MSU/CSE 260 Fall 20091 1 Integer Representations & Base Conversion Section 3.6.

MSU/CSE 260 Fall 2009 11

Integer Representations &

Base ConversionSection 3.6


Object Representation

Objects/Concepts Representation

A one-to-one function is required

.

.


Object Representation…

Need symbols; how many? What are other issues?

Physical cost of representation Representation and algorithms

We’ll concentrate on “number” representation.


ASCII code for characters

CHAR CODE CHAR CODE

A 001000001 a 001100001 B 001000010 b 001100010 C 001000011 c 001100011 + 000101011 . 000101110 * 000101010 % 000100101


Encoding baseball teams

A’s 000 Cardinals 001 Mariners 010 Orioles 011 Tigers 100 Twins 101Which set of teams?011100101000

A’s 1 Cardinals 01 Mariners 0000 Orioles 0001 Tigers 0010 Twin 0011Which set of teams?01100010011

Fixed size 3-bit code Variable size coding


2^n strings of exactly n bits each in length: uniform length code

n=1: {0,1} n=2: {00, 01, 10, 11} n=3: {000,001,010,011,

100,101,110,111} n=8: 256 unique strings (character set) n=10: 1024 strings n=30: over one billion strings n=33: unique string for every person n=64: new IP address size


Physical reps for 0 and 1

Up versus Down 5 volts versus 0 volts Hot versus cold Pit versus no pit on CD surface Red checker versus black checker


CD has many tracks of spots

Shiny spot is 0 Burned spot is 1 5 billion spots total 700 MB total 44,000 x 16 spots

for just 1 second of high fidelity music

Laser reads shiny versus dull spots

CD

CD spins


Example: Numbers Using only one symbol

1

2

3

12

.

.

.

▲

▲ ▲

▲ ▲ ▲

▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲

.

.

.

As long as we have a one-to-one function, we have a correct presentation

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Example: Using two symbols

1

2

3

12

.

.

.

▲

■ ▲

▲ ■

▲ ■ ▲ ▲

.

.

.


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Example: Using two symbols

1

2

3

12

.

.

.

▲

▲ ▲

▲ ▲ ▲

▲ ■ ▲ ▲

.

.

.


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Understanding Number Representation

We usually use {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} as base to write numbers, but we can use any base.

Remember that :

The above is a positional system representation

It has a “compact” manual There are other representations

2 1 0453 4 5 310 10 10

MSU/CSE 260 Fall 2009 1313

Representation of Integers… Theorem: Let b be a positive integer > 1. We

can write every positive integer uniquely in the following form:

where k is a positive integer and a0, a1, …ak {0, 1, 2, …, b – 1} and ak ≠ 0.

Example:

1 1 01 1 0

k kk kn b b ba a aba

2 1 0

2 1 02 1 0

453 4 10 5 10 3 10

n a b a b a b

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Base conversion

To convert a number, say 6543, in base 10 to an equivalent number in another base, say 7, we need to come up with the coefficients in the following expression:

It seems that we have only one equation but many unknowns.

We use the restriction that the coefficients are all integers between 0 and 6.

A series of divisions would do

4 33

1 014

22 06543 7 7 7 7 7a a a a a

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Example: Base conversion

4 3 2 1 04 3 2 0

4 3 2 0

4 3 2 0

4 3 2

1

4 3 2 11

3 21

Re

3 21

24 13 2

6543 7 7 7 7 7

6543 7 7 7 7

6543 ( 7 7 7 )7

6543 934 7 5

934 7 7 7

934 ( 7 7 )7

and so on...

mainderQuotient

a a a a a

a a a a a

a a a a a

a a a a

a a a a

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1 3 2 1 01 3 2 1 0

1 2 2 11 3 2 1 0

2 3 11 3 2 1 0

( )

(( ) )

k kk k

k kk k

k kk k

n a a a a a a

n a a a a a a

n a

b b

a a

b b b b

b b

a a

b b b

b ab b b b

Integer n and base b

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From decimal to another base

