outies r169c4 = 13
21a. must have 4/5 for r6c4 -> r19c4 = 8/9 (no 9)
21b. no 1 in r6c1 (Outies n7 = 10)
21c. no 6 in r7c1

22. 1 in r9 only in n7: 1 locked for n7
22a. no 6 in r6c1
22b. no 4 in r9c4 (Outies n7 = 10)

23. 9(3)n7 = {126/135/234}
23a. 1 in {135} must be in r9c1 -> no 5 in r9c1

24. 17(2)n3 = {89}: both locked for n3 & c9
24a. no 1,2 in 10(2)n9

25. "45" on r9: 2 innies r9c19 = 8 (no 4) = [17/26] = [1/6..]
25a. no 6,7 in r8c9

26. 15(3)r9c2: {168} blocked by h8(2)r9c19
26a. {159} blocked by r9c4 from (678)
26b. -> no 1 in 15(3)

27. r9c1 = 1 (hsingle r9)
27a. r9c9 = 7 (h8(2)r9c19)
27b. r8c9 = 3
27c. r8c12 = {26}: both locked for r8 & n7
Missing clean-up now to race to the end

28. "45" on c1: 3 outies r5c23+r8c2 = 8
28a. min. r5c23 = 4 -> r8c2 = 2 (can't be 6!)
28b. -> r5c23 = 6 = {15}: both locked for r5, n4 and 24(5)r3c1

29. 5(2)n9 = {14}: both locked for c8 & n9

30. "45" on r789: 3 innies r7c179 = 14: must have 3/5 for r7c1 = {239/356}(no 8)
30a. must have 3 -> r7c1 = 3

cracked
Cheers
Ed


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