1. 1 Information Representation in Computer Lecture Nine

2. 2 Outline  Everything is a number (in computer, c.f. Pythagoras)  Binary number system  Negative integers  Floating point numbers (reals)  Accuracy and limitation in “computer numbers”  ASCII code for characters  Code for programs, graphics, music, etc

3. 3 Bit, Byte, and Word 0 0 1 1 01 1 0 0 0 1 1 10 1 1 00 1 1 11 0 0 1 1 1 1 0 0 0 1 1 0 0 0 1 0 0 cont’d A bit is a size that can store 1 digit of a binary number, 0 or 1. A byte is 8 bits, which can store eight 0’s or 1’s. A word is either 32 or 64 bits, depending on computers. Regular PC’s are 32-bit word in size, higher-end workstations are 64-bit. Word size is the size of the registers. What do these bits mean is a matter of interpretation! All information in a computer are represented in a uniform format of bit patterns.

4. 4 Binary Nonnegative Integers Given a 32-bit pattern, e.g., 0000 0000 … 0000 1101 1100, it can represent the integer (if you interpret it that way) Note that if we use 32-bit word, the smallest number is 0, and the largest number is 11111111 11111111 11111111 11111111, which is 2 32 -1=4294967295. Numbers bigger than this cannot be represented. If such things happen in a calculation, we say it overflowed. Such interpretation is called unsigned int in the programming language C. a 0 lowest bit a 31 highest bit

5. 5 Negative Integers  To represent negative numbers, we’ll agree that the highest bit a 31 representing sign, rather than magnitude, 0 for positive, 1 for negative numbers.  More precisely, all modern computers use 2’s complement representations.  The rule to get a negative number representation is: first write out the bit pattern of the corresponding positive number, complement all bits, then add 1.

6. 6 Property of 2’s Complement Numbers  Two’s complement m of a number n is such that adding it together you get zero (m + n = 0, modulo word size)  Thus m is interpreted as negative of n.  The key point is that computer has a finite word length, the last carry is thrown away.

7. 7 2’s Complement, example  1 in bits is 00000000 00000000 00000000 00000001 complementing all bits, we get 11111111 11111111 11111111 11111110 Add 1, we get representation for –1, as 11111111 11111111 11111111 11111111 00000000 00000000 00000000 00000001 + 11111111 11111111 11111111 11111111 1 00000000 00000000 00000000 00000000 Decimal 1 –1 0 Overflow bit not kept by computer

8. 8 2’s Complement, another e.g.  125 in bits is 00000000 00000000 00000000 01111101 complement all bits, we get 11111111 11111111 11111111 10000010 Add 1, we get representation for –125, as 11111111 11111111 11111111 10000011 00000000 00000000 00000000 01111111 + 11111111 11111111 11111111 10000011 1 00000000 00000000 00000000 00000010 Decimal 127 –125 +2 Overflow bit not kept by computer

9. 9 Dual Interpretation of Bit Patterns 0000 0000 = 0 0000 0001 = 1 0000 0010 = 2 0000 0011 = 3 0000 0100 = 4 0000 0101 = 5 0000 0110 = 6... 0111 1111 = 127 1000 0000 = 128 or -128 1000 0001 = 129 or -127 1000 0010 = 130 or -126 1000 0011 = 131 or -125... 1111 1100 = 252 or -4 1111 1101 = 253 or -3 1111 1110 = 254 or -2 1111 1111 = 255 or -1 A byte (or char in C) Unsigned char Signed char

10. 10 Signed and Unsigned Int  If we interpret the number as unsigned, an unsigned integer takes the range 0 to 2 32 -1 (that is 000…000 to 1111….1111)  If we interpret the number as signed value, the range is –2 31 to 2 31 -1 (that is 1000….000 to 1111…111, to 0, to 0111….1111).  Who decide what interpretation to take? What if we need numbers bigger than 2 32 ?

11. 11 Addition, Multiplication, and Division in Binary 0 0 0 1 0 0 1 1 0 0 1 0 0 1 0 1 0 0 1 1 1 0 0 0 + 0 1 1 1 0 1 0 1 0 1 1 1 1 0 0 0 1 1 +  1 0 0 1 1 0 0 0 1 0 0 1 0 1 0 1 0 0 0 1 0 1 0 1 0 0 0 1 0 – – 75=35 74  8 = 9 remainder 2 19+37=56

12. 12 Floating Point Numbers (reals)  To represent numbers like 0.5, 3.1415926, etc, we need to do something else. First, we need to represent them in binary, as E.g. 11.00110 for 2+1+1/8+1/16=3.1875  Next, we need to rewrite in scientific notation, as 1.100110 2 1. That is, the number will be written in the form: 1.xxxxxx…  2 e x = 0 or 1

13. 13 Real Numbers on Computer 0 0.5 1 2 3 4 5 6 7 ε Example for β=2, p=3, e min = -1, e max =2 On modern computers the base β is exclusively 2.

14. 14 IEEE 754 standard  The IEEE standard for floating numbers specifies for 32-bit single precision (float in C) as follows:  The highest 31-th bit a 31 is used for sign, 0 for positive, 1 for negative  8 bits, from position 23 to 30, are used to store an integer, for the biased exponent e+127.  The rest 23 bits are used for the significand. signexponentsignificandbit 0 bit 31

15. 15 IEEE Floating Point signexponentsignificandbit 0 a 31 s -1 s -2 s -3 s -23 e0e0 e1e1 e7e7 Bit 31 (a 31 ) specifies the sign Bit 23 to 30 (e 7,e 6,…,e 1,e 0 ), interpreted as an 8-bit unsigned integer, specifies the biased exponent E = e + 127. Bit 0 to 22 (s -1,s -2,…,s -22,s -23 ) specifies the fractional part of number in scientific notation. The value of the pattern is: However, the number +0 is a string of all zeros.

