algorithms-ch1-Algorithms with numbers

1.1Basic arithmetic

1.1.1addition

-Given two binary numbers x and y, how long does our algorithm take to add them?
-We want the answer expressed as a function of the size of the input: the number of bits of x and y, the number of keystrokes needed to type them in.

Suppose x and y are each n bits long; O(n).

1.1.2multiplication

二進(jìn)制乘法的兩個(gè)算法:


0.0

If x and y are both n bits, then there are n intermediate rows, with lengths of up to 2n bits (taking the shifting into account). The total time taken to add up these rows, doing two numbers at a time, is O(n) + O(n) + · · · + O(n)..(n-1 times): O(n^2)


@.@
function multiply(x, y)
Input: Two n-bit integers x and y, where y ