Is the number of operators always one less than the number of values in an arithmetic Reverse Polish Notation expression with only binary operators?

This question seems very trivial, but I don't know if the statement is necessarily true for all N > 1 where N is the number of values. It seems intuitively true because any expression with more than one binary operator can be expressed as a combination of two different expressions through induction, but I'm not very confident.

  • $\begingroup$ I don't need a proof. This is the wrong sentiment when trying to learn computer science. We're not here to help you with your homework – we're here to help you understand the material, so that you can do your homework yourself next time. $\endgroup$ – Yuval Filmus Mar 4 '16 at 10:53
  • $\begingroup$ I'm really sorry; I was mostly trying to ask if this statement is even true in the first place. I didn't really want to ask for a proof because that will waste people's time since it's probably trivial to everyone here. $\endgroup$ – Sky Mar 4 '16 at 13:18

RPN expressions have stack semantics: a value pushes a number onto the stack, whereas a binary operator pops two values and pushes one back, for a net loss of one element. At the start of an expression the stack is empty, and at the end there is exactly one value – the result.

Now let $V$ be the number of values in the expression, and $B$ be the number of binary operators. Then at the end of the expression, the stack contains $V-B$ values. Since there should be exactly one value remaining, we conclude that $V-B = 1$, or $B = V-1$.

Another way to see this is using structural induction. As you mention, every expression is either a value or is obtained from two expressions combined using a binary operator. Let's prove by induction on the number of values and operators appearing in the expression that $V = B+1$ (in the terminology of the preceding section).

When the expression consists of just a value, $B=0$ and $V=1$, so $V = B+1$. When the expression consists of a binary operator together with two other expressions $E_1,E_2$, we have $V = V(E_1)+V(E_2)$ and $B = B(E_1)+B(E_2)+1$ (I hope the notation is clear). By induction, $V(E_1) = B(E_1)+1$ and $V(E_2) = B(E_2)+1$, so $$ V = V(E_1) + V(E_2) = B(E_1) + 1 + B(E_2) + 1 = B + 1. $$

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A simple inductive argument will work here. Expressions in RPN can be defined recursively by using the grammar:

E: operand
E: E E operator

Where E is an expression, operand is a number and operator is a binary operator.

The base case is when the expression is just a number. In this case the number of operators is clearly one less than the number of operands.

Let E be an expression of length n > 1 and assume that the statement is true for expressions of length < n (n here is the number of operands). The expression is of the form E1 E2 operator. The first two expressions have smaller lengths and so we have:

o1 = n1 - 1.
o2 = n2 - 1.
o = o1 + o2 + 1 = n1 + n2 - 2 + 1 = n - 1 (we add 1 for the `operator` in `E1 E2 operator`).

o stands for number of operators. n stands for number of operands.

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