16. 16 Example of Floating Point Number signexponentsignificand bit 31 bit 0 0 0 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 The sign is positive, the biased exponent E=127=e+127, so the original exponent e is 0. The significand f is 0. Thus the pattern represents: +(1.0 + 0)2 0 = 1.0 signexponentsignificand bit 31 bit 0 1 0 1 1 1 1 1 1 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 The sign is negative, the biased exponent E=126=e+127, so the original exponent e is –1. The significand f =1/2+1/8. Thus the pattern represents: –(1.0 + f)2 -1 = –0.8125.

17. 17 Encoding to IEEE Format  What is the bit pattern of the real number 10.25 in IEEE single precision (32-bit) format?  Ans: first, we need to convert the number to binary. We can consider separately the integer 10, and the fractional part 0.25. For 10, we have 10 = 52 + 0 5 = 22 + 1 2 = 12 + 0 1 = 02 + 1 so 10 10 =1010 2 For 0.25, we multiply by 2 (repeatedly) and take the integer part as the digit: 20.25 = 0.5 -> 0 20.5 = 1.0 -> 1 20.0 = 0.0-> 0 So the fractional part is 0.25 10 =0.010000 2 Thus 10.25 10 =1010.01 =1.010012 3

18. 18 10.25 in bits  10.25 10 =1.01001 2 2 3  Thus sign is positive, f =.01001, e=3, and E=e+127=130=10000010 2. The bit pattern is signexponentsignificand bit 31 bit 0 0 1 0 0 0 0 0 1 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 fE=e+127

19. 19 IEEE 754 (32-bit) Summary  The bit pattern represents  If e = 0: (-1) s  f 2 -126  If 0<e<255: (-1) s (1+f) 2 e-127  If e=255 and f = 0: + or - ∞ and f ≠ 0: NaN. … s e f = b -1 2 -1 + b -2 2 -2 + … + b -23 2 -23 b -1 b -2 b -23

20. 20 Limitations in 32-bit Integer and Floating Point Numbers  Limited range of values (e.g. integers only from –2 31 to 2 31 –1)  Limited resolution for real numbers. E.g., if x is a machine representable value, the next value is x + ε (for some small ε). There is not value in between. This causes “floating point errors” in calculation. The accuracy of a single precision floating point number is about 6 decimal places.

21. 21 Limitations of Single Precision Numbers  Given the representation of the single precision floating point number format, what is the largest magnitude possible? What is the smallest number possible?  With floating point number, it can happen that 1 + ε = 1. What is that largest ε?

22. 22 Double Precision  An IEEE double precision takes twice the number of the bits of the single precision number, 64 bits.  1 bit for the sign  11-bit for exponent  52-bit for significand  The double precision has about 15 decimal places of accuracy

23. 23 Floating Point Rounding Error  Consider 4-bit mantissa floating point addition:  1.0102 2 + 1.1012 -1 1.010000  2 2 0.001101  2 2 1.011101  2 2 1.100  2 2 + Shift exponent to that of the large number Round to 4 bits In decimal, it means 5.0 + 0.8125  6.0

24. 24 Representation of Characters, the ASCII code 0 1 2 3 4 5 6 7 8 9 A B C D E F 0 NUL SOH STX ETX EOT ENQ ACK BEL BS HT LF VT FF CR SO SI 1 DLE DC1 DC2 DC3 DC4 NAK SYN ETB CAN EM SUB ESC FS GS RS US 2 SP ! " # $ % & ' ( ) * +, -. / 3 0 1 2 3 4 5 6 7 8 9 : ; ? 4 @ A B C D E F G H I J K L M N O 5 P Q R S T U V W X Y Z [ \ ] ^ _ 6 ` a b c d e f g h I j k l m n o 7 p q r s t u v w x y z { | } ~ DEL How to read the table: the top line specifies the last digit in hexadecimal, the leftmost column specifies the higher value digit. E.g., at location 41 16 (=0100 0001 2 =65 10 ) is the letter ‘A’.

25. 25 Base-16 Number, or Hexadecimal  Instead of writing out strings of 0’s and 1’s, it is easier to read if we group them in group of 4 bits. A four bit numbers can vary from 0 to 15, we denote them by 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F. binary hexadecimal decimal 0000 0 0 0001 1 1 0010 2 2 0011 3 3 0100 4 4 0101 5 5 0110 6 6 0111 7 7 binary hexadecimal decimal 1000 8 8 1001 9 9 1010 A 10 1011 B 11 1100 C 12 1101 D 13 1110 E 14 1111 F 15

26. 26 Program as Numbers  High level programming language C = A + B;  Assembly language add $5, $10, $3  Machine code (MIPS computer) bit 0 bit 31 0 0 0 0 0 0 0 1 0 1 0 0 0 0 1 1 0 0 1 0 1 0 0 0 0 0 1 0 0 0 0 0 Not used source 1 source 2 destination not used operation $10$3$5add

27. 27 Graphics as Numbers A picture like this is also represented on computer by bits. If you magnify the picture greatly, you’ll see square “pixels” which is either black or white. These can be represented as binary 1’s and 0’s. Color can also be presented with numbers, if we allow more bits per pixel.

28. 28 Music as Numbers – MP3 format  CD music is sampled 44,100 times per second (44.1 kHertz), each sample is 2 bytes long  The digital music signals are compressed, at a rate of 10 to 1 or more, into MP3 format  The compact MP3 file can be played back on computer or MP3 players

29. 29 Summary  All information in computer are represented by bits. These bits encode information. It’s meaning has to be interpreted in a specific way.  We’ve learned how to represent unsigned integer, negative integer, floating pointer number, as well as ASCII characters.