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See also A000375, which is the main entry for this problem.
Comments from Joshua Zucker, Jan 24 2007: (Start)
For some n, there are some longest chains which end up sorted and some which don't, like n = 6, with the following four chains that end sorted and one that does not:
The examples below use 0 through n-1 instead of 1 through n.
((#(4 5 3 0 2 1) #(2 0 3 5 4 1) #(3 0 2 5 4 1) #(5 2 0 3 4 1) #(1 4 3 0 2 5) #(4 1 3 0 2 5) #(2 0 3 1 4 5) #(3 0 2 1 4 5) #(1 2 0 3 4 5) #(2 1 0 3 4 5) #(0 1 2 3 4 5))
(#(3 4 5 1 0 2) #(1 5 4 3 0 2) #(5 1 4 3 0 2) #(2 0 3 4 1 5) #(3 0 2 4 1 5) #(4 2 0 3 1 5) #(1 3 0 2 4 5) #(3 1 0 2 4 5) #(2 0 1 3 4 5) #(1 0 2 3 4 5) #(0 1 2 3 4 5))
(#(3 0 4 1 5 2) #(1 4 0 3 5 2) #(4 1 0 3 5 2) #(5 3 0 1 4 2) #(2 4 1 0 3 5) #(1 4 2 0 3 5) #(4 1 2 0 3 5) #(3 0 2 1 4 5) #(1 2 0 3 4 5) #(2 1 0 3 4 5) #(0 1 2 3 4 5))
(#(2 5 4 0 3 1) #(4 5 2 0 3 1) #(3 0 2 5 4 1) #(5 2 0 3 4 1) #(1 4 3 0 2 5) #(4 1 3 0 2 5) #(2 0 3 1 4 5) #(3 0 2 1 4 5) #(1 2 0 3 4 5) #(2 1 0 3 4 5) #(0 1 2 3 4 5)))
(#(3 0 5 4 1 2) #(4 5 0 3 1 2) #(1 3 0 5 4 2) #(3 1 0 5 4 2) #(5 0 1 3 4 2) #(2 4 3 1 0 5) #(3 4 2 1 0 5) #(1 2 4 3 0 5) #(2 1 4 3 0 5) #(4 1 2 3 0 5) #(0 3 2 1 4 5))
And for some n, e.g. n=12, none of the longest chains end up sorted (and hence A000375 and A000376 are different). I did an exhaustive search with n = 12 working backwards from sorted and found 63 as the longest chain beginning with one of these four:
#(9 10 6 0 2 7 1 8 11 5 3 4)
#(9 10 6 0 1 2 7 8 11 5 3 4)
#(7 8 11 5 0 6 10 9 2 1 3 4)
#(5 0 1 7 10 3 11 8 9 6 2 4)
But 65 is attainable if you don't assume it ends up sorted. (End)
Comment from Quan T. Nguyen, William Fahle (tuongquan.nguyen(AT)utdallas.edu), Oct 21 2010: (Start)
(6 14 9 2 15 8 1 3 4 12 18 5 10 13 16 17 11 7) is the only permutation of length 18 that takes 191 steps (also-called longest winded permutation) of "topswopping moves" before it goes to the identity permutation (i.e. in sorted order).
For n=17, (2 10 15 11 7 14 5 16 6 4 17 13 1 3 8 9 12) is the only longest-winded permutation (or order of cards) that takes 159 steps of "topswopping moves" before it terminates in sorted order, i.e. (1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17). (End)
Comment from Moritz Franckenstein, Apr 16 2011: (Start)
n=18 -> 191
6 14 9 2 15 8 1 3 4 12 18 5 10 13 16 17 11 7
n=19 -> 207
2 3 7 8 4 19 10 15 17 6 1 11 5 18 12 9 13 14 16
3 7 2 8 4 19 10 15 17 6 1 11 5 18 12 9 13 14 16
n=20 -> 231
5 20 6 3 2 15 10 1 16 9 18 14 19 7 12 17 8 11 13 4
3 20 5 6 2 15 10 1 16 9 18 14 19 7 12 17 8 11 13 4
5 20 2 6 3 15 10 1 16 9 18 14 19 7 12 17 8 11 13 4 (End)
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