Difference between revisions of "CSC212 Lab 9 2014"
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::::| largest of array[0] and max( array[1 : N-1] ), where N is the size of the array. | ::::| largest of array[0] and max( array[1 : N-1] ), where N is the size of the array. | ||
<br /> | <br /> | ||
+ | ===Binary Search=== | ||
+ | <br /> | ||
+ | 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=== | ===RPN=== | ||
<br /> | <br /> |
Revision as of 14:50, 22 October 2014
--D. Thiebaut (talk) 15:00, 22 October 2014 (EDT)
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.
- sum( 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.
- sum( 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
- sum( array ) =
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.
- max( 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.