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CS180: Algorithm Design and Analysis
Homeworks assigned noon Tue. and due 7:59am Thu.
Office: 3532C BH
=============
Jan. 5, 2016
=============
Al Khowrazmi --> Algorithm
Serial Model of Computation
----------------------------
+------------------------+ +-----+
| | | |
| +-----------+ +-----+ | | |
| | Registers | | CPU | | | RAM |
| +-----------+ +-----+ | | |
| | | |
+-----------|------------+ +-----+
|
I/O
- CPU can do only basic operations
Example
--------
Add n integers using this model
|
+--> 'n' means a large number in algorithm design
Assume reading from I/O to CPU takes about 1 unit of time
saving from CPU to RAM takes about 1 unit of time
--------------------
read ~ 2n
RAM to CPU ~ n
adding ~ n-1
output ~ 1
--------------------
total ~ 4n
--------------------
T(n) = O(n)
We can either improve the algorithm or prove that we cannot
Example: Famous People
-----------------------
everyone knows
he/she does not know anyone
Model of Computation
---------------------
know
A --------> B takes 1 unit of time
Ask 2(n-1) questions to know if one person is famous or not
Cannot have second famous person or else contradiction
We repeat for each person in class 2(n-1) --+
2(n-1) |
2(n-1) +--> n times
... |
2(n-1) --+
n * 2(n-1) = 2n^2-2n
~ 2n^2
~ n^2
Reduce size of problem, eliminate candidates
Random vs. Arbitrary
Pick 2 arbitrary people
yes
know +-----> A is not famous
A --------> B +
+-----> B is not famous
no
After n-1 questions only 1 candidate
Ask 2(n-1) questions about the last candidate to verify
3(n-1) --> ~ n
Asymptotic Analysis
--------------------
-------------
n = 10
-------------
n 10 --+
2n 20 +---> polynomial
n^2 100 --+
2^n 1014 --+
n! 4M +---> exponential
------------- --+
f(n) = O(g(n)) ----> about or less, order of
if there exists constants n0,c such that
f(n) <= cg(n) n >= n0
f(n) = O(n^2)
n^2/4
5n^2
logn
2^n is not order of n^2
=============
Jan. 7, 2016
=============
Develop general methodologies that can be applied to general cases
---------------------
The Matching Problem
---------------------
In a group, match two people with each other
one-to-one, one person can only match with another single individual
two groups: male and female, each with n people
match one male with one female
M1 W1
M2 W2
M3 W3
M4 W4
assume each person has an ordering list (preference list)
(W', W)
M ---- W
/\
\
\ double arrow as shown means unstable
\
\/
M' --- W'
(M, M')
generic step for inductive algorithm
-------------------------------------
if M is not matched, M will pick the highest woman on his list and propose
if W is not matched, she will accept the proposal
otherwise she will look in her list
if M has higher priority, she will discard current match, and match with M
otherwise, she will not accept the match
M will get worse matches eventually (down priority list)
W will get better matches eventually (up priority list)
how to prove that this is stable?
----------------------------------
assume by contradiction that we get an unstable result as follows
M ---- W
/\
\
\
\
\/
M' --- W'
case 1 : M did not propose to W'
contradiction to algorithm because W' is higher on M's list
so M must have proposed to W' first before matching W
case 2 : M did propose to W'
since M is higher on priority list of W' and M proposed to W'
W' should be matched with M, so contradiction
M" > M > M', so W' is matched to M", not M', so contradiction
How to show perfect match?
