university of hawaii ics141: discrete mathematics …janst/141/lecture/18-integeralgorithms.pdfics...

13-1 ICS 141: Discrete Mathematics I – Fall 2011

University of Hawaii University of Hawaii

ICS141: Discrete Mathematics for Computer Science I

Dept. Information & Computer Sci., University of Hawaii

Jan Stelovsky based on slides by Dr. Baek and Dr. Still

Originals by Dr. M. P. Frank and Dr. J.L. Gross Provided by McGraw-Hill


University of Hawaii University of Hawaii Quiz

1.  gcd(84,96) = 2.  lcm(84,96) = 3.  gcd(84,96) x lcm(84,96)=

n  Hints n  What's the prime factorization of 84? n  What's the prime factorization of 96?

n  Try the primes 2, 3 and 7


University of Hawaii University of Hawaii

Lecture 18

Chapter 3. The Fundamentals 3.6 Integers and Algorithms


University of Hawaii University of Hawaii Review: Greatest Common Divisor n  The greatest common divisor gcd(a,b) of integers

a,b (not both 0) is the largest integer d that is a divisor both of a and of b. d = gcd(a,b) = max(d: d|a ∧ d|b) ⇔ d|a ∧ d|b ∧ ∀e∈Z, (e|a ∧ e|b) → (d ≥ e)

n  If the prime factorizations are written as and , then the GCD is given by:

n  Example: 84 = 22·31·71 and 96 = 25·31·70

n  gcd(84,96) = 22·31·70 = 2·2·3 = 12.

.),gcd( ),min(),min(2

),min(1

2211 nn ban

baba pppba …=

nan

aa pppa …2121=

nbn

bb pppb …2121=


University of Hawaii University of Hawaii Review: Least Common Multiple n  lcm(a,b) of positive integers a, b, is the smallest

positive integer that is a multiple both of a and of b. E.g. lcm(6,10) = 30

m = lcm(a,b) = min(m: a|m ∧ b|m) ⇔ a|m ∧ b|m ∧ ∀n∈Z: (a|n ∧ b|n) → (m ≤ n)

n  If the prime factorizations are written as and , then the LCM is given by:

n  Example: 84 = 22·31·71 and 96 = 25·31·70

n  lcm(84,96) = 25·31·71 = 32·3·7 = 672.

.),(lcm ),max(),max(2

),max(1

2211 nn ban

baba pppba …=

nan

aa pppa …2121=

nbn

bb pppb …2121=


University of Hawaii University of Hawaii GCD and LCM n  Theorem: Let a and b be positive integers. Then

ab = gcd(a,b)·lcm(a,b)

n  Example n  a = 84 = 2·2·3·7 = 22·31·71 n  b = 96 = 2·2·2·2·2·3 = 25·31·70

n  ab = (22·31·71)·(25·31·70) = 22·31·70·25·31·71 = 2min(2,5)·3min(1,1)·7min(1,0)·2max(2,5)·3max(1,1)·7max(1,0)

= gcd(a,b)·lcm(a,b)


University of Hawaii University of Hawaii Integers and Algorithms

n  Topics: n  Base-b representations of integers.

n  Especially: binary, hexadecimal, octal.

n  Algorithms for computer arithmetic: n  Binary addition and multiplication.

n  Euclidean algorithm for finding GCD’s.


University of Hawaii University of Hawaii Base-b Number Systems n  Ordinarily, we write base-10 representations

of numbers, using digits 0-9. n  But, 10 isn’t special! Any base b > 1 will work. n  For any positive integers n and b, there is a

unique sequence akak-1… a1a0 of digits ai < b such that:

∑=

−−

−−

=

+++++=k

i

ii

kk

kk

kk

ba

ababababan

0

01

12

21

1

The “base-b expansion of n”

n  Notation: n = (akak-1… a1a0)b


University of Hawaii University of Hawaii Particular Bases of Interest

n  Base b = 10 (decimal): 10 digits: 0,1,2,3,4,5,6,7,8,9.

n  Base b = 2 (binary): 2 digits: 0,1. (“Bits”=“binary digits.”)

n  Base b = 8 (octal): 8 digits: 0,1,2,3,4,5,6,7.

n  Base b = 16 (hexadecimal): 16 digits: 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F

