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Integer Representations & Base Conversion

Integer Representations & Base Conversion. Section 3.6. 1. Object Representation. Objects/Concepts. Representation. . . A one-to-one function is required. 2. Object Representation… . Need symbols; how many? What are other issues? Physical cost of representation

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Integer Representations & Base Conversion

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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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