1 / 14

Discrete Mathematics

Discrete Mathematics. University of Jazeera College of Information Technology & Design Khulood Ghazal. Integers & Algorithms. Euclidean Algorithm for GCD. Finding GCDs by comparing prime factorizations can be difficult if the prime factors are unknown.

nelson
Download Presentation

Discrete Mathematics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Discrete Mathematics University of Jazeera College of Information Technology & Design Khulood Ghazal Integers & Algorithms

  2. Euclidean Algorithm for GCD • Finding GCDs by comparing prime factorizations can be difficult if the prime factors are unknown. • Euclid discovered: For all integers a, b,gcd(a, b) = gcd((a mod b), b). • Sort a,b so that a>b, and then (given b>1) (a mod b) < a, so problem is simplified.

  3. Euclidean Algorithm Example • gcd(372,164) = gcd(372 mod 164, 164). • 372 mod 164 = 372164372/164 = 372164·2 • = 372328 = 44. • gcd(164,44) = gcd(164 mod 44, 44). • 164 mod 44 = 16444164/44 = 16444·3 = 164132 = 32. • gcd(44,32) = gcd(44 mod 32, 32) = gcd(12, 32) = gcd(32 mod 12, 12) = gcd(8,12) = gcd(12 mod 8, 8) = gcd(4,8) = gcd(8 mod 4, 4) = gcd(0,4) = 4. • Note: See class notes for more Examples .

  4. Euclid’s Algorithm Pseudocode procedure gcd(a, b: positive integers) whileb  0 r := amodb; a := b; b := r return a

  5. Base-b number systems • Ordinarily we write base-10 representations of numbers (using digits 0-9). • 10 isn’t special; any base b>1 will work. • For any positive integers n, b, there is a unique sequence ak ak-1… a1a0of digitsai<b such that

  6. Particular Bases of Interest Used only because we have 10 fingers • Base b=10 (decimal):10 digits: 0,1,2,3,4,5,6,7,8,9. • Base b=2 (binary):2 digits: 0,1. (“Bits”=“binary digits.”) • Base b=8 (octal):8 digits: 0,1,2,3,4,5,6,7. • Base b=16 (hexadecimal):16 digits: 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F Usedinternally in all modern computers Octal digits correspond to groups of 3 bits Hex digits give groups of 4 bits

  7. Converting to Base b (Algorithm, informally stated) • To convert any integer n to any base b>1: • To find the value of the rightmost (lowest-order) digit, simply compute n mod b. • Now replace n with the quotient n/b. • Repeat above two steps to find subsequent digits, until n is gone (=0). Note: See class notes for more Examples .

  8. Addition of Binary Numbers procedureadd(an−1…a0, bn−1…b0: binary representations of non-negative integers a,b) carry := 0 forbitIndex := 0 to n−1 {go through bits} begin bitSum:=abitIndex+bbitIndex+carry{2-bit sum} sbitIndex := bitSummod 2 {low bit of sum} carry := bitSum/ 2 {high bit of sum} end sn := carry returnsn…s0: binary representation of integer s Note: See class notes for more Examples .

  9. Two’s Complement • In binary, negative numbers can be conveniently represented using two’s complement notation. • In this scheme, a string of n bits can represent any integer i such that −2n−1 ≤ i < 2n−1. • The bit in the highest-order bit-position (n−1) represents a coefficient multiplying −2n−1; • The other positions i < n−1 just represent 2i, as before. • The negation of any n-bit two’s complement number a = an−1…a0 is given by an−1…a0 + 1.

  10. Why two’s complement? binary 0111 0110 … 0001 0000 1111 1110 … 1001 1000 Decimal 7 6 … 1 0 -1 -2 … -7 -8 binary 1111 1110 … 0011 0010 0001 0000 Decimal 15 14 … 3 2 1 0

  11. Why two’s complement? binary 0111 0110 … 0001 0000 1111 1110 … 1001 1000 Decimal 7 6 … 1 0 -1 -2 … -7 -8 binary 1111 1110 … 0011 0010 0001 0000 Decimal 15 14 … 3 2 1 0 Positive values 22,21,20 -23 Positive values 23,22,21,20

  12. When you want to find the negative of a number, you take the number, and subtract it from zero No difference between subtracting the number from 0 and subtracting it from 2n to find the negative of an n-bit number, subtract the number from 0 or subtract it from 2n subtract our number from (1111 + 1) rather than 10000. (n=4 case)

  13. Subtraction of Binary Numbers proceduresubtract(an−1…a0, bn−1…b0: binary two’s complement representations of integers a,b) returnadd(a, add(b,1)) { a + (−b) } This fails if either of the adds causes a carry into or out of the n−1 position, since 2n−2+2n−2 ≠ −2n−1, and −2n−1 + (−2n−1) = −2n isn’t representable! Note: See class notes for more Examples .

  14. Multiplication of Binary Numbers proceduremultiply(an−1…a0, bn−1…b0: binary representations of a,bN) product := 0 fori := 0 to n−1 ifbi = 1 then product := add(an−1…a00i, product) returnproduct Note: See class notes for more Examples .

More Related