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Evaluation within If or Which

Alain RIWAN
Posted 10 years ago

For these pieces of code how can I "r" evaluated. When I use If "r" remains unevaluated. When I use Which "r" is only evaluated when the first case is True. Can I get some help? Thanks

Clear[aa, bb, fIf, r];
fIf[u_, v_] := Block[{aa = u + v + w, bb = 2 u - v - w},
  If[bb > aa, r = BB];
  If[aa > bb, r = AA];
  If[aa == bb, r = AB];
  {aa, bb, r}
  ]
{fIf[2, 2], fIf[2, 1], fIf[2, 0]} //. w -> -1
{fIf[2, 2], fIf[2, 1], fIf[2, 0]} //. w -> 0
{fIf[2, 2], fIf[2, 1], fIf[2, 0]} //. w -> 1

Clear[aa, bb, r1, r2, r3];
fWhich[u_, v_] = Block[{aa = u + v + w, bb = 2 u - v - w},
   r1 = Which[aa == bb, AB, aa > bb, AA, bb > aa, BB];
   r2 = Which[aa > bb, AA, aa == bb, AB, bb > aa, BB];
   r3 = Which[bb > aa, BB, aa > bb, AA, aa == bb, AB];
   {aa, bb, r1, r2, r3}
   ];
{{"aa", "bb", "r1", "r2", "r3"}, fWhich[2, 1], fWhich[2, 2],    fWhich[2, 0]} //. w -> -1 // MatrixForm
{{"aa", "bb", "r1", "r2", "r3"}, fWhich[2, 1], fWhich[2, 2],    fWhich[2, 0]} //. w ->  0 // MatrixForm
{{"aa", "bb", "r1", "r2", "r3"}, fWhich[2, 1], fWhich[2, 2],    fWhich[2, 0]} //. w ->  1 // MatrixForm
POSTED BY: Alain RIWAN
10 Replies
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Eric Rimbey
Eric Rimbey
Posted 9 years ago

I'm interpreting your question to be: "Does Mathematica use 'which' where other languages use 'then', and does Mathematica use 'switch' where other languages use 'else'?"

The answer is "no".

Mathematica has an If[...] form, and it is analogous to the if {...} else {...} branching structure of procedural languages like Java.

Mathematica has a Swich[...] form, and it is analogous to the switch {case ...} branching structure of procedural languages like Java.

Mathematica has a Which[...] form, and I think most procedural languages like Java have no analog to it. It is basically the same as cond from Lisp.
POSTED BY: Eric Rimbey

Is which used in place of then and switch used in place of else? Or am I working with these ideas incorrectly. I am new to programming.
POSTED BY: Leighton Cooper
Alain RIWAN
Alain RIWAN
Posted 10 years ago

Thanks. I'm pretty sure I tried it at first and it didn't work!!!
POSTED BY: Alain RIWAN
Eric Rimbey
Eric Rimbey
Posted 10 years ago

Well, I'm not sure where you're going with this, and how it relates to the original post, but the easy way to get "a>b" to evaluate in your example is to replace

Dynamic[a] > Dynamic[b]

with

Dynamic[a > b]
POSTED BY: Eric Rimbey
Alain RIWAN
Alain RIWAN
Posted 10 years ago

In fact I think the core of the problem is how to get the expression "a>b" dynamically evaluated in this example

{Slider[Dynamic[a]], Dynamic[a]} {Slider[Dynamic[b]], Dynamic[b]} Dynamic[a]> Dynamic[b]
POSTED BY: Alain RIWAN
Alain RIWAN
Alain RIWAN
Posted 10 years ago

Thanks for your answer.
POSTED BY: Alain RIWAN
Eric Rimbey
Eric Rimbey
Posted 10 years ago
POSTED BY: Eric Rimbey
POSTED BY: Frank Kampas
Alain RIWAN
Alain RIWAN
Posted 10 years ago
POSTED BY: Alain RIWAN
Eric Rimbey
Eric Rimbey
Posted 10 years ago

Alain,

Your variables aa & bb don't resolve to a number until you do the final replacements of w. Consider one of your If expressions:

If[bb > aa, r = BB]

Assuming u=2 and v=1, this is effectively trying to determine 3+w > 3-w, which cannot be determined, because w is still unknown. Since your assignment to r was inside the If, it never gets evaluated.

In your second example, you are setting r to the Which expression itself, which, upon application of the replacement rules for w, sometimes resolves to one of the values. The same thing would have happened if you used r=If[...] in your first example.

What I would suggest, is that you simply make w an input variable to fIf and fWhich. Like this:

fIf[u_, v_, w_] :=
 Block[
  {aa = u + v + w, bb = 2 u - v - w},
  If[bb > aa, r = BB];
  If[aa > bb, r = AA];
  If[aa == bb, r = AB];
  {aa, bb, r}
  ]
POSTED BY: Eric Rimbey
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