# Prim’s Algorithm A Proof. 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.

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Prim’s Algorithm A Proof

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

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

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

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

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

A B C D E F G H I J K

A B C D E F G H I J K

A B C D E F G H I J K

A B C D E F G H I J K

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

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

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)

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

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

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

Prim’s Algorithm A Proof

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

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

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

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

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

A B C D E F G H I J K

A B C D E F G H I J K

A B C D E F G H I J K

A B C D E F G H I J K

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

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

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)

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

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

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

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

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

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.

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.

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

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?

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.

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