# Computer Science 212 Title: Data Structures and Algorithms

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Computer Science 212 Title: Data Structures and Algorithms
Instructor: Harry Plantinga Text: Algorithm Design: Foundations, Analysis, and Internet Examples by Goodrich and Tamassia

Computer Science 212 Data Structures and Algorithms
The Heart of Computer Science Data structures Study of important algorithms Algorithm analysis Algorithm design techniques Plus Intelligent systems Course Information: Grades on moodle

Why study DS & Algorithms?
Some problems are difficult to solve and good solutions are known Some “solutions” don’t always work Some simple algorithms don’t scale well Data structures and algorithms make good tools for addressing new problems Fun! Joy! Esthetic beauty! Example: successor in a BST? Algorithm example: celebrity problem Algorithms that don’t scale: e.g. using bubble sort in virtual trackball program, Microsoft Word on MB+ files Example of bad “solutions” – making change when you run out of nickels (derive dynamic programming solution) A better algorithm is much to be preferred over a faster computer Story of how my old minister always used to sing the doxology…

Place in the curriculum
The study of interesting areas of computer science: upper-level electives 108, 112: basic tools (like algebra to mathematics) 212: More advanced tools. Introduction to the science of computing. (A little more mathematical than other core CS courses, but not as rigorous as a real math course.) New way of thinking to many of you. But you’ll get used to it.

Example: Search and Replace
Goal: replace one string with another Which version is better? #!/usr/bin/perl my \$a = shift; my \$b = shift; my \$input = ""; while (<STDIN>) { \$input .= \$_; } while (\$input =~ m/\$a/) { \$input =~ s/\$a/\$b/; #!/usr/bin/perl my \$a = shift; my \$b = shift; my \$input = ""; while (<STDIN>) { \$input .= \$_; } \$input =~ s/\$a/\$b/g; Graph data in Excel. Find the curves; fit the functions. Acolyte: public/examples

Empirical Analysis Implement Gather runtime data for various inputs
Analyze runtime vs. size Example: suppose we test swap1 Input sizes: 1, 2, 4, 8, 16, 32, 64, 128, 256 Runtimes: 70, 110, 250, 560, 810, 1750, 6510, 25680, Get data – what now?

swap1 runtime analysis size Swap1 (ms) 1 70 2 110 4 250 8 560 16 810
32 1750 64 6510 128 25680 256 102050 see online: acolyte:public/212/examples/swap/swap1

swap2 runtime analysis size swap2 4 70 8 50 16 32 80 64 100 128 120
256 230 512 380 1024 580 2048 910 4096 1680 8192 3130 16384 6160

Can I solve the "bad algorithm" problem by just getting a faster computer?

How do we find the runtime from the fitted straight line?
Y = x^3 log(y) = 3 log(x) Slope is exponent

Note that slope gives growth rate of runtime Note that first few points are inflated by startup costs

Empirical runtime analysis
What can we say about the runtimes of these two algorithms? Could give runtime as a function of input size, for a particular implementation But would that apply on different hardware? What about a different implementation in C++ instead of Perl? Is there anything we can say that is true in general? Is version 2 better than version 1 in general? Why? Have we done science? I.e. have we proven that version 1is better than version 2, in general, i.e. independent of implementation details?

Empirical analysis: conclusions
Empirical analyses are implementation-dependent You have to use representative sample data You can learn something about the growth rate from the slope or curve of the runtime graph Bad algorithms can’t be fixed with faster computers Big companies (e.g. Microsoft) are not immune to bad algorithms

What’s the runtime? for i  0 to n  1 do for j  0 to n  1 do
if A[i]  A[j] then A[j]  A[i] What's the runtime? How do you know? What is n? What does O(n^2) even mean?

Theoretical Analysis Uses a high-level description of the algorithm instead of an implementation Characterizes running time as a function of the input size, n. Takes into account all possible inputs, often analyzing the worst case Allows us to evaluate the speed of an algorithm independent of the hardware/software environment

Pseudocode (§1.1) Example: find max element of an array
Algorithm arrayMax(A, n) Input array A of n integers Output maximum element of A currentMax  A[0] for i  1 to n  1 do if A[i]  currentMax then currentMax  A[i] return currentMax Example: find max element of an array High-level description of an algorithm More structured than English prose Less detailed than a program Preferred notation for describing algorithms Hides program design issues

Pseudocode Details Control flow Method declaration Method call
if … then … [else …] while … do … repeat … until … for … do … Indentation replaces braces Method declaration Algorithm method (arg [, arg…]) Input … Output … Method call var.method (arg [, arg…]) Return value return expression Expressions Assignment (like  in Java) Equality testing (like  in Java) n2 Superscripts and other mathematical formatting allowed

Questions… Can a program be asymptotically faster on one type of CPU vs another? Do all CPU instructions take equally long?

