RE: Simplify a module
- To: mathgroup at smc.vnet.net
- Subject: [mg39476] RE: [mg39450] Simplify a module
- From: "Wolf, Hartmut" <Hartmut.Wolf at t-systems.com>
- Date: Mon, 17 Feb 2003 18:17:34 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
>-----Original Message----- >From: flip [mailto:flip_alpha at safebunch.com] To: mathgroup at smc.vnet.net >Sent: Sunday, February 16, 2003 12:14 PM >To: mathgroup at smc.vnet.net >Subject: [mg39476] [mg39450] Simplify a module > > >Hi All, > >Can someone recommend a simplification to this module (just a >bunch of if >checks). > >It was specified to be done that way (and I know how to sove >the problem >using PowerMod in one step, so please bear with me here). > >Thanks, Flip > >(* to email me remove "_alpha" *) > >Anyway, here goes. > >dcalc[ein_] := Module[{e = ein, a1 = 0, a2 = 0, a3 = 0}, > {T1, M1, T2, M2, T3, M3} = {2, 392, 3, 147, 47, 24}; > If[Mod[e, 3] == 1, a1 = 1]; If[Mod[e, 3] == 2, a1 = 2]; > If[Mod[e, 8] == 1, a2 = 1]; > If[Mod[e, 8] == 3, a2 = 3]; If[Mod[e, 8] == 5, a2 = 5]; > If[Mod[e, 8] == 7, a2 = 7]; > If[Mod[e, 49] == 1, a3 = 1]; If[Mod[e, 49] == 2, a3 = 25]; > If[Mod[e, 49] == 3, a3 = 33]; If[Mod[e, 49] == 4, a3 = 37]; > If[Mod[e, 49] == 5, a3 = 10]; If[Mod[e, 49] == 6, a3 = 41]; > If[Mod[e, 49] == 8, a3 = 43]; If[Mod[e, 49] == 9, a3 = 11]; > If[Mod[e, 49] == 10, a3 = 5]; If[Mod[e, 49] == 11, a3 = 9]; > If[Mod[e, 49] == 12, a3 = 45]; If[Mod[e, 49] == 13, a3 = 34]; > If[Mod[e, 49] == 15, a3 = 36]; If[Mod[e, 49] == 16, a3 = 46]; > If[Mod[e, 49] == 17, a3 = 26]; If[Mod[e, 49] == 18, a3 = 30]; > If[Mod[e, 49] == 19, a3 = 31]; If[Mod[e, 49] == 20, a3 = 27]; > If[Mod[e, 49] == 22, a3 = 29]; If[Mod[e, 49] == 23, a3 = 32]; > If[Mod[e, 49] == 24, a3 = 47]; If[Mod[e, 49] == 25, a3 = 2]; > If[Mod[e, 49] == 26, a3 = 17]; If[Mod[e, 49] == 27, a3 = 20]; > If[Mod[e, 49] == 29, a3 = 22]; If[Mod[e, 49] == 30, a3 = 18]; > If[Mod[e, 49] == 31, a3 = 19]; If[Mod[e, 49] == 32, a3 = 23]; > If[Mod[e, 49] == 33, a3 = 3]; If[Mod[e, 49] == 34, a3 = 13]; > If[Mod[e, 49] == 36, a3 = 15]; If[Mod[e, 49] == 37, a3 = 4]; > If[Mod[e, 49] == 38, a3 = 40]; If[Mod[e, 49] == 39, a3 = 44]; > If[Mod[e, 49] == 40, a3 = 38]; If[Mod[e, 49] == 41, a3 = 6]; > If[Mod[e, 49] == 43, a3 = 8]; If[Mod[e, 49] == 44, a3 = 39]; > If[Mod[e, 49] == 45, a3 = 12]; If[Mod[e, 49] == 46, a3 = 16]; > If[Mod[e, 49] == 47, a3 = 24]; If[Mod[e, 49] == 48, a3 = 48]; > Return[Mod[a1*T1*M1 + a2*T2*M2 + a3*T3*M3, 1176]]] > > > > > > > > Flip, to "simplify" your module, resort to metaprogramming, i.e. write another program that constructs that module from data you supply from a table. The following is a somewhat foolish example, just to show the idea. To begin with, I cut out part of your coding... In[2]:= Cases[Hold[If[Mod[e, 3] == 1, a1 = 1]; If[Mod[e, 3] == 2, a1 = 2]; If[Mod[e, 8] == 1, a2 = 1]; ......,and so on...... If[Mod[e, 49] == 47, a3 = 24]; If[Mod[e, 49] == 48, a3 = 48];], If[Mod[_, g_] == h_, ax_ = k_] :> {ax, g, h, k}, Infinity]; ...to extract your data In[3]:= Split[%, #1[[1]] === #2[[1]] &]; ...group, and transform it to a handy structure In[4]:= ctab = {#[[1, {1, 2}]], #[[All, {3, 4}]]} & /@ %; Of course in future you'll just start with ctab, or something similar. In[5]:= dcalc2[e_] := Block[{a1, a2, a3}, {T1, M1, T2, M2, T3, M3} = {2, 392, 3, 147, 47, 24}; (Evaluate[#[[1, 1]]] = Switch[Mod[e, #[[1, 2]]], Evaluate[Apply[Sequence, #[[2, All]], {0, 1}]], _, 0 ]) & /@ ctab; Mod[a1*T1*M1 + a2*T2*M2 + a3*T3*M3, 1176]] In[6]:= d100 = dcalc /@ Range[100]; In[7]:= d100x = dcalc2 /@ Range[100]; In[8]:= d100 == d100x Out[8]= True As said, just to get an idea; this is not a suggestion as how to code it, don't use the hard-wired symbols a1, a2, a3 between ctab and dcalc2 (generate them as needed) or thread, avoid global T1, M2, etc. Your exercise. -- Hartmut PS, perhaps something like dcalc3[e_] := Mod[Plus @@ MapThread[ Switch[Mod[e, #1[[1, 2]]], Evaluate[Apply[Sequence, #1[[2, All]], {0, 1}]], _, 0 ]*#2*#3 &, {ctab, {2, 3, 47}, {392, 147, 24}}], 1176] (a1, etc. not used)
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