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

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

3. 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

4. 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
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5. 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
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6. 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
i b.length b
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7. 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; }
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8. 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;
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9. 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.
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10. 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?
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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
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12. Simple idea, but . . .
Method dates back to 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
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13. Algorithm Development
Specification. Store an integer in k to truthify: b[0..k] <= x < b[k+1..b.length-1]
Examples: b x k
<= x >x
k b.length b
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14. Invariant? Store an integer in k to truthify
b[0..k] <= x < b[k+1..b.length-1]
Invariant
<= x >x
k b.length b
<= x >x
k j b.length
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15. Let’s fill in the blanks
j= ;
while ( ) {
int e = ;
// We know that -1 <= k < e < j <= b.length
if ( b[e] <= x )
else }
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16. 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;
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17. 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.
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18. How fast is binary search?
b.length no.of iterations log(b.length+1) =
1 =
3 =
7 =
15 =
32767 = =
j is the “base 2 logarithm of 2j.
m is the base 10 logarithm of 10m.
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19. 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.
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20. Classes redux Suppose we want to manipulate geometric objects
Superclass: geometricObject
Potential subclasses:
Triangle
Square
Rectangle
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21. The Big Picture Triangle sideOne sideTwo sideThree isEquilateral
isRight
area
GeometricObject
color
numSides
area
Rectangle
length
width
area
Square
side
area
isRhombus
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Lecture 11