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*To*: mathgroup at smc.vnet.net*From*: Dr Bob <drbob at bigfoot.com>*Date*: Thu, 23 Oct 2003 07:14:38 -0400 (EDT)*Reply-to*: drbob at bigfoot.com*Sender*: owner-wri-mathgroup at wolfram.com

Here's a simple answer (not necessarily the fastest): overlapping[{a_, b_}, {c_, d_}] := Or[a == c, a < c < b, c < a < d] Count[Flatten@Outer[overlapping, v1, v2, 1], True] More efficient algorithms are possible if you can guarantee the bricks in each list are non-overlapping and exhaustive. Bobby recode procedural algorithm to faster functional module Subject: [mg44097] [mg44079] [mg44034] recode procedural algorithm to faster functional module From: "Prince-Wright, Robert G SEPCO" <robert.prince-wright at shell.com> To: mathgroup at smc.vnet.net Can someone please help me recode this Module so it is less procedural and hopefully runs a lot faster. The Lists V1 and V2 represent two time series with 'bricks' laid out along them. The Bricks have varying length set by v[[i]][[1]] and v[[i]][[2]] and the idea is to count the number of times there is an overlap. I can only see the dumb procedural way of taking each brick and checking if it overlaps in time with another! Concurrence[v1_List, v2_List,nsim_]:=Module [ {k=1,c=0}, Do[ Do[ Which[ v1[[i]][[1]] == v2[[k]][[1]], c=c+1, v1[[i]][[1]] < v2[[k]][[1]] && v1[[i]][[2]] > v2[[k]][[1]], c=c+1, v1[[i]][[1]] > v2[[k]][[1]] && v1[[i]][[1]] < v2[[k]][[2]], c=c+1 ]; (*Write[ strm1, {v1[[i]][[1]],v2[[i]][[1]],c}];*) , {i,1,nsim} ], {k,1,nsim} ]; c ] I've have a PowerPoint illustration if this is unclear and can e-mail it. thanks Robert Prince-Wright