Number conversion Algorithms
CS Principles
This activity introduces the concept that abstractions built upon binary sequences can be used to represent all digital data . It also introduces the concept of an algorithm. It focuses on the following learning objectives:
Introduction
This lesson assumes you have completed the homework on binary and hexadecimal number systems. That homework decribed how the binary, decimal, and hexadecimal number systems work and showed how to convert from one number system to another.
In this lesson we want generalize what we learned their by seeing those number systems as specific examples of a more general concept, a positional number system.
We will develop algorithms that will enable you to perform conversions from one number system to another.
The type of generalization we are doing in this lesson is another example of the abstraction principle in computer science -- here we are focusing on a general pattern that holds true for all positional number systems.
Algorithms and Pseudocode
An algorithm is a step-by-step procedure to perform some computation. For example, the steps you take in the Hello Purr app when the button is clicked is an example of a simple 2-step algorithm:
To help us talk about algorithms we will use pseudocode, a language or notation that has many of the structures of a programming language but is easy to read. Pseudocdes are halfway between natural languages like English and formal programming languages.
Positional Number Systems
Let's review some of the key points that you learned in the Khan Academy videos.
Name | Base | Symbols |
Decimal | 10 | 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 |
Binary | 2 | 0, 1 |
Hexadecimal | 16 | 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F |
Octal | 8 | 0, 1, 2, 3, 4, 5, 6, 7 |
System | Base | Value | Conversion Formula | Decimal Value |
Decimal | 10 | 104 | (1 × 102) + (0 × 101) + (4 × 100) | 100 + 0 + 4 = 104 |
Binary | 2 | 111 | (1 × 22) + (1 × 21) + (1 × 20) | 4 + 2 + 1 = 7 |
Octal | 8 | 104 | (1 × 82) + (0 × 81) + (4 × 80) | 64 + 0 + 4 = 68 |
Hexadecimal | 16 | FEC | (F × 162) + (E × 161) + (C × 80) | 15 × 256 + 14 × 16 + 12 × 1 = 3840 + 224 + 12 = 4076 |
Conversion Algorithms
Let's summarize these conversion formulas by developing an general algorithm that will convert from any base into decimal.
Algorithm to Convert From Any Base to Base 10 Decimal
Let's try it on the binary number 1011.
Let n = 4.
Let b = 2.
Let s = 0.
First digit, 1: n = 3, 1 × bn is 1 × 23 = 8. So s = 8.
Second digit, 0: n = 2, 0 × bn is 0 × 22 = 0. So s = 8.
Third digit, 1: n = 1, 1 × bn is 1 × 21 = 2. So s = 10
Last digit, 1: n = 0, 1 × bn is 1 × 20 = 1. So 10112 = 1110
Digit | n | Value = Digit * bn | Running Total |
1 | 3 | 1 × 23 = 8 | 8 |
0 | 2 | 0 × 22 = 0 | 8 |
1 | 1 | 1 × 21 = 2 | 10 |
1 | 0 | 1 × 20 = 1 | 11 |
Let's try it on the hex number 7E.
Let n = 2.
Let b = 16.
Let s = 0.
First digit, 7: n = 1, 7 × bn is 7 × 161 = 7 × 16 = 112. So s = 112.
Last digit, E: n = 0, 14 × bn is 14 × 160 = 14. So s = 112 + 14 = 126. So 7E16 = 12610
Digit | n | Value = Digit * bn | Running Total |
7 | 1 | 7 × 161 = 112 | 114 |
E | 0 | 14 × 160 = 14 | 126 |
Let's try it on the octal number 124.
Let n = 3.
Let b = 8.
Let s = 0.
First digit, 1: n = 2, 1 × bn is 1 × 82 = 1 × 64 = 64. So s = 64.
Second digit, 2: n = 1, 2 × bn is 2 × 81 = 2 × 8 = 16. So s = 64 + 16 = 80.
Last digit, 4: n = 0, 4 × bn is 4 × 80 = 4. So s = 80 + 4 = 84. So 1248 = 8410
Digit | n | Value = Digit * bn | Running Total |
1 | 2 | 1 × 82 = 64 | 64 |
2 | 1 | 2 × 81 = 16 | 80 |
4 | 0 | 4 × 80 = 4 | 84 |
Algorithm to Convert From Decimal To Another Base
Let's use the algorithm to convert 45 into binary.
Let n = 45.
Let b = 2.
Repeat
45 divided by b is 45/2 = 22 remainder 1. So d=22 and r=1. So m= 1 and the new n is 22.
22 divided by b is 22/2 = 11 remainder 0. So d=11 and r=1. So m= 01 and the new n is 11.
11 divided by b is 11/2 = 5 remainder 1. So d=5 and r=1. So m= 101 and the new n is 5.
5 divided by b is 5/2 = 2 remainder 1. So d=2 and r=1. So m= 1101 and the new n is 2.
2 divided by b is 2/2 = 1 remainder 0. So d=1 and r=0. So m= 01101 and the new n is 1.
1 divided by b is 1/2 = 0 remainder 1. So d=0 and r=1. So m=101101 and the new n is 0. So 4510 = 1011012
Let's use it to convert 99 into binary.
Let n = 99.
Let b = 2.
Repeat
99 divided by b is 99/2 = 49 remainder 1. So d=49 and r=1. So m= 1 and the new n is 49.
49 divided by b is 49/2 = 24 remainder 1. So d=24 and r=1. So m= 11 and the new n is 24.
24 divided by b is 24/2 = 12 remainder 0. So d=12 and r=0. So m= 011 and the new n is 12.
12 divided by b is 12/2 = 6 remainder 0. So d=6 and r=0. So m= 0011 and the new n is 6.
6 divided by b is 6/2 = 3 remainder 0. So d=3 and r=0. So m= 00011 and the new n is 3.
3 divided by b is 3/2 = 1 remainder 1. So d=1 and r=1. So m= 100011 and the new n is 1.
1 divided by b is 1/2 = 0 remainder 1. So d=0 and r=1. So m=1100011 and the new n is 0. So 9910 = 11000112
Let's use it to convert 45 into hexadecimal.
Let n = 45.
Let b = 16.
Repeat
45 divided by b is 45/16 = 2 remainder 13. So d=2 and r=13. So m= D and the new n is 2.
2 divided by b is 2/16 = 0 remainder 2. So d=0 and r=2. So m=2D and the new n is 0. So 4510 = 2D16.
Let's use it to convert 99 into hexadecimal.
Let n = 99.
Let b = 16.
Repeat
99 divided by b is 99/16 = 6 remainder 3. So d=6 and r=3. So m= 3 and the new n is 6.
6 divided by b is 6/16 = 0 remainder 6. So d=0 and r=6. So m=63 and the new n is 0. So 9910 is 6316.