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PostPosted: Fri Jan 10, 2020 9:50 pm 
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Grand Master
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 603
Semi-Symmetrical NC Killer 4

I commented on the players forum that I could not find an NC puzzle that was fully Semi-symmetric (i.e. at least on number is paired with itself in at least one position pair) all I could find were asymmetric ones. Wecoc took this as a challenge and found a set of them, posting a nice vanilla puzzle.

From his puzzle I made a killer which I re-post here. I have solved it a couple of ways but find my solutions mucky, can anyone post a neat solution?



Image

Wecoc's solution:
725849163
948361725
163527948
816273594
594618372
372495816
639752481
257184639
481936257


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PostPosted: Sun Jan 12, 2020 2:47 am 
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Grand Master
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 203
Location: California, out of London
I don't know if this is any less mucky that yours HATMAN - but here's how I started and identified the pairings.
Semi-Symmetric NC Killer 4 - Partial WT:
1. NC -> 7(2) = {16} or {25}
-> Of 7(2)n1 and 7(2)c4 - one of them is {25} and the other is {16}

2. If 7(2)c6 has different values from 7(2)c4 -> one each of (16) pairs with one each of (25)
(SS) But this would require 9(2)n9 (paired with 7(2)n1) to have values from the set (1256) which is impossible.
-> 7(2)c6 has the same values as 7(2)c4

3. Trying 7(2)c4 = {16} puts 8(2)c4 = {35}, 7(2)c6 = {16}, 8(2)c6 = {35}, 7(2)n1 = {25}
Puts {16} as a pair
Also puts (35) in n5 in r456c5
(NC) puts 4 in n5 in r5c46
(NC) puts r46c5 = {35} -> (35) is a pair
But this would require 9(2)n9 to have a 3 or 5. (SS)
But it cannot be {45} (NC) or {36} since (SS) 6 pairs with a 1 and there is no 1 or 6 in 7(2)n1
-> 7(2)c4 = {25}

4. -> 7(2)n1 = {16}, 8(2)c4 = {17}, 7(2)c6 = {25}, 8(2)c6 = [17]
-> (25) is a pair

5. NC -> 13(2)n4 from {58} or {49}
But cannot be {58} since that would require 10(2)n6 to have a 2. (SS)
-> 13(2)n4 = {49}
-> 10(2)n6 cannot have a 2 (since 2 would require a paired 5 in 13(2)n4)
-> 10(2)n6 = {37}
(NC) -> 14(2)n6 = {59}
(SS) -> 9(2)n4 has a 2 or a 5 -> (NC) 9(2)n4 = {27}
-> (79) is a pair
-> (34) is a pair

6. (79) is a pair -> r46c5 = [79]
-> HS 1 in c5 -> r5c5 = 1 (unpaired)
-> Last remaining pair = (68)


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