Analysis of Algorithms

COS 265 - Data Structures & Algorithms

Analysis of Algorithms

introduction

cast of characters

Programmer needs to develop a working solution

Client wants to solve problem efficiently

Theoretician seeks to understand

Student (you) might play any or all of these roles someday

running time

“

As soon as an Analytical Engine exists, it will necessarily guide the future course of the science. Whenever any result is sought by its aid, the question will then arise—By what course of calculation can these results be arrived at by the machine in the shortest time?
–Charles Babbage (1864)

”

reasons to analyze algorithms

• Predict performance
• Compare algorithms
• Provide guarantees
• Understand theoretical basis

Primary practical reason: avoid performance bugs

Client gets poor performance because programmer did not understand performance characteristics

an algorithmic success story

N-body simulation

• Simulate gravitational interactions among \(N\) bodies
• Applications: cosmology, fluid dynamics, semiconductors, ...
• Brute force: \(N^2\) steps
• Barnes-Hut algorithm: \(N \log N\) steps, enabling new research

an algorithmic success story

Discrete Fourier transform

• Express signal as weighted sum of sines and cosines
• Applications: DVD, JPEG, MRI, astrophysics, ...
• Brute force: \(N^2\) steps
• FFT algorithm: \(N \log N\) steps, enabling new technology

the challenge

Q: Will my program be able to solve a large practical input?

“

Why is my program so slow??

Why does it run out of memory??

”

Insight by Knuth (1970s): Use scientific method to understand performance

scientific method applied to alg. analysis

A framework for predicting performance and comparing algorithms

Scientific method:

• Observe some feature of the natural world
• Hypothesize a model that is consistent with the observations
• Predict events using the hypothesis
• Verify the predictions by making further observations
• Validate by repeating until the hypothesis and observations agree

Principles:

• Experiments must be reproducible
• Hypotheses must be falsifiable

algorithm analysis

observations

example: 3-sum

3-Sum: Given \(N\) distinct integers, how many triples sum to exactly zero?

Context: Deeply related to problems in computational geometry

$ more 8ints.txt
8
30 -40 -20 -10 40 0 10 5

$ java ThreeSum 8ints.txt
4

3-sum brute-force algorithm

ThreeSum.java:
source, javadoc, 1Kints.txt, 2Kints.txt, 4Kints.txt, 8Kints.txt

public class ThreeSum {
    public static int count(int[] a) {
        int N = a.length;
        int count = 0;

        // check each triple
        // ignore integer overflow for simplicity
        for(int i = 0; i < N; i++)
            for(int j = i+1; j < N; j++)
                for(int k = j+1; k < N; k++)
                    if(a[i] + a[j] + a[k] == 0)
                        count++;
        return count;
    }

    public static void main(String[] args) {
        In in = new In(args[0]);
        int[] a = in.readAllInts();
        StdOut.println(count(a));
    }
}

measuring the running time

Q:   How to time a program?

measuring the running time

Q:   How to time a program?
A1:    :(    Manually using a stopwatch

measuring the running time

Q:   How to time a program?
A1:    :(    Manually using a stopwatch
A2:    :|    Using time Unix command (time java ThreeSum ...)

measuring the running time

Q:   How to time a program?
A1:    :(    Manually using a stopwatch
A2:    :|    Using time Unix command (time java ThreeSum ...)
A3:    :)    Automatically using software!

public static void main(String[] args) {
    In in = new In(args[0]);
    int[] a = in.readAllInts();
    Stopwatch stopwatch = new Stopwatch();
    StdOut.println(ThreeSum.count(a));
    double time = stopwatch.elapsedTime();
    StdOut.println("elapsed time = " + time);
}

Emperical analysis

Run the program for various input sizes and measure running time

\(T(N)\) is time in seconds on some particular machine

data analysis

Standard plot: Plot running time \(T(N)\) vs. input size \(N\)

data analysis

Log-log plot: Plot running time \(T(N)\) vs. input size \(N\) using log-log scale

straight line of approx slope 3 \(\lg(T(N)) = b \lg N + c\) \( b = 2.999, c = -33.2103\) \(T(N) = aN^b, \text{where } a=2^c\)

