1. May 5, 2015Applied Discrete Mathematics Week 13: Boolean Algebra 1 Dijkstra’s Algorithm procedure Dijkstra(G: weighted connected simple graph with vertices a = v 0, v 1, …, v n = z and positive weights w(v i, v j ), where w(v i, v j ) =  if {v i, v j } is not an edge in G) for i := 1 to n L(v i ) :=  L(a) := 0 S :=  {the labels are now initialized so that the label of a is zero and all other labels are , and the distinguished set of vertices S is empty}

2. May 5, 2015Applied Discrete Mathematics Week 13: Boolean Algebra 2 Dijkstra’s Algorithm while z  S begin u := the vertex not in S with minimal L(u) S := S  {u} for all vertices v not in S if L(u) + w(u, v) < L(v) then L(v) := L(u) + w(u, v) {this adds a vertex to S with minimal label and updates the labels of vertices not in S} end {L(z) = length of shortest path from a to z}

3. May 5, 2015Applied Discrete Mathematics Week 13: Boolean Algebra 3 Dijkstra’s Algorithm Example: abdz ec 4 2 1 5 8 10 2 6 30    Step 0

4. May 5, 2015Applied Discrete Mathematics Week 13: Boolean Algebra 4 Dijkstra’s Algorithm Example: abdz ec 4 2 1 5 8 10 2 6 30    Step 1 4 (a) 2 (a)

5. May 5, 2015Applied Discrete Mathematics Week 13: Boolean Algebra 5 Dijkstra’s Algorithm Example: abdz ec 4 2 1 5 8 10 2 6 30    Step 2 4 (a) 2 (a) 3 (a, c) 10 (a, c) 12 (a, c)

6. May 5, 2015Applied Discrete Mathematics Week 13: Boolean Algebra 6 Dijkstra’s Algorithm Example: abdz ec 4 2 1 5 8 10 2 6 30    Step 3 4 (a) 2 (a) 3 (a, c) 10 (a, c) 12 (a, c) 8 (a, c, b)

7. May 5, 2015Applied Discrete Mathematics Week 13: Boolean Algebra 7 Dijkstra’s Algorithm Example: abdz ec 4 2 1 5 8 10 2 6 30    Step 4 4 (a) 2 (a) 3 (a, c) 10 (a, c) 12 (a, c) 8 (a, c, b) 10 (a, c, b, d) 14 (a, c, b, d)

8. May 5, 2015Applied Discrete Mathematics Week 13: Boolean Algebra 8 Dijkstra’s Algorithm Example: abdz ec 4 2 1 5 8 10 2 6 30    Step 5 4 (a) 2 (a) 3 (a, c) 10 (a, c) 12 (a, c) 8 (a, c, b) 10 (a, c, b, d) 14 (a, c, b, d) 13 (a, c, b, d, e)

9. May 5, 2015Applied Discrete Mathematics Week 13: Boolean Algebra 9 Dijkstra’s Algorithm Example: abdz ec 4 2 1 5 8 10 2 6 30    Step 6 4 (a) 2 (a) 3 (a, c) 10 (a, c) 12 (a, c) 8 (a, c, b) 10 (a, c, b, d) 14 (a, c, b, d) 13 (a, c, b, d, e)

10. May 5, 2015Applied Discrete Mathematics Week 13: Boolean Algebra 10 The Traveling Salesman Problem The traveling salesman problem is one of the classical problems in computer science. A traveling salesman wants to visit a number of cities and then return to his starting point. Of course he wants to save time and energy, so he wants to determine the shortest path for his trip. We can represent the cities and the distances between them by a weighted, complete, undirected graph. The problem then is to find the circuit of minimum total weight that visits each vertex exactly once.

11. May 5, 2015Applied Discrete Mathematics Week 13: Boolean Algebra 11 The Traveling Salesman Problem Example: What path would the traveling salesman take to visit the following cities? ChicagoToronto New York Boston 600 700 200 650550 700 Solution: The shortest path is Boston, New York, Chicago, Toronto, Boston (2,000 miles).

12. May 5, 2015Applied Discrete Mathematics Week 13: Boolean Algebra 12 The Traveling Salesman Problem Question: Given n vertices, how many different cycles C n can we form by connecting these vertices with edges? Solution: We first choose a starting point. Then we have (n – 1) choices for the second vertex in the cycle, (n – 2) for the third one, and so on, so there are (n – 1)! choices for the whole cycle. However, this number includes identical cycles that were constructed in opposite directions. Therefore, the actual number of different cycles C n is (n – 1)!/2.

13. May 5, 2015Applied Discrete Mathematics Week 13: Boolean Algebra 13 The Traveling Salesman Problem Unfortunately, no algorithm solving the traveling salesman problem with polynomial worst-case time complexity has been devised yet. This means that for large numbers of vertices, solving the traveling salesman problem is impractical. In these cases, we can use approximation algorithms that determine a path whose length may be slightly larger than the traveling salesman’s path, but can be computed with polynomial time complexity. For example, artificial neural networks can do such an efficient approximation task.

14. May 5, 2015Applied Discrete Mathematics Week 13: Boolean Algebra 14 Let us talk about… Trees

15. May 5, 2015Applied Discrete Mathematics Week 13: Boolean Algebra 15Trees Definition: A tree is a connected undirected graph with no simple circuits. Since a tree cannot have a simple circuit, a tree cannot contain multiple edges or loops. Therefore, any tree must be a simple graph. Theorem: An undirected graph is a tree if and only if there is a unique simple path between any of its vertices.

