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CS 6515 FINAL EXAM ACTUAL COMPLETE 90 QUESTIONS AND CORRECT DETAILED ANSWERS (VERIFIED ANSWERS) WITH RATIONALES |ALREADY GRADED A+||BRAND NEW VERSION!!.£14.70
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CS 6515 FINAL EXAM ACTUAL COMPLETE 90 QUESTIONS AND CORRECT DETAILED ANSWERS (VERIFIED ANSWERS) WITH RATIONALES |ALREADY GRADED A+||BRAND NEW VERSION!!.
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Module
CS 6515
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CS 6515
CS 6515 FINAL EXAM ACTUAL COMPLETE 90 QUESTIONS AND CORRECT DETAILED ANSWERS (VERIFIED ANSWERS) WITH RATIONALES |ALREADY GRADED A+||BRAND NEW VERSION!!.
CS 6515 FINAL EXAM ACTUAL EXAM COMPLETE 90
QUESTIONS AND CORRECT DETAILED ANSWERS (VERIFIED
ANSWERS) WITH RATIONALES |ALREADY GRADED A+||BRAND
NEW VERSION!!.
Kite Reduction ✔✔CORRECT ANSWER✔✔Prove Kite is in np by verifying that
- check for the clique - O(n^2)
- check for the tail of degree <= 2 nodes O(n)
- check for that the tail is connected to the clique with Explore (O(n)
Reduce Clique to kite by adding a tail of length g to each vertex in graph G
O(n^2)
Transform back to Clique by removing all vertexs of degree <= 2 (prune the
tails)
If there is no solution to kite then there is no clique in G
If there is a solution to kite, removing the tail is then a solution to clique
CMM ✔✔CORRECT ANSWER✔✔Chain Matrix Multiply
If your problem needs a windowing check
If your problem is looking to split an array
Ilooking to split an array in some way to do something
,Start a window size of 1-> n
set i to 1->n-window
Set j to i+window
Calculate T(i,j) using smaller windows between i and j
Let m[i-1] and m[i] be the dimensions of A[i]
Let T(i,j) = minimum cost for computing A[i] * A[i+1] * .. A[j]
T(i,i) = 0 for 1<=i<=n
For 1<=i<j<=n
T(i,j) = Min {T(i,k) + T(k+1,j) + m[i-1] *m[k]*m[j] L i<=k<=j-1}
Quick Sort Runtime ✔✔CORRECT ANSWER✔✔O(n^2)
O(nlogn) with median of medians
Merge Sort Runtime ✔✔CORRECT ANSWER✔✔O(nlogn)
Quick Select Runtime ✔✔CORRECT ANSWER✔✔O(n^2)
O(n) with median of medians (Fast select)
Dijkstra's algorithm ✔✔CORRECT ANSWER✔✔An algorithm for finding the
shortest paths between nodes in a weighted graph. For a given source node
in the graph, the algorithm finds the shortest path between that node and
every other. It can also be used for finding the shortest paths from a single
node to a single destination node by stopping the algorithm once the
,shortest path to the destination node has been determined. Its time
complexity is O(E + VlogV), where E is the number of edges and V is the
number of vertices.
Kruskal's Algorithm ✔✔CORRECT ANSWER✔✔(Minimum Spanning Trees,
O(mlogn) with a union find, which is fast for sparse graphs) Builds up
connected components of vertices, repeatedly considering the lightest
remaining edge and tests whether its two endpoints lie within the same
connected component. If not, insert the edge and merge the two
components into one.
input: undirected G=(V,E) with weights w(e)
1. sort E by incr weight (O(mlogn))
2. set X = empty set
3. for e = (v, w) exists E (go through in order)
if x Ue doesnt have a cycle
then: X = X U e(checks if v and w are in different components: O(log n))
4. return X
, returns MST defined by edges X
Prim's Algorithm for MST ✔✔CORRECT ANSWER✔✔Start with a node and
add edge with the lowest weight. Then from the current existing nodes add
the next smallest edge that exists going out from all the existing nodes. Do
this until all vertices are connected.
- min spanning tree defined by array prev[]
- O(m log n)
Ford-Fulkerson Algorithm ✔✔CORRECT ANSWER✔✔Runtime: O(mC)Input:
Graph with integer edge weights. (Note: Does not work with Infinity)Output:
max flow f*
1. set f_e = 0 for all edges
2. Build residual network G^f for current flow f
3. Check for st-path (cal P) in G^f for current flow f (DFS, BFS)- if no path
exists - output f
4. Given cal P, let c(cal P) = min capacity along cal P in G^f
5. Augment f by c(cal P) units along cal P
6. Repeat 2 until no st-path
Assumes all capacities are integers - C, C = size of max flow
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