Modular Arithmetic for Beginners

Revision en5, by Spheniscine, 2019-12-27 15:51:41

Introduction

If you're new to the world of competitive programming, you may have noticed that some questions, typically combinatorial and probability questions, have this funny habit of asking you to calculate a huge number, then tell you that "because this number can be huge, please output it modulo $$$10^9 + 7$$$". Like, it's not enough that they ask you to calculate a number they know will overflow basic integer data types, but now you need to apply the modulo operation after that? Even worse are those that say you need to calculate a fraction $$$\frac pq$$$ and ask you to output $$$r$$$ where $$$r \cdot q = p \text{ mod } n$$$... not only do you have to calculate a fraction with huge numbers, how in the world are you going to find $$$r$$$?

Actually, the modulo is there to make the calculation easier, not harder. This may sound counterintuitive, but once you know how modular arithmetic works, you'll see why too. Soon you'll be solving these problems like second nature.

Terminology and notation

For convenience, I will define the notation $$$n \text{ mod } m$$$ to mean $$$n - \lfloor \dfrac nm \rfloor \cdot m$$$, where $$$\lfloor x \rfloor$$$ is the largest integer that doesn't exceed $$$x$$$. This may or may not correspond to the expression n % m in your programming language (% is often called the "modulo operator" but in some instances, it's more correct to call it the "remainder operator"). If -8 % 7 == 6, you're fine, but if it is -1, you'll need to adjust it by adding $$$m$$$ to any negative results. If you use Java/Kotlin, the standard library function Math.floorMod(n, m) does what we need.

Also for convenience, I will also define the $$$\text{ mod }$$$ operator to have lower precedence than addition or subtraction, thus $$$ax + b\ \text{ mod } m \Rightarrow (ax + b) \text{ mod } m$$$. This probably does not correspond with the precedence of the % operator.

The value $$$m$$$ after the modulo operator is known as the modulus

"Basic" arithmetic

First off, some important identities about the modulo operator:

$$$(a \text{ mod } m) + (b \text{ mod } m)\ \text{ mod } m = a + b\ \text{ mod } m$$$

$$$(a \text{ mod } m) - (b \text{ mod } m)\ \text{ mod } m = a - b\ \text{ mod } m$$$

$$$(a \text{ mod } m) \cdot (b \text{ mod } m)\ \text{ mod } m = a \cdot b\ \text{ mod } m$$$

These identities have the very important consequence in that you generally don't need to ever store the "true" values of the large numbers you're working with, only their residue $$$\text{ mod } m$$$. You can then add, subtract, and multiply with them as much as you need for your problem, taking the modulo as often as needed to avoid integer overflow. You may even decide to wrap them into their own object class with overloaded operators if your language supports them, though you may have to be careful of any object allocation overhead. If you use Kotlin like I do, consider using the inline class feature.

But what about division and fractions? That's slightly more complicated, and requires a concept called the "modular multiplicative inverse". The modular multiplicative inverse of a number $$$a$$$ is the number $$$a^{-1}$$$ such that $$$a \cdot a^{-1}\ \text{ mod } m = 1$$$

But how do you actually find such a number? Bruteforcing all numbers to a prime number close to a billion will usually cause you to exceed the time limit. There are two faster ways to calculate the inverse: the extended GCD algorithm, and Fermat's little theorem. Though the extended GCD algorithm is more versatile and sometimes slightly faster, the Fermat's little theorem method is often more popular, simply because it's almost "free" once you implement exponentiation, another useful operation to do, so that's what we'll cover here.

