Posted: Sat Jan 20, 2007 3:55 am
by nd
Hm. Just been working through the walkthrough, & I must confess I don't really understand the ordering of steps or the key stage at step 20.
Re: the ordering: I don't understand why you've consistently chosen the most complicated & laborious innie-outies & combinations, & neglected to do all the basic steps first. In particular, fully half the first two dozen steps would be unnecessary if you simply applied the 45 rule to N1 (R3C13 = 11(2)), N3 (R3C79 = 12(2)), C1234 (R16C4 = 7(2)), and C5678 (R16C6 = 12(2)).
Secondly, I just don't follow what you're doing in step 20; probably if I sat down with a combination chart & did the branching myself I could verify the results, but since the point of a walkthrough is to provide an explanation that doesn't need extensive further work from the reader, then I stopped at this point.
Some remarks--
rcbroughton wrote:9. 45 rule on column 7. Innies cells r1234597 equal 27 -> {124578} {123579} {124569} {123678} {134568} {234567}
9a. No valid combination with 5/6/7/8/9 in r2c7
9b. No valid combination with 5/6/7 in r3c7
9c. No valid combination with 5/6/7/8/9 in r9c7
I don’t understand the notation (r1234597) or what you’re doing here. The deductions are correct, though. But the best (clearest, most logical) way to deal with this is via the hidden subset move I mentioned earlier.
12. 45 rule on column 9. Outies r2c7 r2345c8 equal 14
12a Cage 10(2) at r6c8 restricts possible values in r2345c8 {1236} & {1234} not possible
12b r2c7 cannot be 4
12c r2c8 can't now be 8/9
12d r3c8 can't now be 8/9
13. No possible combinations left with 8 in 19(5) at r1c9
14. Must use 2 in cage 19(5) at r1c9 -> cannot have 2 in r3c79
You’re doing things the exceedingly hard way in all these steps. All you need to do is: 45 rule on N3 => R3C79 = 12 = [39|48] => in conjunction with the 14(2) cage in R1C78 (={59|68}) we have a hidden pair on {89} so those candidates are eliminated in all other cells of N3. This places {127} in the 19(5) cage & excludes {89}.
18. 45 rule on N2. Innies r3c5 minus outies r3c3 r3c7 equals -8
18a Max val in r3c5 is 5 so max of r3x3+r3c7 is 13
18b No valid combination with 8
19c Remembering that r3c13 total 11 r3c1 cannot now be 3
Again this is doing things the hard way. {38} is excluded in R3C13 by R3C79 = [39|48].
20. 45 rule on N2. Innies r2c4 r3c4 r2c6 r3c6 r3c5 equal 28
20a r3c13 limits possibilities in r3c456 - no possible combo with 6 at r3c4
This is not at all immediately clear to me from looking at the grid. Please explain this one.
20b.this combo now means 9 cannot be in 17(4) at r1c5, r1c6 or r2c5
Again this step isn’t clear to me at all.
Posted: Sat Jan 20, 2007 6:22 am
by nd
Haven't proofed this yet but here's a walkthrough.
Step 1. 12(4) cage at R4C3 = {1236|1245}. 29(4) cage at R4C6 = {5789}. R78C6 = {12}. R1C78 = {59|68}. 22(3) cage at R2C4+R3C34 = {589|679}.
Step 2. 18(3) cage at R6C7 must have 2 cells containing {56789}. (If it had only one, then the max cage-sum would be 3+4+9 = 16.) So the 18(2) cage forms a hidden quint on {56789} in conjunction with R145C7 => R239C7 = {1..4}.
Step 3. 45 rule on N3 => R3C79 = 12 = [39|48] => R23C6 = {47|56|37}. This means that R45C6 cannot be {57}. 45 rule on C6789 => R16C6 = 12 = {39|48} (again, this cannot be {57} because of R23C6). This means that R45C6 cannot be {89}. So R45C6 must contain exactly one of {57} and one of {89}. This means we have a hidden quad on {5789} within R123456C6!!! So R9C6 = {346}.
Step 4. 45 rule on N1 => R3C13 = 11 = {56} or [29] or [47] ([38] is blocked because R3C79 = [39|48], see step 3).
