Use a circuit to multiply two resistances

Question

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.

Answer 1

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.

Alternatively:

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.

Answer 2

I think it is impossible.

Proof:

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:

$$\deg(s(a,b))=\max(\deg(a),\deg(b))$$
_{(The $=$ can be used rather than $≤$ since $x$ and $y$ can only be used once overall)}

$$\deg(p(a,b))=\deg(a)+\deg(b)-\deg(s(a,b))$$
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.

Answer 3

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 $$ d(s(s(c(s(x,y)),b(c(x))),b(c(y))))=xy. $$

Answer 4

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.