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Understanding Integer Representations and Base Conversions in Computer Science

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This document provides an insightful overview of integer representations and base conversions, focusing on the importance of unique representations and one-to-one functions in encoding numbers. It explores various numeral systems, such as binary, octal, and hexadecimal, highlighting how numbers can be transformed and represented differently. Fundamental concepts, including the physical representation of data and the practical challenges in base conversions, are addressed, along with examples of converting numbers across different bases. Ideal for students and enthusiasts in computer science education.

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Understanding Integer Representations and Base Conversions in Computer Science

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  1. Integer Representations & Base Conversion Section 3.6 MSU/CSE 260 Fall 2009 1

  2. Object Representation Objects/Concepts Representation . . A one-to-one function is required 2 MSU/CSE 260 Fall 2009

  3. Object Representation… • Need symbols; how many? • What are other issues? • Physical cost of representation • Representation and algorithms • We’ll concentrate on “number” representation. 3 MSU/CSE 260 Fall 2009

  4. ASCII code for characters CHAR CODE CHAR CODE A 001000001 a 001100001 B 001000010 b 001100010 C 001000011 c 001100011 + 000101011 . 000101110 * 000101010 % 000100101 MSU/CSE 260 Fall 2009

  5. A’s 000 Cardinals 001 Mariners 010 Orioles 011 Tigers 100 Twins 101 Which set of teams? 011100101000 A’s 1 Cardinals 01 Mariners 0000 Orioles 0001 Tigers 0010 Twin 0011 Which set of teams? 01100010011 Encoding baseball teams Fixed size 3-bit code Variable size coding MSU/CSE 260 Fall 2009

  6. 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 MSU/CSE 260 Fall 2009

  7. 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 MSU/CSE 260 Fall 2009

  8. 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 MSU/CSE 260 Fall 2009

  9. 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 9 MSU/CSE 260 Fall 2009

  10. Example: Using two symbols ▲ ■ ▲ ▲ 1 ■ ▲ 2 ▲ ■ 3 ... ▲ 12 ... As long as we have a one-to-one function, we have a correct presentation 10 MSU/CSE 260 Fall 2009

  11. Example: Using two symbols ▲ 1 ▲ ▲ 2 ▲ ▲ ▲ 3 ... ▲ ■ ▲ ▲ 12 ... As long as we have a one-to-one function, we have a correct presentation 11 MSU/CSE 260 Fall 2009

  12. 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 12 MSU/CSE 260 Fall 2009

  13. 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: 13 MSU/CSE 260 Fall 2009

  14. 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 14 MSU/CSE 260 Fall 2009

  15. Example: Base conversion 15 MSU/CSE 260 Fall 2009

  16. Integer n and base b 16 MSU/CSE 260 Fall 2009

  17. From decimal to another base The base b expansion of n is (ak-1…a1a0)b 17 MSU/CSE 260 Fall 2009

  18. 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 18 MSU/CSE 260 Fall 2009

  19. Conversion from decimal to octal 19 MSU/CSE 260 Fall 2009

  20. 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 20 MSU/CSE 260 Fall 2009

  21. 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 thatA=10, B=11, C=12, D=13, E=14, F=15. 21 MSU/CSE 260 Fall 2009

  22. 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 22 MSU/CSE 260 Fall 2009

  23. 1 5 0 6 4 7 001101 000 11010 0111 13 1 10 7 D 1 A 7 Converting Bases Related by Powers Since 8 = 23 then every 3 binary digits makes an Octal digit. Since 16=24 then every 4 binary digits makes a Hexadecimal digit Octal (base 8) Binary (base 2) 53671 base 10 Hexadecimal (base 16) 23 MSU/CSE 260 Fall 2009

  24. Representation of integers in different bases 24 MSU/CSE 260 Fall 2009

  25. 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. 25 MSU/CSE 260 Fall 2009

  26. 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. 26 MSU/CSE 260 Fall 2009

  27. 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 2nbits may be needed to represent the product of two n-bit numbers. 27 MSU/CSE 260 Fall 2009

  28. More Examples on Base Arithmetic • Compute 222 + 222 in base 3. • Answer: • (1221)3 • Compute 222 + 222 in base 4. • Answer: • (1110)4 • Compute 222 + 222 in base 5. • Answer: • (444)5 28 MSU/CSE 260 Fall 2009

  29. 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); MSU/CSE 260 Fall 2009

  30. results >> decimal2base(983, 5) ans = 1 2 4 1 3 >> decimal2base(453, 5) ans = 3 3 0 3 >> decimal2base(6543, 7) ans = 2 5 0 3 5 >> decimal2base(78, 2) ans = 1 0 0 1 1 1 0 MSU/CSE 260 Fall 2009

  31. 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) MSU/CSE 260 Fall 2009

  32. 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.) MSU/CSE 260 Fall 2009

  33. 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. MSU/CSE 260 Fall 2009

  34. Hash function effort O(n): n = #chars int 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; } MSU/CSE 260 Fall 2009

  35. 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. MSU/CSE 260 Fall 2009

  36. 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; MSU/CSE 260 Fall 2009

  37. Using a polynomial hash function -----+----- Crypto/Hashing Demonstration -----+----- ====== Give TABLESIZE (between 10 and 1000000): 20 Give plain text string = harrycarrie Your digital signature is: 10 Give plain text string = spartyparty Your digital signature is: 3 Give plain text string = sparty Your digital signature is: 13 Give plain text string = party Your digital signature is: 2 Give plain text string = 122345678901234567890!@#$%^&&* Your digital signature is: 13 MSU/CSE 260 Fall 2009

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