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2IL50 Data Structures
Spring 2015
Lecture 1: Introduction
Algorithms
Algorithm
a well-defined computational procedure that takes some value, or a
set of values, as input and produces some value, or a set of values,
as output.
Algorithm
sequence of computational steps that transform the input into the
output.
Algorithms research
design and analysis of algorithms and data structures for
computational problems.
Data structures
Data Structure
a way to store and organize data to facilitate access and
modifications.
Abstract data type
describes functionality (which operations are supported).
Implementation
a way to realize the desired functionality
 how is the data stored (array, linked list, …)
 which algorithms implement the operations
The course
 Design and analysis of efficient algorithms for some basic
computational problems.
 Basic algorithm design techniques and paradigms
 Algorithms analysis: O-notation, recursions, …
 Basic data structures
 Basic graph algorithms
Some administration first
before we really get started …
Organization
Lecturer:
Prof. Dr. Bettina Speckmann, MF 4.105,
[email protected]
I’m here every day but Friday …
Web page:
http://www.win.tue.nl/~speckman/2IL50.html
Book:
T.H. Cormen, C.E. Leiserson, R.L. Rivest and C. Stein.
Introduction to Algorithms (3rd edition)
mandatory
Prerequisites
 Being able to work with basic programming constructs such as
linked lists, arrays, loops …
 Being able to apply standard proving techniques such as proof by
induction, proof by contradiction ...
 Being familiar with sums and logarithms such as discussed in
Chapter 3 and Appendix A of the textbook.
 If you think you might lack any of this knowledge, please come and
talk to me immediately!
Grading scheme 2IL50
1. 6 homework assignments, the best 5 of which each count for 10%
of the final grade.
2. A written exam (closed book) which counts for the remaining 50%
of the final grade.
 If you reach less than 50% of the possible points on the homework
assignments, then you are not allowed to participate in the final exam nor
in the second chance exam. You will fail the course and your next chance
will be next year. Your grade will be the minimum of 5 and the grade you
Do the
assignments!
achieved. If you reach
lesshomework
than 50% of
the points on the final exam, then
you will fail the course, regardless of the points you collected with the
homework assignments. However, you are allowed to participate in the
second chance exam. The grade of the second chance exam replaces the
grade for the first exam, that is, your homework assignments always count
for 50% of your grade.
Homework Assignments
 Posted on web-page on Monday before lecture.
 Due Sundays at 23:59 as .pdf in the electronic mailbox of your
instructor.
 Late assignments will not be accepted.
Only 5 out of 6 assignments count, hence there are no exceptions.
 Must be typeset – use Latex! See example file on web-page.
 Name scheme: Ai-LastName.pdf.
If your name is Anton van Gelderland and you submit the 1st
assignment, then your file must be named A1-vanGelderland.pdf.
 Use the tag [2IL50] in the subject line of your email.
Any questions: Stop by my office whenever you want (except Fridays!)
(send email if you want to make sure that I have time)
Academic Dishonesty
Academic Dishonesty
All class work has to be done independently. You are of course
allowed to discuss the material presented in class, homework
assignments, or general solution strategies with me or your
classmates, but you have to formulate and write up your solutions
by yourself. You must not copy from the internet, your friends, or
other textbooks. Problem solving is an important component of this
course so it is really in your best interest to try and solve all
problems by yourself. If you represent other people's work as your
own then that constitutes fraud and will be dealt with accordingly.
Organization
Components:
1. Lectures
Monday 5+6
Wednesday 3+4
AUD 6
AUD 3
2. Tutorials
Wednesday 1+2
see web-page for rooms
and instructors!
The instructors will explain the solutions to the homework assignments of the previous week and answer any questions that arise.
