writing algorithms using the while-statement

Writing algorithms using the while-statement

Previously discussed

• Syntax of while-statement:

Previously discussed (cont.)

• Statements discussed so far:

• Assignment statement:

variable = expression ;


• Conditional statements:

if ( condition ) statement

if ( condition ) statement1 else statement2


• Loop (while) statements:

while ( condition ) { statement1 statement2 ... }

Combining different types of statements

• Computer Science Fact:

• Every programming language (including Java) has 3 types of statements:

• Assignment statements

• Conditional statements

• Loop statements

Combining different types of statements (cont.)

• The good news is: you have learned all of them now !

• However, you still need to learn how to use them effectively


• Rule of programming languages:

• Whenever you see a statement in a syntax construct, you can use any type of statement !!!

Example: while ( condition ) while ( condition ) { { statement ===> if ( condition2 ) } { statement1 } else { statement2 } }


• Here, we used an if-else (conditional) statement as the statement in the while-body

• In fact, the statement1 and statement2 in the then-part and else-part of the if-else (conditional) statement can themselves be a assignment statement, a conditional statement, or a loop statement !!!

• Therefore, you can create very complex programs


• Advice:

• The point of computer programming is not writing complex programs... but write out an algorithm in simple steps for a dumb machine (computer).

• There are many different ways to write the same computer program, some ways can be very convoluted than others.

• You should keep the structure of computer programs simple


• We will now learn how to use the power of a programming language (Java) by combining (nesting) different statements

Developing computer algorithms

• Pre-requisite to developing a computer algorithm:

• Before you can write a computer program to solve a problem, you yourself must know what you need to do

Because:

• Programming a computer = tell a computer what to do to solve a problem

• If you don't know what to do, you cannot tell someone else (or something else like a computer) what to do.... (A blind cannot lead a blind...)

Developing computer algorithms (cont.)

• Developing a computer algorithm:

• Develop a computer algorithm = Write down the steps that a human must do to solve a problem in such a detail than a dumb machine (computer) can do it !!!

Programming example 1: find all divisors of a number

• Problem description:

• Write a Java program that reads in an integer n...

• and prints all its divisors

Programming example 1: find all divisors of a number (cont.)

• A concrete example:

• Input: n = 12

• Output: 1, 2, 3, 4, 6, 12


• What would you do to solve this problem ?

• Suppose: n = 12

• Check if 12 is divisible by 1


• ...



• Note:

• When the remainder of the division is equal to 0, we print out the divisor

• We do not need to check numbers > 12 because only number ≤ 12 can be divisors !


• Algorithm:

The algorithm contains some abstract steps that need to be fleshed out (work out the details)


• We have learned the programming technique (trick) to test for divisibility previously:

if ( n % a == 0 ) n is divisible by a; else n is not divisible by a;


• Apply this programming technique to obtain a refined algorithm:

Now the algorithm contain only (concrete) steps that can be easily translated into a Java program !!!


• Java program:

import java.util.Scanner; public class Divisors01 { public static void main(String[] args) { Scanner in = new Scanner(System.in); int n; int a; System.out.print("Enter a number n: "); n = in.nextInt(); // Read in number


a = 1; while ( a <= n ) // Run a = 1, 2, ..., n { if ( n % a == 0 ) { // a is a divisor of n System.out.println(a); // Print a (because it's a divisor) } a++; // Make sure we more to the next number !! // or else: infinite loop !!! } } }


• Example Program: (Demo above code) – Prog file:

http://mathcs.emory.edu/~cheung/Courses/170/Syllabus/07/Progs/Divisors01.java

• How to run the program:

• Right click on link and save in a scratch directory

• To compile: javac Divisors01.java

• To run: java Divisors01

The brute force search method: a commonly used solution method in computer algorithms

• The previous algorithm is an example of the brute force search method

• The general form of the Brute force search method:

for every possible value x do { if ( x is a solution ) print x; }

The brute force search method: a commonly used solution method in computer algorithms (cont.)

• Pre-conditions for using the brute force search method:

• The number of possible values that needs to be searched must be finite

(I.e.: a finite search space)

• There is a method to determine if a value x is a solution

Programming example 2: find all common divisors of 2 numbers


• Write a Java program that reads in 2 numbers x and y...

• and prints all numbers that are divisors of both x and y

Programming example 2: find all common divisors of 2 numbers (cont.)