The Random Access Machine (RAM) Model
A CPU An potentially unbounded bank of memory cells, each of which can hold an arbitrary number or character 1 2 What's the purpose of the RAM model? Memory cells are numbered and accessing any cell in memory takes unit time.

Primitive Operations Basic computations performed by an algorithm
Examples: Evaluating an expression Assigning a value to a variable Indexing into an array Calling a method Returning from a method Basic computations performed by an algorithm Identifiable in pseudocode Largely independent from any programming language Exact definition not important (constant number of machine cycles per statement) Assumed to take a constant amount of time in the RAM model

Counting Primitive Operations (§1.1)
By inspecting the pseudocode, we can determine the maximum number of primitive operations executed by an algorithm, as a function of the input size Algorithm arrayMax(A, n) currentMax  A[0] # operations for i  1 to n  1 do n if A[i]  currentMax then 2(n  1) currentMax  A[i] 2(n  1) { increment counter i } 2(n  1) return currentMax Total 7n  1

Estimating Running Time
Algorithm arrayMax executes 7n  1 primitive operations in the worst case. Define: a = Time taken by the fastest primitive operation b = Time taken by the slowest primitive operation Let T(n) be worst-case time of arrayMax. Then a (7n  1)  T(n)  b (7n  1) Hence, the running time T(n) is bounded by two linear functions

Theoretical analysis: conclusion
We can count the number of RAM-equivalent statements executed as a function of input size All remaining implementation dependencies amount to multiplicative constants, which we will ignore (But watch out for statements that take more than constant time, such as \$input =~ s/\$a/\$b/g) Linear, quadratic, etc. runtime is an intrinsic property of an algorithm

What’s the runtime? int n; cin >> n; for (int i=0; i<n; i++)
for (int j=0; j<n; j++) for (int k=0; k<n; k++) { cout << "Hello, "; cout << "greetings, "; cout << "bonjour, "; cout << "guten tag, "; cout << "Здравствуйте, "; cout << "你好 "; cout << " world!"; } You've seen big-O notation. What is the runtime of this function in terms of n? Does the constant matter? Why?

What’s the runtime? int n; cin >> n; if (n<1000)
for (int i=0; i<n; i++) for (int j=0; j<n; j++) for (int k=0; k<n; k++) cout << "Hello\n"; else cout << "world!\n"; What's the runtime here? Isn't n^3 a constant when n is <1000?

Function Growth Rates Growth rates of functions:
Linear  n Quadratic  n2 Cubic  n3 In a log-log chart, the slope of the line corresponds to the growth rate of the function

Constant factors The growth rate is not affected by Examples
constant factors or lower-order terms Examples 102n is a linear function 105n n is a quadratic function

Asymptotic (big-O) Notation (§1.2)
Given functions f(n) and g(n), we say that f(n) is O(g(n)) if there are positive constants c and n0 such that f(n)  cg(n) for n  n0 Example: 2n + 10 is O(n) 2n + 10  cn (c  2) n  10 n  10/(c  2) Pick c = 3 and n0 = 10

Example Example: the function n2 is not O(n) n2  cn n  c
The above inequality cannot be satisfied since c must be a constant

More Big-O Examples 3n3 + 20n2 + 5 3 log n + log log n 7n-2
7n-2 is O(n) need c > 0 and n0  1 such that 7n-2  c•n for n  n0 this is true for c = 7 and n0 = 1 3n3 + 20n2 + 5 3n3 + 20n2 + 5 is O(n3) need c > 0 and n0  1 such that 3n3 + 20n2 + 5  c•n3 for n  n0 this is true for c = 4 and n0 = 21 3 log n + log log n 3 log n + log log n is O(log n) need c > 0 and n0  1 such that 3 log n + log log n  c•log n for n  n0 this is true for c = 4 and n0 = 2

Asymptotic analysis of functions
Asymptotic analysis is equivalent to ignoring multiplicative constants ignoring lower-order terms “for large enough inputs” Big-O and growth rate Big-O gives an upper bound on the growth rate of a function Think of it as <= [asymptotically speaking]

