1. Analysis of Algorithms

2. Time and Space to analyze an algorithm means:
developing a formula for predicting how fast an algorithm is, based on the size of the input
time complexity
and/or
developing a formula for predicting how much memory an algorithm requires, based on the size of the input
space complexity
usually time is our biggest concern
most algorithms require a fixed amount of space
These slides are based on Dr. David Matuszek’s lectures

3. What Does “Size of the Input” Mean?
if we are searching an array, the “size” of the input could be the size of the array
if we are merging two arrays, the “size” could be the sum of the two array sizes
if we are computing the nth Fibonacci number, or the nth factorial, the “size” is n
we choose the “size” to be a parameter that determines the actual time (or space) required
it is usually obvious what this parameter is
sometimes we need two or more parameters

4. Characteristic Operations
in computing time complexity, one good approach is to count characteristic operations
what a “characteristic operation” is depends on the particular problem
if searching, it might be comparing two values
if sorting an array, it might be:
comparing two values
swapping the contents of two array locations
both of the above
sometimes we just look at how many times the innermost loop is executed

5. Exact Values
it is sometimes possible, in assembly language, to compute exact time and space requirements
we know exactly how many bytes and how many cycles each machine instruction takes
for a problem with a known sequence of steps (factorial, Fibonacci), we can determine how many instructions of each type are required
however, often the exact sequence of steps cannot be known in advance
the steps required to sort an array depend on the actual numbers in the array (which we do not know in advance)

6. Higher-Level Languages
in a higher-level language (such as Java), we do not know how long each operation takes
which is faster, x < 10 or x <= 9 ?
we don’t know exactly what the compiler does with this
the compiler almost certainly optimizes the test anyway (replacing the slower version with the faster one)
in a higher-level language we cannot do an exact analysis
our timing analyses will use major oversimplifications
nevertheless, we can get some very useful results

7. Average, Best, and Worst Cases
usually we would like to find the average time to perform an algorithm
however,
sometimes the “average” isn’t well defined
example: Sorting an “average” array
time typically depends on how out of order the array is
how out of order is the “average” unsorted array?
sometimes finding the average is too difficult
often we have to be satisfied with finding the worst (longest) time required
sometimes this is even what we want (say, for time-critical operations)
the best (fastest) case is seldom of interest

8. Constant Time it does not include a loop
constant time means there is some constant k such that this operation always takes k nanoseconds
a Java statement takes constant time if:
it does not include a loop
it does not include calling a method whose time is unknown or is not a constant
if a statement involves a choice (if or switch) among operations, each of which takes constant time, we consider the statement to take constant time
this is consistent with worst-case analysis

9. Linear Time
We may not be able to predict to the nanosecond how long a Java program will take, but do know some things about timing:
for (i = 0, j = 1; i < n; i++) { j = j * i; }
This loop takes time k*n + c, for some constants k and c
k : How long it takes to go through the loop once (the time for j = j * i, plus loop overhead)
n : The number of times through the loop (we can use this as the “size” of the problem)
c : The time it takes to initialize the loop
the total time k*n + c is linear in n

10. Constant Time vs. Linear Time
constant time is (usually) better than linear time
suppose we have two algorithms to solve a task:
algorithm A takes 5000 time units
algorithm B takes 100*n time units
which is better?
clearly, algorithm B is better if our problem size is small, that is, if n < 50
algorithm A is better for larger problems, with n > 50
So B is better on small problems that are quick anyway
but A is better for large problems, where it matters more
we usually care most about very large problems
but not always!

11. The Array subset Problem
Suppose you have two sets, represented as unsorted arrays:
int[] sub = { 7, 1, 3, 2, 5 }; int[] super = { 8, 4, 7, 1, 2, 3, 9 };
and you want to test whether every element of the first set (sub) also occurs in the second set (super):
System.out.println(subset(sub, super));
(the answer in this case should be false, because sub contains the integer 5, and super does not)
we are going to write method subset and compute its time complexity (how fast it is)
let’s start with a helper function, member, to test whether one number is in an array

12. member if x is not in a, the loop executes n times, where n = a.length
static boolean member(int x, int[] a) { int n = a.length; for (int i = 0; i < n; i++) { if (x == a[i]) return true; } return false; }
if x is not in a, the loop executes n times, where n = a.length
this is the worst case
if x is in a, the loop executes n/2 times on average
either way, linear time is required: k*n+c

