RE: Interval arithmetic, of a sort
- To: mathgroup at smc.vnet.net
- Subject: [mg33762] RE: [mg33721] Interval arithmetic, of a sort
- From: F.H.Simons at tue.nl
- Date: Thu, 11 Apr 2002 02:14:23 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
Here is a function that does the job:
visibleintervals[intervals_] := Module[{f, n = Length[intervals] + 1, res},
(* define the function *)
(n = n - 1; f[x_ /; #[[1]] <= x <= #[[2]]] = n) & /@ Reverse[intervals];
(* find all subintervals and the corresponding value of the function *)
res = {f[(#[[1]] + #[[2]])/2], #} & /@
Partition[Union @@ intervals, 2, 1];
(* combine neighbouring subintervals with the same function value *)
{#[[1, 1]], {#[[1, 2, 1]], #[[-1, 2, 2]]}} & /@
Split[res, #1[[1]] == #2[[1]] && #1[[2, 2]] == #2[[2, 1]] &] ]
visibleintervals[{{0, 100}, {2, 3}, {1, 20}, {15, 30}}]
{{1, {0, 1}}, {3, {1, 15}}, {4, {15, 30}}, {1, {30, 100}}}
In the first line of the Module for each interval the function f is defined
only on that interval with value the number of that interval. We start with
the last interval, so our definitions are ordered according to decreasing
function values. Mathematica uses the definitions in this order, so for any
argument in any interval the returned value is the highest interval.
Fred Simons
Eindhoven University of Technology