Writing algorithms using the `for`-statement

Programming example 1: find all divisors of a number

We have seen a program using a while-statement to solve this problem.
Here, we will see that this problem can be solved more naturally using a for-statement.
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

Reminder: 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

This notation can be converted to a for-statement very naturally:

for ( x = 1; x <= 10; x = x + 1 ) { print x; }

Rough algorithm (pseudo code) to find all divisors:
Structure diagram of the algorithm:
Notice it is much easier to write the algorithm using a for-statement (than a while-statement - in this example)

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 for ( x = 1; x <= n; x++ ) // 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) } } } }

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

Programming example 2: compute the sum 1+2+3+...+n

Problem description:
A concrete example:

What would you do to solve this problem ?

Imagine again that you are using a calculator....

I think you would have done this:

Punch clear to set the total to 0.
Press 1, then press add to add to the running total
Press 2, then press add to add to the running total
...
Press 5, then press add to add to the running total
Show the total

Rough algorithm (pseudo code) to compute 1 + 2 + 3 + ... + n:

input: n = some integer number
sum = 0; // Clear the running total ! for ( x = 1, 2, 3, ..., n ) do { Add x to sum } Print sum;

Sample execution (n = 5):

sum = 0; ---> sum = 0 for-statement: add 1 to sum; ---> sum = 1 add 2 to sum; ---> sum = 3 add 3 to sum; ---> sum = 6 add 4 to sum; ---> sum = 10 add 5 to sum; ---> sum = 15 Print sum; ---> Prints 15

Statement to perform the task: "add x to sum"

sum = sum + x;

Example:

Suppose the variable sum contains 6 and x = 4 Then: sum = sum + x; = 6 + 4; = 10; ---> Result, sum = 10 We have added 4 to sum !

Structure diagram of the algorithm:

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, sum; System.out.print("Enter a number n: "); n = in.nextInt(); // Read in number sum = 0; // Clear running total for ( x = 1; x <= n; x++ ) // Run x = 1, 2, ..., n { sum = sum + x; // Add x to sum } System.out.println(sum); // Print final sum } }

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

Programming example 3: compute factorial n!

Problem description:
A concrete example:

We can re-use the previous programming paradigm to construct the algorithm:

Imagine you are using a calculator....

This is how one may compute the factorial:

Punch 1 to set the running product to 1.
(You can't use 0 as starting value for multiplication , because the multiplying with 0 will always produce 0)
Press 1, then press multiply to multiply 1 into the running product
Press 2, then press multiply to multiply 2 into the running product
...
Press 5, then press multiply to multiply 5 into the running product
Show the product

Rough algorithm (pseudo code) to compute n! = 1 × 2 × 3 × ... × n:

input: n = some integer number
product = 1; // Set running product to 1 for ( x = 1, 2, 3, ..., n ) do { Multiply x into product } Print sum;

Sample execution (n = 5):

product = 1; ---> product = 1 for-statement: multiply 1 into product; ---> product = 1 multiply 2 into product; ---> product = 2 multiply 3 into product; ---> product = 6 multiply 4 into product; ---> product = 24 multiply 5 into product; ---> product = 120 Print product; ---> Prints 120

Statement to perform the task: "multiply x into product"

product = product * x;

Example:

Suppose the variable product contains 6 and x = 4 Then: product = product * x; = 6 * 4; = 24; ---> Result, product = 24 We have multiplied 4 into the product !

Structure diagram of the algorithm:

Java program:

import java.util.Scanner; public class Factorial01 { public static void main(String[] args) { Scanner in = new Scanner(System.in); int n; int x, product; System.out.print("Enter a number n: "); n = in.nextInt(); // Read in number product = 1; // Set init. product to 1 for ( x = 1; x <= n; x++ ) // Run x = 1, 2, ..., n { product = product * x; // Multiply x into product } System.out.println(product); // Print final product } }

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

Writing algorithms using the for-statement

Writing algorithms using the `for`-statement