13. subset the loop (and the call to member) will execute:
static boolean subset(int[] sub, int[] super) { int m = sub.length; for (int i = 0; i < m; i++) if (!member(sub[i], super) return false; return true; }
the loop (and the call to member) will execute:
m = sub.length times, if sub is a subset of super
this is the worst case, and the one we’re most interested in
fewer than sub.length times (but we don’t know how many)
we would need to figure this out in order to compute average time complexity
the worst case is a linear number of times through the loop
but the loop body doesn’t take constant time, since it calls member, which takes linear time

14. Analysis of Array subset Algorithm
We have seen that the loop in subset executes m = sub.length times (in the worst case)
also, the loop in subset calls member, which executes in time linear in n = super.length
hence, the execution time of the array subset method is m*n, along with assorted constants
we go through the loop in subset m times, calling member each time
we go through the loop in member n times
If m and n are similar, this is roughly quadratic, i.e., n2

15. What About the Constants?
an added constant, f(n)+ c, becomes less and less important as n gets larger
a constant multiplier, k*f(n), does not get less important, but...
improving k gives a linear speedup (cutting k in half cuts the time required in half)
improving k is usually accomplished by careful code optimization, not by better algorithms
we aren’t that concerned with only linear speedups!
bottom line – forget the constants!

16. Simplifying the Formulae
throwing out the constants is one of two things we do in analysis of algorithms
by throwing out constants, we simplify 12n to just n2
our timing formula is a polynomial, and may have terms of various orders (constant, linear, quadratic, cubic, etc.)
we usually discard all but the highest-order term
we simplify n2 + 3n + 5 to just n2

17. Big O Notation
when we have a polynomial that describes the time requirements of an algorithm, we simplify it by:
throwing out all but the highest-order term
throwing out all the constants
if an algorithm takes 12n3+4n2+8n+35 time, we simplify this formula to just n3
we say the algorithm requires O(n3) time
we call this Big O notation
(more accurately, it’s Big , but we’ll leave that for Turpin’s class)

18. Big O for subset Algorithm
recall that, if n is the size of the set, and m is the size of the (possible) subset:
we go through the loop in subset m times, calling member each time
we go through the loop in member n times
hence, the actual running time should be k*(m*n) + c, for some constants k and c
we say that subset takes O(m*n) time

19. Can We Justify Big O Notation?
big O notation is a huge simplification; can we justify it?
it only makes sense for large problem sizes
for sufficiently large problem sizes, the highest-order term swamps all the rest!
consider R = x2 + 3x + 5 as x varies:
x = 0 x2 = 0 3x = 0 5 = 5 R = 5
x = 10 x2 = 100 3x = 30 5 = 5 R = 135
x = 100 x2 = x = = 5 R = 10,305
x = 1000 x2 = x = = 5 R = 1,003,005
x = 10,000 x2 = 108 3x = 3*104 5 = 5 R = 100,030,005
x = 100,000 x2 = x = 3* = 5 R = 10,000,300,005

20. y = x2 + 3x + 5, for x=1..10

21. y = x2 + 3x + 5, for x=1..20

22. Common Time Complexities
BETTER
WORSE
O(1) constant time
O(log n) log time
O(n) linear time
O(n log n) log linear time
O(n2) quadratic time
O(n3) cubic time
O(2n) exponential time

23. Simple Sorting Algorithms

24. Bubble Sort
compare each element (except the last one) with its neighbor to the right
if they are out of order, swap them
this puts the largest element at the very end
the last element is now in the correct and final place
compare each element (except the last two) with its neighbor to the right
this puts the second largest element next to last
the last two elements are now in their correct and final places
compare each element (except the last three) with its neighbor to the right
continue as above until you have no unsorted elements on the left

25. Example of Bubble Sort 7 2 8 5 4 2 7 5 4 8 2 5 4 7 8 2 4 5 7 8 2 7 8 5 4 2 7 5 4 8 2 5 4 7 8 2 4 5 7 8 2 7 8 5 4 2 5 7 4 8 2 4 5 7 8
(done) 2 7 5 8 4 2 5 4 7 8 2 7 5 4 8

26. Code for Bubble Sort
public static void bubbleSort(int[] a) { int outer, inner; for (outer = a.length - 1; outer > 0; outer--) { // counting down for (inner = 0; inner < outer; inner++) { // bubbling up if (a[inner] > a[inner + 1]) { // if out of order... int temp = a[inner]; // ...then swap a[inner] = a[inner + 1]; a[inner + 1] = temp; } } } }

