Writing algorithms using the `while`-statement

Previously discussed

Syntax of while-statement:

Statements discussed so far:

Assignment statement:

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:
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
The point is:
There are many different ways to write the same computer program
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 a computer algorithm:

Programming example 1: find all divisors of a number

Problem description:
A concrete example:

What would you do to solve this problem ?

Problem:
I think you would have done this:
Note:
When the remainder of the division is equal to 0, we print out the divisor

Computer programming:

Notation:

The notation:

for (x = 1, 2, 3, ...., 10) do { print x; }

means:

The variable x takes on (assumes, gets) the value 1, 2, 3, ..., 10, one at a time
For each individual value taken on by x, you perform the operation "print x" once

The result of:

for (x = 1, 2, 3, ...., 10) do { print x; }

is printing the numbers 1, 2, 3, ..., 10.

We will use this notation to expression some of our algorithms.

Rough algorithm (pseudo code) to find all divisors:
Note:
The algorithm contains some abstract steps that need to be fleshed out (work out the details)

Working out some details:

How to make variable x take on the values 1..n one at a time:

Use a loop statement:

x = 1; // x starts at 1 while ( x <= n ) // Loop will stop when x = n { (other statements omitted for clarity) x++; // Variable x will take on: 1, 2, 3, ..., n }

How to test if n is divisible by x

We have previously learned about this programming technique (trick) to test for divisibility:

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

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

Example Program: (Demo above code)
- Prog file: click here
How to run the program:

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; }

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

Problem description:
A concrete example:

What would you do to solve this problem ?

Suppose: x = 24 and y = 16
Check if 16 and 24 are divisible by 1
Check if 16 and 24 are divisible by 2
...
Check if 16 and 24 are divisible by 16
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 (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 !!! } } }

Example Program: (Demo above code)
- Prog file: click here
How to run the program:

Example 3: find the occurence of a letter in a word

Problem description:

A concrete example:

Input: x = "Abracadabra

Output: 3 5 7 10

because:

       Abracadabra
       01234567890
          ^ ^ ^  ^

Pseudo code:

for ( every possible position i ) { if ( character at position i in string x == 'a ) { position i is a correct solution } }

Starting position to search: 0 !!!
Ending position to search: x.length() − 1 !!!

Algorithm in Java:

import java.util.Scanner; public class FindChar01 { public static void main(String[] args) { Scanner in = new Scanner(System.in); String x; /* =========================================== Read in the word and store it in variable x ============================================ */ System.out.print("Enter a word x = "); x = in.next(); // Read in a word int i; i = 0; while ( i <= x.length() - 1 ) // Run i = 0, 1, ..., x.length()-1 { if ( x.charAt(i) == 'a' ) { // found 'a' at position i System.out.println(i); // Print position i } i++; // Make sure we more to the next position !! // or else: infinite loop !!! } } }

Example Program: (Demo above code)
- Prog file: click here
How to run the program:

Programming example 4: factor a number (into prime factors)

Problem description:
A concrete example:

What would you do to solve this problem ?

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
When the number = 1, we have found completed the factorization algorithm !

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
Iteration 3 through the while-loop: (notice that x = 3 in this iteration !)
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
Iteration 5 through the while-loop: (notice that x = 1 in this iteration !)
The loop-continuation-condition is not true !!!
The while-loop terminates
Result:

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 } } } }

Example Program: (Demo above code)
- Prog file: click here
How to run the program:

Program example 5: division by repeated subtraction

Problem description:

Write a Java program that reads in 2 integers A and B ....
and prints the quotient A/B and remainder A%B using the "dvision by repeated subtraction" method.

The division by repeated subtraction method illustrated:

Divide: 37 by 7 Initially: Quotient = 0 Remainder = 37 Is Remainder >= 7 ? Yes: Quotient = 1, Remainder = 30 Is Remainder >= 7 ? Yes: Quotient = 2, Remainder = 23 Is Remainder >= 7 ? Yes: Quotient = 3, Remainder = 16 Is Remainder >= 7 ? Yes: Quotient = 4, Remainder = 9 Is Remainder >= 7 ? Yes: Quotient = 5, Remainder = 2 Is Remainder >= 7 ? NO Answer: Quotient = 5, Remainder = 2

Algorithm in pseudo code:

Input A, B Q = 0; R = A; as long as ( R >= B ) { Q = Q + 1; R = R - B; } Print Q, R

Java program:

import java.util.Scanner; public class Divide01 { public static void main(String[] args) { Scanner in = new Scanner(System.in); int A, B; int Q, R; System.out.print("Enter A = "); A = in.nextInt(); // Read in number System.out.print("Enter B = "); B = in.nextInt(); // Read in number Q = 0; R = A; while ( R >= B ) { Q = Q + 1; R = R - B; } System.out.println("Quotient = " + Q); System.out.println("Remainder = " + R); } }

Example Program: (Demo above code)
- Prog file: click here
How to run the program:

Writing algorithms using the while-statement

Writing algorithms using the `while`-statement