4 2 10 13 11.1.3 Example: Take N-2, K-3, a constant interest rate r 1 /9. Take P

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4 2 10 13 11.1.3 Example: Take N-2, K-3, a constant interest rate r 1 /9. Take P(w1) = 1/2, P(w2) = 1/4, P(ws) 1/4. Note that here, B,-(1+r). The price process is given by Take u(v)--exp(-v). So (t)-1 (y)-Iny. Also, we use the notation L- P/P. In order to compute P, we solve the system of equations characterizing P The previous system of equations is equivalent to Plan) P(w2) +P(ws) = 1. The unique solution is (1/3,1/3,1/3). We deduce (w) = 2/3,L(w2) = L(ws) = 4/3· We then use the formulae derived above for the optimal solutiorn

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Use two iterators p and q, both initially at the start of the list. Advance p one step at a time, and q two steps at a time. If q reaches the end there is no cycle; otherwise, p and q will eventually catch up to each other in the middle of the cycle

2) Like stacks the linked list and array implementation gives give fast O(1) running times for every opertion so that the linked list is straight forward and left in an exercise.

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