==============
Jan. 12, 2016
==============
For every given problem we do not want a totally new algoritm
so we will study algorithmic paradigms
But let's start with a few examples
-----------------------
The Scheduling Problem
-----------------------
Given set of intervals, want to find a subset
We cannot overlap tasks, and cannot do tasks partially
S E
+-----+ +---+
a b
+-----------+
c
+---+ +-----+
d e
+-----+ +---+
f g
Sa <= Ea
Want to maximize the NUMBER of intervals
Length of interval does not matter
+-----------------+ +---------------+
a d
+---+ +---+
b c
Can pick 'a' or 'b', both have same cardinality |a| = |b|
However, we should pick 'b' and 'c' instead of 'a'
Then we should pick 'b' and 'c' and 'd', which has cardinality 3
exhausive search algorithms vs. greedy algorithms
Solve first using exhausive search
-----------------------------------
Pick every subset
write the solution as a binary number
10110 means that for each bit turned on, we pick corresponding interval
abcde in this case, we pick 'acd'
This algorithm takes o(2^n Tn), which is too slow
Solve using greedy algorithm
-----------------------------
Pick one and stick with it, will not change it later
Pick according to length of time interval (E - S)
Pick the shortest interval, exclude any other that overlap with it
Pick the next, and exclude any other that overlap with it
Repeat the process
This does not work
a b
+--------------------+ +--------------------+
+-----+
c
What if we choose the interval that has the fewest number of conflicts
This does not work either (try to prove it)
Solve using plane sweep
------------------------
Starting point is an event, ending point is an event
Start from -inf to inf, choose first starting event
Whenever we encounter an event, decide what to do
Encounter a Start, include it and exclude any overlapping intervals
Repeat the process until we are out of choices
This does not work
a
+---------------------+
b c
+---+ +---+
What if we choose the first ending event
We cannot find counter example, let's try proving that it works
Proof by Induction
-------------------
Assume that we have an optimal solution all the way up to the 'i'th selection
Base case: Choose the first interval, it is optimal
Inductive Step:
------------+---------------------
i | i + 1
------------+---------------------
|
x | a
+---------+ | +-----+
|
x | b
+---------+ | +-----+
|
Ea <= Eb
Sb >= Ex
Sa >= Ex
Runtime analysis
-----------------
1. sort the endpoints O(nlogn)
2. pick the first ending interval
remove the overlapping ones O(n)
==============
Jan. 14, 2016
==============
Suppose we have integers as follows, would like to sort in non-decreasing order
6 2 4 9 1 5 7 3
Technique: reduce problem size
Famous Problem: reduce problem size from n to n-1
Methodology: Divide and conquer
keep dividing until it is small enough
'glue' back together
Merging
--------
Two lists A and B
A B
+---+ +----+ +----+
->| 2 | ->| 6 | | 2 |
| 4 | | 10 | | 4 |
| 7 | | 11 | ---> | 6 |
| 8 | | 14 | | 7 |
| . | | . | | 8 |
| . | | . | | 10 |
+---+ +----+ | 11 |
s t | 14 |
| .. |
+----+
runtime of merging O(st)
a better runtime analysis is O(s+t)
6 2 4 9 1 5 7 3 ----+
|
6 2 4 9 1 5 7 3 +----> divide
|
6 2 4 9 1 5 7 3 ----+
2 6 4 9 1 5 3 7 ----+
|
2 4 6 9 1 3 5 7 +----> merge
|
1 2 3 4 5 6 7 9 ----+
Runtime analysis of mergesort
------------------------------
T(1) = O(1)
T(n) = 2T(n/2) + O(n)
= 2T(n/2) + cn
= 2[2T(n/4) + cn/2] + cn
= 2^2 * T(n/2^2) + 2cn
= 2^3 * T(n/2^3) + 3cn
= 2^i * T(n/2^i) + icn
find out for what values are n/2^i = 1
-> 2^i = n
-> i = logn
= 2^logn * T(n/2logn) + logn * cn
= n + cnlogn
The algorithm is O(nlogn)
Closest pair problem
---------------------
Assume we have a set of points in a 2-D space
Computing distances takes O(1) time
Runtime
--------
n-1
n-2
.
.
.
