Discrete Structures & Algorithms
Graphs and Trees: II
EECE 320
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Some examples
Questions on graphs and solutions
Ideas for other proofs
Minimal spanning trees
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A spanning tree of a graph G is a tree that touches every node of G and uses only edges from G
Every connected graph has a spanning tree
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Prove: A graph G is 2-colourable iff it contains no cycle of odd length.
What does 2-colorable mean?
• We can color the vertices of the graph with at most two colors such that two adjacent nodes do not get the same color.
Adjacent nodes?
• Two vertices are adjacent if there is an edge connecting the two vertices.
Cycle? Of odd length?
• If there is a path that starts at a vertex and ends at the same vertex without repeating an edge, then the graph has a cycle.
• The number of edges along the cycle is the length of the cycle.
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Prove: A graph G is 2-colourable iff it contains no cycle of odd length.
First: Assume that the graph is 2-colourable and prove that it contains no odd cycle.
• Select a 2-colouring of G. Consider an arbitrary cycle v
1
, v
2
, …, v k
, v
1
.
• The vertices with even subscript must be one colour and the vertices with odd subscripts must be a different colour.
• Since v
1 and v k must be coloured differently, k must be even.
• But, k is the length of the cycle so all cycles must be of even length.
v
4 v
1 v
2 v
3
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Prove: A graph G is 2-colourable iff it contains no cycle of odd length.
Second: Assume that the graph has no cycle of odd length and prove that it must be 2-colourable.
• If we can 2-color every connected component of G, then we can 2-color all of G. Thus, it suffices to show that we can 2-color an arbitrary component of G.
• Consider an arbitrary component H of G.
• Consider some spanning tree T of H.
• Choose any 2-colouring of T. Any tree can be coloured with at most 2 colours because we can think of a tree as having parents and children and we colour each generation separately.
• Let x-y be some edge not in T and consider the path from some vertex v to x. Let z be the last vertex on the path from v to x that also lies on the path from v to y.
• Of the paths from z to x and from z to y , exactly one must have odd length else the these two paths together with the edge x-y would form an odd length cycle.
• Thus x and y belong to different generations wrt z and will have different colours.
w v
H z y
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Problem: If a graph G has 15 edges and its complementary graph G’ has 13 edges, how many vertices does G have?
Complementary graph G’: An edge between vertices x and y exists in G’ iff x and y are not adjacent in G.
v
1 v
1 v
2 v
2
G v
4 v
4
G’ v
3 v
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Problem: If a graph G has 15 edges and its complementary graph G’ has 13 edges, how many vertices does G have?
• The maximum number of edges a graph with n vertices can have is n(n-1)/2.
• The complement of a complete graph has no edges.
• If E is the set of edges in G and E’ the set of edges in G’ then: |E|+|E’| = n(n-
1)/2.
• Solve for n when n(n-1)/2 = 28.
• n = 8.
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Use elementary properties of graphs to construct proofs. Start with properties that you understand easily:
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Trees have many nice properties
(uniqueness of paths, no cycles, etc.)
We may want to compute the “best” tree approximation to a graph
If all we care about is communication , then a tree may be enough. We want a tree with smallest communication link costs
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Problem: Find a minimum spanning tree , that is, a tree that has a node for every node in the graph, such that the sum of the edge weights is minimum
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4
7 8
5
9
6
7
11
8 9
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Create a forest where each node is a separate tree
Make a sorted list of edges S
While S is non-empty:
Remove an edge with minimal weight
If it connects two different trees, add the edge. Otherwise discard it.
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9
7 4
1 5
10
3
9 8
7
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The algorithm outputs a spanning tree T .
[By contradiction] Suppose that it’s not minimal.
(For simplicity, assume all edge weights in graph are distinct)
Then let M be a minimum spanning tree. M≠T
Let e be the first edge chosen by the algorithm that is in T but not in M.
If we add e to M , it creates a cycle. Since this cycle isn’t fully contained in T , it has (at least) an edge f not in T .
N = M+e-f is another spanning tree.
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Proving Correctness of the Algorithm (cont)
N = M+e-f is another spanning tree.
