Add all my notes

imperfect_notes/pragy/Maths 2 - Factorization.md

@@ -0,0 +1,225 @@
+Maths 2 - Factorization
+-----------------------
+
+All Factors
+-----------
+
+> Given number n, find all factors of n
+> 
+
+**Observations:**
+1. largest factor is n itself
+1. all other factors are $\leq n/2$
+
+**Brute Force:**
+So, loop from $1 \ldots \frac n 2$
+
+**Time complexity:** $O(n)$
+
+
+**Observation 3**
+Factors come in pairs
+(1,n), (2, n/2), (x, n/x) .. 
+Need to go only till $\sqrt n$
+Why sqrt? if x = n/x, then $x = \sqrt n$
+
+```python
+for(i = 1; i < sqrt(n); i++)
+    if( n%i == 0 ):
+        factors.append(i)
+        factors.append(n/i)
+if( n % i == 0)
+    factors.append(i) # because i = n/i = sqrt(n)
+```
+
+Complexity: $O(\sqrt n)$
+
+-- --
+
+Open Door question
+------------------
+
+100 doors, all are closed.
+1. Open all doors
+2. Close every 2nd door
+3. Open every 3rd door
+4. Close every 4th door
+
+100. Close every 100th door.
+
+Find the number of doors that are left open.
+
+Approach: Door is toggled by each factor.
+Find the number of factors for each 0 <= i <= 100
+
+Check if the number of factors is even or odd.
+
+**How to find if the number of factors is even or odd?**
+
+
+**Brute Force**: Loop and count
+
+**Observation:** Factors come in pairs. Except for $\sqrt n$
+
+Check if perfect square. If perfect suqre - odd, else even.
+
+Finding $\sqrt n$: Binary search / Newton Raphson
+
+
+-- --
+
+Check Prime
+-----------
+
+**Brute Force** Loop from 1 to $\sqrt n$ and check if the count is 1
+$O(\sqrt n)$
+
+-- --
+
+Generate all primes till n
+--------------------------
+
+Brute force using $\sqrt n$ check prime: $O(n \times \sqrt n)$
+
+**Sieve of Eratosthenes**  (era - tos - thenis)
+
+- Assume all prime
+- cut 1
+- next uncut is prime
+- cut all multiples - because can't be prime
+- go only till sqrt
+
+![3fa0e7ff.png](:storage/3d9f53c5-6f21-4ce2-b310-40e38a17113c/3fa0e7ff.png =400x)
+
+Optimizatiopn
+- start cutting from $i\times i$, because $i \times (i-1,2..)$ is already cut
+- $i$ needs to go only till $sqrt n$
+
+```python
+primes[N] = {1}
+primes[0] <- 0
+primes[1] <- 0
+for(i = 2; i * i <= N; i++)
+    if(primes[i] == 1)
+        for(j = i; i * j <= N; j++)
+            primes[i*j] = 0
+```
+
+Complexity:
+
+2 -> N/2
+3 -> N/3
+4 -> O(1)
+5 -> N/5
+6 -> O(1)
+7 -> N/7
+8 -> O(1)
+9 -> O(1)
+...
+$\sqrt N$ -> 1 or O(1)
+
+total = $N (1/1 + 1/2 + 1/3 + 1/5 + 1/7...)$
+
+$\Large N \sum_p \frac 1 p$
+
+= $O(n \log \log n)$ ([Mertens' theorems - Wikipedia](https://en.wikipedia.org/wiki/Mertens%27_theorems))
+
+Memory = $O(n)$
+
+Simple Upperbound can be $n \log n$ because $\sum_i \frac 1 i = \Theta(\log n)$ (proof by integration)
+
+
+Note: better algos exist for primality testing, as well as sieve.
+
+-- --
+
+Caution: Don't build a sieve if you just want to check one number. $O(\sqrt n)$ vs $O(n \log n)$
+
+
+-- --
+
+Prime Factorization
+-------------------
+
+1. find all factors and check if prime
+
+
+> $24 = 2^3 \cdot 3$
+> Factorize N
+
+```python
+p_factors = []
+for f <- 1.. sqrt(n)
+    while n % f == 0
+        p_factors.append(f)
+        n /= f
+```
+
+Issue:
+> 404 = 2 * 2 * 101
+> our algo gives only 2, 2
+> i.e., if a prime factorization contains a prime number that is greater than the sqrt of n. we have an issue
+
+Fixed:
+
+```python
+p_factors = []
+for f <- 1.. sqrt(n)
+    while n % f == 0
+        p_factors.append(f)
+        n /= f
+if n != 1
+    p_factors.append(n)
+```
+
+Time complexity: $O(\sqrt n)$
+Inner loop doesn't matter since n is getting decreased
+
+
+**Optimization 1:**
+
+We're going over composite numbers as well. Go over primes only.
+
+But sieve is costly. Ignoring sieve, we will take
+O(number of primes < N) time
+
+**Optimization 2:**
+
+If we can get SmallestPrimeFactor(n) in O(1) time, we can solve in logn time.
+
+Because,
+
+6060 - 2
+3030 - 2
+1515 - 3
+505 - 5
+101
+
+Everytime, number if being reduced by atleast half
+
+In sieve itself, we can find the SPF of each number - the first prime to cut that number
+
+Helpful when doing multiple queries
+
+
+-- --
+
+
+Number of factors
+-----------------
+
+If $\Large n = p_1 ^ a p_2 ^ b p_3 ^ c \cdots$
+
+Then number of factors $= \phi(n) = (1+p_1)(1+p_2)(1+p_3)\cdots$
+Sum of factors
+$= (1 + p_1 + p_1^2 + \cdots + p_1^a) \cdot (1 + p_2 + p_2^2 + \cdots + p_2^b) \cdots$
+
+$= \Large \frac{p_1^{a+1} - 1}{p_1 - 1} \cdot \frac{p_2^{b+1} - 1}{p_2 - 1} \cdots$
+can be found using mod-exponentiation
+
+Product of factors = $n^{\phi(n)}$ where $\phi(n)$ is the number of factors
+
+-- --
+
+
+https://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n