Fermat's theorem says that as long as the modulus $$$m$$$ is a prime number and $$$a \text{ mod } m \neq 0$$$, then $$$a^m \text{ mod } m = a \text{ mod } m$$$. Working backwards, $$$a^{m-1} \text{ mod } m = a \cdot a^{m-2}\ \text{ mod } m = 1$$$

Tags modulo, modular arithmetic

History

 
 
 
 
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  Rev. Lang. By When Δ Comment
en33 English Spheniscine 2020-10-24 18:50:23 20
en32 English Spheniscine 2020-04-18 06:33:00 319
en31 English Spheniscine 2020-02-12 04:43:04 101
en30 English Spheniscine 2020-02-12 04:39:24 214 Tiny change: 'efined as $1$](https://' -> 'efined as 1](https://'
en29 English Spheniscine 2020-01-16 12:01:55 73 Tiny change: ' produce a number between' -> ' produce an integer between'
en28 English Spheniscine 2020-01-16 11:40:32 252
en27 English Spheniscine 2020-01-07 15:14:50 14 Tiny change: '$0 \cdot x \text{ mo' -> '$0 \cdot x\ \text{ mo'
en26 English Spheniscine 2020-01-06 22:33:32 8
en25 English Spheniscine 2019-12-29 14:18:37 66
en24 English Spheniscine 2019-12-29 01:11:18 2 Tiny change: 'v p \pmod n$... not o' -> 'v p \pmod m$... not o'
en23 English Spheniscine 2019-12-29 01:10:14 218 Tiny change: ' \equiv p \text{ mod } n$... not o' -> ' \equiv p (\pmod n)$... not o'
en22 English Spheniscine 2019-12-28 13:56:37 90 Tiny change: 'ession $n mod m$ is kno' -> 'ession $n \text{ mod } m$ is kno'
en21 English Spheniscine 2019-12-28 05:46:39 3 Tiny change: 'minator isn't coprime t' -> 'minator is coprime t'
en20 English Spheniscine 2019-12-28 05:16:04 2 Tiny change: 'a fraction, we want a' -> 'a fraction; we want a'
en19 English Spheniscine 2019-12-28 05:15:21 181
en18 English Spheniscine 2019-12-28 04:34:30 32
en17 English Spheniscine 2019-12-28 04:30:00 7 Tiny change: 'r \cdot q = p \text{ ' -> 'r \cdot q \equiv p \text{ '
en16 English Spheniscine 2019-12-28 04:22:39 111
en15 English Spheniscine 2019-12-28 04:21:05 37
en14 English Spheniscine 2019-12-28 04:18:50 325
en13 English Spheniscine 2019-12-28 03:57:50 55
en12 English Spheniscine 2019-12-28 03:44:37 2 Tiny change: 'xt{ mod } 0 = 1$\n\nM' -> 'xt{ mod } m = 1$\n\nM'
en11 English Spheniscine 2019-12-28 03:44:15 174
en10 English Spheniscine 2019-12-27 17:41:43 36 Tiny change: 'orces.com/problemset/problem/935/D)' -> 'orces.com/contest/935/problem/D)'
en9 English Spheniscine 2019-12-27 17:37:21 8 Tiny change: 'division) in a way ' -> 'division) defined in a way '
en8 English Spheniscine 2019-12-27 17:26:44 344 Tiny change: 'actorials are too big ' -> 'actorials can be too big ' (published)
en7 English Spheniscine 2019-12-27 17:06:35 4705 Tiny change: 'method is often more popu' -> 'method is more popu'
en6 English Spheniscine 2019-12-27 15:56:03 270 Tiny change: 'd } m = 1$' -> 'd } m = 1$, therefore the number we need is $a^{m-2} \text{ mod } m$'
en5 English Spheniscine 2019-12-27 15:51:41 790 Tiny change: 'll see why. Soon you' -> 'll see why too. Soon you'
en4 English Spheniscine 2019-12-27 15:29:05 1601 Tiny change: 't{ mod } m' -> 't{ mod } m$'
en3 English Spheniscine 2019-12-27 15:04:10 518 Tiny change: ' b \text{ mod } m \' -> ' b \text{ mod } m \'
en2 English Spheniscine 2019-12-27 14:56:23 257
en1 English Spheniscine 2019-12-27 14:50:55 987 Initial revision (saved to drafts)