Problem Solving Guide to Modular Combinatorics and Exponentiation

Revision en6, by jeqcho, 2020-06-16 10:22:30

Sometimes, you are asked to calculate the combination or permutation modulo a number, for example $^nC_k \mod p$. Here I want to write about a complete method to solve such problems with a good time complexity because it took me a lot of googling and asking to finally have the complete approach. I hope this blog can help other users and save their time when they solve combinatorics problem in Codeforces.

Example Problem

Find the value of $^nC_k$, ($1 \leq n,k \leq 10^6$). As this number can be rather large, print the answer modulo $p$. ($p = 1000000007 = 10^9 + 7$)

Combination (binomial coefficients)

$^nC_k$ means how many ways you can choose $k$ items from an array of $n$ items, also denoted as $\binom{n}{k}$. This is also known as binomial coefficients. The formula for combination is

$^nC_k = \frac{n!}{k!(n-k)!}$

Sometimes, the denominator $k!(n-k)!$ is very large, but we can't modulo it since modulo operations can't be done independently on the denominator. $\frac{n!}{k!(n-k)!} \mod p \neq \frac{n! \mod p}{k!(n-k)! \mod p}$. Now I will introduce the modular multiplicative inverse to solve this problem.

Modular multiplicative inverse

The modular multiplicative inverse $x$ of $a$ modulo $p$ is defined as

$a \cdot x \equiv 1 \pmod p$

Here, I will replace $x$ with $\text{inv}(a)$, so we have

$a \cdot \text{inv}(a) \equiv 1 \pmod p$

Getting back to the formula for combination, we can rearrange so that

$^nC_k = n! \cdot \frac{1}{k!} \cdot \frac{1}{(n-k)!}$

Here, we can use $\text{inv}(a)$ as follows

$^nC_k \equiv n! \cdot \text{inv}(k!) \cdot \text{inv}((n-k)!) \pmod p$

Now we can distribute the modulo to each of the terms by the distributive properties of modulo

$^nC_k \mod p = n! \mod p \cdot \text{inv}(k!) \mod p \cdot \text{inv}((n-k)!) \mod p$

Now I will discuss on how to calculate $\text{inv}(a)$

Fermat's Little Theorem

You can easily remember this theorem. Let $a$ be an integer and $p$ be a prime number,

$a^p \equiv a \pmod p$

It is helpful to know that the $p$ in the problem ($10^9 + 7$) is indeed a prime number! We can divide both sides with $a$ to get

$a^{p-1} \equiv 1 \pmod p$

Looking back at our equation for $\text{inv}(a)$, both equations equate to 1, so we can equate them as

$a \cdot \text{inv}(a) \equiv a^{p-1} \pmod p$

Dividing each side by $a$ we have

$\text{inv}(a) \equiv a^{p-2} \pmod p$

We now have a directly formula for $\text{inv}(a)$. However, we cannot use the pow() function to calculate $a^{p-2}$ because $a$ and $p$ is a large number (Remember $1 \leq n,k \leq 10^6$) ($p = 10^9 + 7$). Fortunately, we can solve this using modular exponentiation.

Modular Exponentiation

To prevent integer overflow, we can carry out modulo operations during the evaluation of our new power function. But instead of using a while loop to calculate $a^{p-2}$ in $O(p)$, we can use a special trick called squaring the powers. Note that if $b$ is an even number

$a^b = (a^2)^{b/2}$

Every time we calculate $a^2$, we reduce the exponent by a factor of 2. We can do this repeatedly until the exponent becomes zero where we stop the loop. This will give us a time complexity of $O(\log p)$ to calculate $a^{p-2}$ because we halve the exponent in each step. For the case when $b$ is odd, we can use the property

$a^b = a^{b-1} \cdot a$

We then store the trailing $a$ into a variable. Then $b-1$ is even and we can proceed as previously stated. We can repeatedly apply these two equations to calculate $a^{p-2}$. Here I will show you the implementation of this modified powmod() function to include modulo operations. ll is defined as long long

ll powmod(ll a, ll b, ll p){
a %= p;
if (a == 0) return 0;
ll product = 1;
while(b > 0){
if (b&1){    // you can also use b % 2 == 1
product *= a;
product %= p;
--b;
}
a *= a;
a %= p;
b /= 2;    // you can also use b >> 1
}
return product;
}


Then we can finally implement the $\text{inv}(a)$ function simply as

ll inv(ll a, ll p){
return powmod(a, p-2, p);
}


Then, finally, we can implement $^nC_k$ as

ll nCk(ll n, ll k, ll p){
return ((fact[n] * inv(fact[k], p) % p) * inv(fact[n-k], p)) % p;
}


We used the dp-approach for factorial where the factorial from 1 to n is pre-computed and stored in an array fact[].

Time complexity

• Pre-computation of factorial: $O(n)$
• Calculation of $^nC_k$, which is dominated by modular exponentiation powmod: $O(\log p)$
• Total: $O(n + \log p)$

Problems for you

Please comment below if you know similar problems.

Stay safe and thank you for reading.

History

Revisions

Rev. Lang. By When Δ Comment
en10 jeqcho 2020-06-16 10:51:53 33
en9 jeqcho 2020-06-16 10:45:59 2 Tiny change: 'e a directly formula f' -> 'e a direct formula f'
en8 jeqcho 2020-06-16 10:37:26 47
en7 jeqcho 2020-06-16 10:34:25 34
en6 jeqcho 2020-06-16 10:22:30 46 Tiny change: ' an array fact[].\n\n#### ' -> ' an array fact[].\n\n#### ' (published)
en5 jeqcho 2020-06-16 10:17:21 100
en4 jeqcho 2020-06-16 10:08:26 129
en3 jeqcho 2020-06-15 19:44:38 7
en2 jeqcho 2020-06-15 19:41:02 3 Tiny change: 'k]) % p;\n~~~~~\n\' -> 'k]) % p;\n}\n~~~~~\n\'
en1 jeqcho 2020-06-15 19:31:13 5463 Initial revision (saved to drafts)