## Euler Phi Extension and Divisor Sum Theorem

Previously we learned about Euler Phi Function. Today we are going to look at two theorems related to Euler Phi that frequently appears in CPPS. I am not sure whether these theorems have any official names, so I just made them up. These allow easy references so I will be using these names from now on.

# Euler Phi Extension Theorem

Theorem: Given a number $N$, let $d$ be a divisor of $N$. Then the number of pairs $\\{a, N \\}$, where $1 \leq a \leq N$ and $gcd(a,N) = d$, is $\phi(\frac{N}{d})$.

# Problem

Given value of $A$ and $M$, find the value of $X$ such that $AX \equiv 1 \ \text{(mod M)}$.

For example, if $A = 2$ and $M = 3$, then $X = 2$, since $2 \times 2 = 4 \equiv 1 \ \text{(mod 3)}$.

We can rewrite the above equation to this:

$AX \equiv 1 \ \text{(mod M)}$
$X \equiv \frac{1}{A} \ \text{(mod M)}$
$X \equiv A^{-1} \ \text{(mod M)}$

Hence, the value $X$ is known as Modular Multiplicative Inverse of $A$ with respect to $M$.

# How to Find Modular Inverse?

First we have to determine whether Modular Inverse even exists for given $A$ and $M$ before we jump to finding the solution. Modular Inverse doesn’t exist for every pair of given value.

## Existence of Modular Inverse

Modular Inverse of $A$ with respect to $M$, that is, $X = A^{-1} \text{(mod M)}$ exists, if and only if $A$ and $M$ are coprime.

Why is that?

$AX \equiv 1 \ \text{(mod M)}$
$AX – 1 \equiv 0 \ \text{(mod M)}$

Therefore, $M$ divides $AX-1$. Since $M$ divides $AX-1$, then a divisor of $M$ will also divide $AX-1$. Now suppose, $A$ and $M$ are not coprime. Let $D$ be a number greater than $1$ which divides both $A$ and $M$. So, $D$ will divide $AX – 1$. Since $D$ already divides $A$, $D$ must divide $1$. But this is not possible. Therefore, the equation is unsolvable when $A$ and $M$ are not coprime.

From here on, we will assume that $A$ and $M$ are coprime unless state otherwise.

## Using Fermat’s Little Theorem

Recall Fermat’s Little Theorem from a previous post, “Euler’s Theorem and Fermat’s Little Theorem“. It stated that, if $A$ and $M$ are coprime and $M$ is a prime, then, $A^{M-1} \equiv 1 \text{(mod M)}$. We can use this equation to find the modular inverse.

$A^{M-1} \equiv 1 \ \text{(mod M)}$ (Divide both side by $A$)
$A^{M-2} \equiv \frac{1}{A} \ \text{(mod M)}$
$A^{M-2} \equiv A^-1 \ \text{(mod M)}$

Therefore, when $M$ is prime, we can find modular inverse by calculating the value of $A^{M-2}$. How do we calculate this? Using Modular Exponentiation.

This is the easiest method, but it doesn’t work for non-prime $M$. But no worries since we have other ways to find the inverse.

## Using Euler’s Theorem

It is possible to use Euler’s Theorem to find the modular inverse. We know that:

$A^{\phi(M)} \equiv 1 \text{(mod M)}$
$\therefore A^{\phi(M)-1} \equiv A^{-1} \text{(mod M)}$

This process works for any $M$ as long as it’s coprime to $A$, but it is rarely used since we have to calculate Euler Phi value of $M$ which requires more processing. There is an easier way.

## Using Extended Euclidean Algorithm

We are trying to solve the congruence, $AX \equiv 1 \text{(mod M)}$. We can convert this to an equation.

$AX \equiv 1 \text{(mod M)}$
$AX + MY = 1$

Here, both $X$ and $Y$ are unknown. This is a linear equation and we want to find integer solution for it. Which means, this is a Linear Diophantine Equation.

Linear Diophantine Equation can be solved using Extended Euclidean Algorithm. Just pass $\text{ext_gcd()}$ the value of $A$ and $M$ and it will provide you with values of $X$ and $Y$. We don’t need $Y$ so we can discard it. Then we simply take the mod value of $X$ as the inverse value of $A$.