27. Analysis of Bubble Sort
for (outer = a.length - 1; outer > 0; outer--) { for (inner = 0; inner < outer; inner++) { if (a[inner] > a[inner + 1]) { // code for swap omitted } } }
let n = a.length = size of the array
the outer loop is executed n-1 times (call it n, that’s close enough)
each time the outer loop is executed, the inner loop is executed
inner loop executes n-1 times at first, linearly dropping to just once
on average, inner loop executes about n/2 times for each execution of the outer loop
in the inner loop, the comparison is always done (constant time), the swap might be done (also constant time)
result is n * n/2 * k, that is, O(n2/2 + k) = O(n2)

28. for all j > outer, if i < j, then a[i] <= a[j]
Loop Invariants
you run a loop in order to change things
oddly enough, what is usually most important in understanding a loop is finding an invariant:
that is, a condition that doesn’t change
in bubble sort, we put the largest elements at the end, and once we put them there, we don’t move them again
the variable outer starts at the last index in the array and decreases to 0
our invariant is: every element to the right of outer is in the correct place
for all j > outer, if i < j, then a[i] <= a[j]
when this is combined with the loop exit test, outer == 0, we know that all elements of the array are in the correct place

29. Selection Sort given an array of length n,
search elements 0 through n-1 and select the smallest
swap it with the element in location 0
search elements 1 through n-1 and select the smallest
swap it with the element in location 1
search elements 2 through n-1 and select the smallest
swap it with the element in location 2
search elements 3 through n-1 and select the smallest
swap it with the element in location 3
continue in this fashion until there’s nothing left to search

30. Example and Analysis of Selection Sort
selection sort might swap an array element with itself – this is harmless, and not worth checking for
analysis:
the outer loop executes n-1 times
the inner loop executes about n/2 times on average (from n to 2 times)
work done in the inner loop is constant (swap two array elements)
time required is roughly (n-1)*(n/2)
you should recognize this as O(n2) 7 2 8 5 4 2 7 8 5 4 2 4 8 5 7 2 4 5 8 7 2 4 5 7 8

31. Code for Selection Sort
public static void selectionSort(int[] a) { int outer, inner, min; for (outer = 0; outer < a.length - 1; outer++) { min = outer; for (inner = outer + 1; inner < a.length; inner++) { if (a[inner] < a[min]) { min = inner; } // Invariant: for all i, // if outer <= i <= inner, then a[min] <= a[i] } // a[min] is least among a[outer]..a[a.length - 1] int temp = a[outer]; a[outer] = a[min]; a[min] = temp; // Invariant: for all i <= outer, // if i< j then a[i] <= a[j] } }

32. Invariants for Selection Sort
for the inner loop:
this loop searches through the array, incrementing inner from its initial value of outer+1 up to a.length-1
as the loop proceeds, min is set to the index of the smallest number found so far
our invariant is:
for all i such that outer <= i <= inner, a[min] <= a[i]
for the outer (enclosing) loop:
the loop counts up from outer = 0
each time through the loop, the minimum remaining value is put in a[outer]
our invariant is: for all i <= outer, if i < j then a[i] <= a[j]

33. Insertion Sort
the outer loop of insertion sort is: for (outer = 1; outer < a.length; outer++) {...}
the invariant is that all the elements to the left of outer are sorted with respect to one another
for all i < outer, j < outer, if i < j then a[i] <= a[j]
this does not mean they are all in their final correct place; the remaining array elements may need to be inserted
when we increase outer, a[outer-1] becomes to its left; we must keep the invariant true by inserting a[outer-1] into its proper place
this means:
finding the element’s proper place
making room for the inserted element (by shifting over other elements)
inserting the element

34. One Step of Insertion Sort3 4 7 12 14 20 21 33 38 10 55 9 23 28 16
sorted
next to be inserted 10
temp
less than 10 12 14 14 20 21 33 38 10 3 4 7 55 9 23 28 16
sorted

35. Analysis of Insertion Sort
we run once through the outer loop, inserting each of n elements; this is a factor of n
on average, there are n/2 elements already sorted
the inner loop looks at (and moves) half of these
this gives a second factor of n/4
hence, the time required for an insertion sort of an array of n elements is proportional to n2/4
discarding constants, we find that insertion sort is O(n2)

36. Summary bubble sort, selection sort, and insertion sort are all O(n2)
we can do much better than this with somewhat more complicated sorting algorithms (wait for Jeff’s class)
within O(n2),
bubble sort is very slow, and should probably never be used for anything
selection sort is intermediate in speed
insertion sort is usually the fastest of the three – in fact, for small arrays (say, 10 or 15 elements), insertion sort is faster than more complicated sorting algorithms
selection sort and insertion sort are good enough for small arrays