Thanks Ruud for another nice puzzle. As you say there are a few presents including some hidden in the branches of the Christmas Tree.
This one felt a bit harder than last week so I'll rate it at 1.25.
Here is my walkthrough. It's late so I haven't had time to check it. I'll do that tomorrow, hope there aren't any mistakes apart from typos.
Edit. Went through again in the morning. I've simplified from step 24 onward. Step 24 had been written before I inserted step 20a. I haven't colour coded the changes this time, hope that's not a problem for anyone.
Prelims
a) R3C67 = {17/26/35}, no 4,8,9
b) R9C34 = {13}, locked for R9
c) R9C67 = {49/58/67}, no 2
d) R123C1 = {389/479/569/578}, no 1,2
e) 21(3) cage at R2C4 = {489/579/678}, no 1,2,3
f) R234C8 = {123}, locked for C8
g) R456C1 = 1{25/34}, 1 locked for C1 and N4
h) 19(3) cage in N6 = {289/379/469/478/568}, no 1
i) R789C5 = {127/145/235} (cannot be {136} which clashes with R9C4), no 6,8,9
1. 45 rule on C1234 1 innie R4C4 = 4
1a. R34C5 = 11 = {29/38/56}, no 1,7
2. 45 rule on C1 3 innies R789C1 = 17
2a. Cage overlap R67C2 = R9C1 + 5
2b. IOU no 5 in R6C2
3. 45 rule on C9 3 innies R789C9 = 12
3a. Cage overlap R67C8 = R9C9 + 4
3b. IOU no 4 in R6C8
4. 45 rule on C12 2 innies R15C2 = 7 = [16]/{25/34}, no 7,8,9, no 6 in R1C2
5. 45 rule on C89 2 innies R15C8 = 17 = {89}, locked for C8
5a. 4 in C5 locked in R789C5, locked for N9, clean-up: no 9 in R9C6
6. 18(3) cage in N9 = {459/468/567} (cannot be {289} because 2,8,9 only in R9C9), no 2
7. 45 rule on C6789 3 innies R245C6 = 9 = {126/135/234}, no 7,8,9
7a. Min R45C6 = 4 (from 21(4) cage) -> no 6 in R2C6
7b. Max R45C6 = 8 -> min R56C5 = 13, no 1,2,3
8. 45 rule on N3 3 outies R13C6 + R4C8 = 10, max R13C6 = 9, no 9 in R1C6
9. 45 rule on N1 2 outies R1C4 + R4C2 = 1 innie R3C3 + 9, min R3C3 = 4 -> min R1C4 + R4C2 = 13, no 1,2,3
10. 16(4) cage at R6C8 = {1267/1357/1456/2347/2356} (cannot be {1249/1258/1348} because R67C8 must contain two of 4,5,6,7), no 8,9
10a. 7 of {1267/1357/2347} must be in R6C8 -> no 7 in R78C9
10b. Cannot be {1456} because R7C8 = 4, R78C9 = {15/16} clash with 18(3) cage
10c. -> 16(4) cage at R6C8 = {1267/1357/2347/2356}
10d. -> R78C9 = {123}, no 5,6
11. R789C9 = 12 (step 3) = {129/138/237} (cannot be {147/156/246} because R78C9 only contain 1,2,3) -> R9C9 = {789}
11a. 18(3) cage in N9 (step 6) = {459/468/567}
11b. R9C9 = {789} -> no 7 in R89C8
12. 45 rule on N4 3 innies R4C2 + R6C23 = 24 = {789}, locked for N4
13. R67C2 = R9C1 + 5 (step 2a), min R67C2 = 8 -> no 2 in R9C1
13a. Max R9C1 = 9 -> max R67C2 = 14, no 7,8,9 in R7C2
14. 45 rule on R9 2 outies R8C28 = 1 innie R9C5 + 7, min R9C5 = 2 -> min R8C28 = 9, no 1,2 in R8C2
15. 45 rule on N6 3 innies R4C8 + R6C78 = 11 = [137/146/236/245/317/326] -> R6C7 = {1234}
15a. Max R6C7 = 4 -> min R6C6 + R7C7 = 13, no 1,2,3
