Given two resistors of unknown resistances and an infinite supply of wires and other resistors, create a circuit using multiple series and parallel combinations such that the effective resistance of the circuit is equal to the product of the two unknown resistances.

Mathematically (for those who don't want to bother about the physics),

$$s(a,b)=a+b$$ $$p(a,b)=\frac{ab}{a+b}$$

Using the two functions given above, two variables $x$ and $y$ and any other numbers, write a mathematical expression that evaluates to give $xy$. You cannot use any other operators or functions, not even division.

Your score is given by the number of times a function appears in the expression, so try to minimize this score.

Example: $s(s(x,p(y,2)),p(3.5,1))$ is a valid expression with a score of 4.

P.S. I'm not 100% sure this is even possible, if not please try giving a proof for the same.

  • $\begingroup$ Your example uses the resistor 'y' twice. $\endgroup$ Commented Apr 6, 2016 at 15:18
  • $\begingroup$ @ChrisTaylor I was unsure whether to allow that or not, anyways now I have disallowed it. Thanks for spotting anyways. $\endgroup$ Commented Apr 6, 2016 at 15:22
  • $\begingroup$ Perhaps this might be useful: $p(a,b)=\frac{1}{s(1/a,1/b)}$ $\endgroup$
    – Roby5
    Commented Apr 6, 2016 at 15:44
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    $\begingroup$ s and p both have units of $\Omega$. No nested set of one inside the other will result in units of $\Omega^2$. $\endgroup$ Commented Apr 7, 2016 at 0:44
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    $\begingroup$ This makes no sense dimensionally. The product of two resistances is not a resistance! $\endgroup$ Commented Apr 7, 2016 at 11:14

4 Answers 4


Another proof that it is impossible.

Consider the case that resistor $x$ has zero resistance. Then the complete circuit must also have zero resistance, so there must be a path with zero resistance; either this path has no resistors or it has only resistor $x$.

If there are no resistors on the path, the circuit will always have a resistance of zero, which is obviously wrong. If the path has only $x$, then the maximum possible resistance of the circuit is $x$, so it can't be correct if $x\ne0$ and $y>1$. Therefore such a circuit is impossible.


Note that for positive $a$ and $b$, $p(a,b)\le s(a,b)=a+b$, so the total resistance is at most the sum of all the resistances in the circuit. In particular, the sum of all the fixed resistances must be at least $xy-x-y$. However, this value increases without bound as $x$ and $y$ increase, so the circuit is impossible.

  • $\begingroup$ Why must the complete circuit have 0 resistance if $x$ has 0 resistance? $\endgroup$
    – Shuri2060
    Commented Apr 6, 2016 at 17:53
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    $\begingroup$ @QuestionAsker Because then the product of $x$ and $y$ is zero. $\endgroup$
    – f''
    Commented Apr 6, 2016 at 17:58
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    $\begingroup$ @RyanO'Hara The same circuit has to work if $x=0$ and also if $x\ne0$. I have shown that if it works for $x=0$, then it does not work for $x\ne0$. $\endgroup$
    – f''
    Commented Apr 6, 2016 at 18:47

I think it is impossible.


Consider the degrees of the expressions which you can use: $x$ and $y$ have a degree of 1. You are also allowed to use constants which have a degree of 0.

Let the function $\deg(A)$ calculate the degree of the expression $A$.

Then under the restrictions of $a$ and $b$ given in the question:

(The $=$ can be used rather than $≤$ since $x$ and $y$ can only be used once overall)

Function $s$ will always have a degree equal to 1 if you use either or both of the variables $x$ or $y$ as the parameters. Otherwise, it'll have a degree of 0 if only constants are used in the parameters.

This still leaves you with expressions of degree 0 or 1 to use.

Function $p$ will always give an expression of degree 0 or 1 if the parameters have degrees 0 or 1.

This still leaves you with expressions of degree 0 or 1 to use.

Hence it is impossible to reach the expression $xy$ which has degree 2.

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    $\begingroup$ Well, it's more fundamental than that. "A resistance of xy" is undefined — it's like cutting a piece of rope with a length of 2 ft². $\endgroup$ Commented Apr 6, 2016 at 15:59
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    $\begingroup$ @PeregrineRook: What you're calling "units" is basically what Question Asker is calling the "degree". His argument is basically that the two operations $p$ and $s$ never change the degree of their inputs (assuming that their inputs are both of degree 1); but this is basically the same as saying that if $x$ and $y$ are both in Ohms, then $p(x,y)$ and $s(x,y)$ are also both in Ohms. $\endgroup$ Commented Apr 6, 2016 at 16:06
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    $\begingroup$ @ Question Asker: Yes. We know that we can map the math onto the circuit and back. If an expression for xy existed, the same expression would fail to multiply resistances measured in a different unit (e.g. deci-ohms). But the expressions p and s give the right physical answers regardless of the choice of unit. $\endgroup$
    – jsocolar
    Commented Apr 6, 2016 at 16:13
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    $\begingroup$ @QuestionAsker: My argument is pretty similar to your argument, except it says, not that it's impossible to achieve the desired result, but that there is no such thing.  For example, if x = 2 ohm and y = 3 ohm, is xy = 6 ohm?  But x = 2,000,000 µohm and y = 3,000,000 µohm, and 6,000,000,000,000 µohm = 6,000,000 ohm. MichaelSeifert: I disagree (pun not intended).  QA is using "degree" to mean dimensionality.  Inch, foot, mile, cm and km are all the same dimensionality — acre and ft² are a different dimensionality. $\endgroup$ Commented Apr 6, 2016 at 16:16
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    $\begingroup$ Give unknown resistors of resistance $x$ ohms and $y$ ohms, it makes sense to ask for a circuit of resistance $xy$ ohms. This is how I interpreted the question (this interpretation is consistent with the purely mathematical formulation of the question). $\endgroup$ Commented Apr 6, 2016 at 17:12