1
sum(i=1,n) i = n(n+1)/2
Thus it takes O(n^2) time
Find minimum value of list
---------------------------
12 4 1 9 11 6 18
keep local variable min = 12
search through each element and compare with min
takes O(n) time, with n-1 comparisons
==============
Jan. 19, 2016
==============
Graphs
-------
G = (V, E)
| |
| edges (pair of vertices)
|
vertices / nodes
Simple Graph (Basic version of graph)
--------------------------------------
Unweighted graphs: no weight on vertices/edges
Undirected graphs: no direction on edge (unordered pair of vertices)
No parallel edges (pair of vertices appearing multiple times)
Using adjacency matrix to represent graph
------------------------------------------
each entry a(i, j) in matrix tells if there is an edge between
vertices i and j, either 1 (there is edge) or 0 (no edge)
for weighted graph: use weight instead of 0 and 1
for directed graph: edge between i and j does not imply edge between j and i
1 2 3
+-----------
1 | [ 0 0 1 ]
2 | [ 1 0 1 ]
3 | [ 1 1 0 ]
|
matrix representation takes space (n^2 entries)
fast for checking if nodes are connected
Using linked list to represent graph
-------------------------------------
1 --> 2 --> 5
2 --> 1
3
does not take as much space as adjacency
basic operations take more time
Path Takes from vertex i to vertex j
simple path: no repeated edges and vertices
Cycle Path that takes vertex i back to i
simple cycle: no repeated edges and vertices (except first)
Connected If we can reach any vertex from any vertex
Maximum number of edges: C(n 2) dense graph
Minimum number of edges: n - 1 sparse graph
Breadth First Search (BFS)
---------------------------
search the close nodes first
use a first in first out queue
runtime: O(|E|) --> also O(n^2) (but not as tight)
|
number of edges
'jump' between components if graph is not connected
then runtime: O(|E|+|V|)
Length of path: number of edges from beginning to end vertex
Bread First Search Tree
Connect the vertices that vertex i is searched through first
Depth First Search (DFS)
-------------------------
search the nodes next first
Edges that allow to see node for first time form a DFS tree
Edges that allow to see node for second or more times are backward edges
- this implies cycles in the graph
Give algorithm to determine if there are cycles in graph
Run depth first search to determine backward edges
runtime: O(|E|)
Majority Problem
-----------------
Majority: person with more than n/2 number of votes
either no majority or exactly one majority
Count number of vote see if anyone has more n/2 votes
Majority candidate must pass middle line
runtime: O(nlogn)
Find O(n) time algorithm to solve this problem
Pick two different elements, get rid of them, until we have 2 left
If they are different, then we have no majority
If they are the same, then it is the majority
In the end we should double check
Keep local 'maj-candidate' variable, and count of votes
2 3 2 2 2 3 2 1
get rid of 2 3 --> 2 2 2 3 2 1
maj-candidate: 2, 3
* What about 1 1 2 2 2 3 3 ?
==============
Jan. 21, 2016
==============
Connected and Directed Graphs
------------------------------
this kind of graph has nodes where each node has a direction
it may or may not have a cycle, which depends on direction
Techniques for BFS and DFS are the same except the definition of 'neighbor' is
different in that we have to follow the direction of the node instead of
simply getting the 'adjacent' nodes.
Directed Acyclic Graph directed graph with no cycles
widely studied, many applications
use DFS to determine if there are cycles
Topological Soring a Graph
---------------------------
must be a DAG
there is no way to topologically sort a graph with a cycle
+-->a --> b --> c --+
| |
+-------------------+
no way to sort this graph topologically
a
/ \
b c
| \ |
| \|
d e
ordering: a c b e d
no precedence relationship between 'b' and 'c'
In degree of vertex ingoing edges of the vertex
Out degree of vertex outgoing edges of the vertex
In-deg = 0 --> Source could have n-1 sources
OUt-deg = 0 --> Sink could have n-1 sinks
Degree of vertex (undirected) number of connected edges
Topological Sorting Algorithm
------------------------------
for a DAG with n nodes, we can have n! (total ordering) orderings
consider a graph with n sources and 1 sink
in an ordering there is no pointer from a later node to an earlier node
building the algorithm high-level
----------------------------------
generate a list of sources
output a source
remove the source and outgoing edges from the outputed source
update the list of sources
output a source
... repeat ...
