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I have expressions like these:

expr1 = -2 a Z[-2] - 2 b Z[-1] - 2 Z[-1] + a Z[-1]
expr2 = -2 a Z[-2] - 2 b Z[-1] - 2 Z[-1] + a Z[-1] + c Z[0]
expr3 = -2 a Z[-2] - 2 c Z[-1] - 2 Z[-1] + a Z[-1] + c Z[0]

and I want to find out whether an expression contains as coefficient a, b and c. So what I want is a function f[expr], such that

f[expr1] (* false because it only contains a and b *)
f[expr2] (* true because it contains a, b and c *)
f[expr3] (* false because it only contains a and c *)

The only way I can think of is converting it to a string and do a string-search, but I'm sure there must be a more clever way. I was trying to use Case[], but I was not able to find a solution yet.

Edit: There are several different solutions, I tested them on speed. Run each of the functions 100.000 times (with one argument only, in case the {a,b,c}-list was a function-argument, it was replaced to be constant - for fair comparison) with the expr1, expr2, expr3 expressions.

Results:

  • Dr. belisarius: 11.3443515 sec
  • Bill: 2.0469818 sec
  • eldo's 1st: 3.6095660 sec
  • eldo's 2nd: 2.4532568 sec
  • Suba Thomas: 6.8753645 sec
  • (Algohi: My Mathmatica9 does not support SubsetQ unfortunatly.)

The fastest solution will get the acceptance-reward, after roughly 24hours of the original questions. Thanks for the infos and the fun :-)

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5 Answers 5

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check[expr_, vars_]:= Not[Or @@ Map[FreeQ[expr, #] &, vars]];
check[expr1, {a, b, c}]
check[expr2, {a, b, c}]
check[expr3, {a, b, c}]

(* False, True, False *)

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f[expr_] := Count[D[expr, {{a, b, c}}], 0] == 0
f /@ {expr1, expr2, expr3}

{False, True, False}

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  • $\begingroup$ wow that is an unusual solution :) $\endgroup$
    – Mario Krenn
    Jan 13, 2016 at 0:24
  • $\begingroup$ Turns out to also have the fewest characters (so far). But not the fastest! $\endgroup$
    – Suba Thomas
    Jan 13, 2016 at 20:14
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f[x_] := Not[Or @@ (PossibleZeroQ /@ Last@CoefficientArrays[x, {a, b, c}])]
f /@ {expr1, expr2, expr3}
(* {False, True, False} *)
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f[expr_, var_List] := SubsetQ[Level[expr, {-1}], var]
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Table[Union@Flatten@Map[Cases[ex, #, Infinity] &, {a, b, c}] == 
 {a, b, c}, {ex, {expr1, expr2, expr3}}]

{False, True, False}

Or

f[ex_] := 
 Union@Flatten@Map[Cases[ex, #, Infinity] &, {a, b, c}] == {a, b, c}

f /@ {expr1, expr2, expr3}

{False, True, False}

Or

 f[ex_] := And @@ Map[MemberQ[ex, #, Infinity] &, {a, b, c}]

 f /@ {expr1, expr2, expr3}

{False, True, False}

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