Big-O Rules If is f(n) a polynomial of degree d, then f(n) is O(nd), i.e., Drop lower-order terms Drop constant factors Use the smallest possible class of functions, if possible Say “2n is O(n)” instead of “2n is O(n2)” (The former is a stronger statement) Use the simplest expression of the class Say “3n + 5 is O(n)” instead of “3n + 5 is O(3n)”

Asymptotic Algorithm Analysis
Asymptotic analysis: determine the runtime in big-O notation To perform the asymptotic analysis Find the worst-case number of primitive operations executed as a function of the input size Express this function with big-O notation Example: We determine that algorithm arrayMax executes at most 7n  1 primitive operations We say that algorithm arrayMax “runs in O(n) time” Since constant factors and lower-order terms are eventually dropped anyway, we can disregard them when counting primitive operations

Computing Prefix Averages
Runtime analysis example: Two algorithms for prefix averages The i-th prefix average of an array X is average of the first (i + 1) elements of X: A[i] = (X[0] + X[1] + … + X[i])/(i+1) Computing the array A of prefix averages of another array X has applications to financial analysis

The following algorithm computes prefix averages in quadratic time by applying the definition Algorithm prefixAverages1(X, n) Input array X of n integers Output array A of prefix averages of X #operations A  new array of n integers n for i  0 to n  1 do n s  X[0] n for j  1 to i do …+ (n  1) s  s + X[j] …+ (n  1) A[i]  s / (i + 1) n return A There's that gauss sum. You know the story about Gauss being asked in elementary school to add … + 100?

Prefix Averages (Linear)
The following algorithm computes prefix averages in linear time by keeping a running sum Algorithm prefixAverages2(X, n) Input array X of n integers Output array A of prefix averages of X #operations A  new array of n integers n s  for i  0 to n  1 do n s  s + X[i] n A[i]  s / (i + 1) n return A Algorithm prefixAverages2 runs in O(n) time

Arithmetic Progression
The running time of prefixAverages1 is O( …+ n) The sum of the first n integers is n(n + 1) / 2 There is a simple visual proof of this fact Thus, algorithm prefixAverages1 runs in O(n2) time

What’s the runtime? 2n3+n2+n+2? O(n3) runtime
int n; cin >> n; for (int i=0; i<n; i++) for (int j=0; j<n; j++) for (int k=0; k<n; k++) cout << “Hello world!\n”; 2n3+n2+n+2? O(n3) runtime Memory leak What if the last line is replaced by: string *s=new string(“Hello world!\n”); O(n3) time and space

What’s the runtime? O(n3) + O(n3) = O(n3)
int n; cin >> n; for (int i=0; i<n; i++) for (int j=0; j<n; j++) for (int k=0; k<n; k++) cout << “Hello world!\n”; O(n3) + O(n3) = O(n3) Statements or blocks in sequence: add

What’s the runtime? Loops: add up cost of each iteration
int n; cin >> n; for (int i=0; i<n; i++) for (int j=n; j>1; j/=2) cout << “Hello world!\n”; Loops: add up cost of each iteration (multiply loop cost by number of iterations if they all take the same time) log n iterations of n steps  O(n log n)

What’s the runtime? Loops: add up cost of each iteration
int n; cin >> n; for (int i=1; i<=n; i++) for (int j=1; j<=i; j++) cout << “Hello world!\n”; Loops: add up cost of each iteration … + n = n(n+1)/2 = Q(n2)

What’s the runtime? template <class Item>
void insert(Item a[], int l, int r) { int i; for (i=r; i>l; i--) compexch(a[i-1],a[i]); for (i=l+2; i<=r; i++) { int j=i; Item v=a[i]; while (v<a[j-1]) { a[j] = a[j-1]; j--; } a[j] = v; }

Math you need to Review Summations (Sec. 1.3.1)
Logarithms and Exponents (Sec ) Proof techniques (Sec ) Basic probability (Sec ) properties of logarithms: logb(xy) = logbx + logby logb (x/y) = logbx - logby Logb xa = a logb x logba = logxa/logxb properties of exponentials: a(b+c) = aba c abc = (ab)c ab /ac = a(b-c) b = a logab bc = a c*logab

