I am trying to check if a list of resistors can be used to achieve a set of equivalent resistances. (Reciprocal of the sum of the reciprocals of the individual resistors is the equivalent resistance) Achieving the exact value of equivalent resistance is very tough, so I would like to allow a tolerance of 5% max. privArrayTest generates a list of all combinations of the resistors and calculates the equivalence. I attempted to implement the tolerance with tolerance[x_], and run that through ContainsAll, but it appears to just be checking the edge values rather than values within the range.

ParallelResist[arrayInTest_, arrayInCompare_] :=
 Module[{privArrayTest, invert, sum, privArrayC, tolerance},
  (*internal functions*)
  invert[x_] := x^-1;
  sum[x_] := Total[x];
  tolerance[x_] := {x - x*.05, x + x*.05};

  (*main body*)
  privArrayTest = Map[invert, N[Map[sum, Subsets[Map[invert, arrayInTest], {1, \[Infinity]}]]]];
  (*1,\[Infinity] prevents the first blank subset from being \generated*)

  privArrayC = Map[tolerance, arrayInCompare];
  Print[privArrayC]; (*Debugging Code*)
  Print["Debug code,privArrayTest", privArrayTest];(*Debugging code*)

  (*compare code, returns true false*)
  ContainsAll[privArrayTest, privArrayC]

Example input:

resistors = {1500, 1000, 2500, 20000};
equivalence = {472.441, 483.871, 895.522};

Example output:


Debug code,privArrayTest{1500.,1000.,2500.,20000.,600.,937.5,1395.35,714.286,952.381,2222.22,483.871,582.524,895.522,689.655,472.441}


1 Answer 1


ContainsAll checks whether expressions contain other expressions on a pattern matching level. It does not check containment in intervals.

Maybe this helps?

tol = 0.05;

possibleresistances = 1/Total[1/Rest@subsets, {2}];
A = Outer[
   {equiv, resistance} \[Function] Between[resistance, (1 + {-1, 1} tol) equiv],

And @@ Or @@@ A


  • 1
    $\begingroup$ Some alternatives: 1. possibleresistances = HarmonicMean[#]/Length[#] & /@ subsets 2. the main testing function can be done as {equiv, resistance} \[Function] Between[resistance, (1 + {-1, 1} tol) equiv]. $\endgroup$ Commented Mar 3, 2019 at 10:00
  • $\begingroup$ Hi @J.M. Good points. However, it appears to me that HarmonicMean is somewhat inefficient (which does matter in this scenario). $\endgroup$ Commented Mar 3, 2019 at 10:12
  • $\begingroup$ Huh, that's interesting. What about 1/Total[1/subsets, {2}]? $\endgroup$ Commented Mar 3, 2019 at 10:14
  • $\begingroup$ That's better! Thank you! $\endgroup$ Commented Mar 3, 2019 at 10:34
  • $\begingroup$ So the main issue is I need the possible resistances contain each member of the equivalence list(+-5%) at least once for the possible resistances to be a successful list. The code I have works, albeit not the best written, except for looking through the ranges of acceptable tolerance. I was able to use ContainsAll when I matched the machine precision (without the tolerance), so I am looking for a replacement command, if one exists. $\endgroup$
    – Sc Bones
    Commented Mar 3, 2019 at 16:39

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