16. Hidden triple {123} in N9 -> R8C7 = {123}
16a. Max R8C7 = 3 -> min R78C6 = 15, R78C6 = {6789}
17. 45 rule on R789 2 innies R7C37 = 2 outies R6C28 + 3, min R6C28 = 12 -> min R7C37 = 15, R7C37 = {6789}
17a. R7C37 = 15,16,17 -> R6C28 = 12,13,14 = [75/76/85/86/95], no 7 in R6C8
18. R7C8 = 7 (hidden single in C8), clean-up: no 6 in R9C6
18a. Naked triple {689} in R7C367, locked for R7
19. R789C9 (step 11) = {129/138}, 1 locked in R78C9 for C9 and N9
20. 1 in N6 locked in R4C8 + R6C7
20a. R4C8 + R6C78 (step 15) = [146] (only remaining permutation)
20b. R6C6 + R7C7 = 13 = [58/76]
20c. R78C9 = {12} (step 10c), locked for C9 and N9, R8C7 = 3, clean-up: no 5 in R3C6
20d. R78C6 = [69/87/96], no 8 in R8C6
21. Naked pair {23} in R23C8, locked for N3, clean-up: no 6 in R3C6
21a. Naked pair {45} in R89C8, locked for N9, R9C9 = 9 (step 11a), clean-up: no 4,8 in R9C6
21b. Naked pair {68} in R79C7, locked for C7, clean-up: no 2 in R3C6
22. Naked quad {5789} in R6C2356, locked for R6 -> R6C9 = 3
22a. 5 locked in R6C56 for N5, clean-up: no 6 in R3C5 (step 1a)
22b. R45C9 = 12 = {57} (only remaining combination), locked for C9 and N6
22c. Naked pair {29} in R45C7, locked for C7 and N6, R5C8 = 8, R1C8 = 9
23. 23(4) cage at R5C4, max R67C3 + R6C4 = 19 -> no 1,2,3 in R5C4
24. 45 rule on R123 1 remaining outie R4C2 = 1 innie R3C5 + 6 -> R3C5 = {23}, R4C2 = {89}, clean-up: no 2,3,6 in R4C5 (step 1)
24a. Naked pair {89} in R4C25, locked for R4 -> R45C7 = [29]
25. Hidden triple {123} in R45C6 + R6C4, no 6 in R45C6 -> R4C6 = 3, R4C1 = 5, R4C3 = 6, R45C9 = [75], clean-up: no 5 in R3C7
25a. R5C6 = {12} -> R2C6 = {45} (step 7)
25b. Min R2C6 = 4 -> max R12C5 = 8, no 8,9
25c. 9 in N2 locked in R23C4, locked for C4 amd 21(3) cage -> no 9 in R3C3
25d. 9 in N8 locked in R78C6 = {69}(step 20d)
26. Naked pair {57} in R69C6, locked for C6 -> R2C6 = 4, R3C67 = [17], R5C6 = 2, R6C4 = 1, R6C1 = 2, R5C1 = 1, R9C34 = [13]
26a. Naked pair {69} in R78C6, locked for C6 and N8 -> R1C6 = 8
27. R45C6 = 5 -> R56C5 = 16 = {79}, locked for C5 and N5 -> R5C4 = 6, R6C6 = 5, R9C6 = 7, R9C7 = 6, R7C7 = 8, R7C3 = 9, R78C6 = [69], R6C3 = 7 (cage sum), R56C5= [79], R6C2 = 8, R4C2 = 9, R4C5 = 8, R3C5 = 3 (cage sum), R23C8 = [32]
28. Naked triple {579} in R123C4, locked for C4 and N2 -> R78C4 = [28], R78C9 = [12]
29. Naked pair {45} in R9C58, locked for R9 -> R9C12 = [82], R8C2 = 7 (cage sum)
30. Naked pair {45} in R8C38, locked for R8 -> R8C1 = 6, R8C5 = 1
31. R6C2 + R8C1 = 14 -> R7C12 = 8 = [35], R79C5 = [45], R89C8 = [54], R8C3 = 4, R5C23 = [43], R23C2 = [16] , R1C2 = 3, R12C7 = [15]
32. 21(3) cage at R2C4 = {579} (only remaining combination) -> R3C3 = 5, R3C4 = 9, R2C4 = 7
and the rest is naked singles
Happy Christmas to everyone on the forum!
Assassin 82
Assassin 82
Last edited by Andrew on Thu Dec 27, 2007 11:58 pm, edited 2 times in total.
