Is the Circuit Possible?

Question

Write a program or function that takes input: all resistors available and a resistance value and outputs a truthy value of whether or not it is possible to get the resistance by using those resistors.

Rules:

Any format for input will do.

There will be at least 1 available resistor and your program should output for at least 10 available resistors.

Resistance of all available resistors and the required resistance will be positive integers.

For available resistors if a fractional value is also possible, the required resistance may be the approximated value.(See example)

Output should be any 2 unique values for Possible and Not possible.

The resistors may be connected in any manner.

Series Resistance: For n resistances in series: Result=R1+R2+R3+....Rn

Parallel Resistance: For n resistances in parallel: Result=1/(1/R1+1/R2+1/R3+....+1/Rn)

The circuit may not require all the resistors to get the required resistance (output True if that is the case).

Winner:

This is code-golf so shortest-code wins.

Examples:

R     List
110   220,220 -> True
440   220,220 -> True
550   400,300 -> False
3000  1000,3000 -> True
750   1000,3000 -> True
333   1000,1000,1000 -> True (1000||1000||1000=333.333)
667   1000,1000,1000 -> True ((1000+1000)||1000=666.6666)
8000  1000,1000,7000 -> True 
190   100,200,333,344,221 -> True 
193   105,200,333,344,221 -> True
400   200,100 -> False

Explanation for the last two examples: https://physics.stackexchange.com/questions/22252/resistor-circuit-that-isnt-parallel-or-series

Answer 1

Python 3, 253 bytes

import functools
def p(l): return functools.reduce(lambda z,x:z+[y+[x]for y in z],l,[[]])
print(l[-1:][0]in[round(a)for a in[sum(a)for a in p([sum(a)for a in p(l)]+[1/sum(c)for c in[[1/b for b in a if b!=0]for a in p(l).copy()]if sum(c)!=0])]])

I take the powerset of all resistor values and then calculate the sums for series and 1/sum(1/values) for paralel and then I take a powerset of these two sets. When you take the sum of all the subsets and put them into a set then this set eiter contains the value or not. -> return True/False

@stephen thanks :)

Answer 2

Japt, 52 bytes

à má

ÈÊ<2?X:X¯1 ïXÅgW @[X+Y1/(1/XÄ/Y]Ãc
mmW c mr øV

Try it!

This was a tough one, and I had to do a couple of strange things to make it work. I can't mathematically prove that this works for everything, but it works for all the test cases as well as my extra proposed test case. Specifically, I know that the function I define called W gives different results depending on the order of the resistors in its input so I run it on each possible ordering of each possible combination of resistors. I also know that it will produce a list of resistances that are all possible to create using the input resistors. I don't know with 100% certainty that those two things together end up with every possible resistance.

Explanation:

à       Get each combination e.g. [1,2,3] -> [[1,2,3],[1,2],[1,3],[2,3],[1],[2],[3]]
  m     For each combination...
   á    Get each order e.g. [1,2,3] -> [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]

    Skip a line to avoid overwriting the target resistance

È                                     Define a function W taking an array input X:
 Ê<2?X                                  If X is only 1 resistor, return it as-is
            XÅgW                        Get the resistances from applying W to the tail of X
       X¯1 ï                            Pair the first resistor with each possible tail resistance
                 @[            ]Ã       For each pair get these two values:
                   X+Y                    Those resistances in series
                      1/(1/XÄ/Y           Those resistances in parallel
                                 c      Collapse all those resistances into a single flat list

mmW            For each combination and order, get the resistances using W
    c          Collapse to a single list of resistances
      mr       Round each one
         øV    Return true if the target resistance is in the list, false otherwise

Answer 3

Ruby, 153 bytes

f=->v,r{[r]-v==[]||r[1]&&[*2..v.size].any?{|n|v.permutation.any?{|l|[l[0,n].sum,(1.0/l[0,n].reduce(0){|s,x|s+1.0/x}).round].any?{|b|f[l[n..-1]+[b],r]}}}}

Try it online!

Brute force. I mean it.