Step 5. 45 rule on N2 => R3C37 = R3C5 + 8 => maximum value of R3C5 = 5 (since max val of R3C37 = 13).
Step 6. 45 rule on N2 => R2C46 + R3C456 = 28(5). Obviously if the 28(5) cage has a 1 or 2 in it, then R3C5 = {12}. If it does not, the only possible combos are {34579} or {34678}, i.e. it must have both 3 and 4 in it. But this is impossible, because R3C7 = {34} "sees" all three cells R2C6 + R3C56. Therefore R3C5 = {12}. All combinations of 28(5) with 1 or 2 in them include a 9 => the 9 is locked in R23C4 within N2/C4/the 22(3) cage.
Step 7. So at this point there is only one square in N8 where the 9 can go (since 11(2) = {29} is blocked by the 3(2) cage). So R7C5 = 9 => 9 is locked in N5/the 29(4) cage in R45C6 => R16C6 = {48}, R45C6 = {(5|7)9}, R45C7 = {(5|7)8}.
Step 8. R78C4 = {48|57}, R89C5 = {38|56} ({47} is blocked by the 12(2) cage). 45 rule on N8 => R9C46 = 10 = [73|46]. Thus R78C4 + R9C4 contain a hidden {47} pair => {47} is excluded in R1..6C4 => R23C4 = {(6|8)9}, R3C3 = {57}, R3C1 = {46}.
Step 9. 45 rule on C1234 => R16C4 = 7 = {16|25}. Therefore the only spot left in C4 for the 3 is R45C4 => the 12(4) cage is {1236} with the 3 in R45C4.
Step 10. This leaves only two possible combinations for the 25(4) cage at R6C4: {1789} or {2689} => R6C6 = 8, R1C6 = 4, and the 4 is locked in C5/N5/the 11(3) cage within R45C5 => R45C5 = {4(6|7)}. This in turn forms a hidden {67} pair with the 25(4) cage => R45C6 = {59}!
Step 11. The rest is mop-up. R45C7 = {78}. R23C6 = {37} (see step 3), R3C7 = 4, R3C9 = 8, R9C6 = 6, R89C5 = {38}, R78C4 = {57}, R9C4 = 4, R23C4 = [89] (because this is the only spot for 8 in C4), R3C3 = 5, R3C1 = 6, R16C4 = [61], R45C4 = {23}, R45C3 = {16}. 45 rule on N2 => R3C5 = 1. R1C78 = {59}, R12C5 = [25], R6C5 = 7, R45C5 = {46}.
Step 12. In C9 the only spot for the 9 is R4C9 (can't be in a 11(3) cage) => R4C89 = [19], R45C6 = [59], R4C12 = {27}, R45C3 = [61], R45C4 = [32], R45C7 = [87], R45C5 = [46]. 45 rule on N4 => R6C23 = 13 = {49}, 45 rule on N6 => R6C78 = {36}, R6C19 = [52], R789C1 = {489}, R5C12 = [38], R1C23 = [38] (only place for 8 to go in N1).
Step 13. 11(3) cage in N9 must have a 1 in it (since it must have 1 and/or 2, & 2 is blocked) => R2C7 = 1, R1C1 = 1. 45 rule on C7 => R19C7 = 7 = [52], R1C8 = 9, R1C9 = 7, R89C8 = {58}. R89C58 form an x-wing on 8 => R789C1 = [849], and you carry on........
Posted: Sat Jan 20, 2007 9:05 am
by rcbroughton
nd wrote:Re: the ordering: I don't understand why you've consistently chosen the most complicated & laborious innie-outies & combinations, & neglected to do all the basic steps first. In particular, fully half the first two dozen steps would be unnecessary if you simply applied the 45 rule to N1 (R3C13 = 11(2)), N3 (R3C79 = 12(2)), C1234 (R16C4 = 7(2)), and C5678 (R16C6 = 12(2)).
true enough - I tend to work more with combinations and permutations - I sometimes find I make simple arithmetical errors trying to tally up cage totals - mental arithmetic not necessarily my strong point. (Plus I use a littel spreadsheet that shows my combinations & permutations for cage totals)
Secondly, I just don't follow what you're doing in step 20; probably if I sat down with a combination chart & did the branching myself I could verify the results, but since the point of a walkthrough is to provide an explanation that doesn't need extensive further work from the reader, then I stopped at this point.
