Transcript Lecture 1 Preliminaries - University of Liverpool

Lectures on
Greedy Algorithms and
Dynamic Programming
COMP 523: Advanced Algorithmic Techniques
Lecturer: Dariusz Kowalski
Greedy Algorithms and Dynamic Programming
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Overview
Previous lectures:
• Algorithms based on recursion - call to the same
procedure to solve the problem for the smallersize sub-input(s)
• Graph algorithms: searching, with applications
These lectures:
• Greedy algorithms
• Dynamic programming
Greedy Algorithms and Dynamic Programming
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Greedy algorithm’s paradigm
Algorithm is greedy if :
• it builds up a solution in small consecutive steps
• it chooses a decision at each step myopically to optimize
some underlying criterion
Analyzing optimal greedy algorithms by showing that:
• in every step it is not worse than any other algorithm, or
• every algorithm can be gradually transformed to the
greedy one without hurting its quality
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Interval scheduling
Input: set of intervals on the line, represented by
pairs of points (ends of intervals)
Output: the largest set of intervals such that
none two of them overlap
Generic greedy solution:
• Consider intervals one after another using some
rule
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Rule 1
Select the interval that starts earliest
(but is not overlapping the already chosen intervals)
Underestimated solution!
optimal
algorithm
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Rule 2
Select the shortest interval
(but not overlapping the already chosen intervals)
Underestimated solution!
optimal
algorithm
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Rule 3
Select the interval intersecting the smallest number of
remaining intervals
(but still is not overlapping the already chosen intervals)
Underestimated solution!
optimal
algorithm
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Rule 4
Select the interval that ends first
(but still is not overlapping the already chosen intervals)
Hurray! Exact solution!
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Analysis - exact solution
Algorithm gives non-overlapping intervals:
obvious, since we always choose an interval which does
not overlap the previously chosen intervals
The solution is exact:
Let:
• A be the set of intervals obtained by the algorithm,
• Opt be the largest set of pairwise non-overlapping
intervals
We show that A must be as large as Opt
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Analysis - exact solution cont.
Let A = {A1,…,Ak} and Opt = {B1,…,Bm} be sorted.
By definition of Opt we have k  m.
Fact: for every i  k, Ai finishes not later than Bi.
Proof: by induction.
For i = 1 by definition of the first step of the algorithm.
From i -1 to i : Suppose that Ai-1 finishes not later than Bi-1.
From the definition of a single step of the algorithm, Ai is the first
interval that finishes after Ai-1 and does not overlap it.
If Bi finished before Ai then it would overlap some of the previous
A1,…, Ai-1 and consequently - by the inductive assumption - it
would overlap or end before Bi-1, which would be a
contradiction.
Bi-1
B
i
Ai-1
Greedy Algorithms and Dynamic Programming
Ai
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Analysis - exact solution cont.
Theorem: A is the exact solution.
Proof: we show that k = m.
Suppose to the contrary that k < m.
We already know that Ak finishes not later than Bk.
Hence we could add Bk+1 to A and obtain a bigger solution
by the algorithm - a contradiction.
Bk-1
Ak-1
Bk
Ak
Bk+1
algorithm finishes selection
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Implementation & time complexity
Efficient implementation:
• Sort intervals according to the right-most ends
• For every consecutive interval:
– If the left-most end is after the right-most end of the
last selected interval then we select this interval
– Otherwise we skip it and go to the next interval
Time complexity: O(n log n + n) = O(n log n)
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Textbook and Exercises
READING:
• Chapter 4 “Greedy Algorithms”, Section 4.1
EXERCISE:
• All Interval Scheduling problem from Section
4.1
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Minimum spanning tree
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Greedy algorithm’s paradigm
Algorithm is greedy if :
• it builds up a solution in small consecutive steps
• it chooses a decision at each step myopically to optimize
some underlying criterion
Analyzing optimal greedy algorithms by showing that:
• in every step it is not worse than any other algorithm, or
• every algorithm can be gradually transformed to the
greedy one without hurting its quality
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Minimum spanning tree
Input: weighted graph G = (V,E)
• every edge in E has its positive weight
Output: spanning tree such that the sum of weights is
not bigger than the sum of weights of any other
spanning tree
Spanning tree: subgraph with
– no cycle, and
– spanning and connected (every two nodes in V are
connected by a path)
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Properties of minimum spanning
trees MST
Properties of spanning trees:
• n nodes
• n - 1 edges
• at least 2 leaves (leaf - a node with only one neighbor)
MST cycle property:
• after adding an edge we obtain exactly one cycle and
each edge from MST in this cycle has no bigger weight
than the weight of the added edge
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1
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cycle
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Crucial observation about MST
Consider sets of nodes A and V - A
• Let F be the set of edges between A and V - A
• Let a be the smallest weight of an edge in F
Theorem:
Every MST must contain at least one edge of weight a
from set F
A
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1
2
A
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3
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Proof of the Theorem
Let e be the edge in F with the smallest weight - for simplicity
assume that such edge is unique. Suppose to the contrary that
e is not in some MST. Consider one such MST.