• Input: x = 24 and y = 16

• Output: 1, 2, 4, 8



• Suppose: x = 24 and y = 16

• The lesser of the values is 16

• Therefore, all divisors are ≤ 16


• Check if 16 and 24 are divisible by 1


• ...


• When the remainder of both divisions are equal to 0, we print out the divisor


• Rough algorithm:

input x, y;

min = min(x, y); // this is the range of the brute force search

for every value a = {1, 2, ...., min} do { if (x and y are divisible by a) { print a; } }


• Algorithm (structured diagram):


• Java program:

import java.util.Scanner; public class Divisors02 { public static void main(String[] args) { Scanner in = new Scanner(System.in); int x, y, a, min = 0; x = in.nextInt(); // Read in x y = in.nextInt(); // Read in y if ( x < y ) // Find min(x,y) min = x; else min = y;


a = 1; while ( a <= min ) // Run a = 1, 2, ..., min(x,y) { if ( x % a == 0 && y % a == 0 ) { // a is a divisor of x and y System.out.println(a); // Print a (because it's a common divisor) } a++; // Make sure we move to the next number !! // or else: infinite loop !!! } } }



http://mathcs.emory.edu/~cheung/Courses/170/Syllabus/07/Progs/Divisors02.java



• To compile: javac Divisors02.java

• To run: java Divisors02

Programming example 3: factor a number (into prime factors)


• Write a Java program that reads in an integer number x....

• and prints the prime factorization of the number x

Programming example 3: factor a number (into prime factors) (cont.)


• Input: x = 420

• Output: 2, 2, 3, 5, 7 (because 2 x 2 x 3 x 5 x 7 = 420 and all factors are prime numbers)



• Suppose: x = 420

• Factorization algorithm taught in High Schools:

The number 420 is divisible by 2. Factor out the number 2:

210 ------ 2 / 420


The number 210 is divisible by 2. Factor out the number 2:

105 ------ 2 / 210


The number 105 is not divisible by 2 ==> try 3 !!! The number 105 is divisible by 3. Factor out the number 3:

35 ------ 3 / 105


The number 35 is not divisible by 3 ==> try 4 !!! The number 35 is not divisible by 4 ==> try 5 !!! The number 35 is divisible by 5. Factor out the number 5:

7 ----- 5 / 35


The number 7 is not divisible by 5 ==> try 6 !!! The number 7 is not divisible by 6 ==> try 7 !!! The number 7 is divisible by 7. Factor out the number 7:

1 ----- 7 / 7


• Rough algorithm:

input x;

a = 2; <---- Try use a = 2 as factor for x

as long as x is not equal to 1 do { if (x is divisible by a) { // a is a factor of x !!!

print a; (Because we have found another factor) Remove factor a from the number x (and continue) } else { Try use a+1 as factor } }


• Algorithm (structured diagram):


• Check for correctness !!!

• This algorithm is complicated enough to warrant a correctness check

• We do so using a small example: suppose x = 12


• Execution of the algorithm:

• Iteration 1 through the while-loop:

It has printed the factor 2 and starts a new loop with x = 6


• Iteration 2 through the while-loop: (notice that x = 6 in this iteration !)

It has printed another factor 2 and starts a new loop with x = 3



Because a = 2 is not a factor of x = 3, the else-part is executed.

A new iteration is started with a = 3 (we are trying a new factor)


• Iteration 4 through the while-loop: (notice that a = 3 in this iteration !)

It has printed the factor 3 and starts a new loop with x = 1



The loop-continuation-condition is not true !!!

The while-loop terminates


Result:

• The program has printed the factors 2, 2, 3

• (Which is correct !!!)


• Java program:

import java.util.Scanner; public class Factor01 { public static void main(String[] args) { Scanner in = new Scanner(System.in); int x; int a; System.out.print("Enter a number x: "); x = in.nextInt(); // Read in number


a = 2; while ( x > 1 ) { if ( x % a == 0 ) { // a is a factor of x System.out.println(a); // Print a (because it's a divisor) x = x / a; // Remove factor a from x } else { a++; // Use next number as factor } } } }



http://mathcs.emory.edu/~cheung/Courses/170/Syllabus/07/Progs/Factor01.java



• To compile: javac Factor01.java

• To run: java Factor01

writing algorithms using the while-statement

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