Algorithms for Primality test & GCD number - naive & efficient algorithms explained. The slides were presented by me during one of the programming sessions conducted by codechef campus chapter,

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2. —Team GCC
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bool isPrime (int n)
{
for (int i=2; i<n; i++)
if (n%i==0)
return false;
return true;
}
( Simply traverse from 2 to 𝑛 − 1 )

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bool isPrime (int n)
{
if (n<=1) return false;
if (n==2) return true;
if (n%2==0) return false;
for (int i=3; i<n; i+=2)
if (n%i==0)
return false;
return true;
}
number not
divisible by 2 is
not divisible by
any even number

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bool isPrime (int n)
{
for (int i=2; i*i<n; i++)
if (n%i==0)
return false;
return true;
}
( Simply traverse from 2 to 𝑛 )

9. ALL prime numbers are of form 6𝑘 ± 1
FACT

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bool isPrime (int n)
{
if (i==2) return true;
if (i==3) return true;
for (int i=6; i*i<n; i+=6)
if (n%(i-1)==0)
return false;
else if (n%(i+1)==0)
return false;
return true;
}

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Basic concept - we pick one prime number and eliminate all the
multiples of that prime number
1. Create an array of 𝑛 + 1 elements (0,1,…,n)
2. Initialise the array with 1 (consider all as prime)
3. Mark 0 & 1 as 0 (!prime)
4. Iterate array till 𝒊 = 𝟏 (prime is found)
5. Mark all consecutive multiple of p as 0 (!prime)
6. Repeat step 4 & 5 till 𝑖 ≠ 𝑛

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void generatePrime (int n)
{
int isPrime[n+1];
for (int i=0; i<n; i++)
isPrime[i] = 1;
isPrime[0] = isPrime[1] = 0;
for (int i=2; i<n; i++)
if (isPrime[i]==1)
for (int j=i+i; j<n; j+=i)
isPrime[i] = 0;
}
Create an array of 𝑛 + 1 elements (0,1,…,n)Initialise the array with 1 (consider all as prime)Mark 0 & 1 as 0 (!prime)Iterate array till 𝑖 = 1 (prime is found)Mark all consecutive multiple of p as 0 (!prime)Repeat step 4 & 5 till 𝑖 ≠ 𝑛

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What is GCD of 54 and 24?
Divisor of 54: 1, 2, 3, 6, 9, 18, 27, 54
Divisor of 24: 1, 2, 3, 4, 6, 8, 12, 24
Common divisor are: 1, 2, 3, 6
GCD (54, 24) = 6

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Homework :D

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https://en.wikipedia.org/wiki/Euclidean_algorithm
[video content removed]

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—pGxplorer
Keep practicing,
It’s a sport worth playing!

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THANK YOU