3. Office hours
Thursday 17:00-18:00
4. Reading and writing mathematical proofs
around MF 4.104b
How to read and write mathematical
proofs
Lecturer:
Dr. Kevin Verbeek, MF 4.106,
[email protected]
Web page:
Part of Data Structures:
http://www.win.tue.nl/~speckman/2IL50.html
Book:
Daniel Solow
How to Read and Do Proofs (5th edition)
not mandatory
Signing up for groups
 Group registration is open from today 02-02-2015 at 18:00 until
tomorrow 03-02-2015 at 18:00.
Some statistics …
Sorting
let’s get started …
The sorting problem
Input: a sequence of n numbers ‹a1, a2, …, an›
Output: a permutation of the input such that ‹ai1 ≤ … ≤ ain›
 The input is typically stored in arrays
 Numbers ≈ Keys
 Additional information (satellite data) may be stored with keys
 We will study several solutions ≈ algorithms for this problem
Describing algorithms
 A complete description of an algorithm consists of three parts:
1. the algorithm
(expressed in whatever way is clearest and most concise,
can be English and / or pseudocode)
2. a proof of the algorithm’s correctness
3. a derivation of the algorithm’s running time
InsertionSort
 Like sorting a hand of playing cards:
 start with empty left hand, cards on table
 remove cards one by one, insert into
correct position
 to find position, compare to cards in hand from right to left
 cards in hand are always sorted
InsertionSort is
 a good algorithm to sort a small number of elements
 an incremental algorithm
Incremental algorithms
process the input elements one-by-one and maintain the solution
for the elements processed so far.
Incremental algorithms
Incremental algorithms
process the input elements one-by-one and maintain the solution
for the elements processed so far.
 In pseudocode:
Check book for more pseudocode conventions
IncAlg(A)
// incremental algorithm which computes the solution of a problem
with input A = {x1,…,xn}
1. initialize: compute the solution for {x1}
2. for j = 2 to n
3.
do compute the solution for {x1,…,xj} using the (already
computed) solution for {x1,…,xj-1}
no “begin - end”, just indentation
InsertionSort
InsertionSort(A)
// incremental algorithm that sorts array A[1..n] in non-decreasing
order
1. initialize: sort A[1]
2. for j = 2 to A.length
3.
do sort A[1..j] using the fact that A[1.. j-1] is already sorted
InsertionSort
InsertionSort(A)
// incremental algorithm that sorts array A[1..n] in non-decreasing
order
InsertionSort is an in place algorithm:
1. initialize: sort A[1]
the numbers are rearranged within the
2. for j = 2 to A.length
array with only constant extra space.
3.
do key = A[j]
4.
i = j -1
5.
while i > 0 and A[i] > key
6.
do A[i+1] = A[i]
7.
i = i -1
8.
A[i +1] = key
1
j
1 3 14 17 28 6
n
…
Correctness proof
 Use a loop invariant to understand why an algorithm gives the
correct answer.
Loop invariant (for InsertionSort)
At the start of each iteration of the “outer” for loop (indexed by j) the
subarray A[1..j-1] consists of the elements originally in A[1..j-1] but
in sorted order.
Correctness proof
 To proof correctness with a loop invariant we need to show three
things:
Initialization
Invariant is true prior to the first iteration of the loop.
Maintenance
If the invariant is true before an iteration of the loop, it remains true
before the next iteration.
Termination
When the loop terminates, the invariant (usually along with the
reason that the loop terminated) gives us a useful property that
helps show that the algorithm is correct.
Correctness proof
InsertionSort(A)
1. initialize: sort A[1]
2. for j = 2 to A.length
3.
do key = A[j]
4.
i = j -1
5.
while i > 0 and A[i] > key
6.
do A[i+1] = A[i]
7.
i = i -1
8.
A[i +1] = key
Loop invariant
At the start of each iteration of the
“outer” for loop (indexed by j) the
subarray A[1..j-1] consists of the
elements originally in A[1..j-1] but
in sorted order.
Initialization
Just before the first iteration, j = 2 ➨ A[1..j-1] = A[1], which is the
element originally in A[1], and it is trivially sorted.
