Algorithms Binary Search recursion ppt BSIT

Building Java Programs
Chapter 12: Recursive public/private pairs
Chapter 13: Searching
reading: 13.3

4
Recursion and cases
 Every recursive algorithm involves at least 2 cases:
 base case: simple problem that can be solved directly.
 recursive case: more complex occurrence of the problem
that cannot be directly answered, but can instead be described
in terms of smaller occurrences of the same problem.
 Some recursive algorithms have more than one base or
recursive case, but all have at least one of each.
 A crucial part of recursive programming is identifying these
cases.

5
Recursion Challenges
 Forgetting a base case
 Infinite recursion resulting in StackOverflowError
 Working away from the base case
 The recursive case must make progress towards the base case
 Infinite recursion resulting in StackOverflowError
 Running out of memory
 Even when making progress to the base case, some inputs
may require too many recursive calls: StackOverflowError
 Recomputing the same subproblem over and over again
 Refining the algorithm could save significant time

6
Exercise
 Write a method crawl accepts a File parameter and prints
information about that file.
 If the File object represents a normal file, just print its name.
 If the File object represents a directory, print its name and
information about every file/directory inside it, indented.
cse143
handouts
syllabus.doc
lecture_schedule.xls
homework
1-tiles
TileMain.java
TileManager.java
index.html
style.css
 recursive data: A directory can contain other directories.

7
File objects
 A File object (from the java.io package) represents
a file or directory on the disk.
Constructor/method Description
File(String) creates File object representing file with given name
canRead() returns whether file is able to be read
delete() removes file from disk
exists() whether this file exists on disk
getName() returns file's name
isDirectory() returns whether this object represents a directory
length() returns number of bytes in file
listFiles() returns a File[] representing files in this directory
renameTo(File) changes name of file

8
Public/private pairs
 We cannot vary the indentation without an extra
parameter:
public static void crawl(File f, String indent) {
 Often the parameters we need for our recursion do not
match those the client will want to pass.
In these cases, we instead write a pair of methods:
1) a public, non-recursive one with parameters the client wants
2) a private, recursive one with the parameters we really need

9
Exercise solution 2
// Prints information about this file,
// and (if it is a directory) any files inside it.
public static void crawl(File f) {
crawl(f, ""); // call private recursive helper
}
// Recursive helper to implement crawl/indent behavior.
private static void crawl(File f, String indent) {
System.out.println(indent + f.getName());
if (f.isDirectory()) {
// recursive case; print contained files/dirs
for (File subFile : f.listFiles()) {
crawl(subFile, indent + " ");
}
}
}

10
Recursive Data
 A file is one of
 A simple file
 A directory containing files
 Directories can be nested to an arbitrary depth
 Iterative code to crawl a directory structure requires data
structures
 In recursive solution, we use the call stack

11
Binary search (13.1)
 binary search: Locates a target value in a sorted
array/list by successively eliminating half of the array from
consideration.
 Can be implemented with a loop or recursively
 Example: Searching the array below for the value 42:
inde
x
0 1 2 3 4 5 6 7 8 9 1
0
1
1
1
2
1
3
1
4
1
5
16
valu
e
-4 2 7 1
0
1
5
2
0
2
2
2
5
3
0
3
6
4
2
5
0
5
6
6
8
8
5
9
2
10
3
min mid ma
x

12
Binary search code
// Returns the index of an occurrence of target in a,
// or a negative number if the target is not found.
// Precondition: elements of a are in sorted order
public static int binarySearch(int[] a, int target) {
int min = 0;
int max = a.length - 1;
while (min <= max) {
int mid = (min + max) / 2;
if (a[mid] < target) {
min = mid + 1;
} else if (a[mid] > target) {
max = mid - 1;
} else {
return mid; // target found
}
}
return -(min + 1); // target not found
}

13
Recursive binary search (13.3)
 Write a recursive binarySearch method.
 If the target value is not found, return its negative insertion
point.
int index = binarySearch(data, 42); // 10
int index2 = binarySearch(data, 66); // -14
inde
x
0 1 2 3 4 5 6 7 8 9 1
0
1
1
1
2
1
3
1
4
1
5
16
valu
e
-4 2 7 1
0
1
5
2
0
2
2
2
5
3
0
3
6
4
2
5
0
5
6
6
8
8
5
9
2
10
3

14
Ordering and objects
 Can we compare Strings?
 Operators like < and > do not work with String objects.
 But we do think of strings as having an alphabetical ordering.
 natural ordering: Rules governing the relative placement
of all values of a given type.
 comparison function: Code that, when given two values
A and B of a given type, decides their relative ordering:
 A < B, A == B, A > B

15
The compareTo method (10.2)
 The standard way for a Java class to define a comparison
function for its objects is to define a compareTo method.
 Example: in the String class, there is a method:
public int compareTo(String other)
 A call of A.compareTo(B) will return:
a value < 0 if A comes "before" B in the ordering,
a value > 0 if A comes "after" B in the ordering,
or 0 if A and B are considered "equal" in the
ordering.

16
Using compareTo
 compareTo can be used as a test in an if statement.
String a = "alice";
String b = "bob";
if (a.compareTo(b) < 0) { // true
...
}
Primitives Objects
if (a < b) { ... if (a.compareTo(b) < 0) { ...
if (a <= b) { ... if (a.compareTo(b) <= 0) { ...
if (a == b) { ... if (a.compareTo(b) == 0) { ...
if (a != b) { ... if (a.compareTo(b) != 0) { ...
if (a >= b) { ... if (a.compareTo(b) >= 0) { ...
if (a > b) { ... if (a.compareTo(b) > 0) { ...

17
Exercise solution
// Returns the index of an occurrence of the given value in
// the given array, or a negative number if not found.
// Precondition: elements of a are in sorted order
public static int binarySearch(int[] a, int target) {
return binarySearch(a, target, 0, a.length - 1);
}
// Recursive helper to implement search behavior.
private static int binarySearch(int[] a, int target,
int min, int max) {
if (min > max) {
return -1; // target not found
} else {
int mid = (min + max) / 2;
if (a[mid] < target) { // too small; go right
return binarySearch(a, target, mid + 1, max);
} else if (a[mid] > target) { // too large; go left
return binarySearch(a, target, min, mid - 1);
} else {
return mid; // target found; a[mid] == target
}
}
}

18
Binary search runtime
 For an array of size N, it eliminates ½ until 1 element
remains.
N, N/2, N/4, N/8, ..., 4, 2, 1
 How many divisions does it take?
 Think of it from the other direction:
 How many times do I have to multiply by 2 to reach N?
1, 2, 4, 8, ..., N/4, N/2, N
 Call this number of multiplications "x".
2x
= N
x = log2 N
 Binary search is in the logarithmic complexity class.

19
Complexity classes
http://recursive-design.com/blog/2010/12/07/comp-sci-101-big-o-notation/ - post about a Google interview

20
 See section 12.4
Recursive Graphics

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