If we have multiple copies of each of the unknown resistors (so we can use the variables $x$ and $y$ more than once in the mathematical formulation), and we allow complex-valued resistors, then this is possible.

We can build the following operators: $$a(x):=s(p(s(x,-1),1),-1)=\frac{-1}{x},$$ $$ b(x):=a(s(a(s(a(s(x,i)),-i)),i))=-x, \hspace{5mm}\text{where $i$ is the imaginary unit}, $$ $$ c(x):=s(b(p(x,s(b(x),1))),x)=x^2, $$ $$ d(x):=a(s(a(s(a(s(x,-\sqrt{2})),-1/\sqrt{2})),-\sqrt{2}))=\frac{x}{2}. $$ Finally, we have $$ d(s(s(c(s(x,y)),b(c(x))),b(c(y))))=xy. $$

  • $\begingroup$ I wonder if it's possible to use only negative-valued resistors and avoid the complex ones. $\endgroup$ Commented Apr 6, 2016 at 21:38
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    $\begingroup$ Negative resistances are generally harder to construct than imaginary ones in any case - all the latter need are inductors and capacitors. $\endgroup$
    – Zandar
    Commented Apr 6, 2016 at 21:48
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    $\begingroup$ You can simplify $d(x)=p(x,x)$. $\endgroup$ Commented Apr 7, 2016 at 0:39
  • $\begingroup$ @2012rcampion Indeed, that's much easier! $\endgroup$ Commented Apr 7, 2016 at 0:41

Here is a purely mathematical version of f''’s answer.

First note that the functions $s(x,y)$ and $p(x,y)$ always take non-negative inputs to positive outputs; hence, so does any operation build up from them. (To make this argument fully rigorous, it’s probably most natural to consider them as total binary operations on the extended non-negative real line $\{ x \in \mathbb{R}\ |\ x \geq 0 \} \cup \{\infty\}$, with $p(x,\infty)$ and $p(\infty,x)$ defined to be $x$; or, even better, on the projective line $\mathbb{P}^1(\mathbb{R}_{>0})$. But I won’t quite be that formal here.)

Now I claim: for any function $f(x,y)$ built up from the operations $s$ and $p$ together with non-negative constants, if $f(0,y) = 0$ for all $y$, then $f(x,y) = 0$ for all $x,y$.

Suppose this failed. Then there would be some minimal counterexample $f(x,y)$ — that is, minimal in its complexity as a formula built from $s$ and $p$, not necessarily minimal in its numerical output values.

It can’t be a constant, so it’s either of the form $s(g(x,y),h(x,y))$ or $p(g(x,y),h(x,y))$, where $g$ and $h$ are simpler expressions, so in particular are not counterexamples to the claim.

First case: $f(x,y) = s(g(x,y),h(x,y))$. So $g(0,y) + h(0,y) = 0$ for all $y$. So, since everything involved is non-negative, $g(0,y) = h(0,y) = 0$, for all $y$; so since $g$ and $h$ were not counterexamples to the claim, $g(x,y) = h(x,y) = 0$ for all $x,y$. So $f(x,y) = 0$ always — contradicting the choice of it as a counterexample!

Second case: $f(x,y) = p(g(x,y),h(x,y))$. (This case gets a bit mathematically involved.) Now $p(a,b) = 0$ exactly if at least one of $a,b$ is 0. So we know that for all $y$, at least one of $g(0,y)$ and $h(0,y)$ is $0$. So either $g(0,y)$ is zero for a dense set of $y$, or $h(0,y)$ is zero on some non-empty open set of $y$’s. But these are rational functions — and if a rational function is 0 either on a dense set or on a non-empty open set, then it must be zero everywhere. So either $g(0,y)$ is always zero, or $h(0,y)$ is. Since $g$ and $h$ aren’t counterexamples, this means that either $g(x,y)$ is always zero, or $h(x,y)$ is. But either of these means that $f(x,y)$ is always zero. Contradiction again!

So no such function is possible.

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    $\begingroup$ This proof is missing f(x,y) = x $\endgroup$
    – ffao
    Commented Apr 6, 2016 at 19:01
  • $\begingroup$ @ffao: good point — and that is a counterexample to the claim as stated (and so are constant multiples of it). So this doesn’t work as given. Hopefully it can be fixed — f''’s argument sounds very convincing physically, so it should be possible to state it mathematically! $\endgroup$ Commented Apr 6, 2016 at 19:17

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