Damn Andrew and his time zone advantage 
. Well being the second one to solve it is not so bad. I liked this assassin since one of my key moves was a Killer triple which is one of my favorite moves besides Killer pairs.
A82 Walkthrough:
1. C6789
a) Innies C89 = 17(2) = {89} locked for C8
b) 6(3) = {123} locked for C8
c) 16(4): R78C9 <> 7,8,9 because R8C67 >= 9
d) 18(3): R9C9 <> 1,2,3,4 because R89C8 <= 13
e) Innies N6 = 11(3): R6C7 <> 7,8,9 because R46C8 >= 5
2. C1234
a) Innies N4 = 24(3) = {789} locked for N4
b) 8(3) = 1{25/34}, 1 locked for C1+N4
c) Innies C12 = 7(2): R1C2 <> 6,7,8,9
d) Innies = 4 = R4C4
e) 15(3) = 4{29/38/56}
f) Outies N1 = 30(3+1) <> 1,2,3
3. C456 !
a) ! Killer triple (235) of 15(3) blocks {235} of 10(3)
b) 10(3) = 1{27/36/45}, 1 locked for C5+N8
c) R9C4 = 3, R9C3 = 1
d) 1 locked in 23(4) @ C4 for N5 -> 23(4) = 19{58/67}
e) 2 locked in 14(3) @ C4 for N8 -> 14(3) = 2{39/48/57}, R8C3 <> 2
f) 10(3) = {145} locked for C5+N8
g) 14(3) must have 3,4 xor 5 and it's only possible @ R8C3 -> R8C3 = (345)
h) 6 locked in R789C6 for C6
i) 15(3) <> 6
4. R789
a) 2 locked in R9C12 for N7
b) 2 locked in 17(3) @ R9 = 2{69/78}
c) 13(2): R9C7 <> 8,9
d) 18(3) @ R7C6: R8C7 <> 6,7,8,9 because R78C6 >= 13
e) Hidden pair (89) locked in R7C7+R9C9 @ N9 -> no other candidates
f) 18(3) @ R8C8 = 4{59/68} because R9C9 = (89), 4 locked for C8+N9
g) 13(2) = [67/76/85]
h) 7 locked in 16(4) @ C8 = 17{26/35} -> 1 locked for C9+N9
i) Hidden triple (123) locked in R78C9+R8C7 @ N9 -> no other candidates
5. C123
a) Innies C1 = 17(3) <> 5 because 5{39/48} blocked by Killer pairs (35,45) of 8(3)
b) 22(4): R7C12+R8C1 <> 9 because R6C2 <> 3,4,6
c) Innies C1 = 17(3) <> 9 because R78C1 <> 2,9
d) 9 locked in 20(3) @ C1 for N1
e) 9 locked in 23(4) @ C3 -> R56C4 <> 9
6. N56 !
a) Innies N6 = 11(3): R6C7 <> 5,6 because R6C8 >= 5
b) 17(3) must have 5 xor 7 and it's only possible @ R6C6 -> R6C6 = (57)
c) Killer pair (57) locked in 21(4) + R6C6 for N5
d) 21(4) = 2{379/568} because R6C6 = (57) blocks {3567} -> 2 locked for N5
e) ! 23(4): R56C4 = 1{6/8} (step 3d) and 23(4) must have 6 xor 8 -> R3C67 <> 6,8
f) 21(4) = {2379} locked for N5 because {2568} blocked by Killer pair (68) of R56C4
g) R4C5 = 8 -> R3C5 = 3
h) 17(3) = 5[39/48] -> R6C6 = 5
i) 23(4) = {1679} -> R67C3 = 7 locked for C3, 6 locked for C4
7. N23
a) 6 locked in 12(3) @ C5 = {246} -> R2C6 = 4, {26} locked for C5+N2
b) Hidden Single: R4C8 = 1 @ N6
c) R3C8 = 2, R2C8 = 3
d) Killer pair (56) of 23(4) blocks {567} of 18(3)
e) 18(3) = 4{59/68} -> 4 locked for C9+N3
f) 7 locked in R123C7 for C7
8. N6
a) 7 locked in 15(3) @ C9 -> 15(3) = 7{26/35}
b) R6C8 = 6
c) 15(3) = {357} locked for C9+N6
d) 19(3) = {289} -> 2 locked for C7
9. C123
a) Hidden Single: R6C2 = 8 @ N4
b) 16(3) must have 7 xor 9 and R4C2 = (79) -> R23C2 <> 7
c) 7,9 locked in 20(3) @ N1 -> 20(3) = {479} locked for C1+N1
d) 22(4) = {3568} -> R7C2 = 5
e) 21(3): R3C3 <> 8 because R23C4 <> 6
10. Rest is singles.
Rating: 1.25. It felt nearly as difficult as A81 but the moves I used here were more advanced.
