Motivation Problem
Given a sequence of positive integers, say a, count number of ordered distinct triplets such that their greatest common divisor is 1.
Naive solution
\mathcal{O}(n^3\log{\text{max }a_{i}})
O(n3logmax ai)
Expected Solution
\mathcal{O(n\log{n})}
O(nlogn)
\mathcal{O}({n\log{\text{max }a_{i}}})
O(nlogmax ai)
Greatest Common Divisor
Euler's Algorithm
\text{Time Complexity: }\mathcal{O}(\log{\text{max }(a, b)})
Time Complexity: O(logmax (a,b))
Multiplicative Function
\text{An arithmetic function } f(n) : \mathbb{N}\rightarrow\mathbb{C} \text{ is multiplicative }
An arithmetic function f(n):N→C is multiplicative
\text{if for any relatively
prime } n, m \in \mathbb{N}:
if for any relatively prime n,m∈N:
f(mn) = f(m)f(n).
f(mn)=f(m)f(n).
One Sieve to rule them all
int primes[N] = {1} // 1 a prime
int lowestPrime[N]
int phi[N]
int mu[N]
for i in [2 .. N]:
if primes[i] is 1:
lowestPrime[i] = i
j = i
while j <= N:
primes[j] = 0
if primes[j] is 1:
lowestPrime[j] = i
primes[j] = 0
j += i
phi[i] = i - 1
mu[i] = -1
else:
if lowestPrime[i] is lowestPrime[i / lowestPrime[i]]:
mu[i] = 0
phi[i] = lowestPrime[i] * phi[i / lowestPrime[i]]
else:
mu[i] = -1 * mu[i / lowestPrime[i]]
phi[i] = (lowestPrime[i] - 1) * phi[i / lowestPrime[i]]Bonus
We can do all of the above in:
\mathcal{O}(n)
O(n)
Copy of deck
By APwhitehat
Copy of deck
- 174
