Sieve of Eratosthenes Find All Primes in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the array that marks prime numbers.

DSA C

int prime[[1]];

Drag options to blanks, or click blank then click option'

An+1

B100

Cn-1

Dn*2

Attempts:

3 left

💡 Hint

Common Mistakes

Using size n instead of n+1 causes out-of-bound errors.

Using a fixed size like 100 limits the range.

We need an array of size n+1 to mark numbers from 0 to n.

2fill in blank

medium

Complete the code to initialize all entries as true (1) in the prime array.

DSA C

for (int i = 0; i <= n; i++) {
    prime[i] = [1];
}

Drag options to blanks, or click blank then click option'

A-1

Bn

C1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Setting to 0 marks all as not prime initially.

Using negative values is incorrect for boolean flags.

We mark all numbers as prime initially by setting to 1 (true).

3fill in blank

hard

Fix the error in the loop condition to run only up to the square root of n.

DSA C

for (int p = 2; p [1]= sqrt(n); p++) {
    if (prime[p] == 1) {
        for (int i = p * p; i <= n; i += p) {
            prime[i] = 0;
        }
    }
}

Drag options to blanks, or click blank then click option'

A<=

B<

C>

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' misses markings when sqrt(n) is integer (perfect square n).

Using '>=' or '>' reverses the loop and causes no iterations.

We loop while p is less than or equal to the square root of n.

4fill in blank

hard

Fill both blanks to print all prime numbers after sieving.

DSA C

for (int i = [1]; i <= n; i++) {
    if (prime[i] == [2]) {
        printf("%d ", i);
    }
}

Drag options to blanks, or click blank then click option'

A2

B1

C0

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Starting from 0 or 1 prints non-prime numbers.

Checking for prime[i] == 0 prints non-primes.

We start from 2 because 0 and 1 are not prime, and print numbers marked as 1 (prime).

5fill in blank

hard

Fill all three blanks to complete the sieve function header and variable declaration.

DSA C

[1] sieve(int [2]) {
    int [3][n+1];
    // rest of the code
}

Drag options to blanks, or click blank then click option'

Avoid

Bn

Cprime

Dint

Attempts:

3 left

💡 Hint

Common Mistakes

Using int as return type without return statement.

Wrong parameter name or type.

Incorrect array name.

The function returns nothing (void), takes int n as input, and declares int prime array.

Practice

1. What is the main purpose of the Sieve of Eratosthenes algorithm?

easy

A. To sort an array of numbers

B. To find all prime numbers up to a given limit

C. To find the greatest common divisor of two numbers

D. To reverse a string

Solution

1. Step 1: Understand the algorithm's goal

   The Sieve of Eratosthenes is designed to identify prime numbers efficiently.

2. Step 2: Recognize the output

   It marks multiples of primes as non-prime to find all primes up to a limit.

3. Final Answer:

   To find all prime numbers up to a given limit -> Option B

4. Quick Check:

   Prime finding = B [OK]

Hint: Sieve marks multiples to find primes fast [OK]

Common Mistakes:

• Confusing prime finding with sorting
• Thinking it finds divisors or reverses strings
• Assuming it works for negative numbers

2. Which of the following is the correct way to initialize the sieve array in C for finding primes up to n?

easy

A. int sieve[n+1]; for(int i=0; i<=n; i++) sieve[i] = 1;

B. int sieve[n]; for(int i=1; i

C. int sieve[n+1]; for(int i=0; i<=n; i++) sieve[i] = 0; sieve[0] = sieve[1] = 1;

D. int sieve[n+1]; for(int i=0; i<=n; i++) sieve[i] = i;

Solution

1. Step 1: Understand sieve initialization

   The sieve array uses 1 to mark potential primes initially.

2. Step 2: Check array size and initialization

   Array size should be n+1 to include number n, and all set to 1 (true) initially.

