In this question, we'll modify the cost of flipping a bit in the k-bit counter discussed in the
lectures. Suppose instead of incurring a cost of 1 to flip a bit, for this question the cost to flip a
bit is dependent on the position of the bit is given by: flipping the ith bit (or equivalently bit i ),
for i0, incurs a cost of i. Assume the bits are labelled from 0 to k1, where bit 0 is the least
significant bit, while bit k1 is the most significant bit. For example, for k=4, going from 0000 to
0001 costs nothing, going from 0001 to 0010 costs 1 , going from 0111 to 1000 incurs a cost of
0+1+2+3=6 because it costs 0 to flip bit 0,1 to flip bit 1 , and 2 to bit 2 , and 3 to flip bit 3. Using
the method of aggregate analysis, with the counter initialized to 000..0 (zero), show that n
increments of the counter incurs a total cost of at most 4n. Hint: It may be helpful to apply one of
the following. n=0nxn1=(1x)21 and n=0nxn=(1x)2x, for x<1.
In this question, well modify the cost of flipping a bit in the k-bi.pdf
1. In this question, we'll modify the cost of flipping a bit in the k-bit counter discussed in the
lectures. Suppose instead of incurring a cost of 1 to flip a bit, for this question the cost to flip a
bit is dependent on the position of the bit is given by: flipping the ith bit (or equivalently bit i ),
for i0, incurs a cost of i. Assume the bits are labelled from 0 to k1, where bit 0 is the least
significant bit, while bit k1 is the most significant bit. For example, for k=4, going from 0000 to
0001 costs nothing, going from 0001 to 0010 costs 1 , going from 0111 to 1000 incurs a cost of
0+1+2+3=6 because it costs 0 to flip bit 0,1 to flip bit 1 , and 2 to bit 2 , and 3 to flip bit 3. Using
the method of aggregate analysis, with the counter initialized to 000..0 (zero), show that n
increments of the counter incurs a total cost of at most 4n. Hint: It may be helpful to apply one of
the following. n=0nxn1=(1x)21 and n=0nxn=(1x)2x, for x<1
