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高等演算法 -Introduction

1. Analysis2. Basic arithmetic3. Modular arithmetic4. GCD5. Primality testing6. Cryptography

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Algorithm Algorithm ： a specified set of simple

instructions to be followed to

solve a problem.

require how much in the way of

resoureces ： time or space.

以決定該 algorithm 是否 useful.

1. Analysis

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Analysis Note that1. Input size 2. the number of basic computer

steps

1. Analysis

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FibonacciFibonacci number: 0, 1, 1, 2, 3, 5, 8

Rule:

0n if 0

1n if 1

1 n if FF

F2-n1-n

n

1. Analysis

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FibonacciExponential

function fib1(n) if n=0: return 0 if n=1: return 1return fib1(n-1) + fib1(n-2)

polynomial

function fib2(n) if n=0: return 0 create an array f[0…n] f[0]=0, f[1]=1 for i=2…n: f[i]=f[i-1]+f[i-2]return f[n]T(n)=T(n-1)+T(n-2)+3

T(n) ≧FnT(n) = n-1

1. Analysis

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Fibonacci

polynomial

T(n) = O(logn)

F

F

1 1

1 0

F

F

.... F

F

1 1

1 0

F

F

1

0

n

1n

n

1

0

2

1

1. Analysis

兩個 22matrix 相乘 4 加法 8 乘法

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Fibonacci

1 1

1 0

1 1

1 0 ....

1 1

1 0

1 1

1 0

1 1

1 0

1 1

1 0

1. Analysis

T(n) = O(logn)

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Fibonacci

1. Analysis

公式解

nn

2

51

5

1

2

51

5

1Fn

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Big-O Def ： T(N)=O(f(N)) if there are positive constant c and n0 such that T(N) cf(N) when N n0

例： T(N)=1000N, f(N)=N2 n0=10 c=100 1000N cN2 ∴1000N=O(N2)

1. Analysis

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Def ：

Def ：

Def ： 　

0nNwhen)N(cf)N(T

))N(f()N(T

))N(f()N(Tand

))N(f(O)N(Tiff))N(f()N(T

))N(f()N(Tand

))N(f(O)N(Tif))N(f(o)N(T

Big-O

1. Analysis

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)(log)(),()(

)()(),()(),()(2)(( 2342

NNTNNT

NONTNONTNONTNNT

)N(Tonboundloweranis)N(f))N(f()N(T

)N(Tonboundupperanis)N(f))N(f(O)N(T

Big-O

o

O

))(()(

))(()(

NfoNT

NfONT

不可寫

1. Analysis

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如何判斷 function 成長速度 1. 基本 function ：

2.

例如： error

Nk322 2NNNNlogNNNlogNlogc

))()(()()(.

))(()),((max()()(.

))(()())(()(

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21

21

NgNfONTNTb

NgONfONTNTathen

NgONTandNfONTif

)()(

)2()(2

2

NONT

NNONT

1. Analysis

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NlogN

NlogN

1. Analysis

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如何判斷 function 成長速度 3. 利用 limit 比較 f(N) 與 g(N) 計算

例如：比較 NlogN 與 N1.5

?)N(g/)N(flimn

))N(f(o)N(g

))N(g()N(f0c

))N(g(o)N(f0

0N

NlogNlim

5.1n

1. Analysis

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2. Basic arithmeticAddition:

Input:x and y are each n bits long;

bit complexity: O(n)

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2. Basic arithmeticMultiplication :

bit complexity: O(n2)

13 × 11; x = 1101 and y = 1011.

If x and y are both n bits, then there are n intermediate rows, with lengths of up to 2n bits.

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2. Basic arithmeticMultiplication :

Khwarizmi’s method 13 × 11

bit complexity: O(n2)

13)13*2)(5(

)13*2)(2

15()13*2)(

2

11(

13*11

偶數

( 高度 n)

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2. Basic arithmeticMultiplication :

Karatsuba’s method

bit complexity: O(n1.585)

Schonhage-Strassen’s method (based on FFT)

bit complexity: O(nlognloglogn)

A = 2n/2A1 + A0 B = 2n/2B1 + B0

A*B = 2nA1B1 + 2n/2(A0B1 + A1B0)+ A0B0

= (2n + 2n/2) A1B1 + 2n/2(A1- A0)(B0- B1)+ (2n +1)A0B0

T(n) 3T(n/2) + c*n

T(n) 3c’nlog23

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2. Basic arithmeticDivision:

x = yq + r and r < y.

