▪ Hexadecimal Addition and Subtraction |
ABACUS: MYSTERY OF THE BEAD |
Hexadecimal on the Soroban
Used in mathematics and computer sciences, hexadecimal or "hex" is a base 16 numerical system. It's main purpose is to represent the binary code in a way that makes it easier for most of us to read. In fact it acts as a kind of shorthand. In hexadecimal the numbers 0-9 are represented normally then; A=ten, B=eleven, C=twelve, D=thirteen, E=fourteen and F=fifteen.
0 hex = 0 dec
1 hex = 1 dec
2 hex = 2 dec
3 hex = 3 dec0 0 0 0
0 0 0 1
0 0 1 0
0 0 1 14 hex = 4 dec
5 hex = 5 dec
6 hex = 6 dec
7 hex = 7 dec0 1 0 0
0 1 0 1
0 1 1 0
0 1 1 18 hex = 8 dec
9 hex = 9 dec
A hex = 10 dec
B hex = 11 dec1 0 0 0
1 0 0 1
1 0 1 0
1 0 1 1C hex = 12 dec
D hex = 13 dec
E hex = 14 dec
F hex = 15 dec1 1 0 0
1 1 0 1
1 1 1 0
1 1 1 1
Conversion TableBinary Codes
A "binary code" is any system that uses only two states (0 or 1, on or off, true or false, etc.) The base 2 or binary number system is one example of a binary code.
How it works for numbers
Take the example 201, which can also be expressed as: 128 + 64 + 8 + 1
128 = 10000000 64 = 1000000 8 = 1000 1 = 1 201 = 11001001So 201 has a binary representation of 11001001. The problem is many of us find binary numbers difficult to understand. Converting 201 to hex makes the number more accessible. (see Example:1)
Convert Base 10 to Hexadecimal
This is an advanced soroban problem using the division techniques taught by Takashi Kojima.
Basically it works like this:
- Designate a unit rod.
- Set the base 10 number onto the soroban.
- Divide by16.
- Remainders on the right form part of the hexadecimal answer.
- If the quotient on the left is 16 or greater divide again to make it less than 16.
- Use the Conversion Table to express quotient and remainders as hex.
Example 1: Convert 201 to its hexadecimal and binary equivalents
Step 1: Choose rod H as the unit rod. Set 201 and 16 onto the soroban. (Fig.1)
Fig.1
Step 1A B C D E F G H I . . . 1 6 0 0 0 2 0 1 0
Step 2 and the answer: Divide 201 by 16. What remains on the soroban is the quotient 12 & the remainder 09. When expressed as hex 12 & 09 become C9 with a binary representation of 11001001 (C = 1100, 9 = 1001). (Fig.2)
Fig.2
Step 2A B C D E F G H I . . . 1 6 0 0 0 2 0 1 0 (1) - 1 6 1 6 0 1 0 0 4 1 0 (2) - 3 2 1 6 0 1 2 0 0 9 0
Example 2: Convert 2989 to its hexadecimal and binary equivalents
Step 1: Choose K as the unit rod. Set 2989 and 16 onto the soroban. (Fig.3)
Fig.3
Step 1A B C D E F G H I J K L M . . . . 1 6 0 0 0 0 0 2 9 8 9 0 0
Step 2: Divide 2989 by 16 leaving the quotient answer 186 on rods FGH. Notice the remainder 13 on rods JK; it forms the last part of the hexadecimal answer. (Fig.4)
Fig.4
Step 2A B C D E F G H I J K L M . . . . 1 6 0 0 0 0 0 2 9 8 9 0 0 (1) - 1 6 1 6 0 0 0 1 0 1 3 8 9 0 0 (8) - 1 2 8 1 6 0 0 0 1 8 0 1 0 9 0 0 (6) - 9 6 1 6 0 0 0 1 8 6 0 1 3 0 0
Step 3 and the answer: Convert 186 to hex by dividing again. With this last step what remains on the soroban is the quotient 11 and the remainders 10 & 13. When expressed as hex 11, 10 & 13 become BAD with a binary representation of 101110101101 (B = 1011, A = 1010, D = 1101). (Fig.5)
Fig.5
Step 3A B C D E F G H I J K L M . . . . 1 6 0 0 0 1 8 6 0 1 3 0 0 (1) - 1 6 1 6 0 0 0 0 2 6 0 1 3 0 0 (1) - 1 6 1 6 0 1 1 0 1 0 0 1 3 0 0
Convert Hexadecimal to Base 10This is an advanced soroban problem using the multiplication techniques taught byTakashi Kojima.
The technique for converting hexadecimal to base 10 is very much as the one above but in reverse. In other words instead of dividing we multiply in the reverse order.
- Where applicable use the Conversion Table to express the hexadecimal in numbers.
- Set numbers onto the soroban making sure to set unit numbers on unit rods.
- Work left to right starting with the left-hand most group.
- Multiply each group setting the product on the right.
- Move to the next group and multiply again.
All this is best explained by example. It's not at all complicated. Here's a simple example first.
Example 1: Convert the hex D0A to Base 10
Step 1: Expressed as a decimal (base 10) number, hexadecimal D0A becomes 13 00 10. Set each group of numbers in such a way that each unit number falls on a unit rod. In this case 13 on rods GH, 00 on JK & 10 on MN. Set the multiplier 16 on rods AB.
Fig.6
Step 1A B C D E F G H I J K L M N O . . . . . 1 6 0 0 0 0 1 3 0 0 0 0 1 0 0
Step 2: Starting with the left most group multiply 3 on H by 16. Add the product 48 to rods JK. Clear 3 from H.
