Exam (elaborations)
CS6515 Exam 2 QUESTIONS AND ANSWERS 2024
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CS6515 Exam 2 QUESTIONS AND ANSWERS 2024
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CS6515 Exam 2Basic Properties of Trees - ANSWERS-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) with |E| = |V| - 1 is a tree
Kruskal's Algorithm - ANSWERS-1. Sort E by increasing weigt
2. Go through edges in order and add an edge to our current tree if it doesn't
create a cycle
Running Time: O(m log n), m = |E|, n = |V|
Is there ever a reason to use cycles in a flow graph? - ANSWERS-No
Flow Network Constraints: Capacity Constraint - ANSWERS-For all edges, the flow
must be larger than zero, but less than the capacity of that edge
,Goal of Flow Problem - ANSWERS-Maximize the flow out of the source (or into the
sink) of maximum size while satisfying the capacity and conservation of flow
constraints.
Flow Network Constraints: Conservation of Flow - ANSWERS-For all vertices (other
than the starting (source) and ending (sink) vertices), the flow into v must equal
the flow out of v.
Ford-Fulkerson Algo - ANSWERS-1. Start with f_e = 0 for all edges
2. Build the residual network for current flow
3. Find st-path in residual network
- if no such path then output f
4. Let c(p) = min(c_e - f_e); this is available capacity along some path
5. Augment f by c(p) along p
- for forward edges increase flow by c(p)
- for backward edge, decrease flow by c(p)
6. Repeat
Residual Network - ANSWERS-For flow network G = (V, E) with c_e for edges and
f_e for flows:
1. If there exists an edge vw where f_vw < c_vw, add vw to residual network with
capacity c_vw - f_aw
, 2. If there exists an edge vw where f_vw > 0, then add wv to residual network
with capacity f_vw
Note: 1 shows available forwards capacities and 2 shows residual backward
capacities.
Ford-Fulkerson Runtime - ANSWERS-Need to assume all capacities are integers.
This will allow flow to always increase by at least 1 unit per round. If C is the
maxflow, then there are at most C rounds. Each round of FF takes O(m), so the
total time is O(mC), which is pseudo-polynomial
What is the time to check whether or not a flow is a max flow - ANSWERS-- O(n +
m)
1. Build the residual graph takes O(n + m) time
2. Checking if there's a path from s to t using DFS, which takes linear time.
Capacity of a Cut - ANSWERS-Sum of capacities (edges) going from cut L to cut R
Max-flow = Min st-cut - ANSWERS-Size of Max-flow = min capacity of a st-cut
Cut Property: Key Ideas - ANSWERS-1. Take a tree T, add an edge e* will create a
cycle. Removing any edge of the cycle we'll get a new tree
2. A minimum weight edge across a cut is part of a MST
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