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Number Bases Informatics INFO I101 February 9, 2004 John C. Paolillo, Instructor

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Items for Today Last week –Digital logic, Boolean algebra, and circuits –Logic gates and truth tables This Week –Numbers and bases –Working with binary

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Number Base Systems

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The Format of a Base System ##### … ##### … b4b4 b3b3 b2b2 b1b1 b0b0 … b -1 b -2 b -3 b -4 b -5 … The number represented is the sum of all the products of the digit values and their respective place values

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Common Bases Decimal (Base 10) Binary (Base 2) Octal (Base 8) Hexadecimal (Base 16)

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Conversion to Base 10 Identify each of the places in the new number base. These will correspond to the powers of the base, for example, with base 2, they are 1, 2, 4, 8, 16, 32, etc. Multiply the value for each place by the value of the digit appearing there; Add the results up, and you have the result in decimal Note that if you divide and add correctly, you can reverse this procedure to convert decimal into another base. It’s harder, because you’re not used to using the appropriate addition and multiplication tables.

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Try out these examples What is 10011 Base 2 in decimal? 1 16+ 0 8 + 0 4 + 1 2 + 1 1 = 19 What is 121 Base 8 in decimal? 1 64 + 2 8 + 1 1 = 81 What is 247 Base 10 in Binary? Here it helps to have a different procedure…

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Converting to Binary 2424 23232 2121 2020 2727 2626 2525 2828 1684211286432 256 2472560 2471281 119641 55321 23161 780 741 321 101110111 111 What we’re converting

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Octal — base 8 Sixteen digits: 0, 1, 2, 3, 4, 5, 6, 7 7 = 111 two = 7 eight Octal values are usually not specially indicated Unix example:chmod 666 myfile.html

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Octal Digits Octal 0 1 2 3 4 5 6 7 Binary 000 001 010 011 100 101 110 111 Decimal 0 1 2 3 4 5 6 7

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Octal Tips each octal digit corresponds to three binary digits (bits) convert binary to octal by parsing each group of three bits into one octal digit convert octal to binary by translating each digit into three bits Examples: 764 eight =111101100 two 011011101 two = 335 eight

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Hexadecimal — base 16 Sixteen digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F 15 = 1111 two = F sixteen Hexadecimal (“hex”) values are usually indicated by a preceding base marker HTML: #FFFFFF JavaScript, C: 0xF1AD

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Hexadecimal Digits Hex 0 1 2 3 4 5 6 7 8 9 A B C D E F Binary 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 Decimal 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

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Hexadecimal Tips each hex digit corresponds to four binary digits (bits) convert binary to hex by parsing each group of four bits into one hex digit convert hex to binary by translating each digit into four bits Two hex digits make up one byte, a very common unit of memory

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IP Numbers IP = “Internet Protocol” 129.79.142.114 dhcp-Memorial–142-114.memorial.indiana.edu IP Number Host Name A Domain Name Server (DNS) has a database that matches IP and host name

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The IP Number Four Fields 0-255 in each field This is really base 256, but we use decimal numbers in each digit 129.79.142.114 Net Subnet Node

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Binary Addition

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Adding in Binary Zero plus any other values leaves that value Identity value for addition No carry is generated One plus one leaves zero and causes a carry (one) to the next digit Each successive digit must accept the carry from the previous

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Try these calculations 01001101 Base 2 + 00011011 Base 2 01111111 Base 2 + 00000001 Base 2 102 Base 8 + 121 Base 8 01001101 Base 2 00000100 Base 2 0000101 Base 2 0000101 Base 2 01111111 Base 2 00011011 Base 2

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Addition: Truth Tables CI 0 1 A01010101A01010101 B00110011B00110011 SCO S01101001S01101001 0 1 0 1

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Addition: Half Adder +0001 00 01 10 S A B XOR01 0 0 1 1 1 0 C The half adder sends a carry, but can’t accept one So we need another for the carry bit 01 0 0 0 1 0 1

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Addition in Binary Two half-adders gives us a full adder –two inputs plus carry Adders are cascaded to permit adding binary numbers –Eight adders allows adding (0...256) + (0...256) in binary numbers –Overflow can happen (200 + 100) Binary adders are used to do other computations as well...

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Subtraction Complement Representations

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Subtraction –0001 00 01 –0100 Subtraction is asymmetrical That makes it harder We have to borrow sometimes 827 Minuend –223 Subtrahend =604 Difference/remainder

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When Subtraction is Easy 456 –123 333 999 –123 876 Subtraction is easy if you don’t have to borrow –i.e. if all the digits of the minuend are greater than (or equal to) all those of the subtrahend This will always be true if the minuend is all 9’s: 999, or 999999, or 9999999999 etc.

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Using Easy Subtraction Subtract the subtrahend from 999 (or whatever we need) (easy) Add the result to the minuend (ordinary addition) Add 1 (easy) Subtract 1000 (drop highest digit) Difference = Minuend + 999 – Subtrahend + 1 – 1000 This works for binary as well as decimal

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Easy Subtraction in Binary Subtract the subtrahend from 111 (or whatever we need) (easy) Add the result to the minuend (ordinary addition) Add 1 (easy) Subtract 1000 (drop highest digit) Difference = Minuend + 111 – Subtrahend + 1 – 1000

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Binary Subtraction Example 10010101 –01101110 ????????? 11111111 –01101110 10010001 This is the same as inverting each bit + 10010101 100100110 +1 100100111 –100000000 00100111 Regular addition Add one Now drop the highest bit (easy: it’s out of range) 00100111

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Subtraction Procedure Invert each bit Regular addition Add one Now drop the highest bit (easy: it’s out of range) Each of these steps is a simple operation we can perform using our logic circuits Bitwise XOR Cascaded Adders Add carry bit Drop the highest bit ( it overflows)

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Negative Numbers Invert each bit Add one These steps make the negative of a number in twos-complement notation Twos complements can be added to other numbers normally Positive numbers cannot use the highest bit (the sign bit) This is the normal representation of negative numbers in binary

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Negative Numbers in Binary 000000000 000000011 000000102 000000113 000001004 000001015 000001106 000001117 000010008 000010019 etc. 11111111–1 11111110–2 11111101–3 11111100–4 11111011–5 11111010–6 11111001–7 11111000–8 11110111–9 11110110–10 etc.

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Representations The number representation you use (encoding) affects the way you need to do arithmetic (procedure) This is true of all codes: encoding (representation) affects procedure (algorithm) Good binary codes make use of properties of binary numbers and digital logic

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A problem A computer program adds 20,000 and 20,000 and instead of 40,000, it reports – 25,566 No errors in encoding, decoding or addition How? Because the result is a negative number in twos-complement notation (highest bit = sign bit)

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How it works 20,000 base ten is 0100111000100000 binary 0100111000100000 0100111000100000 1001110001000000 Highest bit is set, so number is negative in twos complement notation: subtract one and invert to display 1001110001000000 – 1 = 1001110000111111 0110001111000000 = 25,566

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Bottom Line Representations themselves, as we use them, have limits. Interpretation depends on context two procedures (encoding/decoding and addition) may be in and of themselves correct, but conflict in their application to specific examples

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