. Well being the second one to solve it is not so bad. I liked this assassin since one of my key moves was a Killer triple which is one of my favorite moves besides Killer pairs.A82 Walkthrough:
1. C6789
a) Innies C89 = 17(2) = {89} locked for C8
b) 6(3) = {123} locked for C8
c) 16(4): R78C9 <> 7,8,9 because R8C67 >= 9
d) 18(3): R9C9 <> 1,2,3,4 because R89C8 <= 13
e) Innies N6 = 11(3): R6C7 <> 7,8,9 because R46C8 >= 5
2. C1234
a) Innies N4 = 24(3) = {789} locked for N4
b) 8(3) = 1{25/34}, 1 locked for C1+N4
c) Innies C12 = 7(2): R1C2 <> 6,7,8,9
d) Innies = 4 = R4C4
e) 15(3) = 4{29/38/56}
f) Outies N1 = 30(3+1) <> 1,2,3
3. C456 !
a) ! Killer triple (235) of 15(3) blocks {235} of 10(3)
b) 10(3) = 1{27/36/45}, 1 locked for C5+N8
c) R9C4 = 3, R9C3 = 1
d) 1 locked in 23(4) @ C4 for N5 -> 23(4) = 19{58/67}
e) 2 locked in 14(3) @ C4 for N8 -> 14(3) = 2{39/48/57}, R8C3 <> 2
f) 10(3) = {145} locked for C5+N8
g) 14(3) must have 3,4 xor 5 and it's only possible @ R8C3 -> R8C3 = (345)
h) 6 locked in R789C6 for C6
i) 15(3) <> 6
4. R789
a) 2 locked in R9C12 for N7
b) 2 locked in 17(3) @ R9 = 2{69/78}
c) 13(2): R9C7 <> 8,9
d) 18(3) @ R7C6: R8C7 <> 6,7,8,9 because R78C6 >= 13
e) Hidden pair (89) locked in R7C7+R9C9 @ N9 -> no other candidates
f) 18(3) @ R8C8 = 4{59/68} because R9C9 = (89), 4 locked for C8+N9
g) 13(2) = [67/76/85]
h) 7 locked in 16(4) @ C8 = 17{26/35} -> 1 locked for C9+N9
i) Hidden triple (123) locked in R78C9+R8C7 @ N9 -> no other candidates
5. C123
a) Innies C1 = 17(3) <> 5 because 5{39/48} blocked by Killer pairs (35,45) of 8(3)
b) 22(4): R7C12+R8C1 <> 9 because R6C2 <> 3,4,6
c) Innies C1 = 17(3) <> 9 because R78C1 <> 2,9
d) 9 locked in 20(3) @ C1 for N1
e) 9 locked in 23(4) @ C3 -> R56C4 <> 9
6. N56 !
a) Innies N6 = 11(3): R6C7 <> 5,6 because R6C8 >= 5
b) 17(3) must have 5 xor 7 and it's only possible @ R6C6 -> R6C6 = (57)
c) Killer pair (57) locked in 21(4) + R6C6 for N5
d) 21(4) = 2{379/568} because R6C6 = (57) blocks {3567} -> 2 locked for N5
e) ! 23(4): R56C4 = 1{6/8} (step 3d) and 23(4) must have 6 xor 8 -> R3C67 <> 6,8
f) 21(4) = {2379} locked for N5 because {2568} blocked by Killer pair (68) of R56C4
g) R4C5 = 8 -> R3C5 = 3
h) 17(3) = 5[39/48] -> R6C6 = 5
i) 23(4) = {1679} -> R67C3 = 7 locked for C3, 6 locked for C4
7. N23
a) 6 locked in 12(3) @ C5 = {246} -> R2C6 = 4, {26} locked for C5+N2
b) Hidden Single: R4C8 = 1 @ N6
c) R3C8 = 2, R2C8 = 3
d) Killer pair (56) of 23(4) blocks {567} of 18(3)
