Binary Numbers PHY 201 Blum 1 Why Binary

Binary Numbers PHY 201 (Blum) 1

Why Binary? n n n 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 PHY 201 (Blum) 2

Don’t sweat the small stuff n n For decimal numbers, fluctuations must be less than 0. 25 volts For binary numbers, fluctuations must be less than 1. 25 volts Decimal PHY 201 (Blum) Binary 0 volts 3

Range actually split in three High Forbidden range Low PHY 201 (Blum) 4

It doesn’t matter …. n n n Some of the standard voltages coming from a computer’s power are ideally supposed to be 3. 30 volts, 5. 00 volts and 12. 00 volts Typically they are 3. 28 volts, 5. 14 volts or 12. 22 volts or some such value So what, who cares PHY 201 (Blum) 5

How to represent big integers n n Use positional weighting, same as with decimal numbers 205 = 2 102 + 0 101 + 5 100 n n Decimal – powers of ten 11001101 = 1 27 + 1 26 + 0 25 + 0 24 + 1 23 + 1 22 + 0 21 + 1 20 = 128 + 64 + 8 + 4 + 1 = 205 n Binary – powers of two PHY 201 (Blum) 6

Converting 205 to Binary n 205/2 = 102 with a remainder of 1, place the 1 in the least significant digit position 1 n Repeat 102/2 = 51, remainder 0 0 PHY 201 (Blum) 1 7

Iterate n 51/2 = 25, remainder 1 1 n 1 0 1 25/2 = 12, remainder 1 1 n 0 1 12/2 = 6, remainder 0 0 PHY 201 (Blum) 1 1 8

Iterate n 6/2 = 3, remainder 0 0 n 1 1 0 1 0 1 3/2 = 1, remainder 1 1 n 0 0 0 1 1 1/2 = 0, remainder 1 1 PHY 201 (Blum) 1 0 0 1 1 9

Recap 1 1 0 0 1 1 27 + 1 26 + 0 25 + 0 24 + 1 23 + 1 22 + 0 21 + 1 20 205 PHY 201 (Blum) 10

Finite representation n n 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. PHY 201 (Blum) 11

Same number, different representation n n n 5 using 8 bits 0000 0101 5 using 16 bits 0000 0101 5 using 32 bits 0000 0000 0101 PHY 201 (Blum) 12

Adding Binary Numbers n 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 + PHY 201 (Blum) 1 3 3 7 9 5 4 13

Adding Binary Numbers carries 1 + PHY 201 (Blum) 1 1 1 0 1 0 0 0 1 1 1 0 0 1 0 14

Uh oh, overflow* n What if you use a byte (8 bits) to represent an integer 1 n *The 1 1 0 1 0 + 1 1 0 0 1 0 1 1 0 A byte may not be enough to represent the sum of two such numbers. End of the World as We Know It PHY 201 (Blum) 15

Biggest unsigned integers n n n 4 bit: 1111 15 = 24 - 1 8 bit: 1111 255 = 28 – 1 16 bit: 11111111 65535= 216 – 1 32 bit: 1111111111111111 4294967295= 232 – 1 Etc. PHY 201 (Blum) 16

Bigger Numbers n n You can represent larger numbers 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) PHY 201 (Blum) 17

Negative numbers n Negative x is the number that when added to x gives zero 1 1 n 1 0 1 1 1 0 1 0 1 1 0 0 0 0 0 Ignoring overflow the two eight-bit numbers above sum to zero PHY 201 (Blum) 18

Two’s Complement 0 n 1 0 1 0 Step 1: exchange 1’s and 0’s 1 n 0 1 0 1 Step 2: add 1 (to the lowest bit only) 1 PHY 201 (Blum) 1 0 19

Sign bit n n With the two’s complement approach, all positive numbers start with a 0 in the leftmost, 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 PHY 201 (Blum) 20

Add 1’s to the left to get the same negative number using more bits n n n n -5 using 8 bits 11111011 -5 using 16 bits 1111111011 -5 using 32 bits 111111111111111011 When the numbers represented are whole numbers (positive or negative), they are called integers. PHY 201 (Blum) 21

Biggest signed integers n n n 4 bit: 0111 7 = 23 - 1 8 bit: 01111111 127 = 27 – 1 16 bit: 011111111 32767= 215 – 1 32 bit: 01111111111111111 2147483647= 231 – 1 Etc. PHY 201 (Blum) 22