Relatives of Big-Oh big-Omega
f(n) is (g(n)) if there is a constant c > 0 and an integer constant n0  1 such that f(n)  c•g(n) for n  n0 big-Theta f(n) is (g(n)) if there are constants c’ > 0 and c’’ > 0 and an integer constant n0  1 such that c’•g(n)  f(n)  c’’•g(n) for n  n0 little-o f(n) is o(g(n)) if, for any constant c > 0, there is an integer constant n0  0 such that f(n)  c•g(n) for n  n0 little-omega f(n) is (g(n)) if, for any constant c > 0, there is an integer constant n0  0 such that f(n)  c•g(n) for n  n0

Intuition for Asymptotic Notation
Big-Oh f(n) is O(g(n)) if f(n) is asymptotically less than or equal to g(n) big-Omega f(n) is (g(n)) if f(n) is asymptotically greater than or equal to g(n) big-Theta f(n) is (g(n)) if f(n) is asymptotically equal to g(n) little-oh f(n) is o(g(n)) if f(n) is asymptotically strictly less than g(n) little-omega f(n) is (g(n)) if is asymptotically strictly greater than g(n)

Example Uses of the Relatives of Big-Oh
5n2 is (n2) f(n) is (g(n)) if there is a constant c > 0 and an integer constant n0  1 such that f(n)  c•g(n) for n  n0 let c = 5 and n0 = 1 5n2 is (n) f(n) is (g(n)) if there is a constant c > 0 and an integer constant n0  1 such that f(n)  c•g(n) for n  n0 let c = 1 and n0 = 1 5n2 is (n) f(n) is (g(n)) if, for any constant c > 0, there is an integer constant n0  0 such that f(n)  c•g(n) for n  n0 need 5n02  c•n0  given c, the n0 that satisfies this is n0  c/5  0

Asymptotic Analysis: Review
What does it mean to say that an algorithm has runtime O(n log n)? n: Problem size Big-O: upper bound over all inputs of size n “Ignore constant factor” (why?) “as n grows large” O: like <= for functions (asymptotically speaking) W: like >= Q: like =

Asymptotic notation: examples
Asymptotic runtime, in terms of O, W, Q? Suppose the runtime for a function is n2 + 2n log n + 40 n n1.999 n3 + n2 log n n n2 log n 2n+ 100 n2 n+ 100 n97

Asymptotic comparisons
n2 = O( n1.999 )? n = O(n log n)? 1.0001n = O(n943)? lg n = Q(ln n)? (Evaluate the limit of the quotient of the functions) No – a polynomial with a higher power dominates one with a lower power No – all polynomials (n ) dominate any polylog (log n) No – all exponentials dominate any polynomial Yes – different bases are just a constant factor difference

Estimate the runtime Suppose an algorithm has runtime Q(n3)
suppose solving a problem of size 1000 takes 10 seconds. How long to solve a problem of size 10000? Suppose an algorithm has runtime Q(n log n) runtime 10-8 n3; if n=10000, runtime 10000s = 2.7hr runtime 10-3 n lg n; if n=10000, runtime 133 secs

Worst vs. average case You might be interested in worst, best, or average case analysis of an algorithm You can have upper, lower, or tight bounds on each of those functions. Eg. For each n, some problem instances of size n have runtime n and some have runtime n2. Worst case: Best case: Average case: Q(n2), W(n), W(log n), O(n2), O(n3) W(n), W(log n), O(n2), Q(n) W(n), W(log n), O(n2), O(n3) Average case: need to know distribution of inputs

Problem: Extensible Arrays
A data structure for a stack that is… As fast and easy as an array (usually) Not limited in size How? Extensible Array: Usually works like a normal array When it fills, copy the entire contents into a new array of twice the size. Runtime?

Analysis of Extensible Arrays
Worst-case runtime for a push() operation? Is that the most useful way of describing the result? Amortized analysis: averaging out the cost over many operations. Amortized runtime of a push() operation? Total runtime of m push operations: m steps when array doesn’t grow Worst-case cost of grow operations: …+m < 2m Total: O(m) time for m operations. O(1) amortized runtime per operation

Theoretical and empirical analysis
Why do we ignore constants in analyzing algorithms theoretically? Wouldn’t it be better if we knew the actual constants? Theoretical and empirical analysis gives us Theoretical knowledge of asymptotic runtime Empirical knowledge of actual constants for some implementation Best of both worlds!

Analyze this… function fun (int n) if (n<1000)
for (i=1; i<=n; i++) for (j=1; j<=n; j++) for (k=1; k<=n; k++) cout << “Hello world!\n”; else { j=i; while (j >= 1) j = j / 2; if (n%2) } cout << “Have a nice day.\n”;

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