details of algorithm
---------------------
(1) first go through all vertices and edges and update in-deg and out-deg
a ------------> b
in-deg = 0 in-deg = 0
out-deg = 0 out-deg = 0
whenever we find an edge, compute the in-deg and out-deg of the two vertices
in this case, in-deg(a)++ & out-deg(b)++
worst case analysis
O(n^2) every vertex can be compared to (n-1) other nodes
O(|E|) |E| = number of edges (tighter analysis)
'linear' algorithm means linear in the input size, nothing else
in this case we need algorithm linear in |V|+|E| (number of vertices & edges)
(2) go through all vertices (n), check their in-deg and build list of sources
need data structure that inserts and extracts elements in O(1) time
since we have n nodes, this runs in O(n) time
(3) when we look at a source, we decrement the in-deg of the pointed nodes
one single step could take n-1 comparisons with n-1 nodes
so this step would take O(n^2) time
or we could simply say that we make one move per edge, takeing O(1) time
since we have |E| edges, this takes O(|E|) time
--------------------------------------------------------------------
NP-complete problems : nobody knows how to solve it in linear time
NP-hard problems : also very hard (travelling salesman problem)
--------------------------------------------------------------------
Bipartite Graphs
-----------------
a graph is bipartite if I can take a graph and partition it into two groups
and that the edges of the graph go only from group 1 to group 2 such that
there are no edges between two nodes inside the same group
(1) Can we draw any graph as this way? In essense are all graphs bipartite?
This cannot happen. Take the simplest example
*
/ \
*-----*
Any graph with an odd cycle is not bipartite
(2) Can we prove that any graph without an odd cycle is bipartite?
This is the key to this algorithm
Use BFS in this case, and use its runtime analysis
==============
Jan. 26, 2016
===============
f(n) = Omega(g(n))
if there exists c, n0 such that
f(n) >= cg(n) when n > n0
To improve the lower bound we have to increase it
To improve the upper bound we have to decrease it
Ideally we want the lower bound and upper bound to match
f(n) = Theta(g(n))
if there exists c1, c2, n0, n1 such that
c1g(n) <= f(n) <= c2g(n) when n0 < n < n1
Comparison Exchange Model
--------------------------
Decision Tree
a:b
/ \
/ \
/ \
/ \
a > b b > a
/ \
a:c b:c
/ \ / \
a > c a < c b > c b < c
/ \ / \
b:c *cab a:c *cba
/ \ / \
b > c c > b a > c c < a
/ \ / \
*abc *acb *bac *bca
# leaves >= n!
height >= log(n!)
>= Omega(nlogn)
T(n) = Omega(nlogn)
= Omega(n)
Binary Search
--------------
+------------+-+-----------+
| | | |
+------------+-+-----------+
1 x n
find k in the list, which is sorted
Take the middle value x
(1) x = k, done
(2) x > k, get rid of right-half of list
(3) x < k, get rid of left-half of list
T(n) = T(n/2) + C
= O(logn)
T(n) = Omega(1) -> trivial lower bound
Circularly Sorted List
-----------------------
9 11 14 2 5 7 8
targets: O(n^(1/2)) O(logn) O(1)
^
|
bet on this one
compare middle of the list with the head and tail of the list
if middle = target, done
==============
Jan. 28, 2016
==============
Greedy Paradigm
----------------
Pick an element and stick with it, without looking at global solutions
Usually local, not global, which takes more time to prove optimal
Fast and efficient
----------------------------------
Example: time interval scheduling
----------------------------------
a set of time intervals
any number of processors
cannot change start/end time
want to minimize number of processors
Algorithm
----------
1. sort every interval according to start time
At any random step
p3 |-----|
p2 |---| |-----|
p1 |------| |------| <--- assgin to p1 and we're done
^
---------------------|-------------------------------------
|
|------| |------|
|--------| |-------|
this is greedy because we assign a task to a processor and never change it
How to prove this algorithm?
-----------------------------
Assume up to k'th step that the algorithm is correct but at k+1 step we assign
a task to a wrong processor.