Add e to MST - a cycle is obtained, in which e has weight not smaller
than any other weight of edge in this cycle, by the MST cycle property.
Since the two ends of e are in different sets A and V - A,
there is another edge f in the cycle and in F. By definition of e,
such f must have a bigger weight than e, which is a contradiction.
A
1
3
A
1
2
1
2
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2
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Greedy algorithms finding MST
Kruskal’s algorithm:
• Sort all edges according to their weights
• Choose n - 1 edges, one after another, as follows:
– If a new added edge does not create a cycle with previously
selected edges then we keep it in (partial) solution;
otherwise we remove it
Remark: we always have a partial forest
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Greedy algorithms finding MST
Prim’s algorithm:
• Select an arbitrary node as a root
• Choose n - 1 edges, one after another, as follows:
– Consider all edges which are incident to the currently build
(partial) solution and which do not create a cycle in it, and
select one having the smallest weight
Remark: we always have a connected partial tree
root
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2
1
2
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Why the algorithms work?
Follows from the crucial observations:
Kruskal’s algorithm:
• Suppose we add edge {v,w};
• This edge has a smallest weight among edges between the
set of nodes already connected with v (by a path in
already selected subgraph) and other nodes
Prim’s algorithm:
• Always chooses an edge with a smallest weight among
edges between the set of already connected nodes and free
nodes (i.e., non-connected nodes)
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Time complexity
There are implementations using
• Union-find data structure (Kruskal’s algorithm)
• Priority queue (Prim’s algorithm)
achieving time complexity
O(m log n)
where n is the number of nodes and m is the
number of edges in a given graph G
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Textbook and Exercises
READING:
• Chapter 4 “Greedy Algorithms”, Section 4.5
EXERCISES:
• Solved Exercise 3 from Chapter 4
• Generalize the proof of the Theorem to the case
where may be more than one edges of smallest
weight in F
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Priority Queues (PQ)
Implementation of Prim’s algorithm using PQ
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Minimum spanning tree
Input: weighted graph G = (V,E)
• every edge in E has its positive weight
Output: spanning tree such that the sum of weights is
not bigger than the sum of weights of any other
spanning tree
Spanning tree: subgraph with
– no cycle, and
– connected (every two nodes in V are connected by a path)
1
3
2
1
2
1
3
2
1
2
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1
2
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Crucial observation about MST
Consider sets of nodes A and V - A
• Let F be the set of edges between A and V - A
• Let a be the smallest weight of an edge in F
Theorem:
Every MST must contain at least one edge of weight a
from set F
A
1
3
2
1
2
A
1
2
1
3
Greedy Algorithms and Dynamic Programming
2
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Greedy algorithm finding MST
Prim’s algorithm:
• Select an arbitrary node as a root
• Choose n - 1 edges, one after another, as follows:
– Consider all edges which are incident to the currently build
(partial) solution and which do not create a cycle in it, and
select one which has the smallest weight
Remark: we always have a connected partial tree
root
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2
1
2
1
3
2
1
2
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2
1
2
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Priority queue
Set of n elements, each has its priority value (key)
– the smaller key the higher priority the element has
Operations provided in time O(log n):
• Adding new element to PQ
• Removing an element from PQ
• Taking element with the smallest key
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Implementation of PQ based on heaps
Heap: rooted (almost) complete
binary tree, each node has its
• value
• key
• 3 pointers: to the parent and
children (or nil(s) if parent or
child(ren) not available)
Required property:
in each subtree the smallest key
is always in the root
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5 6
2 3 4 7 5 6
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Operations on the heap
PQ operations:
• Add
• Remove
• Take
Additional supporting operation:
• Last leaf:
Updating the pointer to the rigth-most leaf on the lowest
level of the tree, after each operation (take, add, remove)
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Construction of the heap
Construction:
• Start with arbitrary element
• Keep adding next elements using add operation provided
by the heap data structure
(which will be defined in the next slide)
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Implementing operations on heap