Division of a 2n-bit integer by an n-bit integer can be performed using O(n2) bit operations

There is an algorithm to find the quotient q = [x/y], when the 2n-bit integer x is divided by the integer y having no more than n bits, using O(M(n)), where M(n) is the number of bit operations needed to multiply two n-bit integers.

bit complexity: O(nlognloglogn)

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3. Modular arithmetic

x ≡ y (mod N) ⇐⇒ N divides (x − y).

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3. Modular arithmetic

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Modular addition

Since x and y are each in the range 0 to N − 1, their sum is between 0 and 2(N − 1). If the sumexceeds N − 1, we merely need to subtract off N to bring it back into the required range. The overall computation therefore consists of an addition, and possibly a subtraction, of numbers that never exceed 2N. Its running time is linear in the sizes of these numbers, in other words O(n), where n = log N is the size of N;

x+y mod N

3. Modular arithmetic

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Modular multiplication

The product can be as large as (N − 1)2, but this is still at most 2n bits long since log(N − 1)2 = 2 log(N − 1) ≤ 2n.To reduce the answer modulo N, we compute the remainder upon dividing it by N, using our quadratic-time division algorithm. Multiplication thus remains a quadraticoperation. O(n2)

x.y mod N

3. Modular arithmetic

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Modular exponentiation

xy mod N

O(y) modular multiplications

is clearly exponential in the size of y

O(logy) modular multiplications; bit complexity: O(n3)

3. Modular arithmetic

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Modular exponentiation

3. Modular arithmetic

Square and multiply

3723 mod 55 = 53

3710111 = ((((((((371)2)370)2)371)2)371)2)371) mod 55

(..((371)2mod 55)370)2mod 55)371mod 55 )2mod 55 )371mod 55)2mod 55)371) mod 55 =53

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5. Primality testing

But, 341 = 11 · 31 is not prime, and yet 2340 ≡ 1 mod 341 (a=2)

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5. Primality testingCarmichael Numbers

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5. Primality testingLemma: If aN−1 ≡ 1 mod N for some a relatively prime to N, then it must hold forat least half the choices of a < N.

If N is prime, then aN−1 ≡ 1 mod N for all a < N.If N is not prime, then aN−1 ≡ 1 mod N for at most half the values of a < N.

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5. Primality testing

Pr(Algorithm 1.8 returns yes when N is not prime) ≤1/2k

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5. Primality testingGiven a number N, prove that N is a prime.

Sieve of Eratosthenes

Miller 1975 (using Riemann hypothesis)

Adleman, Pomerance and Rumely 1983

Agrawal, Kayal and Saxena 2002

Lenstra and Pomerance 2003

)N(O

))Nlog((O 5

))Nlog((O Nlogloglogc

))Nlog((O 12

))Nlog((O 6

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6. Cryptography

1. 張三選 2 個大質數 p 和 q ( 至少 100 位數 ) ，令 N = p • q

2. 再計算 Ø(N)=(p-1)(q-1) ，並選 1 個與 Ø(N) 互質數 e

Ø(N) 為 Euler‘s Totient 函數，其意為與 N 互質之個數3. (e,N) 即為張三的公開金鑰

加密法為 C = Me mod N

4. 張三選 1 個數 d ，滿足 e • d mod Ø(N) = 15. d 即為張三的解密金鑰 ( 亦稱私有金鑰或祕密金鑰 )

解密法為 M = Cd mod N RSA 之安全性取決於質因數分解之困難度 要將很大的 N 因數分解成 P 跟 Q 之相乘，是很困難的

RSA

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6. Cryptography

1. 張三選 p=5 ， q=11 ; 此時 N = p • q = 5 x 11 = 552. 張三選出 1 個與 ( p-1 ) x ( q-1 ) = ( 5-1 )( 11-1 )

= 4 x 10 = 40互質數 e=7

3. ( e, N) = (7,55) 即為張三的公開金鑰4. 張三選 1 個數 d=7 當作解密金鑰，

滿足 e • d 1 mod 40 ( 7 x 23 1 mod 40 )

令明文 M = 53加密 : C = Me mod N = 537 mod 55 = 37解密 : M = Cd mod N = 3723 mod 55 = 53