# CSE 421 Algorithms - PowerPoint PPT Presentation

PPT – CSE 421 Algorithms PowerPoint presentation | free to download - id: 9f571-MzFiN

The Adobe Flash plugin is needed to view this content

Get the plugin now

View by Category
Title:

## CSE 421 Algorithms

Description:

### Algorithms with polynomial run time have the property that increasing the ... polynomial time seems to capture the algorithms which are efficient in practice ... – PowerPoint PPT presentation

Number of Views:58
Avg rating:3.0/5.0
Slides: 42
Provided by: csWash
Category:
Tags:
Transcript and Presenter's Notes

Title: CSE 421 Algorithms

1
CSE 421 Algorithms
• Richard Anderson
• Lecture 4

2
What does it mean for an algorithm to be
efficient?

3
Definitions of efficiency
• Fast in practice
• Qualitatively better worst case performance than
a brute force algorithm

4
Polynomial time efficiency
• An algorithm is efficient if it has a polynomial
run time
• Run time as a function of problem size
• Run time count number of instructions executed
on an underlying model of computation
• T(n) maximum run time for all problems of size
at most n

5
Polynomial Time
• Algorithms with polynomial run time have the
property that increasing the problem size by a
constant factor increases the run time by at most
a constant factor (depending on the algorithm)

6
Why Polynomial Time?
• Generally, polynomial time seems to capture the
algorithms which are efficient in practice
• The class of polynomial time algorithms has many
good, mathematical properties

7
Polynomial vs. Exponential Complexity
• Suppose you have an algorithm which takes n!
steps on a problem of size n
• If the algorithm takes one second for a problem
of size 10, estimate the run time for the
following problems sizes

12 14 16
18 20
10 1 second 12 2 minutes 14 6 hours 16 2
months 18 50 years 20 20K years
8
Ignoring constant factors
• Express run time as O(f(n))
• Emphasize algorithms with slower growth rates
• Fundamental idea in the study of algorithms
• Basis of Tarjan/Hopcroft Turing Award

9
Why ignore constant factors?
• Constant factors are arbitrary
• Depend on the implementation
• Depend on the details of the model
• Determining the constant factors is tedious and
provides little insight

10
Why emphasize growth rates?
• The algorithm with the lower growth rate will be
faster for all but a finite number of cases
• Performance is most important for larger problem
size
• As memory prices continue to fall, bigger problem
sizes become feasible
• Improving growth rate often requires new
techniques

11
Formalizing growth rates
• T(n) is O(f(n)) T Z ? R
• If n is sufficiently large, T(n) is bounded by a
constant multiple of f(n)
• Exist c, n0, such that for n gt n0, T(n) lt c f(n)
• T(n) is O(f(n)) will be written as
T(n) O(f(n))
• Be careful with this notation

12
Prove 3n2 5n 20 is O(n2)
Let c Let n0
Choose c 6, n0 5
T(n) is O(f(n)) if there exist c, n0, such that
for n gt n0, T(n) lt c f(n)
13
Order the following functions in increasing order
by their growth rate
• n log4n
• 2n2 10n
• 2n/100
• 1000n log8 n
• n100
• 3n
• 1000 log10n
• n1/2

14
Lower bounds
• T(n) is W(f(n))
• T(n) is at least a constant multiple of f(n)
• There exists an n0, and e gt 0 such that
T(n) gt ef(n) for all n gt n0
• Warning definitions of W vary
• T(n) is Q(f(n)) if T(n) is O(f(n)) and
T(n) is W(f(n))

15
Useful Theorems
• If lim (f(n) / g(n)) c for c gt 0 then
f(n) Q(g(n))
• If f(n) is O(g(n)) and g(n) is O(h(n)) then
f(n) is O(h(n))
• If f(n) is O(h(n)) and g(n) is O(h(n)) then f(n)
g(n) is O(h(n))

16
Ordering growth rates
• For b gt 1 and x gt 0
• logbn is O(nx)
• For r gt 1 and d gt 0
• nd is O(rn)

17
Formalizing growth rates
• T(n) is O(f(n)) T Z ? R
• If n is sufficiently large, T(n) is bounded by a
constant multiple of f(n)
• Exist c, n0, such that for n gt n0, T(n) lt c f(n)
• T(n) is O(f(n)) will be written as
T(n) O(f(n))
• Be careful with this notation

