Nice puzzle Ruud. Usually select for Diagonal Killers these days but couldn't resist the prospect of using 'overlap' on r5 and c1. Didn't end up needing it - but still very challenging. Had to redo it a few times before getting it out - when the sum chart comes out, the mistakes multiply!
Did anyone not need to use a sum chart (much)? Any chance of a walk-through? Always learn lots from others' approaches.
Doesn't worry me about symmetry - a good challenge is the main thing. Thanks.
Assassin Weekly Killer 16 June
Ed--here's my version--tell me if this is clear/if I've made any typos. No extensive use of combination charts &c.
*
Step 1. R1C89 = {13}, R34C9 = {79}, R1C12 = {48|57}, R5C23 = {89}, 11(4) cage beginning in N1 = {1235}.
Step 2. 45 rule on C1 -> R14C2 = 10, so R1C2 = {478}, R4C2 = {236}. 45 rule on R1234 -> R4C12 = 10 -> R56C1 = 5 = {14|23}. 45 rule on R5 -> R6C9 = R5C1 = {1234}.
Step 3. 20(3) cage in R4 = {389}, {569} or {578}, i.e. must contain {79} -> complex pair with {79} in R4C9 -> {79} eliminated in the rest of R4. So R4C12 = [46], R1C12 = [84], R56C1 = R6C9 = {23}, R23C1 = {17}, R789C1 = {569}, R4C3 = {15}, R6C23 = {(1|5)7}.
[Step 4 deleted]
Step 5. In R4 the 2 can't go in R4C5 because this would make R3C45 = {79}, contradicting R3C9. Only place left is R4C4 = 2 -> R4C3 = 1, R7C23 = [13], R3C23 = [35], R6C23 = [57].
Step 6. The 20(3) cage in R4 has now been narrowed down to {389} or {578}, i.e. it must contain an 8. So R4C5 = {35}, and the only remaining combos for the 12(3) cage in N5 = {147} or {156}. So 1 is locked in R5 in N5 and R6C8 = 1 (can't be in 19(3) cage obviously because this would make the remaining two cells > 17) -> R1C89= [31]. Subtraction combo on remaining cells in R6 ({4689}) -> R6C4 = 8, R6C567 = {469}.
Step 7. 45 on R7 -> R8C6 = R7C1 - 5, i.e. R7C1 = {69}, R8C6 = {14}. R7C89 = [78] or [96]. R7C45 = {29|47|56}. So we have a complex triple on {679} in R7! So R7C67 = {45}, R8C6 = 1, R7C45 = [92], R7C1 = 6, R7C89 = [78], R9C7 = 1.
[Footnote: I'll leave this step as it is but I see that it's far more complex than necessary: the simplest thing is to note that in R7 the 8 must go in R7C89. So R7C89 = [78]; in turn R7C67 must have the 4 in them, so the 10(3) cage has {145} in it. No innie-outie differences or complex triples required!]
Step 8. 45 on N9 -> R7C67 = [54], R9C56 = {48}, R89C9 = {25}, R6C9 = R5C1 = 3, R6C1 = 2, R8C7 = R9C4 = 3, R8C45 = {67}, R8C3 = 4, R89C8 = [96], R89C1 = [59], R89C9 = [25], R8C2 = 8, R9C23 = [72], R5C23 = [98], R2C2 = 2.
Step 9. In N6 the 2 and 4 must be in 16(4). So the remaining cell of the 16(4) = 7, which goes in R5C7. R4C9 = 9, R4C6 = 7, R4C78 = {58}, R4C5 = 3, R5C6 = 6, R5C89 = [24], R2C9 = 6, R23C8 = [54]. 45 rule on R1 -> R2C57 = 9 = [18]. And you carry on....
*
Step 1. R1C89 = {13}, R34C9 = {79}, R1C12 = {48|57}, R5C23 = {89}, 11(4) cage beginning in N1 = {1235}.
Step 2. 45 rule on C1 -> R14C2 = 10, so R1C2 = {478}, R4C2 = {236}. 45 rule on R1234 -> R4C12 = 10 -> R56C1 = 5 = {14|23}. 45 rule on R5 -> R6C9 = R5C1 = {1234}.
Step 3. 20(3) cage in R4 = {389}, {569} or {578}, i.e. must contain {79} -> complex pair with {79} in R4C9 -> {79} eliminated in the rest of R4. So R4C12 = [46], R1C12 = [84], R56C1 = R6C9 = {23}, R23C1 = {17}, R789C1 = {569}, R4C3 = {15}, R6C23 = {(1|5)7}.
[Step 4 deleted]
Step 5. In R4 the 2 can't go in R4C5 because this would make R3C45 = {79}, contradicting R3C9. Only place left is R4C4 = 2 -> R4C3 = 1, R7C23 = [13], R3C23 = [35], R6C23 = [57].
Step 6. The 20(3) cage in R4 has now been narrowed down to {389} or {578}, i.e. it must contain an 8. So R4C5 = {35}, and the only remaining combos for the 12(3) cage in N5 = {147} or {156}. So 1 is locked in R5 in N5 and R6C8 = 1 (can't be in 19(3) cage obviously because this would make the remaining two cells > 17) -> R1C89= [31]. Subtraction combo on remaining cells in R6 ({4689}) -> R6C4 = 8, R6C567 = {469}.
Step 7. 45 on R7 -> R8C6 = R7C1 - 5, i.e. R7C1 = {69}, R8C6 = {14}. R7C89 = [78] or [96]. R7C45 = {29|47|56}. So we have a complex triple on {679} in R7! So R7C67 = {45}, R8C6 = 1, R7C45 = [92], R7C1 = 6, R7C89 = [78], R9C7 = 1.