Correctness proof
InsertionSort(A)
1. initialize: sort A[1]
2. for j = 2 to A.length
3.
do key = A[j]
4.
i = j -1
5.
while i > 0 and A[i] > key
6.
do A[i+1] = A[i]
7.
i = i -1
8.
A[i +1] = key
Loop invariant
At the start of each iteration of the
“outer” for loop (indexed by j) the
subarray A[1..j-1] consists of the
elements originally in A[1..j-1] but
in sorted order.
Maintenance
Strictly speaking need to prove loop invariant for “inner” while loop.
Instead, note that body of while loop moves A[j-1], A[j-2], A[j-3],
and so on, by one position to the right until proper position of key is
found (which has value of A[j]) ➨ invariant maintained.
Correctness proof
InsertionSort(A)
1. initialize: sort A[1]
2. for j = 2 to A.length
3.
do key = A[j]
4.
i = j -1
5.
while i > 0 and A[i] > key
6.
do A[i+1] = A[i]
7.
i = i -1
8.
A[i +1] = key
Loop invariant
At the start of each iteration of the
“outer” for loop (indexed by j) the
subarray A[1..j-1] consists of the
elements originally in A[1..j-1] but
in sorted order.
Termination
The outer for loop ends when j > n; this is when j = n+1 ➨ j-1 = n.
Plug n for j-1 in the loop invariant ➨ the subarray A[1..n] consists
of the elements originally in A[1..n] in sorted order.
Another sorting algorithm
using a different paradigm …
MergeSort
 A divide-and-conquer sorting algorithm.
Divide-and-conquer
break the problem into two or more subproblems, solve the
subproblems recursively, and then combine these solutions to
create a solution to the original problem.
Divide-and-conquer
D&CAlg(A)
// divide-and-conquer algorithm that computes the solution of a
problem with input A = {x1,…,xn}
1. if # elements of A is small enough (for example 1)
2.
then compute Sol (the solution for A) brute-force
3.
else
4.
split A in, for example, 2 non-empty subsets A1 and A2
5.
Sol1 = D&CAlg(A1)
6.
Sol2 = D&CAlg(A2)
7.
compute Sol (the solution for A) from Sol1 and Sol2
8. return Sol
MergeSort
MergeSort(A)
// divide-and-conquer algorithm that sorts array A[1..n]
1. if A.length == 1
2.
then compute Sol (the solution for A) brute-force
3.
else
4.
split A in 2 non-empty subsets A1 and A2
5.
Sol1 = MergeSort(A1)
6.
Sol2 = MergeSort(A2)
7.
compute Sol (the solution for A) from Sol1 en Sol2
MergeSort
MergeSort(A)
// divide-and-conquer algorithm that sorts array A[1..n]
1. if A.length == 1
2.
then skip
3.
else
4.
n = A.length ; n1 = n/2 ; n2 = n/2 ;
copy A[1.. n1] to auxiliary array A1[1.. n1]
copy A[n1+1..n] to auxiliary array A2[1.. n2]
5.
MergeSort(A1)
6.
MergeSort(A2)
7.
Merge(A, A1, A2)
MergeSort
3 14 1 28 17 8 21 7 4 35
1 3 4 7 8 14 17 21 28 35
3
3 14 1 28 17
8 21 7 4 35
1 3 14 17 28
4 7 8 21 35
3 14
1 28 17
3 14
1 17 28
14
MergeSort
 Merging
A1
1 3 14 17 28
A
A2
4 7 8 21 35
1 3 4 7 8 14 17 21 28 35
MergeSort: correctness proof
Induction on n (# of input elements)

proof that the base case (n small) is solved correctly

proof that if all subproblems are solved correctly, then the
complete problem is solved correctly
MergeSort: correctness proof
MergeSort(A)
1. if length[A] = 1
2.
then skip
3.
else
4.
n = A.length ; n1 = n/2 ; n2 = n/2 ;
copy A[1.. n1] to auxiliary array A1[1.. n1]
copy A[n1+1..n] to auxiliary array A2[1.. n2]
5.
