1. Prim’s Algorithm A Proof

2. Suppose we have a tree T that is the minimal spanning tree for a graph. Let P be the spanning tree from Prim’s alg. and Then there is an arc, a, in P that is not in T. If arc a connects node x to node y, then there is a path within the tree of T that also connects x to y: x y P T

3. Case 1: The weight of P is smaller than one of the edges on the path from the tree T Then we remove the edge with highest weight from T and replace it with the edge from P. This gives a smaller spanning tree than T – and contradicts the fact that T is the minimal spanning tree. x y P T

4. Case 2: The weight of P is greater than the weight of all the edges in the path from T. Then consider the step – in Prim’s algorithm- when we add edge P. P is the shortest edge remaining which can be added to the graph. But we would add xs before P (or st or tu or uy – whichever is still available to add. So we never add P – a contradiction! x P T y s t u

5. Depth-first search Algorithm DF(G,n) –Start with a given node, n. For all arcs a connected to n Do –DF(G\{n}, a) End - Do

6. Depth-first example A B C D E F G H I J K

7. A B C D E F G H I J K

8. A B C D E F G H I J K

9. A B C D E F G H I J K

10. A B C D E F G H I J K

11. A B C D E F G H I J K Continue: ABEDCHIJK (BACKTRACK to D) GF

12. Depth-first example-from K A B C D E F G H I J K

13. Breadth-First Algorithm BF(G,n) –Start with a given node, n. For all arcs a connected to n Do –Add nodes at end of each vertex. For all arcs a connected to n Do - BF(G\{arcs}, a)

14. Breadth-first example A B C D E F G H I J K

15. Add arcs connected to a A B C D E F G H I J K ABCDE

16. Add arcs connected to b A B C D E F G H I J K ABCDEF

18. Prim’s Algorithm A Proof

19. Suppose we have a tree T that is the minimal spanning tree for a graph. Let P be the spanning tree from Prim’s alg. and Then there is an arc, a, in P that is not in T. If arc a connects node x to node y, then there is a path within the tree of T that also connects x to y: x y P T

20. Case 1: The weight of P is smaller than one of the edges on the path from the tree T Then we remove the edge with highest weight from T and replace it with the edge from P. This gives a smaller spanning tree than T – and contradicts the fact that T is the minimal spanning tree. x y P T

21. Case 2: The weight of P is greater than the weight of all the edges in the path from T. Then consider the step – in Prim’s algorithm- when we add edge P. P is the shortest edge remaining which can be added to the graph. But we would add xs before P (or st or tu or uy – whichever is still available to add. So we never add P – a contradiction! x P T y s t u

22. Depth-first search Algorithm DF(G,n) –Start with a given node, n. For all arcs a connected to n Do –DF(G\{n}, a) End - Do

23. Depth-first example A B C D E F G H I J K

24. A B C D E F G H I J K

25. A B C D E F G H I J K

26. A B C D E F G H I J K

27. A B C D E F G H I J K

28. A B C D E F G H I J K Continue: ABEDCHIJK (BACKTRACK to D) GF

29. Depth-first example-from K A B C D E F G H I J K

30. Breadth-First Algorithm BF(G,n) –Start with a given node, n. For all arcs a connected to n Do –Add nodes at end of each vertex. For all arcs a connected to n Do - BF(G\{arcs}, a)

31. Breadth-first example A B C D E F G H I J K

32. Add arcs connected to a A B C D E F G H I J K ABCDE

33. Add arcs connected to b A B C D E F G H I J K ABCDEF

34. Continue: A B C D E F G H I J K ABCDEFHGJIK

35. Do breadth-first from D A B C D E F G H I J K

36. Be able to: Do depth first search and list nodes Do breadth-first search and list nodes Do either search and give the search tree.

37. Boolean Algebras Named after mathematician George Boole Created Boolean-algebras around 1850 He saw a connection between logic and certain algebraic properties He turned logic into algebra. Used in computers about 100 years after its invention – pure mathematics becomes applied mathematics.

38. Definition – Boolean Algebra A Boolean algebra is a set B, with two binary operations, + and x, a unary operator ‘ and in which there are special elements 0 and 1 such that x+y=y+xx y = y x (x+y)+z=x+(y+z) (x y) z = x (y z) x+(y z) = (x+y) x (x + z)x (y+z)=xy+xz x+0 = xx 1 = 1 x + x’= 1x x’ = 0

39. Examples 1. Sets with union (+), intersection (x) and set complement. What are 0 and 1? 2. Logic statements with or(+) and (x) and not(‘). Again, what are zero and one? 3. B={0,1} with usual multiplication and addition, except with 1+1 = 0. What is complement?

40. Check that every Boolean algebra has the idempotent property Show x+x=x. Proof (x+x)=(x+x) x 1 = (x+x) (x+x’) = x + x x’ (dist prop) X + 0 = x. Look at each example in terms of the idempotent property.