# Code

$A$ and $M$ need to be coprime. Otherwise, no solution exists. The following codes do not check if $A$ and $M$ are coprime. The checking is left of the readers to implement.

## When $M$ is Prime

We will use Fermat’s Little Theorem here. Just call the $bigmod()$ function from where you need the value.

int x = bigmod( a, m - 2, m ); // (ax)%m = 1


Here $x$ is the modular inverse of $a$ which is passed to $bigmod()$ function.

## When $M$ is not Prime

For this, we have to use a new function.

int modInv ( int a, int m ) {
int x, y;
ext_gcd( a, m, &x, &y );

// Process x so that it is between 0 and m-1
x %= m;
if ( x < 0 ) x += m;
return x;
}



I wrote this function since after using $\text{ext_gcd()}$ we need to process $x$ so that it's value is between $0$ and $M-1$. Instead of doing that manually, I decided to write a function.

So, if we want to find the modular inverse of $A$ with respect to $M$, then the result will be $X = modInv ( A, M )$.

# Complexity

Repeated Squaring method has a complexity of $O(logP)$, so the first code has complexity of $O(logM)$, whereas Extended Euclidean has complexity of $O(log_{10}A+log_{10}B)$ so second code has complexity $O(log_{10}A + log_{10}M)$.

# Why Do We Need Modular Inverse?

We need Modular Inverse to handle division during Modular Arithmetic. Suppose we are trying to find the value of the following equations:

$\frac{4}{2} \ \% \ 3$ - This is simple. We just simplify the equation and apply normal modular operation. That is, it becomes $\frac{4}{2} \ \% \ 3 = 2 \ \% \ 3 = 2$.

Then what happens when we try to do same with $\frac{12}{9}\ \% \ 5$? First we simply. $\frac{12}{9}\ \% \ 5 = \frac{4}{3}\ \% \ 5$. Now we are facing an irreducible fraction. Should we simply perform the modular operation with numerator and denominator? That doesn't help since both of them are smaller than $5$.

This is where Modular Inverse comes to the rescue. Let us solve the equation $X \equiv 3^{-1} \ \text{(mod 5)}$. How do we find the value of $X$? We will see that on the later part of the post. For now, just assume that we know the value of $X$.

Now, we can rewrite the above equation in the following manner:

$\frac{12}{9}\ \% \ 5$
$\frac{4}{3}\ \% \ 5$
$(4 \times 3^{-1})\ \% \ 5$
$( (4\ \% \ 5) \times (3^{-1}\ \% \ 5) ) \ \% \ 5$
$\therefore 4X \ \% \ 5$

So, now we can easily find the value of $\frac{A}{B} \ \% \ M$ by simply calculating the value of $(A \times B^{-1}) \ \% \ M$.

# Conclusion

Modular Inverse is a small topic but look at the amount of background knowledge it requires to understand it! Euler's Theorem, Euler Phi, Modular Exponentiation, Linear Diophantine Equation, Extended Euclidian Algorithm and other small bits of information. We covered them all before, so we can proceed without any hitch.

Hopefully, you understood how Modular Inverse works. If not, make sure to revise the articles in the "Reference" section below.

# Reference

1. Wiki - Modular Multiplicative Inverse
2. forthright48 - Euler's Theorem and Fermat's Little Theorem
3. forthright48 - Modular Exponentiation
4. forthright48 - Euler Phi
5. forthright48 - Linear Diophantine Equation
6. forthright48 - Extended Euclidean Algorithm

## Repeated Squaring Method for Modular Exponentiation

Previously on Modular Exponentiation we learned about Divide and Conquer approach to finding the value of $B^P \ \% \ M$. In that article, which is recursive. I also mentioned about an iterative algorithm that finds the same value in same complexity, only faster due to the absence of recursion overhead. We will be looking into that faster algorithm on this post today.

Make sure you know about Bit Manipulation before proceeding.

# Problem

Given three positive integers $B, P$ and $M$, find the value of $B^P \ \% \ M$.