0

while ( 0)

mod

1

end while

k

q n

k

q

a q b

qq

b

k k

is to be converted

is desired new base

is th digit of number

( goes from right to left)k

n

b

a

k

k

The base b expansion of n is

(ak-1…a1a0)b

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Example: Decimal 50 to base 2

50 = 25*2 + 0

25 = 12*2 + 1

12 = 6*2 + 0

6 = 3*2 + 0

3 = 1*2 + 1

1 = 0*2 + 1

So, the base 2 representation is 110010

MSU/CSE 260 Fall 2009 1919

Conversion from decimal to octal

10

0

1

2

10 8

What is 229 in base 8?

229 mod 8 5 (229 8 28 5) 5

229 /8 28

28 mod 8 4 (28 8 3 4) 4

28 /8 3

3 mod 8 3 3

So, 229 345

a

a

a

MSU/CSE 260 Fall 2009 2020

Example: Decimal 229 to base 8

229 = 28*8 + 5

28 = 3*8 + 4

3 = 0*8 + 3

So, decimal 229 is 345 in base 8

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Bases used in Computer Science Base 2 (Binary),

bits in computer, has two symbols {0,1} Base 8 (Octal),

has eight symbols {0,1,2,3,4,5,6,7} Base 16 (Hexadecimal)

has sixteen symbols {0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F}.

Note that A=10, B=11, C=12, D=13, E=14, F=15.

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Exercise

Convert 6.96 to binary, up to 5 places First we convert 6 to binary, which is 110 0.96 ×2 = 1.92, so the first digit after binary

point is 1 0.92 ×2 = 1.84, so the 2nd digit after binary point

is 1 0.84 ×2 = 1.68, so 3rd digit after binary point is 1 0.68 ×2 = 1.36, so the 4th digit after binary point

is 1 0.36 ×2 = 0.72, so the 5th digit after binary point

is 0 …. So, 6.96 is 110.11110…. in binary

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Converting Bases Related by Powers

001101 000 11010 0111

13 1 10 7

D 1 A 7

1 5 0 6 4 7Octal (base 8)

Binary (base 2)

Hexadecimal (base 16)

53671 base 10

Since 8 = 23 then every 3 binary digits makes an Octal digit.

Since 16=24 then every 4 binary digits makes a Hexadecimal digit

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Representation of integers in different bases

Hexadecimal, Octal and Binary Representation of Integers

Base Representation

10 0 1 2 3 4 5 6 7 8 910

11

1213

14

15

16 0 1 2 3 4 5 6 7 8 9 A B C D E F

8 0 1 2 3 4 5 6 7 1011

12

13

1415

16

17

2 0 1 10 11 100 101 110 111 1000100

1101

0101

11100

1101

1110

1111

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Algorithms for Integer Operation Computers use binary numbers to

perform operations in integers. In order to handle integer operations in

binary system first we should convert the decimal numbers into integers and then do the operations using binary numbers.

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Addition of Integers in Base 2 A procedure to perform addition is based on the

usual method for adding numbers with pencil and paper.

Example: 1 1 carry bits

1 1 1 0 (14)10

+ 1 0 1 1 (11)10

= 1 1 0 0 1 (25)10

An overflow may happen when adding two numbers.

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Multiplying Integers in Base 2 Example

1 1 0 (6)10

1 0 1 (5)10

1 1 0 + 0 0 0 1 1 0___ = 1 1 1 1 0 (30)10

Up to 2n bits may be needed to represent the product of two n-bit numbers.

