C-SCAN (circular SCAN) in Operating Systems - Time & Space Complexity

Time Complexity: C-SCAN (circular SCAN)

O(n log n)

Understanding Time Complexity

Analyzing time complexity helps us understand how the C-SCAN disk scheduling algorithm performs as the number of disk requests grows.

We want to know how the total work changes when more requests arrive.

Scenario Under Consideration

Analyze the time complexity of the following C-SCAN scheduling process.

function cscan(requests, head, disk_size) {
  requests.sort()
  let total_movement = 0
  let index = 0;
  while (index < requests.length && requests[index] < head) {
    index++;
  }

  for (let i = index; i < requests.length; i++) {
    total_movement += Math.abs(requests[i] - head)
    head = requests[i]
  }

  total_movement += Math.abs(disk_size - 1 - head) + (disk_size - 1)
  head = 0

  for (let i = 0; i < index; i++) {
    total_movement += Math.abs(requests[i] - head)
    head = requests[i]
  }

  return total_movement
}
    

This code moves the disk head from its current position to the end, then jumps to the start, and services all requests in a circular manner.

Identify Repeating Operations

Look for loops or repeated steps that process requests.

• Primary operation: Two loops each scanning part of the sorted request list.
• How many times: Each request is visited exactly once in total.

How Execution Grows With Input

As the number of requests increases, the algorithm processes each request once in order.

Input Size (n)	Approx. Operations
10	About 10 steps to move and service requests
100	About 100 steps
1000	About 1000 steps

Pattern observation: The work grows roughly in direct proportion to the number of requests.

Final Time Complexity

Time Complexity: O(n log n)

This means the time grows a bit faster than the number of requests because sorting is needed, but each request is only handled once.

Common Mistake

[X] Wrong: "The C-SCAN algorithm runs in linear time because it just moves through requests once."

[OK] Correct: Sorting the requests first takes extra time, which grows faster than just scanning, so the total time is not purely linear.

Interview Connect

Understanding how C-SCAN scales helps you explain disk scheduling efficiency clearly, showing you grasp both algorithm steps and their cost.

Self-Check

"What if the requests were already sorted? How would that affect the time complexity of C-SCAN?"

Practice

1. What is the main characteristic of the C-SCAN disk scheduling algorithm?

easy

A. The disk arm moves back and forth serving requests in both directions.

B. The disk arm moves in one direction and jumps back to the start without serving requests on return.

C. The disk arm serves requests randomly without any specific order.

D. The disk arm serves only the nearest request to its current position.

Solution

1. Step 1: Understand C-SCAN movement

   C-SCAN moves the disk arm in one direction only, serving requests along the way.

2. Step 2: Identify behavior at the end of the disk

   After reaching the last track, it jumps back to the first track without serving requests during the return.

3. Final Answer:

   The disk arm moves in one direction and jumps back to the start without serving requests on return. -> Option B

4. Quick Check:

   C-SCAN moves one way and jumps back = A [OK]

Hint: Remember: C-SCAN moves one way, then jumps back [OK]

Common Mistakes:

• Thinking C-SCAN moves back and forth like SCAN
• Assuming requests are served during the return jump
• Confusing C-SCAN with random or nearest request scheduling

2. Which of the following is the correct way to describe the jump in C-SCAN?

easy

A. The disk arm jumps back to the first track without serving any requests during the jump.

B. The disk arm stops and waits for new requests before moving.

C. The disk arm reverses direction and serves requests immediately.

D. The disk arm jumps back to the first track and serves requests on the way.

Solution

1. Step 1: Analyze the jump behavior in C-SCAN

   C-SCAN jumps back to the first track after reaching the last track.

2. Step 2: Check if requests are served during the jump

   Requests are not served during this jump; the arm moves quickly without processing.

3. Final Answer:

   The disk arm jumps back to the first track without serving any requests during the jump. -> Option A

4. Quick Check:

   Jump without serving requests = D [OK]

Hint: Jump skips requests, no service during return [OK]

Common Mistakes:

• Assuming requests are served during the jump
• Confusing jump with reversing direction
• Thinking the arm waits before jumping

3. Consider a disk with tracks numbered 0 to 199. The disk arm is currently at track 50 and requests are at tracks 95, 180, 34, and 119. Using C-SCAN, which track will the arm serve immediately after 50?

medium

A. 95

B. 34

C. 119

D. 180

Solution

1. Step 1: Identify requests ahead of current position

   Requests are at 34, 95, 119, 180. Current position is 50. C-SCAN moves forward only, so it serves requests with track numbers greater than 50 first: 95, 119, 180.

2. Step 2: Find the nearest request ahead

   The nearest request ahead is at track 95, so it will be served immediately after 50.

3. Final Answer:

   95 -> Option A

4. Quick Check:

   Next request after 50 going forward = 95 [OK]

Hint: Serve next higher track number after current position [OK]

Common Mistakes:

• Serving requests behind current position before forward ones
• Choosing the closest track number regardless of direction
• Confusing C-SCAN with SCAN which moves back and forth

4. A student wrote that in C-SCAN, the disk arm serves requests while moving back to the first track after reaching the last track. What is wrong with this statement?

medium

A. C-SCAN does not move back to the first track; it reverses direction instead.

B. C-SCAN serves requests randomly, so direction does not matter.

C. C-SCAN serves requests only while moving forward, not during the jump back.

D. C-SCAN serves requests only at the last track.

Solution

1. Step 1: Review C-SCAN jump behavior

   C-SCAN moves forward serving requests, then jumps back to the first track without serving requests during the jump.

2. Step 2: Identify the error in the student's statement

   The student incorrectly said requests are served during the jump back, which is false.

3. Final Answer:

   C-SCAN serves requests only while moving forward, not during the jump back. -> Option C

4. Quick Check:

   Requests served only forward, not on jump = C [OK]

Hint: Requests served only forward, jump is skip [OK]

Common Mistakes:

• Thinking C-SCAN reverses direction instead of jumping
• Assuming requests are served during the jump
• Confusing C-SCAN with SCAN algorithm

5. Given a disk with tracks 0 to 199, the disk arm is at track 150. Requests are at tracks 10, 170, 15, 190, and 30. Using C-SCAN, what is the order in which the disk arm will serve these requests?

hard

A. 30, 15, 10, 190, 170

B. 10, 15, 30, 170, 190

C. 190, 170, 30, 15, 10

D. 170, 190, 10, 15, 30

Solution

1. Step 1: Identify requests ahead of current position 150

   Requests ahead are 170 and 190. C-SCAN serves these first moving forward.

2. Step 2: After reaching last track, jump to start and serve remaining requests

   After 190, the arm jumps to track 0 and serves requests at 10, 15, and 30 in ascending order.

3. Final Answer:

   170, 190, 10, 15, 30 -> Option D

4. Quick Check:

   Serve forward requests, jump, then serve from start = A [OK]

Hint: Serve forward requests first, then jump and serve from start [OK]

Common Mistakes:

• Serving requests behind current position before forward ones
• Serving requests in descending order
• Ignoring the jump and serving requests in wrong order