Difference between revisions of "CSC212 Lab 9 2014"

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(Binary Search)
(Binary Search)
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Implement ''Binary Search'' on a sorted array.  The algorithm is simple:
 
Implement ''Binary Search'' on a sorted array.  The algorithm is simple:
# Main Program:
+
* Main Program:
:# Input: A sorted array of size N
+
:* Input: A sorted array of size N
:# a key, of the same type as the elements of A
+
:* a key, of the same type as the elements of A
:# index = binSearch( key, A, 0, N-1 ) // search the key in A, ranging from Index 0 to Index N-1
+
:* index = binSearch( key, A, 0, N-1 ) // search the key in A, ranging from Index 0 to Index N-1
# binSearch( key, A, low, high )
+
* binSearch( key, A, low, high )
:# if low > high, end of search.  Key not found.  Return -1
+
:* if low > high, end of search.  Key not found.  Return -1
:# mid = (high+low) / 2
+
:* mid = (high+low) / 2
:# if key in A[mid], then key found: return mid as index of key
+
:* if key in A[mid], then key found: return mid as index of key
:# if key < A[mid], then key must be in lower part of array, from low to mid-1
+
:* if key < A[mid], then key must be in lower part of array, from low to mid-1
 
::: index = binSearch( key, A, low, mid-1 )
 
::: index = binSearch( key, A, low, mid-1 )
:# otherwise key > A[mid], then key must be in upper part of array, from mid+1 to high
+
:* otherwise key > A[mid], then key must be in upper part of array, from mid+1 to high
 
::: index = binSearch( key, A, mid+1, high )
 
::: index = binSearch( key, A, mid+1, high )
:# return index
+
:* return index
  
 
===RPN===
 
===RPN===

Revision as of 14:51, 22 October 2014

--D. Thiebaut (talk) 15:00, 22 October 2014 (EDT)



...

RecursionFactory.png

Recursion


  • Use the skeleton program below to write different recursive functions.


public class RecurseSkeleton {

	private static int[] initSortedArray( int N ) {
		int[] array = new int[N];
		array[0] = 3;
		for ( int i=1; i<N; i++ ) 
			array[ i ] = array[i-1] + (i*11)*7;
		
		return array;
	}
	
	private static int[] initArray( int N ) {
		int[] array = new int[N];
		array[0] = 3;
		for ( int i=1; i<N; i++ ) 
			array[ i ] = (i+1) * 101 % 23;
		
		return array;
	}
	
	private static int fact( int n ) {
		// stopping condition
		if ( n==1 ) 
			return 1;
		
		// recursive step
		return n * fact( n-1 );
	}
	
	public static void main(String[] args) {
		int N = 20; 
		int A[] = initArray( N );

		int x, y;
		x = 5;
		
		System.out.println( "fact( " + x + " ) = " + fact( x ) );
	}
}


Problem 1


Sum


  • Write a recursive function that will compute the sum of all the elements of the array, following this definition, where A[1:N] indicates the elements of the array going from Index 1 to Index N, included:
sum( array ) =
| 0 if length( array ) == 0
| array[0] if length( array )== 1
| array[0] + sum( array[1 : N-1] ), where N is the size of the array.
  • Write a recursive function that will compute the sum of all the elements of the array, following this definition:
sum( array ) =
| 0 if length( array ) == 0
| array[0] if length( array )== 1
| array[N-1] + sum( array[0 to N-2] ), where N is the size of the array.
  • Write a recursive function that will compute the sum of all the elements of the array, following this definition:
sum( array ) =
| 0 if length( array ) == 0
| array[0] if length( array )== 1
| sum( array[0 to mid] ) + sum( array[mid+1 to N-1] ), where N is the size of the array, and mid = N/2


Max


  • Write a recursive function that finds the largest element of an array recursively:
max( array ) =
| array[0] if length( array )== 1
| largest of array[0] and max( array[1 : N-1] ), where N is the size of the array.


Binary Search


Implement Binary Search on a sorted array. The algorithm is simple:

  • Main Program:
  • Input: A sorted array of size N
  • a key, of the same type as the elements of A
  • index = binSearch( key, A, 0, N-1 ) // search the key in A, ranging from Index 0 to Index N-1
  • binSearch( key, A, low, high )
  • if low > high, end of search. Key not found. Return -1
  • mid = (high+low) / 2
  • if key in A[mid], then key found: return mid as index of key
  • if key < A[mid], then key must be in lower part of array, from low to mid-1
index = binSearch( key, A, low, mid-1 )
  • otherwise key > A[mid], then key must be in upper part of array, from mid+1 to high
index = binSearch( key, A, mid+1, high )
  • return index

RPN


  • Here's a skeleton for a recursive Reverse-Polish-Notation solver:


import java.awt.List;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Stack;

public class RecursiveRPN {
	
	private static int recurseSolve( ArrayList tokens, Stack<Integer> stack ) {
		// STOPPING CONDITION
                // if the size of the token list is 0, then empty expression: return top of stack
                
                //===> put your code here...

	        // get the next token from the array, the first one, and remove it from the array...	

                //===> put your code here...


 		// DO SOME WORK
                // if the token we got above is a "+", then pop 2 ints from stack, add them up, push sum back in stack.
                // similarly, if token is "-", pop 2, subtract, push difference
                // if token is "*", pop 2, multiply them together, push product
                // if token is "/", pop 2, divide 2 by first, push integer quotient back in stack...

                //===> put your code here...

                // and if token was not a +, -, *, or /, then it must be an int.  push it in the stack...
                // (and be careful, because a token is a string, and the stack holds integers only...)

		
                // RECURSIVE STEP: we've done some work, recurse and pass the remaining tokens and stack
                // to recurseSolve() 

                //===> put your code here...

                // return result...
                //  whatever we get back from recurseSolve, in the recursive step, we pass back...

                //===> put your code here...
	}

	
	static public void main( String[] args ) {
		ArrayList<String> tokens = new ArrayList<String>();
		
		// if user specifies expression on command line, then uses that
		if ( args.length != 0 ) {
			for ( int i=0; i<args.length; i++ )
				tokens.add( args[i].trim() );
		}
		// else use the expression below
		else {
			String expression = "3 5 + 7 3 - *";
			String[] tokenStrings = expression.split( " " );
			for ( int i=0; i<tokenStrings.length; i++ )
				tokens.add( tokenStrings[i] );
		}
		
		// display the tokens, for debugging purposes
		System.out.println( "tokens = " + tokens );
		
		// create a stack
		Stack<Integer> stack = new Stack<Integer>();
		
		// solve expression recursively
		int x = recurseSolve( tokens, stack );
		
		// display result.
		System.out.println( "result = " + x );
	}
}


  • Study the comments left in recurseSolve(), and replace the "//===> put your code here" comments by java code implementing all the operations described.
  • Verify that your program works and outputs 32.
  • Modify the expression in main() and solve something a bit more complicated. Say, "12 34 56 + + 3 - 5 9 + * 10 /", which should evaluate to 138.