Amortized analysis: Consider the dynamic table problem (with both inserting and deleting operations). Doubling the table size when it is full. But instead of contracting the table by halving its size when its load factor drops below 1/4, we contract it by multiplying its size by 2/3 when its load factor drops below 1/3. Define a (tight) potential function so that the amortized cost of each operation is bounded above by a constant (as small as possible). (1) Write a tight potential function. (No need to explain.) (2) Give the (smallest) constant that the amortized cost of each operation is bounded above by. No need to prove it. (3) Give the amortized cost for the following four cases. No need to explain. The load factor is full 1, and the next operation is insertion. The load factor is 1/3, and the next operation is insertion. The load factor is 1/3, and the next operation is deletion. The load factor is strictly between 1/3 and 1/2, and the next operation is deletion.

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2.
Consider the dynamic table problem (with both inserting and deleting operations). Doubling the table
size when it is full. But instead of contracting the table by halving its size when its load factor drops
below 1/4, we contract it by multiplying its size by 2/3 when its load factor drops below 1/3. Define a
(tight) potential function so that the amortized cost of each operation is bounded above by a constant
(as small as possible).
(1) Write a tight potential function. (No need to explain.)
(2) Give the (smallest) constant that the amortized cost of each operation is bounded above by.
No need to prove it.
(3) Give the amortized cost for the following four cases. No need to explain.
The load factor is full 1, and the next operation is insertion.
The load factor is 1/3, and the next operation is insertion.
The load factor is 1/3, and the next operation is deletion.
The load factor is strictly between 1/3 and 1/2, and the next operation is deletion.
Amortized analysis:
Transcribed Image Text:2. Consider the dynamic table problem (with both inserting and deleting operations). Doubling the table size when it is full. But instead of contracting the table by halving its size when its load factor drops below 1/4, we contract it by multiplying its size by 2/3 when its load factor drops below 1/3. Define a (tight) potential function so that the amortized cost of each operation is bounded above by a constant (as small as possible). (1) Write a tight potential function. (No need to explain.) (2) Give the (smallest) constant that the amortized cost of each operation is bounded above by. No need to prove it. (3) Give the amortized cost for the following four cases. No need to explain. The load factor is full 1, and the next operation is insertion. The load factor is 1/3, and the next operation is insertion. The load factor is 1/3, and the next operation is deletion. The load factor is strictly between 1/3 and 1/2, and the next operation is deletion. Amortized analysis:
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