CS6515 EXAM 2 | QUESTIONS AND ANSWERS | WITH COMPLETE SOLUTION
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) with |E| = |V| - 1 is a tree
Kruskal's Algorithm - (ANSWER)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? - (ANSWER)No
Flow Network Constraints: Capacity Constraint - (ANSWER)For all edges, the flow must be larger than
zero, but less than the capacity of that edge
Goal of Flow Problem - (ANSWER)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 - (ANSWER)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 - (ANSWER)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
, CS6515 EXAM 2 | QUESTIONS AND ANSWERS | WITH COMPLETE SOLUTION
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 - (ANSWER)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 - (ANSWER)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 - (ANSWER)- 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 - (ANSWER)Sum of capacities (edges) going from cut L to cut R
Max-flow = Min st-cut - (ANSWER)Size of Max-flow = min capacity of a st-cut
Cut Property: Key Ideas - (ANSWER)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
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) with |E| = |V| - 1 is a tree
Kruskal's Algorithm - (ANSWER)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? - (ANSWER)No
Flow Network Constraints: Capacity Constraint - (ANSWER)For all edges, the flow must be larger than
zero, but less than the capacity of that edge
Goal of Flow Problem - (ANSWER)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 - (ANSWER)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 - (ANSWER)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
, CS6515 EXAM 2 | QUESTIONS AND ANSWERS | WITH COMPLETE SOLUTION
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 - (ANSWER)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 - (ANSWER)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 - (ANSWER)- 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 - (ANSWER)Sum of capacities (edges) going from cut L to cut R
Max-flow = Min st-cut - (ANSWER)Size of Max-flow = min capacity of a st-cut
Cut Property: Key Ideas - (ANSWER)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