Claim: e < f , and therefore N < M
Suppose not: e > f
Then f would have been visited before e by the algorithm, but not added, because adding it would have formed a cycle.
Then N < M this M is not a minimal spanning tree.
Contradiction!
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The greedy algorithm, by adding the least costly edges in each stage, succeeds in finding an MST
Obviously greedy approaches do not work for all problems …
Some examples where they do work?
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Given: a number of cities and the costs of traveling from any city to any other city,
Q: What is the cheapest round-trip route that visits each city exactly once and then returns to the starting city?
The problem is in NP
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Modified TSP: We assume that all pairs of nodes are connected, and edge weights satisfy the triangle inequality d(x,y)  d(x,z) + d(z,y)
 Complexity still NP
We can use an MST to derive a TSP tour that is no more expensive than twice the optimal tour.
Idea: walk “around” the MST and take shortcuts if a node has already been visited.
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Shortcuts only decrease the cost, so
Cost(Greedy Tour)  2 Cost(MST)
 2 Cost(Optimal Tour)
This is a 2-competitive algorithm
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Input data:
Directed graph: G= (V, E)
Start and end node: s, t
Length l e for each edge e
Output : shortest path from
to
[cost of path = sum of all edges]
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Dijkstra’s algorithm
Idea: Maintain a set of explored nodes S for which we have determined the shortest path distance d(u) from s to u.
 Initialize S = {s}, d(s) = 0.
 Repeatedly choose unexplored node v to minimize add v to S, and set d(v) = π(v).
 Until all nodes are chosen
Shortest path from a node in u to any other unexplored node
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Dijkstra’s algorithm
Idea: Maintain a set of explored nodes S for which we have determined the shortest path distance d(u) from s to u.
 Initialize S = {s}, d(s) = 0.
 Repeatedly choose unexplored node v to minimize add v to S, and set d(v) = π(v).
 Until all nodes are chosen
Shortest path from a node in u to any other unexplored node
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 Initialize S = {s}, d(s) = 0.
 Repeatedly choose unexplored node v to minimize add v to S, and set d(v) = π(v).
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A graph is bipartite if the nodes can be partitioned into two sets V
1 and V
2 all edges go only between V
1 edges go from V
1 to V
1 such that and V
2 or from V
2
(no to V
2
)
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A group of 100 boys and girls attend a dance. Every boy knows 5 girls, and every girl knows 5 boys. Can they be matched into dance partners so that each pair knows each other?
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Perfect Matching: A matching that covers all nodes in the graph with one edge going to each node.
Theorem: If every node in a bipartite graph has the same degree d  1, then the graph has a perfect matching.
Note: if degrees are the same then |A| = |B|, where A is the set of nodes “on the left” and
B is the set of nodes “on the right”
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Claim: If all nodes have the same degree then |A| = |B|
Proof:
If there are m boys, there are md edges
If there are n girls, there are nd edges
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•
•
Theorem: A bipartite graph has a perfect matching if and only if
|A| = |B| and for all k  [1,n] : for any subset of k nodes of A there are at least k nodes of B that are connected to at least one of them.
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For any subset of (say) k nodes of A there are at least k nodes of B that are connected to at least one of them
The condition fails for this graph
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At least k k
At most n-k n-k
The condition of the theorem still holds if we swap the roles of A and B: If we pick any k nodes in B, they are connected to at least k nodes in A
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Call a bipartite graph “matchable” if it has the same number of nodes on left and right, and any k nodes on the left are connected to at least k on the right
Strategy: Break up the graph into two matchable parts, and recursively partition each of these into two matchable parts, etc., until each part has only two nodes
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Select two nodes a  A and b  B connected by an edge
Idea: Take G
1
= (a,b) and G
2
= everything else
Problem: G neighbors.
2 need not be matchable. There could be a set of k nodes that has only k-1
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k a b k-1
The only way this could fail is if one of the missing nodes is b
Add this in to form
G
1
, and take G
2 everything else.
to be
This is a matchable partition!
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Here’s What
You Need to
Know…
Minimum Spanning Tree
- Definition
Kruskal’s Algorithm
- Definition
- Proof of Correctness
Traveling Salesman Problem
- Definition
- Using MST to get an approximate solution
The Marriage Theorem
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