Cs6515 - Study guides, Class notes & Summaries
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CS6515 - Algorithms- Exam 1
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CS6515 - Algorithms- Exam 1
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CS6515 - Exam 2 Algorithms with complete solution
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CS6515 - Exam 2 Algorithms with complete solution
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CS6515 EXAM 3 NEWEST EXAM 2024 WITH ACTUAL QUESTIONS AND CORRECT VERIFIED ANSWERS|ALREADY GRADED A+
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CS6515 EXAM 3 NEWEST EXAM 2024 WITH 
ACTUAL QUESTIONS AND CORRECT VERIFIED 
ANSWERS|ALREADY GRADED A+
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CS6515 - Exam 2 Algorithms with complete solution
- Exam (elaborations) • 14 pages • 2024
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CS6515 - Exam 2 Algorithms with complete solution
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CS6515 Exam 2 with complete solution
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CS6515 Exam 2 with complete solution
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CS6515 - Algorithms- Exam 1 Complete Questions And Solutions latest
- Exam (elaborations) • 24 pages • 2024
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How do you tell if a graph has negative edges? - ANSWER-when fitting graph on 
a table, if the number of moves decreases the w() from edge to edge, then there 
is a negative edge; 
check from 1 to n 
Why are all pairs Dist(y,z) n^2? - ANSWER-Because it builds a two dim table! 
What is the run time of bellman ford algorithm? 
How about if you had to do it for all edges? - ANSWER-O(nm) 
O(n^2m) 
Floyd-Warshall run time? - ANSWER-O(n^3) 
What is the base case for the bellman ford algorithm? - ANSWE...
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CS6515 - ALGORITHMS- EXAM 1 QUESTIONS AND ANSWERS 2024
- Exam (elaborations) • 20 pages • 2024
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CS6515 - ALGORITHMS- EXAM 1 QUESTIONS AND ANSWERS 2024
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CS6515 - Exam 2 Algorithms questions and answers
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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)...
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CS6515 - Algorithms- Exam bundle
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CS6515 - Algorithms- Exam bundle
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CS6515 Exam 2 with complete solution
- Exam (elaborations) • 11 pages • 2024
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CS6515 Exam 2 with complete solution
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