6212
Lecture {
}
CSCI 1
Design and Analysis of Algorithms
Hoeteck Wee [e-mail]
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Algorithmic ideas are pervasive
applications.
Economics, auctions and game theory
Biology, genes and genomes
Combinatorial optimization
Internet search and massive data sets
this course.
techniques for the design and analysis of algorithms
also, implementation of these algorithms in python
.
Administration
topics.
1. graph algorithms, e.g. graph traversal via breadth-rst search
2. greedy algorithms, e.g. shortest paths and minimum spanning tree
3. divide and conquer algorithms, e.g. mergesort
4. dynamic programming, e.g. sequence alignment
5. network ow & NP-completeness
midterms. tentatively Oct 5 (15%), Nov 2 (30%), Nov 30/Dec 7 (30%).
no rescheduling.
homework & in-class exercises. written, programming assignments (25%).
no late submissions for homework
programming assignments submitted via black-board
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Homework 0
tasks.
1. create a python program that prints your email address 212 times.
2. submit the program as a le named hw0.py via blackboard.
deadline. sep 6 (tue) at 11.59 pm
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Python overview
basics.
interpreted language (i.e. no need to compile)
get help in interpreter using help(<object>) or dir()
properties.
dynamically, implicitly typed (i.e. you dont have to declare variables)
case sensitive
object-oriented (i.e. everything is an object)
.
Python overview
java.
public static void hello1(int n) {
for (int i=0; i<n; i++) {
System.out.println("Hello");
}
}
python.
for i in range(n): ## range(n) returns array [0,1,2,...,n-1]
print "Hello"
syntax.
no mandatory statement termination characters (vs ; in java)
blocks are specied by indentation (vs { } in java)
.
a sample of algorithms
.
Computing the maximum
problem. given a list a of n numbers, compute the maximum.
algorithm.
ans = a[0]
## indices of a from 0 to n-1
for i in range(1, n):
## for i=1,2,3,..,n-1
if a[i] > ans:
ans = a[i]
return ans
running time. linear time O(n),
i.e. running time is at most a constant factor times input size
.
Merging sorted lists
problem. combine two sorted lists a, b into a single sorted list.
algorithm.
i, j, out = 0, 0, []
while both lists are non-empty:
if a[i] <= b[j]:
append a[i] to out and increment i
else:
append to b[j] out and increment j
append remainder of nonempty list to out
return out
running time. linear time O(n)
.
Sorting a list
problem. sort a list a of n numbers.
algorithm. divide-and-conquer
def mergesort(a):
if len(a) <= 1:
return a
else:
mid = len(a)/2
return merge(mergesort(a[:mid]), mergesort(a[mid:]))
running time. O(n log n)
.
Closest pair of points
problem. given a list of n points in the plane (x1 , y1 ), . . . , (xn , yn ), nd the pair
that is closest.
algorithm. try all pairs of points
min = (x1 x2 ) + (y1 y2 )
2
2
for i in range(1, n+1):
for j in range(i+1, n+1):
d = (xi xj ) + (yi yj )
2
2
if d < min: min = d
return min
running time. quadratic time O(n2 )
.
Large contiguous sums
problem. Given an array a of n integers positive and negative, nd the largest sum
of contiguous integers.
example. a = [1, 3, 2, 7, 4, 2, 2, 3, 1], output: 4 + 2 2 + 3 = 7.
algorithm. Compute sums of all sequences of contiguous integers
maxsum = 0
for i in range(n):
for j in range(i+1, n):
sum = 0
## compute a[i]+...+a[j-1]
for k in range(i,j):
sum += a[k]
if sum > maxsum: maxsum = sum
return maxsum
running time. O(n3 )
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Binary search
problem. given a sorted list a of n numbers, determine if item is in a.
algorithm. binary search
def bsearch(a, item):
if len(a) == 0: return False
else:
mid = len(a)/2
if a[mid]==item: return True
else:
if item < a[mid]: return bsearch(a[:mid],item)
else: return bsearch(a[mid+1:],item)
running time. O(log n), due to repeated halving.
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Integer square root
problem. given an n-bit integer A, compute the integer square root of A.
attempt 1. try 1, 2, 3, 4, ...
takes time O(2n/2 n2 )
algorithm. use binary search to nd the square root
input 25
25 1
mid-point: 13
13 1
mid-point: 7
7
1
mid-point: 4
7
4
mid-point: 5
output 5
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Integer square root
problem. given an n-bit integer A, compute the integer square root of A.
algorithm. use binary search to nd the square root
def isqrt(A):
lower, upper = 1, A
while upper-lower>1:
mid = (lower+upper)/2
sq = mid*mid
if sq > A:
upper = mid
else:
lower = mid
if (lower*lower == n):
return lower
else:
return -1
running time. O(n3 )
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Next...
1. Mathematical formulation of an abstract problem
2. Present an algorithm
3. Analysis: running time and correctness
4. Discuss implementation
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