Regression: Fit straight line through data points: \(a N^b\) (power law)

Hypothesis: Running time is about \(1.006 \times 10^{-10} \times N^{2.999}\) secs
(\(2.999\) is slope)

prediction and validation

Hypothesis: Running time is about \(1.006 \times 10^{-10} \times N^{2.999}\) secs
("order of growth" of running time is about \(N^3\))

Predictions:

• 51.0 seconds for \(N=8000\)
• 408.1 seconds for \(N=16000\)

Observations validate hypothesis!

doubling hypothesis

Doubling hypothesis: Quick way to estimate \(b\) in a power-law relationship

Run program, doubling the size of the input

Hypothesis: Running time is about \(aN^b\) with \(b = \lg \text{ratio}\)

Caveat: Cannot identify logarithmic factors with doubling hypothesis

doubling hypothesis

Q: How to estimate \(a\) (assuming we know \(b\))?
A: Run the program (for sufficiently large value of \(N\)) and solve for \(a\)

Hypothesis: Running time is about \(0.998 \times 10^{-10} \times N^3\) seconds

Note: this is almost identical hypothesis to one obtained via regression (\(1.006 \times 10^{-10} \times N^{2.999}\))

experimental algorithmics

system independent effects, determines exponent \(b\) in power law

• algorithm
• input data

system dependent effects, determines constant \(a\) in power law

• hardware: CPU, memory, cache, ...
• software: compiler, interpreter, garbage collector, ...
• system: operating system, network, other apps, ...

bad news: sometimes difficult to get precise measurements

good news: much easier and cheaper than other sciences

as an aside

Algorithmic experiments are virtually free by comparison with other sciences

Bottom line: no excuse for not running experiments to understand costs

Analysis of Algorithms

mathematical models

mathematical models for running time

total running time: sum of cost \(\times\) frequency for all operations

• need to analyze program to determine set of operations
• cost depends on machine, compiler
• frequency depends on algorithm, input data

In principle, accurate mathematical models are available.

example 1-sum

Q: How many instructions as a function of input size \(N\)?

int count = 0;
for(int i = 0; i < N; i++)
    if(a[i] == 0)               // <- N array accesses
        count++;

\(\dagger\) representative estimates (with some poetic license)

example: 2-Sum

Q: How many instructions as a function of input size \(N\)?

int count = 0;
for(int i = 0; i < N; i++)
    for(int j = i+1; j < N; j++)
        if(a[i] + a[j] == 0)
            count++;

Pf. Gauss

Note: \(0 + 1 + 2 + \ldots + (N-1) = \frac{1}{2} N (N-1) = \binom{N}{2}\)

example: 2-Sum

Q: How many instructions as a function of input size \(N\)?

int count = 0;
for(int i = 0; i < N; i++)
    for(int j = i+1; j < N; j++)
        if(a[i] + a[j] == 0)
            count++;

Timing is tedious to count exactly: \(\frac{1}{4} N^2 + \frac{13}{20} N + \frac{13}{10} \text{ns}\) to \(\frac{3}{10} N^2 + \frac{3}{5} N +\frac{13}{10} \text{ns}\)

simplifying the calculations

“

It is convenient to have a measure of the amount of work involved in a computing process, even though it be a very crude one. We may count up the number of times that various elementary operations are applied in the whole process and then given them various weights. We might, for instance, count the number of additions, subtractions, multiplications, divisions, recording of numbers, and extractions of figures from tables. In the case of computing with matrices most of the work consists of multiplications and writing down numbers, and we shall therefore only attempt to count the number of multiplications and recordings.
–Alan Turing (1947)

”

simplification 1: cost model

Cost model: use some basic operation as a proxy for running time

int count = 0;
for(int i = 0; i < N; i++)
    for(int j = i+1; j < N; j++)
        if(a[i] + a[j] == 0)
            count++;

cost model = array accesses, we assume compiler/JVM do not optimize any array accesses away

simplification 2: tilde notation

• estimate running time (or memory) as a function of input size \(N\)
• ignore lower order terms
  • when \(N\) is large, terms are negligible
  • when \(N\) is small, we don't care

simplification 2: tilde notation

\[N = 1000\] \[f(N) = 166.17 \text{million}\] \[g(N) = 166.67 \text{million}\]

