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Example

In our example, $\Phi(D_i) \;=\; b_i$ , the number of 1s in D_i (counter).

Let t_i = number of bits reset to 0 on the ith call to Increment.

Thus, the actual cost c_i is at most (cost of reset) + (cost to set one) = $t_i \;+\; 1$ .

We know that $b_i \;\leq\; b_{i-1} \;-\; t_i \;+\; 1$

Thus, $\Phi(D_i) \;-\; \Phi(D_{i-1}) \;\leq\; (b_{i-1} \;-\; t_i \;+\; 1) \;-\; b_{i-1}$

$=\; 1 \;-\; t_i$

$\hat{c_i} \;=\; c_i \;+\; \Phi(D_i) \;-\; \Phi(D_{i-1})$

$\leq\; t_i \;+\; 1 \;+\; (1 \;-\; t_i)$

= 2

If $\Phi(D_0) \;=\; 0$ , then $\Phi(D_i) \;\geq\; 0$ for all i, and the total amortized cost is an upper bound to the actual cost.

The counter starts at zero, $\Phi(D_0) \;=\; 0$ .

$\hat{c_i}$ = O(2)

n calls is O(n)

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