Let's assume that we assign task i to p1 instead of p3
Then for p1, the following assignments can be exchanged into p3 and will work
Thus we have a contradiction and the algorithm is optimal
----------------------------------
Example: shortest path in a graph
----------------------------------
Suppose we have a weighted undirected connected graph with positive or
negative numbers as edge weights
Assume no negative cycles, because shortest path will be -INF
Assume all weights are positive
algorithm is greedy so we want to pick a path without changing it
S ------
| \
| \
| \
| \
Of all neighbors of S, we pick the shortest edge adjacent to S
At any given time based on edges and vertices we have observed so far
pick the minimum value
Then we claim that this is the smallest
Number of links in this algorithm is irrelevant
time complexity
----------------
min t -> x n
fix its distance 1
move neighbors of x to frontier n-1
-------------------------------------------
total time complexity O(n^2)
Minimum Spanning Tree
----------------------
A tree has no cycles
Remove edges from a graph to build a tree (span tree)
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Feb. 2, 2016
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Minimum spanning tree
----------------------
all edges connected and all edges weighted
At least one edge from set L to set R, containing vertices
Prim's Algorithm (Vertex Centered)
-----------------------------------
(1) L has arbitrary vertex V1 in it and everything other vertex is in R
(2) If there is an edge from V1 to a vertex V2 in R, initialize V2[E], where E
is the weight, other edges become Vx[inf]
(3) Choose the minimum edge and move the vertex A to L, and we do not change
our original initialization
(4) Look at all edges adjacent to A, and label those with their edge weights
(5) Repeat this N-1 times, and all vertices will be in L
Runtime : SUM(deg_i) ---> 2e ---> O(elogn)
Kruskal's Algorithm (Edge Centered)
------------------------------------
(0) Set each vertex to a different group
(1) Choose one vertex and its minimum edge, put the two vertices in the edge
in one partition and the others in another partition
(2) Repeat the process for different vertices
(3) When we try to group two vertices in the same group, we quit and try a
different two vertices
Sorting takes O(eloge) time
Takes O(elogn) time
Both time complexity are the same order for a connected graph, so we can choose
O(eloge) or O(elogn), since loge and logn are the same order, O(loge) = O(logn)
Union Find Algorithm
----------------------
Given 3 groups with elements in them
G1 G2 G3
FIND operation (find if they belong in the same group)
UNION opeartion (merge two group)
Use a rooted tree for the elements z --> y --> x
d --> c --> b --> a
O(N) : FIND (x,y) ---> Root(y) == Root(z)
O(1) : UNION (D,Z) ---> d --> c --> b --> a --> z --> y --> x
Alternative to improve FIND
w
/ | \
x y z
Now FIND takes O(1) time and MERGE takes O(N) times
This becomes the reverse
Assume we have 2 trees
UNION
* <--------- *
/ \ / \
* - * * *
height of tree = logN
Hnew = max(Hleft, Hright)
UNION point smaller tree to larger tree
FIND takes O(logn) time
UNION takes O(1) time
Midterm Material
-----------------
UP TO CURRENT LECTURE [!]
Sections 1.1 1.2
2.2 2.4
3.x
4.1 4.4 4.5
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Feb. 4, 2016
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Suppose we have a set of intervals
a |------------------| | programming
b |-----------| | econ takes the least time (lol)
c |-------------------------| | algorithms
deadlines
Schedule all intervals such that we meet deadline
If we do not meet deadline we will be penalized, but we still must finish
1. Sort the deadlines
2. Schedule based on deadlines
(1) First we prove that if we have an optimal schedule, then we will always
have a schedule without any spaces that is also optimal
This lemma would reduce our problem from a scheduling to an ordering
(2) Prove that there is always one in-order solution (prove by contradiction)
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Feb. 16, 2016
==============
Divide and Conquer
-------------------
partition problem into two or more subsets recursively
subsets should be balanced
solve the subsets and combine back up into the original problem
Problem: inverted pairs
------------------------
3 1 2
given a permutation of n numbers
look at every pair of numbers, if not in order, count as 1, otherwise 0
1 + 1 + 0 = 2
Have an obvious O(n^2) algorithm: look at each pair
this algorithm is not necessary optimal if we simply want the count
1 2 3
\ \ /
\ /\
/\ \
3 1 2
count the crossings
divide and conquer algorithm
-----------------------------
partitioning into two parts