Used internally in all modern computers

Octal digits correspond to groups of 3 bits

Hex digits give groups of 4 bits

Used only because we have 10 fingers


University of Hawaii University of Hawaii Examples n  Example 1: Decimal expansion of the integer with

binary expansion (101011111)2? n  (101011111)2

= 1·28 + 0·27 + 1·26 + 0·25 + 1·24 + 1·23 + 1·22 + 1·2 + 1 = (351)10

n  Example 2: Decimal expansion of the integer with hexadecimal expansion (2AE0B)16? n  (2AE0B)16 = 2·164 + 10·163 + 14·162 + 0·16 + 11

= (175627)10


University of Hawaii University of Hawaii Converting to Base b (An algorithm, informally stated.) n  To convert any integer n to any base b > 1: n  To find the value of the rightmost (lowest-

order) digit, simply compute n mod b. n  Now, replace n with the quotient. n  Repeat above two steps to find subsequent

digits, until n is gone (= 0).

Exercise: Write this out in pseudocode…


University of Hawaii University of Hawaii Converting to Base b

baaaaabababa

abaababbabaabqbabaabqb

abaqbaabqb

abqn

)(

)0(

)(

)(

0123

012

23

3

0122

33

0122

23

01222

0112

011

00

=

+++=

++++⋅=

+++=

+++=

++=

++=

+=


University of Hawaii University of Hawaii Examples n  Example 3: Find the base 8, i.e. octal,

expansion of (12345)10 n  12345 = 8·1543 + 1 n  1543 = 8·192 + 7 n  192 = 8·24 + 0 n  24 = 8·3 + 0 n  3 = 8·0 + 3 n  Therefore, (12345)10 = (30071)8


University of Hawaii University of Hawaii Binary ↔ Hexadecimal

n  Hexadecimal expansion of (11 1110 1011 1100)2

∴ (11 1110 1011 1100)2 = (3EBC)16

n  Binary expansion of (A8D)16 (A)16 = (1010)2, (8)16 = (1000)2, (D)16 = (1101)2

∴ (A8D)16 = (1010 1000 1101)2

E B C 0011 = 3


University of Hawaii University of Hawaii Addition of Binary Numbers Carry:

10111 + 11100 n  Carry = ⎣bitSum / 2⎦ n  sbitIndex = bitSum mod 2 = bitSum – 2⋅carry

1

0 0

1

0

0

1

0

1

1

1

1


University of Hawaii University of Hawaii Addition of Binary Numbers procedure add(an−1…a0, bn−1…b0: binary representations of

non-negative integers a, b) carry := 0 for bitIndex := 0 to n−1 begin {go through bits} bitSum := abitIndex+bbitIndex+carry {2-bit sum} carry := ⎣bitSum / 2⎦ {high bit of sum} sbitIndex := bitSum – 2⋅carry {low bit of sum} end sn := carry return sn…s0: binary representation of integer s


University of Hawaii University of Hawaii Multiplication of Binary Numbers

n  ab = a(b0·20 + b1·21 + ··· + bn-1·2n-1) = a(b0·20 ) + a(b1·21 ) + ··· + a(bn-1·2n-1)

110 a x 101 b

110 000

110 11110

shift 1 bit to the left, i.e. append 1 extra 0-bit 0 shift 2 bit to the left, i.e. append 2 extra 0-bits 00


University of Hawaii University of Hawaii Multiplication of Binary Numbers n  ab = a(b0·20 + b1·21 + ··· + bn-1·2n-1)

= a(b0·20 ) + a(b1·21 ) + ··· + a(bn-1·2n-1) procedure multiply(an−1…a0, bn−1…b0: binary

representations of positive integers a,b) product := 0 for i := 0 to n−1 if bi = 1 then product := add(an−1…a00···0, product) return product

i extra 0-bits appended after the digits of a

i times


University of Hawaii University of Hawaii Division with Remainder procedure div-mod(a∈Z,d∈Z+)

{quotient & remainder of a/d} q := 0 r := |a| while r ≥ d begin r := r – d q := q + 1 end if a < 0 and r > 0 then begin {a is a negative}

r := d – r q := –(q + 1) end {q = a div d (quotient), r = a mod d (remainder)}


University of Hawaii University of Hawaii Euclid’s Algorithm for GCD n  Finding GCDs by comparing prime

factorizations can be difficult when the prime factors are not known!

n  Euclid discovered: Let a = bq +r, where a, b, q, and r are integers. Then gcd(a,b) = gcd(b,r) (i.e. gcd(a,b) = gcd(b, (a mod b))) n  Example: gcd(36, 24) = gcd(24, 12)

n  Sort a, b so that a > b, and then (given b > 1) (a mod b) < b, so problem is simplified.


