To show that if a DECREMENT operation were included in the k-bit counter example, n operations could cost as much as Θ(nk) time, we need to analyze the cost of both INCREMENT and DECREMENT operations in a k-bit binary counter.

1. **Binary Counter Basics**:
- A k-bit binary counter can represent values from 0 to 2^k - 1.
- Each bit in the counter can be either 0 or 1.

2. **INCREMENT Operation**:
- When incrementing the counter, we start from the least significant bit (LSB) and flip bits until we find a 0 to flip to 1.
- The cost of an INCREMENT operation is proportional to the number of bits flipped.
- In the worst case, all k bits are flipped (e.g., from 111...111 to 000...000).

3. **DECREMENT Operation**:
- When decrementing the counter, we start from the LSB and flip bits until we find a 1 to flip to 0.
- The cost of a DECREMENT operation is also proportional to the number of bits flipped.
- In the worst case, all k bits are flipped (e.g., from 000...000 to 111...111).

4. **Amortized Analysis**:
- To find the amortized cost, we consider a sequence of n operations (both INCREMENT and DECREMENT).
- Each bit in the counter can be flipped from 0 to 1 or from 1 to 0 at most once per operation.
- Therefore, each bit can be flipped at most 2n times in n operations.

5. **Total Cost**:
- Since there are k bits, and each bit can be flipped at most 2n times, the total number of bit flips in n operations is at most 2nk.
- Thus, the total cost of n operations is O(nk).

6. **Conclusion**:
- The amortized cost per operation is O(k).
- Therefore, n operations can cost as much as Θ(nk) time.

This demonstrates that if a DECREMENT operation were included in the k-bit counter example, n operations could indeed cost as much as Θ(nk) time.

Question

To show that if a DECREMENT operation were included in the k-bit counter example, n operations could cost as much as Θ(nk) time, we need to analyze the cost of both INCREMENT and DECREMENT operations in a k-bit binary counter.

1. **Binary Counter Basics**:
   - A k-bit binary counter can represent values from 0 to 2^k - 1.
   - Each bit in the counter can be either 0 or 1.

2. **INCREMENT Operation**:
   - When incrementing the counter, we start from the least significant bit (LSB) and flip bits until we find a 0 to flip to 1.
   - The cost of an INCREMENT operation is proportional to the number of bits flipped.
   - In the worst case, all k bits are flipped (e.g., from 111...111 to 000...000).

3. **DECREMENT Operation**:
   - When decrementing the counter, we start from the LSB and flip bits until we find a 1 to flip to 0.
   - The cost of a DECREMENT operation is also proportional to the number of bits flipped.
   - In the worst case, all k bits are flipped (e.g., from 000...000 to 111...111).

4. **Amortized Analysis**:
   - To find the amortized cost, we consider a sequence of n operations (both INCREMENT and DECREMENT).
   - Each bit in the counter can be flipped from 0 to 1 or from 1 to 0 at most once per operation.
   - Therefore, each bit can be flipped at most 2n times in n operations.

5. **Total Cost**:
   - Since there are k bits, and each bit can be flipped at most 2n times, the total number of bit flips in n operations is at most 2nk.
   - Thus, the total cost of n operations is O(nk).

6. **Conclusion**:
   - The amortized cost per operation is O(k).
   - Therefore, n operations can cost as much as Θ(nk) time.

This demonstrates that if a DECREMENT operation were included in the k-bit counter example, n operations could indeed cost as much as Θ(nk) time.

Knowee AI · Accepted Answer