Functions 2
Reversing an Integer

Before showing you the function that we came up with to reverse a number, we have to define two functions. First, $\text{len}_b(x)$, which gives the number of digits in $x$ when written in base $b$.

$$\text{len}_b(x) = \big\lceil \log_b(x + 1) \big\rceil$$

Then, we have $\text{at}_b(x, i)$, which gives the digit at index $i$ of $x$ when written in base $b$. This is zero-indexed and $i=0$ refers to the least significant digit. For example $\text{at}_{10}(123, 0) = 3$

$$\text{at}_b(x, i) = \left\lfloor \frac{\lvert x \rvert}{b^i} \right\rfloor \text{ mod } b$$

Here, $\text{mod}$ is an operator, not the equivalence class. Now, here's our definition of reverse,

$$\text{rev}_b(x) = \sum_{i = 0}^{\text{len}_b(x) - 1} \text{at}_b(x, i) \cdot b^{\text{len}_b(x) - i - 1}$$

Let's break this down. The $\Sigma$ can be described as a "loop". We loop through all of the valid indices $i$ of the digits in $x$. For each index $i$, we take the digit in $x$ at that index and "place" it in a new number at a new index $\text{len}_b(x) - i - 1$. Notice, this new index is exactly the index of this digit in the reversed number.

What's even cooler is that we can translate our definition of $\text{rev}$ back into code. Here's a snippet using Python,

from math import *

def len(x, b):
return ceil(log(x + 1, b))

def at(x, i, b):
return floor(abs(x) / (b ** i)) % b

def rev(x, b):
return sum(at(x, i, b) * (b ** (len(x, b) - i - 1))
for i in range(len(x, b)))

Let's test it out here to verify that it it works:

>>> rev(1234510)
54321
>>> rev(0o370018== 0o10073
True
Awesome! The first expression shows that $12345$ reversed when written in base $10$ comes out to $54321$, which is what we expected! Then we see that $\text{rev}(37001_8, 8) = 10073_8$, as expected. Finally we see that $\text{rev}(\text{deadbeef}_{16}, 16) = \text{feebdaed}_{16}$.