University of Hawaii University of Hawaii Euclid’s Algorithm Example n  gcd(372, 164) = gcd(164, 372 mod 164)

n  372 mod 164 = 372 - 164⎣372/164⎦ = 372 - 164·2 = 372 - 328 = 44

n  gcd(164, 44) = gcd(44, 164 mod 44) n  164 mod 44 = 164 - 44⎣164/44⎦

= 164 - 44·3 = 164 - 132 = 32

n  gcd(44, 32) = gcd(32, 44 mod 32) = gcd(32, 12) = gcd(12, 32 mod 12) = gcd(12, 8) = gcd(8, 12 mod 8) = gcd(8, 4) = gcd(4, 8 mod 4) = gcd(4, 0) = 4


University of Hawaii University of Hawaii Euclid’s Algorithm Pseudocode procedure gcd(a, b: positive integers)

x : = a y : = b while y ≠ 0 begin r := x mod y; x := y; y := r; end return x {x = gcd(a, b)}


University of Hawaii University of Hawaii Two’s Complement n  In binary, negative numbers can be

conveniently represented using two’s complement notation.

n  In this scheme, a string of n bits can represent any integer i such that −2n−1 ≤ i < 2n−1.

n  The leftmost bit is used to represent the sign (0:positive, 1:negative integer)

n  The negation of any n-bit two’s complement number a = an−1…a0 is given by an−1…a0 + 1.

The bitwise logical complement of the n-bit string an−1…a0.


University of Hawaii University of Hawaii Two’s Complement Example

value 3-bit pattern 3 0 1 1 2 0 1 0 1 0 0 1 0 0 0 0

–1 1 1 1 –2 1 1 0 –3 1 0 1 –4 1 0 0

n  To obtain the results for –4 ≤ n ≤ –1, consider |n|, then n  In the binary representation of

|n|, replace each 0 by 1, and each 1 by 0, This is the one’s complement of n.

n  Add 1 (i.e. 001) to the result from the previous step. This is the two’s complement of n.

n  Example n  –3: 011 -> 100

+ 001 = 101

−22 ≤ i < 22

(−2n−1 ≤ i < 2n−1)


University of Hawaii University of Hawaii Subtraction of Binary Numbers

procedure subtract(an−1…a0, bn−1…b0: binary two’s complement reps. of integers a, b) return add(a, add(b,1)) { a + (−b) }

n  Note that 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! We call this an overflow.


University of Hawaii University of Hawaii Modular Exponentiation n  Problem: Given large integers b (base),

n (exponent), and m (modulus), efficiently compute bn mod m. n  Note that bn itself may be completely

infeasible to compute and store directly. n  E.g. if n is a 1,000-bit number, then bn itself

will have far more digits than there are atoms in the universe!

n  Yet, this is a type of calculation that is commonly required in modern cryptographic algorithms!


University of Hawaii University of Hawaii Algorithm Concept

00

22

11

00

22

11

)()()( 222

222

nnn

nnnn

bbb

bb

kk

kk

kk

kk

×××=

=

−−

−−

−−

−− ⋅++⋅+⋅

…n  Note that:

n  We can compute b to various powers of 2 by repeated squaring. n  Then multiply them into the partial product, or not,

depending on whether the corresponding ni bit is 1. n  Crucially, we can do the mod m operations as we go

along, because of the various identity laws of modular arithmetic.

n  All the numbers stay small.

The binary expansion of n

= b1 = b


University of Hawaii University of Hawaii Modular Exponentiation procedure modularExponentiation(b: integer,

n = (nk−1…n0)2, m: positive integers) x := 1 {accumulates the result} b2i := b mod m { mod m; i=0 initially} for i := 0 to k−1 begin {go thru all k bits of n} if ni = 1 then x := (x·b2i) mod m

b2i := (b2i·b2i) mod m end return x {x equals bn mod m}

i

b2

)()( 22222 1 iiii

bbbb ⋅== ⋅+

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