Re: Sort into chain

*To*: mathgroup at smc.vnet.net*Subject*: [mg45660] Re: Sort into chain*From*: bobhanlon at aol.com (Bob Hanlon)*Date*: Sat, 17 Jan 2004 02:34:29 -0500 (EST)*References*: <bu8ia8$aqj$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Use ReplaceRepeated: a //. {s___, {x_, y_}, m__, {y_, z_}, e___} :> {s, {x, y}, {y, z}, m, e} Bob Hanlon In article <bu8ia8$aqj$1 at smc.vnet.net>, Goyder Dr HGD <h.g.d.goyder at cranfield.ac.uk> wrote: << How do I sort a list of pairs, for example, a = {{3, 1}, {2, 6}, {7, 3}, {8, 7}, {10, 8}, {6, 11}, {12, 10}, {11, 14}, {15, 12}, {14, 18}, {19, 15}, {18, 20}, {20, 19}, {1, 2}}; Into a list where the second entry in each pair is the same as the first entry in the next pair. Thus the effect of the sort will give {{18, 20}, {20, 19}, {19, 15}, {15, 12}, {12, 10}, {10, 8}, {8, 7}, {7, 3}, {3, 1}, {1, 2}, {2, 6}, {6, 11}, {11, 14}, {14, 18}} The first pair is not important because the pairs form a circular chain. I have tried Sort[a, (#1[[2]] == #2[[1]]) &] but the answer is {{1, 2}, {20, 19}, {14, 18}, {18, 20}, {19, 15}, {15, 12}, {6, 11}, {11, 14}, {12, 10}, {10, 8}, {8, 7}, {7, 3}, {2, 6}, {3, 1}} which has a few correct entries but is not finished. I wondered if the difficulty was due to the circular nature of the chain but if you break the chain by dropping one pair it still does not work.