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MathGroup Archive 2002

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RE: Can anyone simplify this?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg36477] RE: [mg36461] Can anyone simplify this?
  • From: "DrBob" <drbob at bigfoot.com>
  • Date: Sun, 8 Sep 2002 03:31:02 -0400 (EDT)
  • Reply-to: <drbob at bigfoot.com>
  • Sender: owner-wri-mathgroup at wolfram.com

If I haven't missed a step, the following should work nicely -- best
done before defining any of the symbols that appear:

Attributes[dummyIf] = HoldAll;
Attributes[dummyWhich] = HoldAll;
expr = {If[Aa == Ab, a Sa + b Sb, If[Ca == 0 || Cb == 0, a Sa + b Sb,
SS]], \
If[Aa == Ab, a Ca + b Cb, If[Ca == 0 || Cb == 0, a Ca + b 
        Cb, CC]], If[Ca == 0 && Cb == 0, "Nil", If[Z1 == Z2, If[Ca < 
            0, Aa, If[Cb < 0, Ab, 
      If[Ca > 0, If[Aa > 90, Aa - 90, Aa + 90], If[Cb > 0, 
        If[Ab > 90, Ab - 90, Ab + 90]]]]], If[Aa == Ab,
           Aa, If[Ca == 0 && Cb == 0, "Nil", If[
      Ca == 0, Ab, If[Cb == 0, Aa, If[Y == 0, "Nil", If[X > 180, 
        X - 180, If[X < 0, X + 180, X]]]]]]]]]} /. {If -> dummyIf};
rule1 = dummyIf[a_, b_, 
              dummyIf[c_, d_, e_]] -> dummyWhich[a, b, c, d, True, e];
rule2 = dummyIf[a_, dummyIf[
                b_, c_, d_], e_] -> dummyWhich[a && b, c, a, d, True,
e];
rule3 = dummyIf[a_, dummyIf[b_, c_, d_]] -> dummyWhich[a && b, c, a, d];
rule4 = dummyWhich[a__, b_, dummyIf[c_,
               d_, e_], f___] -> dummyWhich[a, b && c, d, b, e, f];
rule5 = dummyWhich[a__,
                   b_, dummyWhich[c_, 
                    d_, e___], f___] -> dummyWhich[a, b && 
                      c, d, b, dummyWhich[e], f];
rule6 = dummyWhich[] :> Null;
rule7 = HoldPattern[True && a_] :> a;
expr //. {rule1, rule2, rule3, rule4, rule5, rule6, 
    rule7} /. {dummyWhich -> Which

The result of that last line is:

{Which[Aa == Ab, a Sa + b Sb, Ca == 0 || Cb == 0, a Sa + b 
    Sb, True, SS], Which[Aa == Ab, a Ca + b Cb, Ca == 0 || Cb == 0, a 
    Ca + b Cb, True, CC],
   Which[Ca == 0 && Cb == 0, "Nil", Z1 == Z2 && Ca < 0, Aa, 
    Z1 == Z2 && Cb < 0, Ab, (Z1 == Z2 && Ca > 0) && Aa > 90, Aa - 90, 
    Z1 == Z2 && Ca > 0, Aa + 90, Z1 == 
    Z2 && (Cb > 0 && Ab > 90), Ab - 90, Z1 == Z2 && Cb > 0,
     Ab + 90, Z1 == Z2, Null, Aa == Ab, Aa, Ca == 
    0 && Cb == 0, "Nil", Ca == 0, Ab, Cb == 0, Aa, 
    Y == 0, "Nil", X > 180, X - 180, X < 0, X + 180, True, X]}

I've used Null where that would be the result of your original logic --
maybe you want it to be "Nil".  Or maybe you want to use Null instead of
"Nil".  Your choice.  Wherever you see Null, there's possibly a case you
haven't covered.

Bobby Treat

-----Original Message-----
From: Moranresearch at aol.com [mailto:Moranresearch at aol.com] 
To: mathgroup at smc.vnet.net
Subject: [mg36477] [mg36461] Can anyone simplify this?


I have substituted the letters SS, CC, X, Y and Z1 and Z2 for some 
complicated expressions just to illustrate the form of the function.
This 
function works and all the conditions are necessary but I am sure a more

elegant programming solution perhaps using While could be found. Any 
suggestions to tighten this up? Thank you.

f[{Sa_, Ca_, Aa_, Sb_, Cb_, Ab_, a_, b_}] := {If[Aa == Ab, a Sa + b Sb, 
      If[Ca == 0 || Cb == 0, a Sa + b Sb, SS]], 
    If[Aa == Ab, a Ca + b Cb, If[Ca == 0 || Cb == 0, a Ca + b Cb, CC]], 
    If[Ca == 0 && Cb == 0, "Nil", If[Z1 == Z2, If[Ca < 0, Aa, If[Cb < 0,
Ab,
            If[Ca > 0,
              If[Aa > 90, Aa - 90, Aa + 90],
              If[Cb > 0,
                If[Ab > 90, Ab - 90, Ab + 90]]]]],
        If[Aa == Ab, Aa,
          If[Ca == 0 && Cb == 0, "Nil",
            If[Ca == 0, Ab,
              If[Cb == 0, Aa,
                If[Y == 0, "Nil",
                  If[X > 180, X - 180, If[X < 0, X + 180, X]]]]]]]]]}






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