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MathGroup Archive 2003

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Re: Simplify a module

  • To: mathgroup at smc.vnet.net
  • Subject: [mg39490] Re: Simplify a module
  • From: Dr Bob <drbob at bigfoot.com>
  • Date: Wed, 19 Feb 2003 04:42:20 -0500 (EST)
  • References: <200302172317.SAA14748@smc.vnet.net>
  • Reply-to: drbob at bigfoot.com
  • Sender: owner-wri-mathgroup at wolfram.com

Here's a marginally simpler solution:

Cases[Hold[
      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];], If[
      Mod[_, g_] == h_, ax_ = k_] :> {ax, g, h, k}, Infinity];
split = Split[%, #1[[1]] === #2[[1]] &];
mytab = {#[[1, {1, 2}]], Flatten[#[[All, {3, 4}]]]} & /@ split;
dcalc3[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[Sequence @@ #[[2]]], _, 0]) & /@ mytab;

dcalc3@# == dcalc2@# & /@ Range[1176];
And @@ %

True

Could someone explain why something like the following doesn't work?  I've 
tried several versions and haven't been able to use a named function like 
sw in place of the identical in-line function.

sw[{{a_, base_}, seq_}] := (Evaluate[a] = Switch[Mod[e, base], 
Evaluate[Sequence @@ seq], _, 0])
dcalc4[e_] := Block[{a1, a2,
         a3}, {T1, M1, T2, M2, T3, M3} = {2, 392, 3, 147, 47, 24};
    Evaluate[sw /@ mytab];
    Mod[a1*T1*M1 + a2*T2*M2 + a3*T3*M3, 1176]]

Bobby

On Mon, 17 Feb 2003 18:17:34 -0500 (EST), Wolf, Hartmut <Hartmut.Wolf@t- 
systems.com> wrote:

>
>> -----Original Message-----
>> From: flip [mailto:flip_alpha at safebunch.com]
To: mathgroup at smc.vnet.net
> To: mathgroup at smc.vnet.net
>> Sent: Sunday, February 16, 2003 12:14 PM
>> To: mathgroup at smc.vnet.net
>> Subject: [mg39490]  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)
>
>
>



-- 
majort at cox-internet.com
Bobby R. Treat



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