mirror of
https://github.com/dholerobin/Lecture_Notes.git
synced 2025-03-15 21:59:56 +00:00
102 lines
3.3 KiB
Markdown
102 lines
3.3 KiB
Markdown
# Calculating ${n \choose r} \% m$ like a pro
|
|
|
|
Let's begin with some prerequisites.
|
|
|
|
## Modular arithmetic
|
|
|
|
$(a \bmod m)$, also represented as $a \% m =$ remainder left when $a$ is divided by $m$.
|
|
Example: $10 \% 7 = 3$
|
|
|
|
### Properties of $\%$
|
|
#### Distritbutive over addition & substraction $(\pm)$
|
|
$$(a\pm b) \% m = \Bigl[ (a \% m ) \pm (b \% m) \Bigr] \% m$$
|
|
$(8+13)\%7 = \Bigl[(8 \% 7) + (13 \% 7)\Bigr]\%7$
|
|
$21\%7 = [1 + 6]\%7$
|
|
$0 = 7 \% 7$
|
|
$0 = 0$
|
|
#### Distritbutive over multiplication $(\times)$
|
|
$$(a \times b) \% m = \Bigl[ (a \% m ) \times (b \% m) \Bigr] \% m$$
|
|
$(8 \times 13)\%7 = \Bigl[(8 \% 7) \times (13 \% 7)\Bigr]\%7$
|
|
$104 \% 7 = [1 \times 6]\%7$
|
|
$6 = 6 \% 7$
|
|
$6 = 6$
|
|
|
|
- **NOT distributive over $\div$**
|
|
Unlike $+, -, \times$, the modulo operator does not distribute over division.
|
|
So, $\left (\dfrac ab \right )\%m \;\;\red \neq\;\; \left[\dfrac{a\%m}{b\%m}\right] \% m$.
|
|
However, the concept of **Modular Inverse** (discussed later) helps us simplify these calculations.
|
|
|
|
## Calculating $(n! \bmod m)$ in $\mathcal O(n)$ time
|
|
|
|
Since $\%$ distributes over $\times$, we can say that
|
|
|
|
$$(n!) \% m = \Bigl[n (n-1)(n-2) \cdots 1\Bigr] \% m \\[.5em]
|
|
=\overbrace{\underbrace{{\underbrace{\overbrace{n \cdot (n-1)}^{\%m} \cdot (n-2)}_{\%m} \cdots} \cdot 2}_{\% m} \cdot 1}^{\%m}$$
|
|
|
|
So, we can simply calculate factorial, by taking $\%$ after every multiplication.
|
|
```python
|
|
def fact_mod(n, mod):
|
|
ans = 1
|
|
for i in [1 .. n]
|
|
ans = (ans * i) % mod
|
|
return ans
|
|
```
|
|
|
|
## Binary Exponentiation: calculating $(n^p \bmod m)$ in $\mathcal O(\log_2 p)$ time
|
|
```python
|
|
def binary_exponentiation(n, p, mod):
|
|
if n == 0: return 0
|
|
if p == 0: return 1
|
|
ans = binary_exponentiation(n, p//2, mod) # note: integer division
|
|
ans = ans * ans
|
|
if p is even:
|
|
return ans
|
|
else:
|
|
return ans * n
|
|
```
|
|
|
|
|
|
|
|
## Fermat's Little Theorem
|
|
If $p$ is prime, then
|
|
$$a^p \equiv a \pmod p$$
|
|
|
|
## Euler's generalization to Fermat's Little theorem
|
|
|
|
Iff $a$ and $m$ are co-prime, then
|
|
$$\boxed{a^{\phi(m)} \equiv 1 \pmod m}$$
|
|
|
|
**Let's break it down:**
|
|
- $a$ is coprime with $m$ iff $gcd(a, m) = 1$, that is, $a$ and $m$ do not have any common factor.
|
|
- If $a \bmod m = b \bmod m$, we say that $a \equiv b \pmod m$
|
|
- $\phi(m)$ is the Euler's Totient Function for $m$
|
|
|
|
## Euler's Totient Function $\phi(m)$
|
|
The Euler's Totient Function $\phi(m)$ counts the number whole numbers smaller than $m$ which are coprime with $m$
|
|
|
|
For example, if $m = 30$, then what numbers are coprime with $m$?
|
|
```
|
|
1 2 3 4 5 6 7 8 9 10
|
|
11 12 13 14 15 16 17 18 19 20
|
|
21 22 23 24 25 26 27 28 29 30
|
|
-----------------------------
|
|
1 7
|
|
11 13 17 19
|
|
23 29
|
|
-----------------------------
|
|
count = 8
|
|
```
|
|
|
|
1. If $m$ is prime,
|
|
$$\boxed{\phi(m) = m - 1}$$
|
|
2. If $m$ is not prime, then let us factorize $m$
|
|
Let $m = p_1^{\alpha_1} \cdot p_2^{\alpha_2} \cdot p_3^{\alpha_3} \ldots$
|
|
Then,
|
|
$$\boxed{\phi(m) = m \left (1 - \dfrac1 p_1 \right )\left (1 - \dfrac1 p_2 \right )\left (1 - \dfrac1 p_3 \right ) \cdots}$$
|
|
|
|
Example,
|
|
- if $m = 17$, $\phi(17) = 16$
|
|
- if $m = 30 = 2^1 \cdot 3^1 \cdot 5^1$, then
|
|
$\phi(30) = 30 \left( 1 - \dfrac 1 2 \right )\left( 1 - \dfrac 1 3 \right )\left( 1 - \dfrac 1 5 \right )$
|
|
$= 30 \cdot \dfrac12 \cdot \dfrac23 \cdot \dfrac45 = 8$
|