3. Final Answer:

   int sieve[n+1]; for(int i=0; i<=n; i++) sieve[i] = 1; -> Option A

4. Quick Check:

   Initialize all to 1 for primes [OK]

Hint: Initialize all to 1 (true) for prime candidates [OK]

Common Mistakes:

• Using size n instead of n+1
• Initializing with zeros instead of ones
• Marking 0 and 1 as prime initially

3. What will be the output of the following C code snippet using Sieve of Eratosthenes to find primes up to 10?

int n = 10;
int sieve[11];
for(int i=0; i<=n; i++) sieve[i] = 1;
sieve[0] = sieve[1] = 0;
for(int i=2; i*i<=n; i++) {
  if(sieve[i]) {
    for(int j=i*i; j<=n; j+=i) sieve[j] = 0;
  }
}
for(int i=2; i<=n; i++) {
  if(sieve[i]) printf("%d ", i);
}

medium

A. 2 3 4 5 7

B. 1 2 3 5 7

C. 2 3 5 7 9

D. 2 3 5 7

Solution

1. Step 1: Initialize sieve and mark 0,1 as non-prime

   Array sieve marks 0 and 1 as 0 (not prime), others as 1 (prime candidate).

2. Step 2: Mark multiples of primes as non-prime

   For i=2, mark 4,6,8,10 as 0; for i=3, mark 9 as 0.

3. Step 3: Print all indices with sieve[i] = 1

   Primes are 2, 3, 5, 7 only.

4. Final Answer:

   2 3 5 7 -> Option D

5. Quick Check:

   Primes up to 10 = 2 3 5 7 [OK]

Hint: Mark multiples starting from i*i to avoid repeats [OK]

Common Mistakes:

• Printing 1 as prime
• Including non-primes like 4 or 9
• Starting marking multiples from i instead of i*i

4. Identify the error in this Sieve of Eratosthenes code snippet and choose the correct fix:

int n = 20;
int sieve[21];
for(int i=0; i<=n; i++) sieve[i] = 1;
sieve[0] = sieve[1] = 0;
for(int i=2; i<=n; i++) {
  if(sieve[i]) {
    for(int j=2*i; j<=n; j+=i) sieve[j] = 0;
  }
}

medium

A. Remove sieve[0] and sieve[1] assignments

B. Initialize sieve array with zeros instead of ones

C. Change outer loop condition to i*i <= n

D. Start inner loop from j = i instead of 2*i

Solution

1. Step 1: Analyze outer loop condition

   Outer loop runs from 2 to n, but it should run only up to sqrt(n) for efficiency.

2. Step 2: Understand inner loop start

   Inner loop correctly starts from 2*i, but starting from i*i is more optimal.

3. Step 3: Correct the outer loop condition

   Change outer loop to for(i=2; i*i<=n; i++) to avoid unnecessary iterations.

4. Final Answer:

   Change outer loop condition to i*i <= n -> Option C

5. Quick Check:

   Outer loop till sqrt(n) improves sieve [OK]

Hint: Outer loop runs only to sqrt(n) for marking multiples [OK]

Common Mistakes:

• Running outer loop till n wastes time
• Starting inner loop from 2*i instead of i*i (less efficient)
• Not marking 0 and 1 as non-prime

5. You want to find all prime numbers up to 50 using the Sieve of Eratosthenes. After running the sieve, how many numbers will remain marked as prime?

hard

A. 15

B. 16

C. 14

D. 13

Solution

1. Step 1: List primes up to 50

   Primes are: 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47 (total 15).

2. Step 2: Count primes carefully

   Counting again: 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47 = 15 primes.

3. Step 3: Check if 1 is counted

   1 is not prime, so total primes = 15.

4. Final Answer:

   15 -> Option A

5. Quick Check:

   Count primes up to 50 = 15 [OK]

Hint: Count known primes or use sieve output count [OK]

Common Mistakes:

• Including 1 as prime
• Miscounting primes between 40 and 50
• Forgetting to exclude non-primes