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**Analysis of Algorithms**

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**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

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**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

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**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

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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)

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**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

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**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

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**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

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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

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**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!

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**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

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**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

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**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

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**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

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**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!

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**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

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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)

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**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

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**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

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

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

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**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

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**Simple Sorting Algorithms**

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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

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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

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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; } } } }

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**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)

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**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

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**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

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**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

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**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] } }

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**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]

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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

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**One Step of Insertion Sort**

3 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

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**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)

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**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

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