16. May 5, 2015Applied Discrete Mathematics Week 13: Boolean Algebra 16Trees Example: Are the following graphs trees? No. Yes. Yes. No.

17. May 5, 2015Applied Discrete Mathematics Week 13: Boolean Algebra 17Trees Definition: An undirected graph that does not contain simple circuits and is not necessarily connected is called a forest. In general, we use trees to represent hierarchical structures. We often designate a particular vertex of a tree as the root. Since there is a unique path from the root to each vertex of the graph, we direct each edge away from the root. Thus, a tree together with its root produces a directed graph called a rooted tree.

18. May 5, 2015Applied Discrete Mathematics Week 13: Boolean Algebra 18 Tree Terminology If v is a vertex in a rooted tree other than the root, the parent of v is the unique vertex u such that there is a directed edge from u to v. When u is the parent of v, v is called the child of u. Vertices with the same parent are called siblings. The ancestors of a vertex other than the root are the vertices in the path from the root to this vertex, excluding the vertex itself and including the root.

19. May 5, 2015Applied Discrete Mathematics Week 13: Boolean Algebra 19 Tree Terminology The descendants of a vertex v are those vertices that have v as an ancestor. A vertex of a tree is called a leaf if it has no children. Vertices that have children are called internal vertices. If a is a vertex in a tree, then the subtree with a as its root is the subgraph of the tree consisting of a and its descendants and all edges incident to these descendants.

20. May 5, 2015Applied Discrete Mathematics Week 13: Boolean Algebra 20 Tree Terminology The level of a vertex v in a rooted tree is the length of the unique path from the root to this vertex. The level of the root is defined to be zero. The height of a rooted tree is the maximum of the levels of vertices.

21. May 5, 2015Applied Discrete Mathematics Week 13: Boolean Algebra 21Trees Example I: Family tree James ChristineBob FrankJoycePetra

22. May 5, 2015Applied Discrete Mathematics Week 13: Boolean Algebra 22Trees Example II: File system / usrtemp binspoolls bin

23. May 5, 2015Applied Discrete Mathematics Week 13: Boolean Algebra 23Trees Example III: Arithmetic expressions  +- yzxy This tree represents the expression (y + z)  (x - y).

24. May 5, 2015Applied Discrete Mathematics Week 13: Boolean Algebra 24Trees Definition: A rooted tree is called an m-ary tree if every internal vertex has no more than m children. The tree is called a full m-ary tree if every internal vertex has exactly m children. An m-ary tree with m = 2 is called a binary tree. Theorem: A tree with n vertices has (n – 1) edges. Theorem: A full m-ary tree with i internal vertices contains n = mi + 1 vertices.

25. May 5, 2015Applied Discrete Mathematics Week 13: Boolean Algebra 25 Binary Search Trees If we want to perform a large number of searches in a particular list of items, it can be worthwhile to arrange these items in a binary search tree to facilitate the subsequent searches. A binary search tree is a binary tree in which each child of a vertex is designated as a right or left child, and each vertex is labeled with a key, which is one of the items. When we construct the tree, vertices are assigned keys so that the key of a vertex is both larger than the keys of all vertices in its left subtree and smaller than the keys of all vertices in its right subtree.

26. May 5, 2015Applied Discrete Mathematics Week 13: Boolean Algebra 26 Binary Search Trees Example: Construct a binary search tree for the strings math, computer, power, north, zoo, dentist, book. math

27. May 5, 2015Applied Discrete Mathematics Week 13: Boolean Algebra 27 Binary Search Trees Example: Construct a binary search tree for the strings math, computer, power, north, zoo, dentist, book. math computer

28. May 5, 2015Applied Discrete Mathematics Week 13: Boolean Algebra 28 Binary Search Trees Example: Construct a binary search tree for the strings math, computer, power, north, zoo, dentist, book. math computer power

29. May 5, 2015Applied Discrete Mathematics Week 13: Boolean Algebra 29 Binary Search Trees Example: Construct a binary search tree for the strings math, computer, power, north, zoo, dentist, book. math computer power north

30. May 5, 2015Applied Discrete Mathematics Week 13: Boolean Algebra 30 Binary Search Trees Example: Construct a binary search tree for the strings math, computer, power, north, zoo, dentist, book. math computer power northzoo

31. May 5, 2015Applied Discrete Mathematics Week 13: Boolean Algebra 31 Binary Search Trees Example: Construct a binary search tree for the strings math, computer, power, north, zoo, dentist, book. math computer power northzoodentist

32. May 5, 2015Applied Discrete Mathematics Week 13: Boolean Algebra 32 Binary Search Trees Example: Construct a binary search tree for the strings math, computer, power, north, zoo, dentist, book. math computer power northzoodentistbook

33. May 5, 2015Applied Discrete Mathematics Week 13: Boolean Algebra 33 Binary Search Trees To perform a search in such a tree for an item x, we can start at the root and compare its key to x. If x is less than the key, we proceed to the left child of the current vertex, and if x is greater than the key, we proceed to the right one. This procedure is repeated until we either found the item we were looking for, or we cannot proceed any further. In a balanced tree representing a list of n items, search can be performed with a maximum of  log(n + 1)  steps (compare with binary search).