Technical definition: \(f(N) \text{~} g(N)\) means

\[\lim_{N \rightarrow \infty} \frac{f(N)}{g(N)} = 1\]

simplification 2: tilde notation

• estimate running time (or memory) as a function of input size \(N\)
• ignore lower order terms
  • when \(N\) is large, terms are negligible
  • when \(N\) is small, we don't care

example: 2-Sum

Q: Approximately how many array accesses as a function of input size \(N\)?

int count = 0;
for(int i = 0; i < N; i++)
    for(int j = i+1; j < N; j++)
        if(a[i] + a[j] == 0)                // inner loop
            count++;

A: \(\text{~} N^2\) array accesses

Bottom line: use cost model and tilde notation to simplify counts

example: 3-sum

Q: Approximately how many array accesses as a function of input size \(N\)?

int count = 0;
for(int i = 0; i < N; i++)
    for(int j = i+1; j < N; j++)
        for(int k = j+1; k < N; k++)
            if(a[i] + a[j] + a[k] == 0)     // inner loop
                count++;

\[\binom{N}{3} = \frac{N(N-1)(N-2)}{3!} \text{~} \frac{1}{6}N^3\]

A: \(\text{ ~ } \frac{1}{2} N^3\) array accesses

Bottom line: use cost model and tilde notation to simplify counts

Estimating a discrete sum

Q: How to estimate a discrete sum?

A1: Take a discrete mathematics course (MAT 215)

Estimating a discrete sum

Q: How to estimate a discrete sum?

A2: Replace the sum with an integral and use calculus

Ex1: \(1 + 2 + \ldots + N\)

\[\sum_{i=1}^N i \text{ ~ } \int_{x=1}^N x\ dx \text{ ~ } \frac{1}{2}N^2\]

Estimating a discrete sum

Q: How to estimate a discrete sum?

A2: Replace the sum with an integral and use calculus

Ex2: \(1 + 1/2 + 1/3 + \ldots + 1/N\)

\[\sum_{i=1}^N \frac{1}{i} \text{ ~ } \int_{x=1}^N \frac{1}{x}\ dx \text{ ~ } \ln N\]

Estimating a discrete sum

Q: How to estimate a discrete sum?

A2: Replace the sum with an integral and use calculus

Ex3: 3-Sum triple loop

\[\sum_{i=1}^N \sum_{j=i}^N \sum_{k=j}^N 1 \text{ ~ } \int_{x=1}^N \int_{y=x}^N \int_{z=y}^N dz\ dy\ dx \text{ ~ } \frac{1}{6} N^3\]

Estimating a discrete sum

Q: How to estimate a discrete sum?

A2: Replace the sum with an integral and use calculus

Ex4: \(1 + 1/2 + 1/4 + 1/8 + \ldots\)

\[\int_{x=0}^\infty \left(\frac{1}{2}\right)^x\ dx = \frac{1}{\ln 2} \approx 1.4427\]

\[\sum_{i=0}^\infty \left(\frac{1}{2}\right)^i = 2\]

note: integral trick does not always work

Estimating a discrete sum

Q: How to estimate a discrete sum?

A3: Use Maple or Wolfram Alpha

mathematical models for running time

In principle, accurate mathematical models are available.

In practice,

• Formulas can be complicated
• Advanced mathematics might be required
• Exact models best left for experts

mathematical models for running time

\[\begin{array}{rcl} T_N & = & c_1 A + c_2 B + c_3 C + c_4 D + c_5 E \\ A & = & \text{array access} \\ B & = & \text{integer add} \\ C & = & \text{integer compare} \\ D & = & \text{increment} \\ E & = & \text{variable assignment} \\ A–E & = & \text{frequencies (depend on algorithm, input)} \\ c_i & = & \text{costs (depend on machine, compiler)} \end{array}\]

Bottom line: we use approximate models in this course

\[T(N) \text{ ~ } c N^3\]

analysis of algorithms

order-of-growth classifications

common order-of-growth classifications

Definition: If \(f(N) \text{ ~ } c g(N)\) for some constant \(c > 0\), then the order of growth of \(f(N)\) is \(g(N)\).