MergeSort(A1)
Lemma
6.
MergeSort(A2)
MergeSort
7.
Merge(A, A1, A2)
sorts the array A[1..n] correctly.
Proof (by induction on n)
Base case: n = 1, trivial ✔
Inductive step: assume n > 1. Note that n1 < n and n2 < n.
Inductive hypothesis ➨ arrays A1 and A2 are sorted correctly
Remains to show: Merge(A, A1, A2) correctly constructs a sorted
array A out of the sorted arrays A1 and A2 …
etc.
■
QuickSort
another divide-and-conquer sorting
algorithm…
QuickSort
QuickSort(A)
// divide-and-conquer algorithm that sorts array A[1..n]
1. if length[A] ≤ 1
2.
then skip
3.
else
4.
pivot = A[1]
5.
move all A[i] with A[i] < pivot into auxiliary array A1
6.
move all A[i] with A[i] > pivot into auxiliary array A2
7.
move all A[i] with A[i] = pivot into auxiliary array A3
8.
QuickSort(A1)
9.
QuickSort(A2)
10.
A = “A1 followed by A3 followed by A2”
Analysis of algorithms
some informal thoughts – for now …
Analysis of algorithms
 Can we say something about the running time of an algorithm
without implementing and testing it?
InsertionSort(A)
1. initialize: sort A[1]
2. for j = 2 to A.length
3.
do key = A[j]
4.
i = j -1
5.
while i > 0 and A[i] > key
6.
do A[i+1] = A[i]
7.
i = i -1
8.
A[i +1] = key
Analysis of algorithms
 Analyze the running time as a function of n (# of input elements)
 best case
 average case
 worst case
An algorithm has worst case running time T(n) if for any
input of size n the maximal number of elementary operations
executed is T(n).
elementary operations
add, subtract, multiply, divide, load, store, copy, conditional and
unconditional branch, return …
Analysis of algorithms: example
InsertionSort:
15 n2 + 7n – 2
MergeSort:
300 n lg n + 50 n
InsertionSort
6 x faster
n=10
n=100
n=1000
1568
150698
1.5 x 107
10466
204316
3.0 x 106
InsertionSort
1.35 x faster
MergeSort
5 x faster
n = 1,000,000
InsertionSort 1.5 x 1013
MergeSort
6 x 109
2500 x faster !
Analysis of algorithms
 It is extremely important to have efficient algorithms for large inputs
 The rate of growth (or order of growth) of the running time is far
more important than constants
InsertionSort: Θ(n2)
MergeSort: Θ(n log n)
Θ-notation
Intuition: concentrate on the leading term, ignore constants
19 n3 + 17 n2 - 3n
becomes
Θ(n3)
2 n lg n + 5 n1.1 - 5
becomes
Θ(n1.1)
n - ¾ n √n
becomes
---
(precise definition next lecture…)
Some rules and notation

log n denotes log2 n

We have for a, b, c > 0 :
1. logc (ab) =
logc a + logc b
2. logc (ab) =
b logc a
3. loga b =
logc b / logc a
Find the leading term
 lg35n vs. √n ?
 logarithmic functions grow slower than polynomial functions
 lga n grows slower than nb for all constants a > 0 and b > 0
 n100 vs. 2
n
?
 polynomial functions grow slower than exponential functions
 na grows slower than bn for all constants a > 0 and b > 1
Announcements
Today 7+8
 Reading and writing mathematical proofs
Today 7+8+9
 Emergency Latex office hour in MF 11
Today 18:00
 Group registration opens on OASE
Wednesday 1+2
 tutorials, discuss solutions to homework A1 from 2014
(see web-page)

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