Most negative signed integers n n n 4 bit: 1000 -8 = - 23 8 bit: 10000000 - 128 = - 27 16 bit: 100000000 -32768= - 215 32 bit: 10000000000000000 -2147483648= - 231 Etc. PHY 201 (Blum) 23

Riddle 1 n n n 1 0 1 1 0 Is it 214? Or is it – 42? Or is it Ö? Or is it …? It’s a matter of interpretation n How was it declared? PHY 201 (Blum) 24

3 -bit unsigned and signed 7 6 5 4 3 2 1 0 1 1 0 0 PHY 201 (Blum) 1 1 0 0 1 0 1 0 3 2 1 0 -1 -2 -3 -4 0 0 1 1 1 0 0 1 0 1 0 Think of an odometer reading 999999 and the car travels one more mile. 25

Fractions n Similar to what we’re used to with decimal numbers 3. 14159 = 3 · 100 + 1 · 10 -1 + 4 · 10 -2 + 1 · 10 -3 + 5 · 10 -4 + 9 · 10 -5 11. 001001 = 1 · 21 + 1 · 20 + 0 · 2 -1 + 0 · 2 -2 + 1 · 2 -3 + 0 · 2 -4 + 0 · 2 -5 + 1 · 2 -6 (11. 001001 3. 140625) PHY 201 (Blum) 26

Places n 11. 001001 Two’s place One’s place Half’s place Fourth’s place Eighth’s place Sixteenth’s place PHY 201 (Blum) 27

Decimal to binary n 98. 61 n Integer part n n n n 98 / 2 49 / 2 24 / 2 12 / 2 6/2 3/2 1/2 = 49 = 24 = 12 = 6 = 3 = 1 = 0 remainder remainder 0 1 0 0 0 1 1 1100010 PHY 201 (Blum) 28

Decimal to binary n 98. 61 n Fractional part n n n n 0. 61 0. 22 0. 44 0. 88 0. 76 0. 52 2 2 2 = = = 1. 22 0. 44 0. 88 1. 76 1. 52 1. 04 . 100111 PHY 201 (Blum) 29

Decimal to binary n n Put together the integral and fractional parts 98. 61 1100010. 100111 PHY 201 (Blum) 30

Another Example (Whole number part) n 123. 456 n Integer part n n n n 123 / 2 61 / 2 30 / 2 15 / 2 7/2 3/2 1/2 = 61 = 30 = 15 = 7 = 3 = 1 = 0 remainder remainder 1 1 0 1 1 1111011 PHY 201 (Blum) 31

Checking: Go to All Programs/Accessories/Calculator PHY 201 (Blum) 32

Put the calculator in Programmer view PHY 201 (Blum) 33

Enter number, put into binary mode PHY 201 (Blum) 34

Another Example (fractional part) n 123. 456 n Fractional part n n n n n 0. 456 0. 912 0. 824 0. 648 0. 296 0. 592 0. 184 … 2 2 2 2 = = = = 0. 912 1. 824 1. 648 1. 296 0. 592 1. 184 0. 368 . 0111010… PHY 201 (Blum) 35

Checking fractional part: Enter digits found in binary mode Note that the leading zero does not display. PHY 201 (Blum) 36

Convert to decimal mode, then PHY 201 (Blum) 37

Edit/Copy result. Switch to Scientific View. Edit/Paste PHY 201 (Blum) 38

Divide by 2 raised to the number of digits (in this case 7, including leading zero) 1 2 3 4 PHY 201 (Blum) 39

Finally hit the equal sign. In most cases it will not be exact PHY 201 (Blum) 40

Other way around n n n Multiply fraction by 2 raised to the desired number of digits in the fractional part. For example 7 = 58. 368 n. 456 2 Throw away the fractional part and represent the whole number n 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. (Also the fraction is less than. 5 so there’s a zero in the ½’s place. ) n 0111010 PHY 201 (Blum) 41

Limits of the fixed point approach n n n 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 = 15. 9375 The smallest, non-zero number would be 0000. 0001 =. 0625 PHY 201 (Blum) 42

Floating point representation n Floating point representation allows one to represent a wider range of numbers using the same number of bits. It is like scientific notation. We’ll do this later in the semester. PHY 201 (Blum) 43

Hexadecimal Numbers n n n 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 PHY 201 (Blum) 44

