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Binary Numbers. Material on Data Representation can be found in Chapter 2 of Computer Architecture (Nicholas Carter). Can’t get away from the noise. Why Binary?. Maximal distinction among values m inimal corruption from noise.
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Binary Numbers Material on Data Representation can be found in Chapter 2 of Computer Architecture (Nicholas Carter)
Why Binary? • Maximal distinction among values minimal corruption from noise. • Imagine taking the same physical attribute of a circuit, e.g. a voltage lying between 0 and 5 volts, to represent a number. • The overall range can be divided into any number of regions.
Don’t sweat the small stuff • For decimal numbers, fluctuations must be less than 0.25 volts. • For binary numbers, fluctuations must be less than 1.25 volts. 5 volts 0 volts Decimal Binary
Range actually split in three High Forbidden range Low
It doesn’t matter …. • Two of the standard voltages coming from a computer’s power supply are ideally supposed to be 5.00 volts and 12.00 volts • Measurements often reveal values that are slightly off – e.g. 5.14 volts or 12.22 volts or some such value. • So what, who cares.
How to represent big integers • Use positional weighting, same as with decimal numbers • 205 = 2102 + 0101 + 5100 • 11001101 = 127 + 126 + 025 + 024 + 123 + 122 + 021 + 120 = 128 + 64 + 8 + 4 + 1 = 205
Converting 205 to Binary • 205/2 = 102 with a remainder of 1, place the 1 in the least significant digit position • Repeat 102/2 = 51, remainder 0
Iterate • 51/2 = 25, remainder 1 • 25/2 = 12, remainder 1 • 12/2 = 6, remainder 0
Iterate • 6/2 = 3, remainder 0 • 3/2 = 1, remainder 1 • 1/2 = 0, remainder 1
Recap 127 + 126 + 025 + 024 + 123 + 122 + 021 + 120 205
Finite representation • Typically we just think computers do binary math. • But an important distinction between binary math in the abstract and what computers do is that computers are finite. • There are only so many flip-flops or logic gates in the computer. • When we declare a variable, we set aside a certain number of flip-flops (bits of memory) to hold the value of the variable. And this limits the values the variable can have.
Same number, different representation • 5 using 8 bits • 0000 0101 • 5 using 16 bits • 0000 0000 0000 0101 • 5 using 32 bits • 0000 0000 0000 0000 0000 0000 0000 0101
Adding Binary Numbers • Same as decimal; if the sum of digits in a given position exceeds the base (10 for decimal, 2 for binary) then there is a carry into the next higher position
Adding Binary Numbers carries 39 35 74
Uh oh, overflow • What if you use a byte (8 bits) to represent an integer • A byte may not be enough to represent the sum of two such numbers. 170 204 118???
Biggest unsigned* integers • 4 bit: 1111 15 = 24 - 1 • 8 bit: 11111111 255 = 28 – 1 • 16 bit: 1111111111111111 65535= 216 – 1 • 32 bit: 11111111111111111111111111111111 4294967295= 232 – 1 • Etc. *If one uses all of the bits available to represent only positive counting numbers, one is said to be working with unsigned integers.
Bigger Numbers • High-level languages often offer a hierarchy of types that differ in the number of bits used. • You can represent larger numbers than allowed by the highest type in the hierarchy by using more words. • You just have to keep track of the overflows to know how the lower numbers (less significant words) are affecting the larger numbers (more significant words).
Negative numbers • Negative x is the number that when added to x gives zero • Ignoring overflow the two eight-bit numbers above add up to zero x -x
Two’s Complement: a two-step procedure for finding -x from x • Step 1: exchange 1’s and 0’s • Step 2: add 1 (to the lowest bit only) x -x
Sign bit • With the two’s complement approach, all positive numbers start with a 0 in the left-most, most-significant bit and all negative numbers start with 1. • So the first bit is called the sign bit. • But note you have to work harder than just strip away the first bit. • 10000001 IS NOT the 8-bit version of –1
Add 1’s to the left to get the same negative number using more bits • -5 using 8 bits • 11111011 • -5 using 16 bits • 1111111111111011 • -5 using 32 bits • 11111111111111111111111111111011 • When the numbers represented are whole numbers (positive or negative), they are called just integers or signed.