[Footnote: I'll leave this step as it is but I see that it's far more complex than necessary: the simplest thing is to note that in R7 the 8 must go in R7C89. So R7C89 = [78]; in turn R7C67 must have the 4 in them, so the 10(3) cage has {145} in it. No innie-outie differences or complex triples required!]
Step 8. 45 on N9 -> R7C67 = [54], R9C56 = {48}, R89C9 = {25}, R6C9 = R5C1 = 3, R6C1 = 2, R8C7 = R9C4 = 3, R8C45 = {67}, R8C3 = 4, R89C8 = [96], R89C1 = [59], R89C9 = [25], R8C2 = 8, R9C23 = [72], R5C23 = [98], R2C2 = 2.
Step 9. In N6 the 2 and 4 must be in 16(4). So the remaining cell of the 16(4) = 7, which goes in R5C7. R4C9 = 9, R4C6 = 7, R4C78 = {58}, R4C5 = 3, R5C6 = 6, R5C89 = [24], R2C9 = 6, R23C8 = [54]. 45 rule on R1 -> R2C57 = 9 = [18]. And you carry on....
Last edited by nd on Thu Aug 10, 2006 10:47 pm, edited 4 times in total.
Thanks nd for the walkthru. You definately made it easier than the way(s!) I took.
The main thing I missed was Step 7 – I focussed on the 3 innies in r7 instead of your much more helpful 1 outie, 1 innie.
I like your 'subtraction combo' in step 6 – another opportunity to ditch the combo chart -
(BTW – I've discovered (the very hard way!) a few mistakes in Miyuki Misawa's “Sum's- all Combinations” chart – in 17(4) and 40(8) – any others I've missed?).
Now, about your step 7 – I can't 'see' the complex triple on {678}. Must need more practice at understanding them so I can spot 'em. Got any suggestions? What I can see is that r7c89 can't be [69] since 6 or 9 is needed in r7c1 -> r7c89=[78] etc, ie exactly the same result as you get with the (too!) complex (for me!) triple.
Also, can you explain from Step 8 how r8c7=r9c4? Is that just coincidence since they both = 3, or is there a logical connection?
I like your step 5 too, eliminating the 2 from r4c5 – missed that.
Now, a bit of nit-picky stuff – only because you asked!
Step 3 – in r4c12 you include an [82] option – your step 1 has already (implicitly) eliminated 8 from there because r5c23 = {89}. So that makes your step 4 (mostly) redundant.
Step 6 – What happened to the 8 in r4c5? Presumably it was eliminated after step 3? – because the 20(3) cage in r4 has become 8{39/57}?
Step 7 - “r8c6 = {69}, r7c1= {14}” - should be the other way around?
Step 9 - “r5c7 = 6” should be r6. “45 rule on r2” should be on r1
Anyway, as usual, I've learnt some more. Thanks ND!
Now, time to get a couple hours sleep before the Aussies cause the upset of the century by beating the Italians. Starts at 1 am local time! Hopefully we get the World Cup here in 2014. My condolences to any Italians reading this!
(If you're Italian Ruud – please don't delete this post – I promise not to gloat!)
The main thing I missed was Step 7 – I focussed on the 3 innies in r7 instead of your much more helpful 1 outie, 1 innie.
I like your 'subtraction combo' in step 6 – another opportunity to ditch the combo chart -
(BTW – I've discovered (the very hard way!) a few mistakes in Miyuki Misawa's “Sum's- all Combinations” chart – in 17(4) and 40(8) – any others I've missed?).
Now, about your step 7 – I can't 'see' the complex triple on {678}. Must need more practice at understanding them so I can spot 'em. Got any suggestions? What I can see is that r7c89 can't be [69] since 6 or 9 is needed in r7c1 -> r7c89=[78] etc, ie exactly the same result as you get with the (too!) complex (for me!) triple.
Also, can you explain from Step 8 how r8c7=r9c4? Is that just coincidence since they both = 3, or is there a logical connection?
I like your step 5 too, eliminating the 2 from r4c5 – missed that.
Now, a bit of nit-picky stuff – only because you asked!
Step 3 – in r4c12 you include an [82] option – your step 1 has already (implicitly) eliminated 8 from there because r5c23 = {89}. So that makes your step 4 (mostly) redundant.
Step 6 – What happened to the 8 in r4c5? Presumably it was eliminated after step 3? – because the 20(3) cage in r4 has become 8{39/57}?
Step 7 - “r8c6 = {69}, r7c1= {14}” - should be the other way around?
Step 9 - “r5c7 = 6” should be r6. “45 rule on r2” should be on r1
Anyway, as usual, I've learnt some more. Thanks ND!
Now, time to get a couple hours sleep before the Aussies cause the upset of the century by beating the Italians. Starts at 1 am local time! Hopefully we get the World Cup here in 2014. My condolences to any Italians reading this!
(If you're Italian Ruud – please don't delete this post – I promise not to gloat!)
Ed--thanks for catching a few typos & roundabout paths! I've deleted step 4 but left the numbering intact to avoid confusion in reference to your comments.
Re: the complex triple in R7 in step 7: R7C1 = {69}, the pair R7C45 must contain {679}, and R7C8 = {79}.
In step 8: well I could have just said R8C7 = 3 (this is simple as there is nowhere else for it to go in N9) -> R9C4 = 3 but was trying to use shorthand.
re: combinations: I never use charts -- not all that reliable. I usually just work out combinations for myself on paper if needed.
Re: the complex triple in R7 in step 7: R7C1 = {69}, the pair R7C45 must contain {679}, and R7C8 = {79}.
In step 8: well I could have just said R8C7 = 3 (this is simple as there is nowhere else for it to go in N9) -> R9C4 = 3 but was trying to use shorthand.
re: combinations: I never use charts -- not all that reliable. I usually just work out combinations for myself on paper if needed.