In this problem we consider two stacks a and b manipulated using the following operations (n denotes the size of a and m the size of b):
• pusha(x): push element x on stack a.
• pushb(x): push element x on stack b.
• multipopa(k): pop min{k, n} elements from a.
• multipopb(k): pop min{k, m} elements from b.
• transfer(k): repeatedly pop an element from a and push it on b, until either k elements have been moved or a is empty.
assume that a and b are implemented using doubly-linked lists such that pusha and pushb, as well as a single pop from a or b, can be performed in o(1) time worst-case.
(a) what is the worst-case running time of the operations multipopa, multipopb and transfer?
(b) define a potential function φ(n, m) and use it to prove that the operations have amortized running time o(1).

22

step-by-step explanation:

87%

step-by-step explanation:

apex answer

6.5

step-by-step explanation:

you said it was in the question

~courtney

so firstly, we need to isolate the y variables to be able to solve these inequalities. firstly, add 0.5x on both sides of the first inequality and subtract 1.5x on both sides of the second inequality:

now since the slope is positive for the first inequality, this means that the line going upwards belongs to the first inequality, and the line going downwards belongs to the second inequality.

next, since y ≥ in the first inequality, this means that the shaded region will be above the first inequality's line, thus shading regions a and b.

next, since y ≤ in the second inequality, this means that the shaded region is going to be below this line, thus shading regions a and c.

since both lines shade region a, this means that region a is the solution.

step-by-step explanation:

mark brainliest!