3-bit signed and unsigned Think of driving a brand new car in reverse. What would happen to the odometer?
Biggest signed integers • 4 bit: 0111 7 = 23 - 1 • 8 bit: 01111111 127 = 27 – 1 • 16 bit: 0111111111111111 32767= 215 – 1 • 32 bit: 01111111111111111111111111111111 2147483647= 231 – 1 • Etc.
Most negative signed integers • 4 bit: 1000 -8 = - 23 • 8 bit: 10000000 - 128 = - 27 • 16 bit: 1000000000000000 -32768= - 215 • 32 bit: 10000000000000000000000000000000 -2147483648= - 231 • Etc.
Riddle • Is it 214? • Or is it – 42? • Or is it Ö? • Or is it …? • It’s a matter of interpretation • How was it declared?
Hexadecimal Numbers • Even moderately sized decimal numbers end up as long strings in binary. • Hexadecimal numbers (base 16) are often used because the strings are shorter and the conversion to binary is easier. • There are 16 digits: 0-9 and A-F.
0 0000 0 1 0001 1 2 0010 2 3 0011 3 4 0100 4 5 0101 5 6 0110 6 7 0111 7 8 1000 8 9 1001 9 10 1010 A 11 1011 B 12 1100 C 13 1101 D 14 1110 E 15 1111 F Decimal Binary Hex
Binary to Hex • Break a binary string into groups of four bits (nibbles). • Convert each nibble separately.
Numbers from Logic • All of the numerical operations we have talked about are really just combinations of logical operations. • E.g. the adding operation is just a particular combination of logic operations • Possibilities for adding two bits • 0+0=0 (with no carry) • 0+1=1 (with no carry) • 1+0=1 (with no carry) • 1+1=0 (with a carry)
Multiplication: Shift and add shift shift
Fractions • Similar to what we’re used to with decimal numbers
Converting decimal to binary II • 98.61 • Integer part • 98 / 2 = 49 remainder 0 • 49 / 2 = 24 remainder 1 • 24 / 2 = 12 remainder 0 • 12 / 2 = 6 remainder 0 • 6 / 2 = 3 remainder 0 • 3 / 2 = 1 remainder 1 • 1 / 2 = 0 remainder 1 • 1100010
Converting decimal to binary III • 98.61 • Fractional part • 0.61 2 = 1.22 • 0.22 2 = 0.44 • 0.44 2 = 0.88 • 0.88 2 = 1.76 • 0.76 2 = 1.52 • 0.52 2 = 1.04 • .100111
Another Example (Whole number part) • 123.456 • Integer part • 123 / 2 = 61 remainder 1 • 61 / 2 = 30 remainder 1 • 30 / 2 = 15 remainder 0 • 15 / 2 = 7 remainder 1 • 7 / 2 = 3 remainder 1 • 3 / 2 = 1 remainder 1 • 1 / 2 = 0 remainder 1 • 1111011
Another Example (fractional part) • 123.456 • Fractional part • 0.456 2 = 0.912 • 0.912 2 = 1.824 • 0.824 2 = 1.648 • 0.648 2 = 1.296 • 0.296 2 = 0.592 • 0.592 2 = 1.184 • 0.184 2 = 0.368 • … • .0111010…
Divide by 2 raised to the number of digits (in this case 7, including leading zero) 1 2 3 4
Finally hit the equal sign. In most cases it will not be exact
Other way around • Multiply fraction by 2 raised to the desired number of digits in the fractional part. For example • .456 27 = 58.368 • Throw away the fractional part and represent the whole number • 58 111010 • But note that we specified 7 digits and the result above uses only 6. Therefore we need to put in the leading 0 • 0111010
Fixed point • If one has a set number of bits reserved for representing the whole number part and another set number of bits reserved for representing the fractional part of a number, then one is said to be using fixed point representation. • The point dividing whole number from fraction has an unchanging (fixed) place in the number.
Limits of the fixed point approach • Suppose you use 4 bits for the whole number part and 4 bits for the fractional part (ignoring sign for now). • The largest number would be 1111.1111 = 15.9375 • The smallest, non-zero number would be 0000.0001 = .0625