Cs6515 - Study guides, Class notes & Summaries

CS6515 Exam 2 Questions And Complete Solutions

CS6515-ALGORITHMS EXAM 1 NEWEST VERSION 2024 COMPLETE 104+ QUESTIONS AND CORRECT DETAILED ANSWERS (VERIFIED ANSWERS) |ALREADY GRADED A+.

CS6515 EXAM 3 NEWEST VERSION 2024 COMPLETE 46+ QUESTIONS AND CORRECT DETAILED ANSWERS (VERIFIED ANSWERS) |ALREADY GRADED A+.

CS6515 EXAM 2 NEWEST VERSION 2024 COMPLETE 46+ QUESTIONS AND CORRECT DETAILED ANSWERS (VERIFIED ANSWERS) |ALREADY GRADED A+.

CS6515 -EXAM 3 VERSION C LATEST VERSION 2024 COMPLETE 46+ QUESTIONS AND CORRECT DETAILED ANSWERS (VERIFIED ANSWERS) |ALREADY GRADED A+.

CS6515 - Algorithms- Exam 1 Complete Questions And Solutions

CS6515 - Algorithms- Exam 1 | Questions and Verified Answers | Latest Update 2024/2025 | Graded A+ 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...

CS6515 Exam 2 QUESTIONS AND ANSWERS 2024

CS6515 Exam 2 | Questions and Correct Answers | Latest Update 2024/2025 | Graded A+ Basic Properties of Trees - Answer -Tree's are undirected, connected and acyclic that connect all nodes. 1. Tree on n vertices has (n-1) edges -> would have a cycle otherwise (more than n-1 edges means cycle) 2. In tree exactly one path between every pair of vertices (otherwise it's not connected) - More than 1 path implies cycle - less than 1 path implies not connected 3. Any connected G(V, E) wi...

Equivalence - Answer-"x ≡ y (mod N) means that x/N and y/N have the same remainder a ≡ b (mod N) and c ≡ d (mod N) then: a + c ≡ a + d ≡ b + c ≡ b + d (mod N) a - c ≡ a - d ≡ b - c ≡ b - d (mod N) a ** c ≡ a ** d ≡ b ** c ≡ b ** d (mod N) ka ≡ kb (mod N) for any integer k ak ≡ bk (mod N) for any natural number k a + k ≡ b + k (mod N) for any integer k a + b = c, then a (mod N) + b (mod N) ≡ c (mod N) a ** b = c, then a (mod N) ** b (mod N) ≡ c (mod N)...