20. 45 rule on N2. Innies r2c4 r3c4 r2c6 r3c6 r3c5 equal 28
20a r3c13 limits possibilities in r3c456 - no possible combo with 6 at r3c4
This is not at all immediately clear to me from looking at the grid. Please explain this one.
OK - at this point we know r2c4 r3c4 r2c6 r3c6 r3c5 equal 28 and we have{6/7/8/9} {7/8/9} {3/4/5/6/7} {3/4/5/6/7} and {1/2/4/5} in those cells
The 17(4) in n2 must use either a 2 or a 7 - so we can't have any combinations with both. That leaves us with {13789}{14689} {15679} {23689} {24589} {34579} {34678}
Looking at the possibilities for placing the 6 - we have combos {14689}{15679} and {34678}
for {14689} the 1 must be in r3c5 - r23c4 can only be {89}
for {34678} the 8 must be in r23c4, the 3 in r23c6 and the 4 in r3c5 - but this leaves us with nothing for r3c7 - so combination cn't be used
for {15679} the 1 must be in r3c5, the 5 in r23c6 and the 9 in r23c4 - the 14(2) cage can only therefore be {563} - so r23c4 can only be {79}
therefore no 6 in r2c4
(a bit long-winded, but I do like playing with permutations!)
20b.this combo now means 9 cannot be in 17(4) at r1c5, r1c6 or r2c5
Again this step isn’t clear to me at all.
now we've eliminated 6 from r2c4, the 22(3) cage can only be {589}/{679} - so 9 must be in r23c4
The point of my post was more to see if I was looking at the same puzzle as everyone else. The comment had been that it was going to be very difficult and needed trial & error. Although there were a few laborious steps - I didn't see this one as particularly difficult - didn't need to use any advanced techniques like x-wing, chains, ALS, etc - and just wanted to check I wasn't being a bit slow and missing something. But I didn't just want to say it was solvable by siple logic without at least showing my working - however inelegant it might be !!
Posted: Wed Jan 28, 2009 7:56 am
by Andrew
As I commented in this thread and the A71 thread, I didn't manage to solve three of Ruud's Assassins when they first appeared. Now having caught up with my backlog of other walkthroughs I'm having another go at them.
I originally used a contradiction move for the key breakthrough but, after I'd finished I had another look and found an alternative more direct way; I've given both in step 17.
Here is my walkthrough.
Prelims
a) R1C78 = {59/68}
b) R1C23 = {29/38/47} (cannot be {56} which clashes with R1C78), no 1,5,6
c) R67C2 = {19/28/37/46}, no 5
d) R67C8 = {19/28/37/46}, no 5
e) R78C4 = {39/48/57}, no 1,2,6
f) R78C6 = {12}, locked for C6 and N8
g) R89C5 = {38/47/56}, no 1,9
h) 22(3) cage at R2C4 = {589/679}, CPE no 9 in R3C56
i) R345C5 = {128/137/146/236/245}, no 9
j) 11(3) cage in N6 = {128/137/146/236/245}, no 9
k) R789C1 = {489/579/678}, no 1,2,3
l) R789C9 = {128/137/146/236/245}, no 9
m) 12(4) cage at R4C3 = {1236/1245}, no 7,8,9
n) 29(4) cage at R4C6 = {5789}
o) 19(5) cage in N3 = {12349/12358/12367/12457/13456}, 1 locked for N3
1. 45 rule on N1 2 innies R3C13 = 11 = [29/38/47/56/65], no 1,7,8,9 in R3C1
2. Min R23C6 = 7 -> max R3C7 = 7