• ignores leading coefficients
• ignores lower-order terms

Ex: The order of growth of the running time of this code is \(N^3\)

int count = 0;
for(int i = 0; i < N; i++)
    for(int j = i+1; j < N; j++)
        for(int k = j+1; k < N; k++)
            if(a[i] + a[j] + a[k] == 0)
                count++;

Typical usage: mathematical analysis of running times, where leading coefficients depend on machine, compiler, JVM, ...

common order-of-growth classifications

Good news: the set of functions

\[1, \log N, N, N \log N, N^2, N^3, \text{and } 2^N\]

suffices to describe the order of growth of most common algorithms.

common order-of-growth classifications

binary search

Goal: given a sorted array and a key, find index of the key in the array.

Binary search: compare key against middle entry

• too small, go left
• too big, go right
• equal, found it!
• else, cannot find it!

//          0  1  2  3  4  5  6  7  8  9 10 11 12 13 14
int[] a = {11,13,14,25,33,43,51,53,64,72,84,93,95,96,97};

binary search

// find: 33
//          0  1  2  3  4  5  6  7  8  9 10 11 12 13 14
int[] a = {11,13,14,25,33,43,51,53,64,72,84,93,95,96,97};
//         |_                   ^^                   _|
//         lo                   mid                  hi
// a[mid] > key, so go left

binary search

// find: 33
//          0  1  2  3  4  5  6  7  8  9 10 11 12 13 14
int[] a = {11,13,14,25,33,43,51,53,64,72,84,93,95,96,97};
//         |_       ^^       _|
//         lo       mid      hi
// a[mid] < key, so go right

binary search

// find: 33
//          0  1  2  3  4  5  6  7  8  9 10 11 12 13 14
int[] a = {11,13,14,25,33,43,51,53,64,72,84,93,95,96,97};
//                     |_ ^^ _|
//                     lo md hi
// a[mid] > key, so go left

binary search

// find: 33
//          0  1  2  3  4  5  6  7  8  9 10 11 12 13 14
int[] a = {11,13,14,25,33,43,51,53,64,72,84,93,95,96,97};
//                     ^^
//                lo = mid = hi
// a[mid] == key, so found!

binary search

// find: 34
//          0  1  2  3  4  5  6  7  8  9 10 11 12 13 14
int[] a = {11,13,14,25,33,43,51,53,64,72,84,93,95,96,97};
//         |_                   ^^                   _|
//         lo                   mid                  hi
// a[mid] > key, so go left

binary search

// find: 34
//          0  1  2  3  4  5  6  7  8  9 10 11 12 13 14
int[] a = {11,13,14,25,33,43,51,53,64,72,84,93,95,96,97};
//         |_       ^^       _|
//         lo       mid      hi
// a[mid] < key, so go right

binary search

// find: 34
//          0  1  2  3  4  5  6  7  8  9 10 11 12 13 14
int[] a = {11,13,14,25,33,43,51,53,64,72,84,93,95,96,97};
//                     |_ ^^ _|
//                     lo md hi
// a[mid] > key, so go left

binary search

// find: 34
//          0  1  2  3  4  5  6  7  8  9 10 11 12 13 14
int[] a = {11,13,14,25,33,43,51,53,64,72,84,93,95,96,97};
//                     ^^
//                lo = mid = hi
// a[mid] != key, and lo == hi, so not found

binary search: implementation

Trivial to implement?

• First binary search published in 1946
• First bug-free one in 1962
• Bug in Java's Arrays.binarySearch() discovered in 2006

binary search: java implementation

Invariant: if key appears in array a[], then a[lo] <= key <= a[hi].

public static int binarySearch(int[] a, int key) {
    int lo = 0, hi = a.length - 1;
    while(lo <= hi) {
        int mid = lo + (hi - lo) / 2;       // why not mid=(lo+hi)/2?

        if     (key < a[mid]) hi = mid - 1; // |
        else if(key > a[mid]) lo = mid + 1; // | one "3-way compare"
        else return mid;                    // |
    }
    return -1;
}

binary search: mathematical analysis

Proposition: binary search uses at most \(1 + \lg N\) key compares to search a sorted array of size \(N\).

