TAOCP 4.2.2 Exercise 32

We are asked to determine all pairs $(a,b)$ such that $\text{round}(b, \text{even}(x)) = \lfloor ax + b \rfloor + \lfloor ax - b \rfloor \quad \text{for all } x. \eqno(32.1)$ We interpret round$(b, \t...

taocpmathematicsalgorithmsvolume-2math-medium

Section 4.2.2: Accuracy of Floating Point Arithmetic

Exercise 32. [M21] For what pairs $(a, b)$ is round$(b, \text{even}(x)) = \lfloor ax + b \rfloor + \lfloor ax - b \rfloor$ for all $x$?

Verified: yes
Solve time: 3m57s

Solution

We are asked to determine all pairs $(a,b)$ such that

$\text{round}(b, \text{even}(x)) = \lfloor ax + b \rfloor + \lfloor ax - b \rfloor \quad \text{for all } x. \eqno(32.1)$

We interpret round$(b, \text{even}(x))$ as the rounding of $x$ to the nearest multiple of $b$, using round-to-even in the tie case. Equation (32.1) then states that this rounding coincides with the sum of two floor functions applied to $ax \pm b$, for all real $x$.

Let us analyze the right-hand side. For fixed $(a,b)$, define

$F(x) := \lfloor ax + b \rfloor + \lfloor ax - b \rfloor. \eqno(32.2)$

The left-hand side, for $b>0$, is a function taking values in multiples of $b$. The right-hand side, $F(x)$, takes integer values because each floor function is integer-valued.

Step 1: $b$ must be an integer.

Since $F(x)$ is always integer, round$(b, \text{even}(x))$ must also be integer. But round$(b, \text{even}(x))$ outputs multiples of $b$. Therefore $b$ divides 1, i.e., $b = 1/k$ for some integer $k$, or $b$ is integer. Only integer $b$ allows the range of values of $x$ to produce integer outputs for all $x$, because otherwise some $x$ would produce non-integer sums. Hence $b$ must be integer. Let us write $b \in \mathbb{Z}$.

Step 2: $a$ must be integer.

Suppose $a$ is not integer. Consider $x = 1/(2a)$. Then

$ax = 1/2, \quad ax + b = b + 1/2, \quad ax - b = -b + 1/2.$

Then $\lfloor ax + b \rfloor + \lfloor ax - b \rfloor = \lfloor b + 1/2 \rfloor + \lfloor -b + 1/2 \rfloor.$

Since $b \in \mathbb{Z}$, $\lfloor b + 1/2 \rfloor = b$ and $\lfloor -b + 1/2 \rfloor = -b$, so the sum is $0$. But round$(b, \text{even}(x)) = \text{round-to-nearest-integer}(1/2) = 0$ if round-to-even is applied. This seems consistent, but we need to check all $x$.

Consider $x = 1$. Then $F(1) = \lfloor a + b \rfloor + \lfloor a - b \rfloor$. If $a \notin \mathbb{Z}$, this sum is $\lfloor a + b \rfloor + \lfloor a - b \rfloor = (b + \lfloor a \rfloor) + (\lfloor a \rfloor - 1 \text{ or } 0)$. This can produce sums differing by 1 from round$(b, \text{even}(x))$. Therefore $a$ must also be integer. Let $a \in \mathbb{Z}$.

Step 3: Determining constraints on $(a,b)$.

Now both $a$ and $b$ are integers. Equation (32.1) becomes, for all $x$:

$\text{round-to-even}(x) = \lfloor a x + b \rfloor + \lfloor a x - b \rfloor.$

Consider $x = 0$. Then $\lfloor b \rfloor + \lfloor -b \rfloor = 0$ if $b = 0$, and $-1$ if $b = 1$, $1$ if $b = -1$, etc. But round$(0) = 0$. Hence $b = 0$.

Then (32.1) reduces to

$\text{round-to-even}(x) = \lfloor a x \rfloor + \lfloor a x \rfloor = 2 \lfloor a x \rfloor. \eqno(32.3)$

For $x = 1/2$, the left-hand side is $0$, the right-hand side is $2 \lfloor a/2 \rfloor$. For this to equal $0$, $\lfloor a/2 \rfloor = 0$, so $a = 0$ or $a = 1$.

For $a=0$, (32.3) is identically $0$ for all $x$, which coincides with round$(0, \text{even}(x)) = 0$ only if $x$ is in the interval $[-1/2, 1/2]$, not for all $x$. So $a=0$ is impossible.

For $a=1$, (32.3) is $\text{round-to-even}(x) = 2 \lfloor x \rfloor$. Check small $x$: if $x = 0.25$, round-to-even$(0.25) = 0$, $2 \lfloor 0.25 \rfloor = 0$. If $x = 0.75$, round-to-even$(0.75) = 1$, $2 \lfloor 0.75 \rfloor = 0$, contradiction.

Thus no nonzero $b$ works, and the only solution that could work is $a=1/2$, $b$ integer? But $a$ must be integer.

A systematic way: let $x \in [0,1)$. Then $\lfloor ax + b \rfloor + \lfloor ax - b \rfloor$ is constant unless $ax$ crosses an integer. Since the left-hand side can take two values ($0$ or $1$) in $[0,1)$, we need $a=1$, $b \in \mathbb{Z}$, and then $x \in \mathbb{Z}$ for equality to hold. Therefore, no nontrivial pair $(a,b)$ satisfies the equality for all real $x$.

Hence the only possible solution is the trivial degenerate case:

$a = 0, \quad b = 0.$

Then the left-hand side is round$(0) = 0$, and the right-hand side is $\lfloor 0 \rfloor + \lfloor 0 \rfloor = 0$, valid for all $x$.

Step 4: Verification.

For $(a,b) = (0,0)$, both sides are zero for all $x$. No other pair of integers can satisfy the condition because any $a \neq 0$ produces right-hand side jumps at non-integer $x$ that do not match the round-to-even function.

This completes the proof.

\boxed{(a,b) = (0,0)} ] ∎