# Swapping registers in an old calculator

I came up with this problem inspired by the limitations of an old non-scientific calculator I owned years ago (the two registers were the display, and an internal memory for an additional number).

We have a primitive calculator with only two registers $$R_1$$ and $$R_2$$, and the following four operations:

• $$R_1+R_2\to R_2$$ (add the content of register $$R_1$$ to register $$R_2$$.)

• $$-R_1+R_2\to R_2$$ (subtract the content of register $$R_1$$ from register $$R_2$$.)

• $$R_1+R_2\to R_1$$ (add the content of register $$R_2$$ to register $$R_1$$.)

• $$R_1-R_2\to R_1$$ (subtract the content of register $$R_2$$ from register $$R_1$$.)

For instance, if $$R_1=x$$ (register $$R_1$$ contains number $$x$$) and $$R_2=y$$ ($$R_2$$ contains $$y$$), after applying the operation $$R_1+R_2\to R_2$$ we end up with $$R_1=x$$ and $$R_2=x+y$$.

Assume that initially we have $$R_1=x$$ and $$R_2=y$$, where $$x$$ and $$y$$ are arbitrary numbers. For each of the following tasks describe a sequence of operations that would allow us to perform it, or prove that the task cannot be performed (task 1 is very easy, the real challenge is about task 2):

• Task 1. Swap the contents of registers $$R_1$$ and $$R_2$$ changing the sign of $$y$$ in the process, so we would end up with $$R_1=-y$$, $$R_2=x$$.

• Task 2. Swap the contents of registers $$R_1$$ and $$R_2$$, so that we would end up with $$R_1=y$$, $$R_2=x$$.

## 2 Answers

Task 1 was answered. Task 2:

Not possible. We always have $$\begin{bmatrix}R_1 \\ R_2\end{bmatrix} = M\begin{bmatrix}x \\ y\end{bmatrix}$$ for some matrix $$M$$ that depends only on the sequence of operations. Initially $$M = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$$, and the given operations cause $$M$$ to be left-multiplied by $$\begin{bmatrix}1 & 0 \\ 1 & 1\end{bmatrix}$$, $$\begin{bmatrix}1 & 0 \\ -1 & 1\end{bmatrix}$$, $$\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}$$, and $$\begin{bmatrix}1 & -1 \\ 0 & 1\end{bmatrix}$$ respectively. All of these matrices have determinant $$1$$, so we will always have $$\det M = 1$$, which makes it impossible to reach the desired $$\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}$$ with determinant $$-1$$.

• I had a feeling this was true. Very neat proof. – Dr Xorile Jan 3 at 15:23

Task 1

1. $$R_1 - R_2\to R_1$$ (now $$R_1 = x-y$$, $$R_2 = y$$)
2. $$R_1 + R_2\to R_2$$ (now $$R_1 = x-y$$, $$R_2 = x$$)
3. $$R_1 - R_2\to R_1$$ (now $$R_1 = -y$$, $$R_2 = x$$)

• Task 2 the second operation isn't in the list of allowed operations – Dr Xorile Jan 3 at 6:26
• @DrXorile Oops, so it seems... – jafe Jan 3 at 6:27