Cs6515 exam 1 - Study guides, Class notes & Summaries

Knapsack without repetition - ANSWER-k(0) = 0 for w = 1 to W: if w_j >w: k(w,j) = k(w, j - 1) else: K(w,j) = max{K(w, j -1),K(w - w_j, j -1) + v_i} knapsack with repetition - ANSWER-knapsack repeat(w_i....w_n, w_i... w_n, B) k(0) = 0 for i = 1 to n if w_i <= b & k(b) <v_i + K(b-w_i) then k(b) = v_i + K(b-w_i) Longest Increasing Subsequence - ANSWER-LIS(a_1.... a_n) for i = 1 to n L(i) = 1 for j = 1 to n -1 if a_j < a_i & L(i) < 1 + L(j) L(i) = 1 + L(j) max = 1 for ...

Steps to solve a Dynamic Programming Problem - ANSWER-1. Define the Input and Output. 2. Define entries in table, i.e. T(i) or T(i, j) is... 3. Define a Recurrence relationship - Based on a subproblem to the main problem. (hint: use a prefix of the original input 1 < i < n). 4. Define the Pseudocode. 5. Define the Runtime of the algorithm. Use Time Function notation here => T(n) = T(n/2) + 1... DP: Types of Subproblems (4) - ANSWER-Input = x1, x2, ..., xn 1) Subproblem = x1, x2...

What is the running time for Longest Increasing Subsequence (LIS) - ANSWER-O(n^2) What is the recurrence for Longest Increasing Subsequence (LIS)? - ANSWER-L(i) = 1 + max{ L(j) | xj < xi} This reads as the answer to index I is 1 + the maximum over all j's between 1 and i where xj is less than xi What is the recurrence for Longest Common Subsequence (LCS) - ANSWER-L(i,j) = 1 + L(i-1, j-1) if xi = yj L(i,j) = max(L(i-1,j),L(i,j-1)) otherwise What is the running time for Longest Common...

CS6515 Exam 2 Questions And Answers