Smallest key element: trivially read from the root
Adding new element:
• find the next last leaf location in the heap
• put the new element as the last leaf
• recursively compare it with its parent’s key:
– if the element has the smaller key then swap the element and its parent and
continue;
otherwise stop
Remark: finding the next last leaf may require to search through the
path up and then down (exercise)
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Implementing operations on heap
Removing element:
• remove it from the tree
• move the value from last leaf on its place
• update the last leaf
• compare the moved element recursively either
– “up” if its value is smaller than its current parent:
swap the elements and continue going up until reaching smaller parent or the
root,
or
– “down” if its value is bigger than its current parent:
swap it with the smallest of its children and continue going down until
reaching a node with no smaller child or a leaf
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Examples - adding
2
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5 6
1
add 1 at the end
7
2 3 4 7 5 6 1
1
5 6
4
swap 1 and 4
2 3 1 7 5 6
4
1
3
2
1 3 2 7 5 6
swap 1 and 2
4
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5 6
4
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Examples - removing
2
3
6
4
3
7
5 6
removing 2
3
4
6
7
5
swap 2 and last element
2 3 4 7 5 6
4
7
5
remove 2 and
swap 6 and 3
6 3 4 7 5
3 6 4 7 5
3
5
4
swap 6 and 5
3 5 4 7 6
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6
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Heap operations - time complexity
• Taking minimum: O(1)
• Adding:
– Updating last leaf: O(log n)
– Going up with swaps through (almost) complete binary
tree: O(log n)
• Removing:
– Updating last leaf: O(log n)
– Going up or down (only once direction is selected)
doing swaps through (almost) complete binary tree:
O(log n) Greedy Algorithms and Dynamic Programming
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Prim’s algorithm - time complexity
Input: graph is given as an adjacency list
• Select a root node as an initial partial tree
• Construct PQ with all edges incident to the root (weights are keys)
• Repeat until PQ is empty
– Take the smallest edge from PQ and remove it
– If exactly one end of the edge is in the partial tree then
• Add this edge and its other end to the partial tree
• Add to PQ all edges, one after another, which are incident to the new node and
remove all their copies from graph representation
Time complexity: O(m log n)
where n is the number of nodes, m is the number of edges
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Textbook and Exercises
READING:
• Chapters 2 and 4, Sections 2.5 and 4.5
EXERCISES:
• Solved Exercises 1 and 2 from Chapter 4
• Prove that a spanning tree of an n - node graph has
n - 1 edges
• Prove that an n - node connected graph has at least
n - 1 edges
• Show how to implement the update of the last leaf in
time O(log n)
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Dynamic programming
Two problems:
• Weighted interval scheduling
• Sequence alignment
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Dynamic Programming paradigm
Dynamic Programming (DP):
• Decompose the problem into series of sub-problems
• Build up correct solutions to larger and larger subproblems
Similar to:
• Recursive programming vs. DP: in DP sub-problems may
strongly overlap
• Exhaustive search vs. DP: in DP we try to find
redundancies and reduce the space for searching
• Greedy algorithms vs. DP: sometimes DP orders subproblems and processes them one after another
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(Weighted) Interval scheduling
(Weighted) Interval scheduling:
Input: set of intervals (with weights) on the line,
represented by pairs of points - ends of intervals
Output: the largest (maximum sum of weights) set
of intervals such that none two of them overlap
Greedy algorithm doesn’t work for weighted case!
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Example
Greedy algorithm:
• Repeatedly select the interval that ends first (but still not
overlapping the already chosen intervals)
Exact solution of unweighted case.
weight 1
weight 3
weight 1
Greedy algorithm gives total weight 2 instead of optimal 3
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Basic structure and definition
• Sort the intervals according to their right ends
• Define function p as follows:
– p(1) = 0
– p(i) is the number of intervals which finish before ith interval
starts
p(1)=0
weight 1
p(2)=1
weight 3
p(3)=0
weight 2
weight 1
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p(4)=2
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Basic property
• Let wj be the weight of jth interval
• Optimal solution for the set of first j intervals satisfies
OPT(j) = max{ wj + OPT(p(j)) , OPT(j-1) }
Proof:
If jth interval is in the optimal solution O then the other intervals in
O are among intervals 1,…,p(j).
Otherwise search for solution among first j-1 intervals.
