Sieve of Eratosthenes Find All Primes in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Sieve for n=10

What is the output list of primes printed by this Sieve of Eratosthenes code for n=10?

DSA C

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

A [2, 3, 5, 6, 7]

B [2, 3, 4, 5, 7]

C [2, 3, 5, 7, 9]

D [2, 3, 5, 7]

💡 Hint

Remember that the sieve marks multiples of primes as not prime.

The sieve marks 4, 6, 8, 9, 10 as not prime. Only 2, 3, 5, 7 remain.

❓ Predict Output

intermediate

Count of primes up to 20

How many primes does the Sieve of Eratosthenes find up to n=20?

DSA C

int n = 20;
int primes[20];
for(int i = 0; i < n; i++) primes[i] = 1;
primes[0] = 0; primes[1] = 0;
for(int i = 2; i*i < n; i++) {
    if(primes[i]) {
        for(int j = i*i; j < n; j += i) primes[j] = 0;
    }
}
int count = 0;
for(int i = 2; i < n; i++) {
    if(primes[i]) count++;
}
printf("%d", count);

A 8

B 9

C 7

D 10

💡 Hint

Count primes: 2,3,5,7,11,13,17,19

There are 8 primes less than 20: 2,3,5,7,11,13,17,19.

🔧 Debug

advanced

Identify the error in Sieve implementation

What error does this Sieve of Eratosthenes code produce when run for n=15?

DSA C

int n = 15;
int primes[15];
for(int i = 0; i < n; i++) primes[i] = 1;
primes[0] = 0; primes[1] = 0;
for(int i = 2; i*i < n; i++) {
    if(primes[i]) {
        for(int j = i*i; j < n; j += i) primes[j] = 0;
    }
}
for(int i = 2; i < n; i++) {
    if(primes[i]) printf("%d ", i);
}

A Array index out of bounds error

B Compilation error due to missing semicolon

C No error, prints primes correctly

D Infinite loop

💡 Hint

Check the loop that initializes the primes array.

The loop runs from i=0 to i<=n which accesses primes[n], out of bounds for array size n.

❓ Predict Output

advanced

Output of Sieve for n=30 with print format

What is the printed output of primes for n=30 using this code?

DSA C

int n = 30;
int primes[30];
for(int i = 0; i < n; i++) primes[i] = 1;
primes[0] = 0; primes[1] = 0;
for(int i = 2; i*i < n; i++) {
    if(primes[i]) {
        for(int j = i*i; j < n; j += i) primes[j] = 0;
    }
}
for(int i = 2; i < n; i++) {
    if(primes[i]) printf("%d->", i);
}
printf("null");

A 2->3->5->7->11->13->17->19->23->29->null

B 2->3->5->7->9->11->13->17->19->23->29->null

C 2->3->5->7->11->13->17->19->23->27->null

D 2->3->5->7->11->13->17->19->23->29

💡 Hint

Remember 9 and 27 are not primes.

The sieve removes multiples of primes, so 9 and 27 are excluded.

🧠 Conceptual

expert

Why does the inner loop start at i*i in Sieve?

In the Sieve of Eratosthenes, why does the inner loop start marking multiples from i*i instead of 2*i?

A Because i*i is the smallest prime number

B Because all smaller multiples of i were already marked by smaller primes

C Because starting at 2*i would miss some primes

D Because starting at i*i reduces memory usage

💡 Hint

Think about when multiples less than i*i get marked.

Multiples less than i*i are already marked by smaller primes, so starting at i*i avoids redundant work.

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