RE: A Functional Programming Question

*To*: mathgroup at smc.vnet.net*Subject*: [mg24734] RE: [mg24705] A Functional Programming Question*From*: "David Park" <djmp at earthlink.net>*Date*: Wed, 9 Aug 2000 02:31:35 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

Thanks everybody for the replies. I zeroed in on two of them, the one by Johannes Ludsteck and the one by Allan Hayes because they seemed to be adaptable to a more general parallel manipulation of equations. Here is a somewhat more general set of parallel equations: eqns = {(a1 x + b1)/c1 == 0, (a2 y + b2)/c2 == 0, (a3 z + b3)/c3 == 0}; Suppose that we wish to solve for the square roots of x, y and z. Here is a routine which is an adaptation of Johannes method: parallelmap[eqns_, op_, parms_] := MapThread[ Function[parm, Function[eq, op[#, parm] & /@ eq]][#2][#1] &, {eqns, parms}] We can now manipulate the equations in parallel, step-by-step. parallelmap[eqns, #1#2 &, {c1, c2, c3}] parallelmap[%, #1 - #2 &, {b1, b2, b3}] parallelmap[%, #1/#2 &, {a1, a2, a3}] parallelmap[%, Sqrt[#] &, {d, d, d}] {b1 + a1 x == 0, b2 + a2 y == 0, b3 + a3 z == 0} {a1 x == -b1, a2 y == -b2, a3 z == -b3} {x == -(b1/a1), y == -(b2/a2), z == -(b3/a3)} {Sqrt[x] == Sqrt[-(b1/a1)], Sqrt[y] == Sqrt[-(b2/a2)], Sqrt[z] == Sqrt[-(b3/a3)]} The last step works because the function only looks for one argument and we can just put in dummy values for the second argument. Using Sqrt instead of Sqrt[#]& will not work. Working with Allan Hayes method, I came up with this routine: parallelmap2[eqns_, op_, parms_] := Equal @@@ op[List @@@ eqns, parms] parallelmap2[eqns2, #1*#2 & , {c1, c2, c3}] parallelmap2[%, #1 - #2 & , {b1, b2, b3}] parallelmap2[%, #1/#2 & , {a1, a2, a3}] parallelmap2[%, Sqrt[#1] & , {d, d, d}] gives the same answers. David Park djmp at earthlink.net http://home.earthlink.net/~djmp/