2a. 45 rule on N3 2 innies R3C79 = 12 = [39/48/57/75], no 2,6, no 3,4 in R3C9
3. 45 rule on R6789 2 innies R6C19 = 7 = {16/25/34}, no 7,8,9
4. 45 rule on C1234 2 innies R16C4 = 7 = {16/25/34}, no 7,8,9
5. 45 rule on C6789 2 innies R16C6 = 12 = {39/48/57}, no 6
6. R678C7 = {189/279/369/378/459/468/567}, must contain at least two of 5,6,7,8,9
6a. Killer quint 5,6,7,8,9 in R145C7 and 18(3) cage in N69, locked for C7, clean-up: no 5,7 in R3C9 (step 2a)
6b. R678C7 can only contain two of 5,6,7 = {189/279/369/378/459/468}
7. Killer pair 8,9 in R1C78 and R3C9, locked for N3
8. 14(3) cage at R2C6 = {347/356}, no 8,9, CPE no 3 in R3C5
8a. R16C6 (step 6) = {39/48} (cannot be {57} which clashes with R23C6), no 5,7
8b. Killer quad 5,7,8,9 in R16C6, R23C6 and R45C6, locked for C6
9. Hidden killer pair 8,9 in R16C6 and R45C6 for C6 -> R45C6 must contain one of 8,9 -> R45C7 must contain one of 8,9
9a. R678C7 (step 6b) = {279/369/378/459/468} (cannot be {189} which clashes with R45C7), no 1
9b. Killer pair 8,9 in R45C7 and R678C7, locked for C7, clean-up: no 5,6 in R1C8
9c. 45 rule on N3 2 remaining innies R13C7 = 9 = [54/63]
9d. R678C7 = {279/369/378/459} (cannot be {468} which clashes with R13C7)
10. 45 rule on N2 2 outies R3C37 = 1 innie R3C5 + 8, IOU no 8 in R3C3, clean-up: no 3 in R3C1 (step 1)
10a. Max R3C37 = 13 -> max R3C5 = 5
11. 22(3) cage at R2C4 = {589/679}
11a. 5 of {589} must be in R3C34 (R3C34 cannot be {89} which clashes with R3C9), no 5 in R2C4
12. 45 rule on C89 3 innies R189C8 = 1 outie R2C7 + 21
12a. Min R189C8 = 22, no 1,2,3,4, 9 locked for C8, clean-up: no 1 in R67C8
12b. Max R189C8 = 24 -> max R2C7 = 3
13. 45 rule on C9 2 innies R12C9 = 2 outies R45C8 + 5
13a. Max R12C9 = 13 -> max R45C8 = 8, no 8
13b. Max R3C9 + R4C8 = 16 -> min R4C9 = 2
14. 45 rule on R123 3 innies R3C159 = 15 = {159/168/249/258} (cannot be {267/456} because R3C9 only contains 8,9)
14a. R3C13 = 11 (step 1) -> R3C19 cannot be 11 (I’ll call that an overlap IOU) -> no 4 in R3C5
15. 45 rule on N8 3 innies R7C5 + R9C46 = 19 = {379/469/478/568}
15a. 3 of {379} must be in R9C6 -> no 3 in R7C5 + R9C4
16. 25(4) cage at R6C4 = {1789/2689/3589/3679/4579/4678}
16a. 1 of {1789} must be in R6C4 -> no 1 in R6C5
My original step 17
17. 17(4) cage in N2 = {1268/1349/1358/2348/2456} (cannot be {1259} which clashes with R3C5, cannot be {1367/1457/2357} which clash with R23C6), no 7
17a. Cannot be {1268}, here’s how
17(4) cage = {1268} => 14(3) cage at R2C6 = {347} => R23C4 = [95] is impossible because no 8 in R3C3
17b. 17(4) cage = {1349/1358/2348/2456}
17c. Hidden killer pair 1,2 in 17(4) cage and R3C5 for N2 -> R3C5 = {12}
Alternative step 17 which avoids using a contradiction move
17. 17(4) cage in N2 = {1268/1349/1358/2348/2456} (cannot be {1259} which clashes with R3C5, cannot be {1367/1457/2357} which clash with R23C6), no 7
17a. 45 rule on N2 (from result of step 17) 5 innies R2C46 + R3C456 = 28 = {13789/15679/24679/25678} (cannot be {34579} which clashes with 14(3) cage at R2C6)
