Announcements P2 is due tomorrow Prelim on Monday Review Session on Friday in class Quiz today CS100 Lecture 11

Today’s Topics Review Search Algorithms Linear Search Binary Search Lecture 11

Review of Arrays How to declare an array? What do we use arrays for? What is the subscript in g[3*x] ? Initializer lists Syntax choices CS100 Lecture 11

The Search Problem Linear Search: Find (the value of) x in array b. This description is too vague, for several reasons: How are we to indicate that x appears in b? What if it appears more than once in b? What is to be done if x is not in b? In general, when approaching a new programming job, the first task is to make the specification of the job as precise as possible. CS100 Lecture 11

More specifically. . . One way to specify the task: Write a method that returns the index of the first element of b that equals x. If x does not occur in b, return b.length. Example: b = {1, 2, 8, 5, 2, 9} (so b.length = 6) If x is 1, return 0 If x is 2, return 1 If x is 5, return 3 if x is 3, return 6 CS100 Lecture 11

How do we construct the algorithm? // return value i that satisfies: 0<=i<=b.length, // x not in b[0..i-1], // (i = b.length or x=b[i]) public static int linearSearch (int [ ] b, int x) Need a loop. It will compare elements of b with x, beginning with b[0], b[1], … Invariant: 0<=i<=b.length and x not in b[0..i-1] x is not here 0 i b.length b CS100 Lecture 11

One linear search solution // return the value i that satisfies: 0<=i<=b.length, // x not in b[0..i-1], and (i = b.length or x=b[i]) public static int linearSearch (int [ ] b, int x) { int i= 0; // Inv: 0<=i<=b.length and x not in b[0..i-1], while (i < b.length && x != b[i]) i= i+1; return i; } CS100 Lecture 11

A second linear search solution int i= 0; // Inv: 0<=i<=b.length and x not in b[0..i-1], while (i < b.length) { if (x == b[i]) return i; i= i+1; } return i; CS100 Lecture 11

Efficiency What does it mean for an algorithm to be efficient? One way is to think about much compute-time is being spent. How many comparisons are made, for example? There is a whole field of computer science that looks at analyzing the cost of algorithms We’ll just touch on it briefly. CS100 Lecture 11

Cost of Linear Search Worst case: We have to make b.length comparisons Can we do better than this? Suppose we’re looking for someone’s name in a NYC phonebook -- would you use linear search? CS100 Lecture 11

Binary Search Assume that the array b is sorted in ascending order Compare the item your interested in to the center of the list If it’s greater, repeat search on one half, if smaller, repeat on the other half -- and continue. . . Each comparison thus eliminates half of the remaining possibilities CS100 Lecture 11

Simple idea, but . . . Method dates back to 1946 -- it wasn’t until 1962 that a bug-free version appeared! One study (albeit in ‘83) showed that 90% of programmers fail to code up binary search correctly. So, we’ll play with some formal algorithmic stuff for awhile before looking at Java CS100 Lecture 11

Algorithm Development Specification. Store an integer in k to truthify: b[0..k] <= x < b[k+1..b.length-1] Examples: b x 1 2 3 4 5 6 7 8 9 10 k -1 0 3 3 4 5 5 7 8 8 <= x >x 0 k b.length b 2 3 3 3 5 6 8 8 9 0 1 2 3 4 5 6 7 8 CS100 Lecture 11

Invariant? Store an integer in k to truthify b[0..k] <= x < b[k+1..b.length-1] Invariant <= x >x 0 k b.length b <= x >x 0 k j b.length CS100 Lecture 11

Let’s fill in the blanks j= ; while ( ) { int e = ; // We know that -1 <= k < e < j <= b.length if ( b[e] <= x ) else } CS100 Lecture 11

Binary Search in Java // Given sorted b, return the integer k that satisfies // b[0..k] <= x < b[k+1..b.length-1] public static int binarySearch( int [ ] b) { int k= -1; int j= b.length; // Invariant: -1 <= k < j <= b.length and // b[0..k] <= x < b[k+1..b.length] while (k+1 != j) { int e = (k+j)/2; // We know -1 <= k < e < j <= b.length if ( b[e] <= x ) k= e; else j= e; } return k; CS100 Lecture 11

About this algorithm It works when the array is empty (b.length = 0) --return -1. If x is in b, finds its rightmost occurrence If x is not in b, finds position where it belongs In general, it’s faster than a binary search that stops as soon as x is found. The latter kind of binary search needs an extra test (x = b[k]) in the loop body, which makes each iteration slower; however, stopping as soon as x is found can be shown to save only one iteration (on the average). It’s easy to verify correctness of the algorithm --to see that it works. CS100 Lecture 11

How fast is binary search? b.length no.of iterations log(b.length+1) 0 = 20 - 1 0 0 1 = 21 - 1 1 1 3 = 22 - 1 2 2 7 = 23 - 1 3 3 15 = 24 - 1 4 4 32767 = 215 -1 15 15 1048575= 220 -1 20 20 j is the “base 2 logarithm of 2j. m is the base 10 logarithm of 10m. CS100 Lecture 11

So, how fast? To search an array of a million entries using linear search may require a million iterations of the loop. To search an array of a million entries using binary search requires only 20 iteration! Linear search is a linear algorithm; the time it takes is proportional in the worst case to the size of the array. Binary search is a logarithmic algoririthm; the time it takes is proportional in the worst case to the log of the size of the array. CS100 Lecture 11

Classes redux Suppose we want to manipulate geometric objects Superclass: geometricObject Potential subclasses: Triangle Square Rectangle CS100 Lecture 11

The Big Picture Triangle sideOne sideTwo sideThree isEquilateral isRight area GeometricObject color numSides area Rectangle length width area Square side area isRhombus CS100 Lecture 11