lists of pairs

*To*: mathgroup at smc.vnet.net*Subject*: [mg8486] lists of pairs*From*: Russell Towle <rustybel at foothill.net>*Date*: Tue, 2 Sep 1997 16:15:27 -0400*Sender*: owner-wri-mathgroup at wolfram.com

Hi all, I have a problem in discarding duplicates from a list of integer pairs, where if pairs of the form {a, b} and {b,a} exist in the list, I wish to discard (or select) only one, it doesn't matter which. The method I use below works, but is slow. First, here is a sample list of integer pairs: edges = {{16, 10}, {10, 4}, {4, 9}, {9, 15}, {15, 16}, {16, 15}, {15, 18}, {18, 20}, {20, 19}, {19, 16}, {16, 19}, {19, 14}, {14, 8}, {8, 10}, {10, 16}, {10, 8}, {8, 3}, {3, 1}, {1, 4}, {4, 10}, {4, 1}, {1, 2}, {2, 7}, {7, 9}, {9, 4}, {9, 15}, {15, 18}, {18, 13}, {13, 7}, {7, 9}, {19, 20}, {20, 17}, {17, 12}, {12, 14}, {14, 19}, {8, 14}, {14, 12}, {12, 6}, {6, 3}, {3, 8}, {1, 3}, {3, 6}, {6, 5}, {5, 2}, {2, 1}, {7, 2}, {2, 5}, {5, 11}, {11, 13}, {13, 7}, {18, 13}, {13, 11}, {11, 17}, {17, 20}, {20, 18}, {12, 17}, {17, 11}, {11, 5}, {5, 6}, {6, 12}} It is this long: In[35]:= Length [ edges ] Out[35]= 60 However, this is twice as long as I want, for every pair {a, b} has an unwanted opposite {b, a}. This is my present method: (*find duplicates (indices in reverse order)*) (*7.60 seconds*) c=Table[ Position[edges, x_ /; x == edges[[i]] || x == Reverse[edges[[i]]] ], {i,Length[edges]}]; Now I have the positions of all duplicates, twice over, so I use Union and Take the First elements. f=Map[Take[#,1]&, Union[c]] And, this is the final list I wanted from the beginning: f=Flatten[f,2] Suggestions? It is desired to minimize the time it takes to do this. Russell Towle Giant Gap Press: books on California history, digital topographic maps P.O. Box 141 Dutch Flat, California 95714 ------------------------------ Voice: (916) 389-2872 e-mail: rustybel at foothill.net ------------------------------