p(1)=0
weight 1
p(2)=1
weight 3
p(3)=0
weight 2
weight 1
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p(4)=2
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Sketch of the algorithm
• Additional array M[0…n] initialized by 0,p(1),…,p(n)
( intuitively M[j] stores optimal solution OPT(j) )
Algorithm
• For j = 1,…,n do
– Read p(j) = M[j]
– Set M[j] := max{ wj + M[p(j)] , M[j-1] }
p(1)=0
weight 1
p(2)=1
weight 3
p(3)=0
weight 2
weight 1
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p(4)=2
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Complexity of solution
Time: O(n log n)
• Sorting: O(n log n)
• Initialization of M[0…n] by 0,p(1),…,p(n): O(n log n)
• Algorithm: n iterations, each takes constant time, total
O(n)
Memory: O(n) - additional array M
p(1)=0
weight 1
p(2)=1
weight 3
p(3)=0
weight 2
weight 1
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p(4)=2
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Sequence alignment problem
Popular problem from word processing and
computational biology
• Input: two words X = x1x2…xn and Y = y1y2…ym
• Output: largest alignment
Alignment A:
set of pairs (i1,j1),…,(ik,jk) such that
• If (i,j) in A then xi = yj
• If (i,j) is before (i’,j’) in A then i < i’ and j < j’ (no
crossing matches)
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Example
• Input: X = c t t t c t c c Y = t c t t c c
Alignment A:
X=ctttct cc
|
|| | |
Y= t
cttcc
Another largest alignment A:
X= ctttctcc
||| | |
Y=tctt c c
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Finding the size of max alignment
Optimal alignment OPT(i,j) for prefixes of X and Y of lengths i
and j respectively:
OPT(i,j) = max{ ij + OPT(i-1,j-1) , OPT(i,j-1) , OPT(i-1,j) }
where ij equals 1 if xi = yj, otherwise is equal to -
Proof:
If xi = yj in the optimal solution O then the optimal alignment
contains one match (xi , yj) and the optimal solution for prefixes
of length i-1 and j-1 respectively.
Otherwise at most one end is matched. It follows that either
x1x2…xi-1 is matched only with letters from y1y2…ym or
y1y2…yj-1 is matched only with letters from x1x2…xn. Hence the
optimal solution is either the same as for OPT(i-1,j) or for
OPT(i,j-1).
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Algorithm finding max alignment
• Initialize matrix M[0..n,0..m] into zeros
Algorithm
• For i = 1,…,n do
– For j = 1,…,m do
• Compute ij
• Set M[i,j] : =
max{ ij + M[i-1,j-1] , M[i,j-1] , M[i-1,j] }
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Complexity
Time: O(nm)
• Initialization of matrix M[0..n,0..m]: O(nm)
• Algorithm: O(nm)
Memory: O(nm)
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Reconstruction of optimal alignment
Input: matrix M[0..n,0..m] containing OPT
values
Algorithm
• Set i = n, j = m
• While both i,j > 0 do
• Compute ij
• If M[i,j] = ij + M[i-1,j-1] then match xi and yj and
set i = i - 1, j = j - 1; else
• If M[i,j] = M[i,j-1] then set j = j - 1 (skip letter yj ); else
• If M[i,j] = M[i-1,j] then set i = i - 1 (skip letter xi )
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Distance between words
Generalization of alignment problem
• Input:
– two words X = x1x2…xn and Y = y1y2…ym
– mismatch costs pq, for every pair of letters p and q
– gap penalty 
• Output:
– (smallest) distance between words X and Y
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Example
• Input: X = c t t t c t c c Y = t c t t c c
Alignment A: (4 gaps of cost  each, 1 mismatch of cost ct)
X=c t t t c t c c
|
| | ^ |
Y= t
c t t c c
Largest alignment A: (4 gaps)
X= c t t t c t c c
| | |
| |
Y= t c t t
c c
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Finding the distance between words
Optimal alignment OPT(i,j) for prefixes of X and Y of lengths i and j
respectively:
OPT(i,j) = min{ ij + OPT(i-1,j-1) ,  + OPT(i,j-1) ,  + OPT(i-1,j) }
Proof:
If xi and yj are (mis)matched in the optimal solution O then the
optimal alignment contains one (mis)match (xi , yj) of cost ij and
the optimal solution for prefixes of length i-1 and j-1 respectively.
Otherwise at most one end is (mis)matched. It follows that either
x1x2…xi-1 is (mis)matched only with letters from y1y2…ym or
y1y2…yj-1 is (mis)matched only with letters from x1x2…xn. Hence
the optimal solution is either the same as counted for OPT(i-1,j) or
for OPT(i,j-1), plus the penalty gap .
Algorithm and complexity remain the same.
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Textbook and Exercises
READING:
• Chapter 6 “Dynamic Programming”, Sections
6.1 and 6.6
EXERCISES:
• All Shortest Paths problem, Section 6.8
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Conclusions
• Greedy algorithms: algorithms constructing solutions
step after step by using a local rule
• Exact greedy algorithm for interval selection problem in time O(n log n) illustrating “greedy stays ahead” rule
• Greedy algorithms for finding minimum spanning tree
in a graph
– Kruskal’s algorithm
– Prim’s algorithm
• Priority Queues
– greedy Prim’s algorithms for finding a minimum spanning tree
in a graph in time O(m log n)
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Conclusions cont.
• Dynamic programming
– Weighted interval scheduling in time O(n log n)
– Sequence alignment in time O(nm)
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