Lecture 1 Introduction
Lecture 1 Introduction
Monday, 25 July 2016 10:02 AM
Reduction Reducing interval scheduling to independent set problem Each interval becomes a node Node is connected if interval overlaps Reducing bipartite matching to independent set problem Each edge becomes a node Node is connected if the end shares an end Polynomial time If you double the input size, algorithm slows down by a constant factor An algorithm is efficient if it's worst case run time is polynomial. Asymptotic Order of Growth Upper bounds
T(n) is guaranteed to be smaller Gives worst case run time
Lower bounds
T(n) is guaranteed to be larger Gives best case run time
Tight bounds
T(n) is both upper and lower bound Gives best approximation of running time
Transitivity and Additivity
O
Order -- from shortest to longest running Logarithms < Polynomials < Exponentials
Lab 01
Wednesday, 27 July 2016 8:52 AM
Which asymptotic growth is faster? Put the two you want to compare in a ratio Find the limit as n approaches infinity
Lecture 2 Graphs
Monday, 1 August 2016 10:03 AM
Undirected and directed graphs Graph representation 1. Adjacency matrix 2. Adjacency list Adjacency matrix n by n matrix Two cells to list an edge Checking for an specific edge on a node is O(1) Checking for all edges is O(n^2) Space is O(n^2) o Not very good for large graphs
Adjacency list Node indexed as an array Checking for a specific edge on a node is O(deg(v)) Checking for all edges is O(n+m) Space is O(2m+n) = O(n+m) Definitions Path
A sequence of nodes such that each consecutive pair is joined by an edge
Simple path
All nodes in the path are distinct
Connected
There is a path between every pair of nodes
Cycle
A path starting and ending on the same node -- all node are distinct
Trees. Any two statements will imply the third: G is connected G does not contain a cycle G has n-1 edges Graph traversal -- BFS and DFS BFS produces a tree rooted at the start vertex on a set of nodes reachable from the start node BFS
Runs O(n+m) using a adjacency list o Each edge is looked at twice, once for each endpoint o O(n+2m) = O(n+m) Runs O(n^2) o To find the neighbours, look through the vector
Transitive closure DFS or BFS on each node Runs O(nm) using an adjacency list From O(n^2 + nm) = O(nm), because nm > n^2 DFS
Runs O(n+m) using an adjacency list
Testing bipartiteness Lemma: If a graph is bipartite, it cannot contain an odd length cycle
Proof? You can pick each layer to be alternating colours: red and blue If we have an edge connecting two nodes of the same layer, then there exists an odd length cycle
Monday, 25 July 2016 10:02 AM
Reduction Reducing interval scheduling to independent set problem Each interval becomes a node Node is connected if interval overlaps Reducing bipartite matching to independent set problem Each edge becomes a node Node is connected if the end shares an end Polynomial time If you double the input size, algorithm slows down by a constant factor An algorithm is efficient if it's worst case run time is polynomial. Asymptotic Order of Growth Upper bounds
T(n) is guaranteed to be smaller Gives worst case run time
Lower bounds
T(n) is guaranteed to be larger Gives best case run time
Tight bounds
T(n) is both upper and lower bound Gives best approximation of running time
Transitivity and Additivity
O
Order -- from shortest to longest running Logarithms < Polynomials < Exponentials
Lab 01
Wednesday, 27 July 2016 8:52 AM
Which asymptotic growth is faster? Put the two you want to compare in a ratio Find the limit as n approaches infinity
Lecture 2 Graphs
Monday, 1 August 2016 10:03 AM
Undirected and directed graphs Graph representation 1. Adjacency matrix 2. Adjacency list Adjacency matrix n by n matrix Two cells to list an edge Checking for an specific edge on a node is O(1) Checking for all edges is O(n^2) Space is O(n^2) o Not very good for large graphs
Adjacency list Node indexed as an array Checking for a specific edge on a node is O(deg(v)) Checking for all edges is O(n+m) Space is O(2m+n) = O(n+m) Definitions Path
A sequence of nodes such that each consecutive pair is joined by an edge
Simple path
All nodes in the path are distinct
Connected
There is a path between every pair of nodes
Cycle
A path starting and ending on the same node -- all node are distinct
Trees. Any two statements will imply the third: G is connected G does not contain a cycle G has n-1 edges Graph traversal -- BFS and DFS BFS produces a tree rooted at the start vertex on a set of nodes reachable from the start node BFS
Runs O(n+m) using a adjacency list o Each edge is looked at twice, once for each endpoint o O(n+2m) = O(n+m) Runs O(n^2) o To find the neighbours, look through the vector
Transitive closure DFS or BFS on each node Runs O(nm) using an adjacency list From O(n^2 + nm) = O(nm), because nm > n^2 DFS
Runs O(n+m) using an adjacency list
Testing bipartiteness Lemma: If a graph is bipartite, it cannot contain an odd length cycle
Proof? You can pick each layer to be alternating colours: red and blue If we have an edge connecting two nodes of the same layer, then there exists an odd length cycle
