SudokuSolver Forum

A forum for Sudoku enthusiasts to share puzzles, techniques and software
It is currently Tue Sep 29, 2020 2:55 pm

All times are UTC




Post new topic Reply to topic  [ 4 posts ] 
Author Message
PostPosted: Fri Jan 10, 2020 9:50 pm 
Offline
Grand Master
Grand Master

Joined: Wed Apr 30, 2008 9:45 pm
Posts: 639
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


Top
 Profile  
Reply with quote  
PostPosted: Sun Jan 12, 2020 2:47 am 
Offline
Grand Master
Grand Master

Joined: Tue Jun 16, 2009 9:31 pm
Posts: 215
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)


Top
 Profile  
Reply with quote  
PostPosted: Wed Jan 22, 2020 9:50 pm 
Offline
Grand Master
Grand Master

Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1728
Location: Lethbridge, Alberta, Canada
I started the same way as wellbeback but then took a very different path. I also got the paired numbers fairly early; the rest is a lot of detailed steps to finish the puzzle.

Here is my walkthrough for Semi-Symmetric 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?”

Prelims, including effect of NC

a) R23C6 = {17/26/35}
b) R3C12 = {16/25} (cannot be {34})
c) R34C4 = {16/25} (cannot be {34})
d) R4C78 = {59/68}
e) R5C23 = {49/58} (cannot be {67})
f) R5C78 = {19/28/37/46}
g) R6C23 = {18/27/36} (cannot be {45})
h) R67C6 = {16/25} (cannot be {34})
i) R78C4 = {17/26/35}
j) R7C89 = {18/27/36} (cannot be {45})

NC only used as stated.

1a. R23C6 = {17/35} (cannot be {26} which clashes with R67C6)
1b. Killer pair 1,5 in R23C6 and R67C6, locked for C6
1c. R78C4 = {17/35} (cannot be {26} which clashes with R34C4)
1d. Killer pair 1,5 in R34C4 and R78C4, locked for C4

2a. R3C12 = {16/25} must have different combination from R34C4
2b. R3C12 = {16/25} corresponds with R7C89 = {18/27/36} -> {16} cannot correspond with {25} -> R34C4 must have the same combination as R67C6
[Possibly slightly simplified; see also wellbeback’s start.]
2c. R4C4 cannot be the same as R6C6 -> R34C4 and R67C6 must have the same vertical order

3a. R4C78 cannot be {68} because R5C78 = {19} (cannot be {37}, NC), R5C23 = {58}, R6C23 = {27/36}, R4C78 corresponds with R6C23, R5C23 corresponds with R5C78 but 8 cannot correspond both with one of 1,9 and one of 2,3,6,7
3b. R4C78 = {59}, locked for R4 and N6, clean-up: no 2 in R3C4, no 1 in R6C78
3c. R34C4 = {16}/[52] -> R67C6 = {16}[52] (steps 2b and 2c), no 2 in R6C6, no 5 in R7C6

4. R3C12 corresponds with R7C89, R4C78 corresponds with R6C23, R6C23 and R7C89 are both {18/27/36} neither containing 5, R4C78 = {59} corresponds with one of the combinations in R6C23
-> R3C12 = {16} (cannot be {25} because {25} and {59} cannot correspond with the same combination), locked for R3 and N1
4a. R3C45 = [52] -> R67C6 = [52] (step 2c), R78C4 = {17}, locked for C4 and N8, R23C6 = [17], clean-up: no 7 in R7C89
4b. R4C4 corresponds with R6C6 -> 2,5 paired
4c. R4C78 = {59} corresponds with R6C23 -> R6C23 = {27}, locked for R6 and N4
4d. 7 paired with 9
4e. R5C23 corresponds with R5C78, R5C23 contains one of 5,9 -> R5C78 must contain one of 2,7 but {58} cannot correspond with {28} since they are in the same row -> R5C23 = {49}, locked for R5 and N4, R5C78 = {37}, locked for R5 and N6
4f. R6C6 = 5 -> R5C46 = [68] (NC) -> R5C5 = 1, R5C19 = [52], R5C23 = [94] (NC), R5C78 = [37] (NC)
4g. 3,4 paired, 6,8 paired
4h. R4C5 = 7 (hidden single in N5)
4i. 1 can only correspond with itself
4j. R3C12 corresponds with R7C89, R3C12 = {16} -> R7C89 = {18} (cannot be {36} because 3 is paired with 4), locked for R7 and N9 -> R78C4 = [71]