Decimal Binary Hex n n n n 0 1 2 3 4 5 6 7 PHY 201 (Blum) 0000 0001 0010 0011 0100 0101 0110 0111 0 1 2 3 4 5 6 7 n n n n 8 1000 8 9 1001 9 10 1010 A 11 1011 B 12 1100 C 13 1101 D 14 1110 E 15 1111 F 45

Binary to Hex n n Break a binary string into groups of four bits (nibbles) Convert each nibble separately 1 1 1 0 0 1 E PHY 201 (Blum) C 9 46

Addresses n n With user friendly computers, one rarely encounters binary, but we sometimes see hex, especially with addresses To enable the computer to distinguish various parts, each is assigned an address, a number n n n Distinguish Distinguish PHY 201 (Blum) among computers on a network keyboard and mouse among files among statements in a program among characters in a string 47

How many? n n One bit can have two states and thus distinguish between two things Two bits can be in four states and … Three bits can be in eight states, … N bits can be in 2 N states 0 0 PHY 201 (Blum) 0 0 1 1 1 0 0 1 1 0 1 48

IP(v 4) Addresses n n n An IP(v 4) address is used to identify a network and a host on the Internet It is 32 bits long How many distinct IP addresses are there? PHY 201 (Blum) 49

Characters n n n We need to represent characters using numbers ASCII (American Standard Code for Information Interchange) is a common way A string of eight bits (a byte) is used to correspond to a character n n Thus 28=256 possible characters can be represented Actually ASCII only uses 7 bits, which is 128 characters; the other 128 characters are not “standard” PHY 201 (Blum) 50

Unicode n n Unicode uses 16 bits, how many characters can be represented? Enough for English, Chinese, Arabic and then some. PHY 201 (Blum) 51

ASCII n n n n n 0 00110000 1 00110001 … A 01000001 B 01000010 … a 01100001 b 01100010 … PHY 201 (Blum) (48) (49) (65) (66) (97) (98) 52

Booleans n n n A Boolean variable is something that is true or false Booleans have two states and could be represented by a single bit (1 for true and 0 for false) Booleans appearing in a program will take up a whole word in memory PHY 201 (Blum) 53

Boolean Operators n n n A. k. a. logical operators Have Boolean input and Boolean output Standard: n n n AND OR NOT XOR (either or but not both) NOR = NOT(OR) NAND = NOT(AND) PHY 201 (Blum) 54

Truth Tables n AND INPUT A 0 0 1 1 PHY 201 (Blum) B 0 1 OUTPUT A AND B 0 0 0 1 55

Truth Tables (Cont. ) n OR INPUT A 0 0 1 1 PHY 201 (Blum) B 0 1 OUTPUT A OR B 0 1 1 1 56

Truth Tables (Cont. ) n XOR (Excluded OR) INPUT A 0 0 1 1 PHY 201 (Blum) B 0 1 OUTPUT A XOR B 0 1 1 0 57

Numbers from Logic n n n 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 n n 0+0=0 0+1=1 1+0=1 1+1=0 PHY 201 (Blum) (with no carry) a carry) 58

Addition Truth Table INPUT OUTPUT Sum Carry A XOR B A AND B A B 0 0 0 1 1 0 1 0 1 PHY 201 (Blum) 59

All is NAND n n n Actually you can use one logic gate (the NAND) and a few tricks (like De Morgan’s theorem) to build all of the “combinatorial” circuitry (the circuitry that doesn’t involve memory) NORs work too But we tend to think in ANDs, ORs and NOTs PHY 201 (Blum) 60

Bit manipulation You can use an AND to select out part of a word (where s is a 1 or 0, etc) n s t u v w x y z 0 0 0 0 AND 1 1 gives s PHY 201 (Blum) t u v 0 61

IP Addresses Revisted n n La. Salle’s IP address is what’s called a Class B IP address Of the 32 bits the first two bits are 10 (this identifies us as Class B) The remaining 14 bits of the first two bytes identify us as La. Salle The remaining 2 bytes are for our internals use (to assign computers within La. Salle) PHY 201 (Blum) 62

In or Out n n n To see if an address is local to La. Salle, you would restrict your attention to the first two bytes. HOW? AND it with FFFF 0000 PHY 201 (Blum) 63

Subnets n n A network (like La. Salle’s) can be divided further into sub-networks Then subnet masks are used to determine whether or not another computer is on the same subnet PHY 201 (Blum) 64