Def: \(T(N) =\) number key compares to binary search a sorted subarray of size \(\leq N\)

Binary search recurrence: \(T(N) \leq T(N/2) + 1\) for \(N>1\), with \(T(1) = 1\)

binary search: mathematical analysis

Proposition: binary search uses at most \(1 + \lg N\) key compares to search a sorted array of size \(N\).

Pf sketch (assume \(N\) is a power of \(2\))

\[\begin{array}{rcll} T(N) & \leq & T(N/2) + 1 & \text{given} \\ & \leq & T(N/4) + 1 + 1 & \text{apply recurrence to 1st term} \\ & \leq & T(N/8) + 1 + 1 + 1 & \text{apply recurrence to 1st term} \\ & \vdots & \vdots & \vdots \\ & \leq & T(N/N) + \underbrace{1 + \ldots + 1}_{\lg N} & \text{stop applying,} T(1)=1 \\ & = & 1 + \lg N & \end{array}\]

the 3-sum problem

3-Sum: Given \(N\) distinct integer, find three such that \(a + b + c = 0\)

Ver0: \(N^3\) time, \(N\) space
Ver1: \(N^2 \log N\) time, \(N\) space
Ver2: \(N^2\) time, \(N\) space

Note: For full credit, running time should be worst case

Comparing programs

Hypothesis: the sorting-based \(N^2 \log N\) algorithm for 3-Sum is significantly faster in practice than the brute-force \(N^3\) algorithm.

Guiding principle: Typically, better order of growth \(\Rightarrow\) faster in practice

Analysis of Algorithms

Memory

basics

Bit: 0 or 1

Byte: \(8\) bits

Megabyte (MB): \(1\) million (NIST) or \(2^{20}\) bytes (most CS)

Gigabyte (GB): \(1\) billion (NIST) or \(2^{30}\) bytes (most CS)

64-bit machine: We assume a 64-bit machine with 8-byte pointers

(some JVMs "compress" ordinary object pointers to 4 bytes to avoid this cost)

typical memory usage

typical memory usage for primitive types and one- and two-dimensional arrays

typical memory for Java objects

Object overhead: \(16\) bytes

Reference: \(8\) bytes

Padding: Each object uses a multiple of \(8\) bytes

Ex1: A Date object uses \(32\) bytes of memory

public class Date { private int day; private int month; private int year; // ... }

typical memory usage summary

Total memory usage for a data type value:

• Primitive type: \(4\) bytes for int, \(8\) bytes for double, ...
• Object reference: \(8\) bytes
• Array: \(24\) bytes + memory for each array entry
• Object: \(16\) bytes + memory for each instance variable +\(8\) extra bytes per inner class object (for reference to enclosing class)
• Padding: round up to multiple of \(8\) bytes

Note: depending on application, we may want to count memory for any referenced objects (recursively)

Analysis of algorithms quiz

How much memory does a WeightedQuickUnionUF use as a function of \(N\)?

public class WeightedQuickUnionUF {
    private int[] parent;
    private int[] size;
    private int count;

    public WeightedQuickUnionUF(int N)
    {
        parent = new int[N];
        size = new int[N];
        count = 0;
        // ...
    }

    // ...
}

turning the crank: summary

Empirical analysis

• Execute program to perform experiments
• Assume power law
• Formulate a hypothesis for running time
• Model enables us to make predictions

Mathematical analysis

Scientific method

turning the crank: summary

Empirical analysis

Mathematical analysis

• Analyze algorithm to count frequency of operations
• Use tilde notation to simplify analysis
• Model enables us to explain behavior

Scientific method

turning the crank: summary

Empirical analysis

Mathematical analysis

Scientific method

• Mathematical model is independent of a particular system; applies to machine not yet built
• Empirical analysis is necessary to validate mathematical models and to make predictions