[The rest will be very dependent on NC and corresponding cells.]
5a. R5C1 = 5 -> no 6 in R46C1 (NC)
5b. 6 in N4 only in R4C23, locked for R4
5c. R5C9 = 2 -> no 1 in R46C9 (NC)
5d. R46C1 correspond with R46C9, no 1 in R46C9 -> no 1 in R46C1
5e. Naked pair {38} in R46C1, locked for C1 and N4
5f. R46C1 corresponds with R46C9, 3 in R46C1 -> R46C9 must contain 4, locked for C9 and N6
5g. R5C8 = 7 -> R6C8 = 1 (NC), R7C89 = [81]
5h. R3C1 corresponds with R7C9 -> R3C12 = [16], R4C23 = [16]
5i. R1C7 = 1 (hidden single in N3)
5j. R9C3 = 1 (hidden single in N7)
5k. R4C5 corresponds with R6C5, R4C5 = 7 -> R6C5 = 9
5l. R6C6 = 5 -> R6C7 = 8 (NC), R46C9 = [46], R4C6 = 3, R46C1 = [83], R6C4 = 4, R4C78 = [59] (NC), R6C23 = [72] (NC)
5m. R3C4 = 5 -> no 4 in R3C5 (NC)
5n. R4C7 = 5 -> no 4 in R3C7 (NC)
5o. R3C8 = 4 (hidden single in R3)
5p. R6C3 = 2 -> no 3 in R7C3 (NC)
5q. R7C6 = 2 -> no 3 in R7C5 (NC)
5r. R7C2 = 3 (hidden single in R7)
5s. R7C8 = 8 -> no 9 in R7C7 (NC)
5t. 9 in R7 only in R7C13, locked for N7
5u. R3C3 corresponds with R7C7, R7C7 = {46} -> R3C3 = {38}
5v. 9 in R3 only in R3C79, locked for N3

6a. R12C4 = {38/39} (cannot be {89}, NC), 3 locked for C4 and N2
6b. R3C6 = 7 -> R3C5 = 2 (NC), R3C7 = 9
6c. R3C8 = 4 -> R3C9 = 8 (NC)
6d. R3C3 corresponds with R7C7, R3C3 = 3 -> R7C7 = 4
6e. R3C5 corresponds with R7C5, R3C5 = 2 -> R7C5 = 5, R7C13 = [69]
6f. R89C7 = {26/27} (cannot be {67}, NC), 2 locked for C7 and N9
6g. R89C8 = {35/36} (cannot be {56}, NC), 3 locked for C8 and N9
6h. R1C7 = 1 -> no 2 in R1C8 (NC)
6i. R2C8 = 2 (hidden single in N3)
6j. R7C2 = 3 -> no 2,4 in R8C2 (NC)
6k. R2C8 corresponds with R8C2, R2C8 = 2 -> R8C2 = 5
6l. R2C1 corresponds with R8C9, R8C9 = {79} -> R2C1 = {79}
6m. R2C1 = {79} -> R2C2 = 4 (NC)
6n. R2C2 corresponds with R8C8, R2C2 = 4 -> R8C8 = 3
6o. R7C5 = 5 -> R28C5 = [68] (NC), R2C7 = 7
6p. R2C2 = 4 -> R2C3 = 8 (NC)
6q. R8C8 = 3 -> R8C7 = 6 (NC)

and the rest is naked singles, without using corresponding pairs.


Top
 Profile  
Reply with quote  
PostPosted: Thu Jan 23, 2020 4:36 pm 
Offline
Grand Master
Grand Master

Joined: Wed Apr 30, 2008 9:45 pm
Posts: 639
Thank you both, you delivered two solutions similar to mine but neater.

Wellbeback’s {16} elimination was like my first attempt, but of course neater and shorter. On my second attempt I tried to avoid this long elimination and started on the 14/2 so similar to Andrew’s but much muckier.

I particularly liked Andrew’s 4 but I would have phrased it slightly differently:


4. R3C12 corresponds with R7C89, R4C78 corresponds with R6C23, R6C23 and R7C89 are both {18/27/36} neither containing 5
R4C78 = {59}: If R3C12 = {25} then the fives dictate they must both be with the same combination but the two and nine dictate that they cannot be; hence R3C12 = {16} etc.


Top
 Profile  
Reply with quote  
Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 4 posts ] 

All times are UTC


Who is online

Users browsing this forum: Bing [Bot] and 9 guests


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Search for:
Jump to:  
cron
Powered by phpBB® Forum Software © phpBB Group