17b. 1,2 only in R3C5 -> R3C5 = {12}
17c. 17(4) cage = {1349/1358/2348/2456}
18. R3C159 (step 14) = {159/168/249/258}
18a. R3C5 = {12} -> no 2 in R3C1, clean-up: no 9 in R3C3 (step 1)
18b. 9 in 22(3) cage at R2C4 locked in R23C4, locked for C4 and N2, clean-up: no 3 in R6C6 (step 5), no 3 in R78C4
19. R7C5 = 9 (hidden single in N8), clean-up: no 3 in R1C6 (step 5), no 1 in R6C2
19a. 45 rule on N8 2 remaining innies R9C46 = 10 = [46/64/73], no 5,8 in R9C4
19b. R89C5 = {38/56} (cannot be {47} which clashes with R78C4), no 4,7
20. Naked pair {48} in R16C6, locked for C6, clean-up: no 6 in R9C4 (step 19a)
21. 9 in C6 locked in R45C6, locked for 29(4) cage at R4C6
21a. 8 in 29(4) cage at R4C6 locked in R45C7, locked for C7 and N6, clean-up: no 2 in R7C8
22. 22(3) cage at R2C4 = {589/679}
22a. 5 of {589} must be in R3C3 -> no 5 in R3C4
23. 17(4) cage (original step 17b or alternative step 17c) = {1358/2348/2456}
23a. 1 of {1358} must be in R1C45 (R1C456 cannot be {358} which clashes with R1C78), no 1 in R2C5
24. 45 rule on C5 3 remaining innies R126C5 = 14 = {167/248/257/347} (cannot be {158/356} which clash with R89C5)
24a. 1 of {167} must be in R1C5 -> no 6 in R1C5
24b. 7 of {167/257/347} must be in R6C5 -> no 3,5,6 in R6C5
25. 25(4) cage at R6C4 = {1789/2689} (cannot be {3589/3679} because 3,5,6 only in R6C4, cannot be {4579} which clashes with R45C6), no 3,4,5 -> R6C6 = 8, R1C6 = 4, clean-up: no 7 in R1C23, no 2,3 in R1C4 (step 4), no 1 in R1C45 (step 23), no 6 in R6C4 (step 4)
25a. 25(4) cage at R6C4 = {1789} (only remaining combination) -> R6C45 = [17], R1C4 = 6 (step 4), R1C7 = 5, R1C8 = 9, R3C9 = 8, R3C7 = 4 (step 2a), clean-up: no 2 in R1C23, no 3,8 in R12C5 (step 23), no 7 in R3C3 (step 1), no 6 in R6C19 (step 3), no 2,3 in R7C2, no 3 in R7C8
25b. R123C5 = [251], clean-up: no 6 in R89C5
25c. R3C37 = R3C5 + 8 (step 10), R3C5 = 1, R3C7 = 4 -> R3C3 = 5, R3C1 = 6
26. R9C6 = 6 (hidden single in C6), R9C4 = 4 (step 19a), clean-up: no 8 in R78C4
26a. Naked pair {57} in R78C4, locked for C4 -> R23C4 = [89]
26b. Naked pair {23} in R45C4, locked for N5 and 12(4) cage at R4C3
26c. 12(4) cage at R4C3 = {1236} (only remaining combination), no 4
26d. Naked pair {16} in R45C3, locked for C3 and N4, clean-up: no 4 in R7C2
26e. Naked pair {78} in R45C7, locked for C7 and N6
27. R678C7 (step 9d) = {369} (only remaining combination), locked for C7
28. 21(4) cage at R8C8 = {2568} (only remaining combination) -> R9C7 = 2, R2C7 = 1, R89C8 = {58}, locked for C8 and N9, clean-up: no 2 in R6C8
29. R4C9 = 9 (hidden single in C9), R4C8 = 1 (cage sum), R45C3 = [61], R45C5 = [46], R45C6 = [59]
30. R3C9 = 6 -> R4C12 = 9 = {27}, locked for R4 and N4 -> R45C4 = [32], R45C7 = [87], clean-up: no 8 in R7C2
31. R6C9 = 2 (hidden single in R6), R5C89 = 9 = [45], R6C1 = 5 (step 3), clean-up: no 6 in R67C8
31a. R67C8 = [37], R678C7 = [639], R78C4 = [57], R9C9 = 1
32. Naked pair {48} in R78C1, locked for C1 and N7, R7C3 = 2, R7C6 = 1, R7C2 = 6, R6C2 = 4
and the rest is naked singles