2a: Multiply 1 on rod G by 16. Add the product 16 to rods IJ. Clear 1 from rod G. (Fig.7)
Fig.7
Steps 2 & 2aA B C D E F G H I J K L M N O . . . . . 1 6 0 0 0 0 1 3 0 0 0 0 1 0 0 + 4 8 clear (-3) 1 6 0 0 0 0 1 0 0 4 8 0 0 0 0 + 1 6 clear (-1) 1 6 0 0 0 0 0 0 2 0 8 0 1 0 0
Step 3: Now move right to the next group and multiply 8 on K by 16. Add the product 128 to rods LMN. Clear 8 from K.
3a and the answer: Multiply 2 on I by 16. Add the product 32 to rods KL. Clear 2 from I and that completes the problem. Hexadecimal D0A converts to its base 10 number 3338. (Fig.8)
Fig.8
Steps 3 & 3aA B C D E F G H I J K L M N O . . . . . 1 6 0 0 0 0 0 0 2 0 8 0 1 0 0 + 1 2 8 clear (-8) 1 6 0 0 0 2 0 0 2 0 0 1 3 8 0 + 3 2 clear (-2) 1 6 0 0 0 0 0 0 0 0 3 3 3 8 0
Example 2: Convert the hex B4E8 to Base 10
Step 1: Expressed as a decimal (base 10) number, hexadecimal B4E8 becomes 11 04 14 08. As before set each group so that each unit number falls on a unit rod; in this case 11 on DE, 04 on GH, 14 on JK & 08 on MN. (Fig.9)
Fig.9
Step 1A B C D E F G H I J K L M N O . . . . . 1 6 0 1 1 0 0 4 0 1 4 0 0 8 0
Step 2: Starting with the left most multiply 1 on E by16. Add the product 16 to rods GH. Clear 1 from rod E.
2a: Multiply 1 on D by 16. Add the product 16 to rods FG. Clear 1 from rod D. (Fig.10)
Fig.10
Steps 2 & 2aA B C D E F G H I J K L M N O . . . . . 1 6 0 1 1 0 0 4 0 1 4 0 0 8 0 + 1 6 clear (-1) 1 6 0 1 0 0 2 0 0 1 4 0 0 8 0 + 1 6 clear(1) 1 6 0 0 0 1 8 0 0 1 4 0 0 8 0
Step 3: Now move right to the next group to and multiply 8 on Gt by 16 Add the product 128 to rods HIJ. Clear 8 from G.
3a: Multiply 1 on rod F by 16. Add the product 16 to rods HI. Clear 1 from rod F. (Fig.11)
Fig.11
Steps 3 & 3aA B C D E F G H I J K L M N O . . . . . 1 6 0 0 0 1 8 0 0 1 4 0 0 8 0 + 1 2 8 clear (-8) 1 6 0 0 0 1 0 1 2 9 4 0 0 8 0 + 1 6 clear (-1) 1 6 0 0 0 0 0 2 8 9 4 0 0 8 0
Step 4: Move to the next group and multiply 4 on rod K by 16. Add the product 64 to rods MN. Clear 4 from K.
4a: Multiply 9 on J by 16 and add the product 144 to KLM. Clear 9 from J.
4b: Multiply 8 on rod I by 16, add the product 128 to JKL. Clear 8 from I
4c and the answer: Multiply 2 on on H by16 and add the product 32 to rods IJ. Clear 2 from H and that completes the problem. Hexadecimal B4E8 converts to its base 10 number 46312.
Fig.12
Steps 4 4a 4b & 4cA B C D E F G H I J K L M N O . . . . . 1 6 0 0 0 0 0 2 8 9 4 0 0 8 0 +6 4 clear (-4) 1 6 0 0 0 0 0 2 8 9 0 0 7 2 0 + 1 4 4 clear (-9) 1 6 0 0 0 0 0 2 8 0 1 5 1 2 0 + 1 2 8 clear (-8) 1 6 0 0 0 0 0 2 0 1 4 3 1 2 0 + 3 2 clear (-2) 1 6 0 0 0 0 0 0 0 4 6 3 1 2 0
Here are two more exciting ways to convert decimal to binary
▪ Decimal/Binary Conversion on the Soroban
▪ Conversion from Decimal to BinaryHere are two more exciting ways to convert hex to decimal
▪ Convert Hex to Decimal by Dividing by Hex A
▪ Decimal Hexadecimal Conversion
A word of thanksI hadn't really thought much about converting back and forth between base 10 and hex before last September when one day Shane Baggs shared his techniques in a post to members of the Yahoo Soroban/Abacus group. Since then with a little study and working closely with Shane's, Steve's and Marcos' techniques, I've learned more about converting hex, base 10 and binary than I ever thought possible. I love new stuff.
Sincere thanks, Totton Heffelfinger
REFERENCES:Baggs, Shane
Swift Division (Hexadecimals)
Yahoo Soroban/Abacus group
September, 2007Østergaard, Erik
Hexadecimal Number SystemWikipedia:
Hexadecimal
Binary Numeral SystemConversion Tools on the Web:
Hexadecimal conversion
Decimal/ Binary Conversion Tool
Hexa to Decimal and Binary ConverterKojima, Takashi.
The Japanese Abacus: Its Use and Theory
Tokyo: Charles E. Tuttle, 1954Print Page (.PDF Format, 108kb)
▪ Adding Hexadecimal numbers
▪ Abacus: Mystery of the Bead
▪ Advanced Abacus Techniques© November, 2007
Totton Heffelfinger Toronto Ontario Canada
totton[at]idirect[dot]com