
Pairings
Posted:
Jul 17, 1996 1:41 AM


I want a function that finds, given n, all pairings of Range[n] excluding {1,2}, {2,3}, {n2,n1}, and {n1,n}. One approach would be to find the appropriate permutations of Range[n] and then use Map[Partition[#,2],%]&. For example, I would want specialpairings[4] to return {{{1,3},{2,4}}} and specialpairings[6] to return {{{1,3},{2,5},{4,6}}, {{1,4},{2,5},{3,6}}, {{1,4},{2,6},{3,5}}, {{1,5},{2,4},{3,6}}, {{1,5},{2,6},{3,4}}, {{1,6},{2,4},{3,5}}, {{1,6},{2,5},{3,4}}}. I have worked out a formula for the number of special pairings as a function of n, and Length[specialpairings[8]] should be 57.
The wellknown and easily derived formula for the number of pairings of Range[n] is
numberofpairings[n_]:=n!/((2^(n/2))(n/2)!)
I can show that the number of special pairings is given by
numberofspecialpairings[n_]:=((n^28n+19)/((n1)(n3))) numberofpairings[n]
Since these formulas are asymptotic to each other as n gets large, it's completely reasonable to generate all pairings and throw out those which contain any of the four "bad" pairs {1,2}, {2,3}, {n2,n1}, and {n1,n}.
The following naive method returns the desired results:
specialpairings[n_Integer]:= Select[Union[ Map[Sort[#] &, Map[Partition[#,2] &, Permutations[Range[n]]],2]], Intersection[#,{{1,2},{2,3},{n2,n1},{n1,n}}] == {} &]
Of course, this function is VERY slow since there is a LOT of duplication. I would like to use some sort of backtracking technique to make the function more efficient. Does anyone have any suggestions?
Rob Pratt Department of Mathematics The University of North Carolina at Chapel Hill CB# 3250, 331 Phillips Hall Chapel Hill, NC 275993250
rpratt@math.unc.edu