e) 18(3) = 4{59/68} -> 4 locked for C9+N3
f) 7 locked in R123C7 for C7
8. N6
a) 7 locked in 15(3) @ C9 -> 15(3) = 7{26/35}
b) R6C8 = 6
c) 15(3) = {357} locked for C9+N6
d) 19(3) = {289} -> 2 locked for C7
9. C123
a) Hidden Single: R6C2 = 8 @ N4
b) 16(3) must have 7 xor 9 and R4C2 = (79) -> R23C2 <> 7
c) 7,9 locked in 20(3) @ N1 -> 20(3) = {479} locked for C1+N1
d) 22(4) = {3568} -> R7C2 = 5
e) 21(3): R3C3 <> 8 because R23C4 <> 6
10. Rest is singles.
Rating: 1.25. It felt nearly as difficult as A81 but the moves I used here were more advanced.
Last edited by Afmob on Thu Jan 10, 2008 8:54 pm, edited 1 time in total.
The time zone advantage only works occasionally for me. Caida, Frank and Susan have also used it a few times but it's more often that those in Europe get to post the first walkthrough.Afmob wrote:Damn Andrew and his time zone advantage
I'm a fairly slow solver and Thursday is often a good TV night for me. However yesterday evening there was only a repeat of something that I'd seen fairly recently and my wife decided to watch something else so I had a go at A82 and just kept going.
I see that Afmob's walkthrough is very different from mine. More interactions between cage combinations, such as the one mentioned in the introduction to Afmob's walkthrough. Maybe I don't spot those as easily because the only combinations I see are those I can work out by mental arithmetic, those I get by using Ruud's combination calculator and those I've already typed into my walkthrough, if I still remember them or look at them.
Last edited by Andrew on Sat Dec 22, 2007 12:36 am, edited 1 time in total.
Maybe I was lucky but this one turned out to be one of the easier ones for me..took about 45 mins to complete.I'ld rate it as 1.0 max.
1.I c1-4-> r4c4=4 -> r34c5=11 =56/38/29 so 10(3) cage N8 <>235
2.So 10(3) cage N8 contains a 1.So r9c3=1 r9c4=3.
3.4 outies N1=30 so r123c4=21-24 ie no 1/2/3 so in c4 1 at r56.So 2 cannot also be at r56c4 so in c4 2 at r78.10(3) n8={145}
4.r1256c5=24 and r12 is max 11 so r56c5 min 13 so no 2/3.In N5 2 and 3 in c6.In N2 2 and 3 in c5.
5.O r123 -> r4c2458=22 -> r4c5=8/9 so r3c5=2/3.
6.O N3 -> R1C6+R3C6+R4C8=10.So r1c6<>4 (if so -> r4c8=1 [2/3 impossible combos] r3c6=5 and then an impossible 3 at r3c7).Thus r2c6=4 -> r12c5={26} -> r3c5=3 r4c5=8 r3c67=17.
7.In N8 6 must be in c6.So r56c4={16}.All over now.
Surprisingly didn't really need to use innies of n4={789} or innies c89 -> r15c8={89}.
Regards
Gary
p.s and a Happy Christmas and a prosperous new year to everyone!!
1.I c1-4-> r4c4=4 -> r34c5=11 =56/38/29 so 10(3) cage N8 <>235
2.So 10(3) cage N8 contains a 1.So r9c3=1 r9c4=3.
3.4 outies N1=30 so r123c4=21-24 ie no 1/2/3 so in c4 1 at r56.So 2 cannot also be at r56c4 so in c4 2 at r78.10(3) n8={145}
4.r1256c5=24 and r12 is max 11 so r56c5 min 13 so no 2/3.In N5 2 and 3 in c6.In N2 2 and 3 in c5.
5.O r123 -> r4c2458=22 -> r4c5=8/9 so r3c5=2/3.
6.O N3 -> R1C6+R3C6+R4C8=10.So r1c6<>4 (if so -> r4c8=1 [2/3 impossible combos] r3c6=5 and then an impossible 3 at r3c7).Thus r2c6=4 -> r12c5={26} -> r3c5=3 r4c5=8 r3c67=17.
7.In N8 6 must be in c6.So r56c4={16}.All over now.
Surprisingly didn't really need to use innies of n4={789} or innies c89 -> r15c8={89}.
Regards
Gary
p.s and a Happy Christmas and a prosperous new year to everyone!!