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More Examples on Base Arithmetic

Compute 222 + 222 in base 3. Answer:

(1221)3


(1110)4


(444)5


A MATLAB function for dec to base

function rep = decimal2base ( decimal, base )%% Convert input decimal number to rep in base.% Output is an array of digits 0,1, ... base-1.%position = 1;rep(1) = 0;while decimal>0 digitValue = mod(decimal,base); rep(position) = digitValue; % stack it in array position = position+1; decimal = floor(decimal/base); end% Flip the output array so higher order digit are at "left".rep = fliplr(rep);


results

>> decimal2base(983, 5)

ans = 1 2 4 1 3


ans = 3 3 0 3


ans = 2 5 0 3 5


ans = 1 0 0 1 1 1 0


Horner’s rule for computing polynomial Consider the compiler processing the

STRING x = 983; Compute number value as characters are

read val = 0; val = val*10 + 9 val = val*10 + 8 val = val*10 + 3 O(N) * ops and O(N) + ops Computing all the powers separately would

require O(N2) * ops (n+(n-1)+ … + 2+1)


Factoring the polynomial p(x)

P(x) = anxn + an-1xn-1 + … + a2x2 + a1x + a0

P(x) = (anxn-1 + an-1xn-2 + … + a2x + a1) x + a0

P(x) = ((anxn-2 + an-1xn-3 + … + a2) x + a1) x + a0

P(X) = ((…((anx + an-1) x + an-2 ) x … + a1) x + a0

This shows a simple repetition of n multiplies and n adds (Horner’s rule).

It could take n multiplies just to compute the term anxn if a simple loop were used for this. (If we have a fixed x, such as a base, we could store its powers in an array and not compute them each time.)


The following example shows a hash function using this polynomial form

The key is a string of characters – any length.

The characters are the coefficients of the polynomial.

The value of x, or the base value, is 32, which makes multiplication very fast via shifting left 5 bits

For speed, the logical and ‘∧’ is used instead of ‘*’ and also for the addition!

Many tests have shown this function to be fast and uniform over the table size.


Hash function effort O(n): n = #charsint TABLESIZE = 10000; // default size

// Hash function from figure 19.2 of old Weiss text (page 611).// This function mixes the bits of the key to produce a pseudo// random integer between 0 and TABLESIZE-1

unsigned int Hash(const string& KEY, const int tableSize){ unsigned int HashValue = 0; // compute value of polynomial P(X), where KEY[i] are the coefficients // and X=32; multiplying by X is done by a left shift of 5 bits // and arithmetic is "abbreviated" using bit operations.

for( int i=0; i<KEY.length(); i++ ) // use every character of key // Horner's rule. Also, use the bitwise OR instead of + op // for speed and overflow suppression { HashValue = ( HashValue << 5 ) ^ KEY[i] ^ HashValue; } return HashValue % tableSize;}


C++ details: http://docs.hp.com/en/B3901-90013/ch05s04.html

exp1 << exp2 Left shifts (logical shift) the bits in exp1 by exp2 positions.

exp1 >> exp2 Right shifts (logical or arithmetic shift) the bits in exp1 by exp2 positions.

exp1 & exp2 Performs a bitwise AND operation.

exp1 ^ exp2 Performs a bitwise OR operation.

exp1 | exp2 Performs a bitwise inclusive OR operation.

~exp1 Performs a bitwise negation (one's complement) operation.


Program to call hash function

int main(){ string plainText; int digitalSignature;

cout << "\n-----+----- Crypto/Hashing Demonstration -----+-----\n"; cout << "\n====== Give TABLESIZE (between 10 and 1000000): "; cin >> TABLESIZE;

while ( 1 ) { cout << "\nGive plain text string = " ; cin >> plainText; if ( plainText == "quit") break;

digitalSignature = Hash(plainText, TABLESIZE); cout << "\nYour digital signature is: " << digitalSignature << endl; } return 0;


Using a polynomial hash function-----+----- Crypto/Hashing Demonstration -----+-----

====== Give TABLESIZE (between 10 and 1000000): 20

Give plain text string = harrycarrieYour digital signature is: 10

Give plain text string = spartypartyYour digital signature is: 3

Give plain text string = spartyYour digital signature is: 13

Give plain text string = partyYour digital signature is: 2

Give plain text string = 122345678901234567890!@#$%^&&*Your digital signature is: 13

MSU/CSE 260 Fall 20091 1 Integer Representations & Base Conversion Section 3.6.

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Transcript of MSU/CSE 260 Fall 20091 1 Integer Representations & Base Conversion Section 3.6.