18
Graph Theory
Explain that there will be some review from 326
• G (V, E)
• V vertices
• E edges
• Undirected graphs
• Edges sets of two vertices u, v
• Directed graphs
• Edges ordered pairs (u, v)
• Many other flavors
• Edge / vertices weights
• Parallel edges
• Self loops

By default V n and E m
19
Definitions
• Path v1, v2, , vk, with (vi, vi1) in E
• Simple Path
• Cycle
• Simple Cycle
• Distance
• Connectivity
• Undirected
• Directed (strong connectivity)
• Trees
• Rooted
• Unrooted

20
Graph search
• Find a path from s to t

S s While there exists (u, v) in E with u in
S and v not in S Predv u Add v to S if (v
t) then path found
21
• Explore vertices in layers
• s in layer 1
• Neighbors of s in layer 2
• Neighbors of layer 2 in layer 3 . . .

s
22
Key observation
• All edges go between vertices on the same layer

1
2
3
7
6
5
4
8
23
Bipartite
• A graph V is bipartite if V can be partitioned
into V1, V2 such that all edges go between V1 and
V2
• A graph is bipartite if it can be two colored

Two color this graph
24
Testing Bipartiteness
• If a graph contains an odd cycle, it is not
bipartite

25
Algorithm
• Run BFS
• Color odd layers red, even layers blue
• If no edges between the same layer, the graph is
bipartite
• If edge between two vertices of the same layer,
then there is an odd cycle, and the graph is not
bipartite

26
Bipartite
• A graph is bipartite if its vertices can be
partitioned into two sets V1 and V2 such that all
edges go between V1 and V2
• A graph is bipartite if it can be two colored

27
Theorem A graph is bipartite if and only if it
has no odd cycles
If a graph has an odd cycle it is not bipartite
If a graph does not have an odd cycle, it is
bipartite
28
Lemma 1
• If a graph contains an odd cycle, it is not
bipartite

29
Lemma 2
• If a BFS tree has an intra-level edge, then the
graph has an odd length cycle

Intra-level edge both end points are in the same
level
30
Lemma 3
• If a graph has no odd length cycles, then it is
bipartite

No odd length cycles implies no intra-level edges
No intra-level edges implies two colorability
31
Connected Components
• Undirected Graphs

32
Computing Connected Components in O(nm) time
• A search algorithm from a vertex v can find all
vertices in vs component
• While there is an unvisited vertex v, search from
v to find a new component

33
Directed Graphs
• A Strongly Connected Component is a subset of the
vertices with paths between every pair of
vertices.

34
Identify the Strongly Connected Components
There is an O(nm) algorithm that we will not be
covering
35
Strongly connected components can be found in
O(nm) time
• But its tricky!
• Simpler problem given a vertex v, compute the
vertices in vs scc in O(nm) time

36
Topological Sort
• Given a set of tasks with precedence constraints,
find a linear order of the tasks

321
322
401
142
143
341
326
421
370
431
378
37
Find a topological order for the following graph
H
E
I
A
D
G
J
C
F
K
B
L
38
If a graph has a cycle, there is no topological
sort
• Consider the first vertex on the cycle in the
topological sort
• It must have an incoming edge

A
F
B
E
D
C
39
Lemma If a graph is acyclic, it has a vertex
with in degree 0
• Proof
• Pick a vertex v1, if it has in-degree 0 then done
• If not, let (v2, v1) be an edge, if v2 has
in-degree 0 then done
• If not, let (v3, v2) be an edge . . .
• If this process continues for more than n steps,
we have a repeated vertex, so we have a cycle

40
Topological Sort Algorithm
While there exists a vertex v with in-degree
0 Output vertex v Delete the vertex v and all
out going edges
H
E
I
A
D
G
J
C
F
K
B
L
41
Details for O(nm) implementation
• Maintain a list of vertices of in-degree 0
• Each vertex keeps track of its in-degree
• Update in-degrees and list when edges are removed
• m edge removals at O(1) cost each