For example, $B=2$, $P=5$ and $M=7$, then $B^P \ \% \ M = 2^5 \ \% \ 7 = 32 \ \% \ 7 = 4$.

# Repeated Squaring Method

Repeated Squaring Method (RSM) takes the advantage of the fact that $A^x \times A^y = A^{x+y}$.

Now, we know that any number can be written as the sum of powers of $2$. Just convert the number to Binary Number System. Now for each position $i$ for which binary number has $1$ in it, add $2^i$ to the sum.

For example, $(19)_{10} = (10011)_2 = ( 2^4 + 2^1 + 2^0)_{10} = (16 + 2 + 1 )_{10}$

Therefore, in equation $B^P$, we can decompose $P$ to the sum of powers of $2$.

So let’s say, $P = 19$, then $B^{19} = B^{2^4 + 2^1 + 2^0} = B^{16+2+1} = B^{16} \times B^2 \times B^1$.

This is the main concept for repeated squaring method. We decompose the value $P$ to binary, and then for each position $i$ (we start from 0 and loop till the highest position at binary form of $P$) for which binary of $P$ has $1$ in $i_{th}$ position, we multiply $B^{2^i}$ to result.

# Code

Here is the code for RSM. The Explanation is below the code.

int bigmod ( int b, int p, int m ) {
int res = 1 % m, x = b % m;
while ( p ) {
if ( p & 1 ) res = ( res * x ) % m;
x = ( x * x ) % m;
p >>= 1;
}
return res;
}


## Explanation

At line $1$, we have the parameters. We simply send the value of $B$, $P$ and $M$ to this function, and it will return the required result.

At line $2$, we initiate some variables. $res$ is the variable that will hold our result. It contains the value $1$ initially. We will multiply $b^{2^i}$ with result to find $b^p$. $x$ is temporary variable that initially contains the value $b^{2^0} = b^1 = b$.

Now, from line $3$ the loop starts. This loop runs until $p$ becomes $0$. Huh? Why is that? Keep reading.

At line $4$ we first check whether the first bit of $p$ is on or not. If it is on, then it means that we have to multiply $b^{2^i}$ to our result. $x$ contains that value, so we multiply $x$ to $res$.

Now line $5$ and $6$ are crucial to the algorithm. Right now, $x$ contains the value of $b^{2^0}$ and we are just checking the $0_{th}$ position of $p$ at each step. We need to update our variables such that they keep working for positions other than $0$.

First, we update the value of $x$. $x$ contains the value of $b^{2^i}$. On next iteration, we will be working with position $i+1$. So we need to update $x$ to hold $b^{2^{i+1}}$.

$b^{2^{i+1}} = b^{2^i \times 2^1} = b ^ {2^i \times 2} = b^{2^i + 2^i} = b^{2^i} \times b^{2^i} = x \times x$.

Hence, new value of $x$ is $x \times x$. We make this update at line $5$.

Now, at each step we are checking the $0_{th}$ position of $p$. But next we need to check the $1_{st}$ position of $p$ in binary. Instead of checking $1_{st}$ position of $p$, what we can do is shift $p$ $1$ time towards right. Now, checking $0_{th}$ position of $p$ is same as checking $1_{st}$ position. We do this update at line $6$.

These two lines ensures that our algorithm works on each iteration. When $p$ becomes $0$, it means there is no more bit to check, so the loop ends.

## Complexity

Since there cannot be more than $log_{2}(P)$ bits in $P$, the loop at line $3$ runs at most $log_{2}(P)$ times. So the complexity is $log_{2}(P)$.

# Conclusion

RSM is significantly faster than D&C approach due to lack of recursion overhead. Hence, I always use this method when I have to find Modular Exponentiation.

The code may seem a little confusing, so feel free to ask questions.

When I first got my hands on this code, I had no idea how it worked. I found it in a forum with a title, “Faster Approach to Modular Exponentiation”. Since then I have been using this code.

# Resources

1. forthrigth48 – Modular Exponentiation
2. forthright48 – Bit Manipulation
3. algorithmist – Repeated Squaring
4. Wiki – Exponentiation by Squaring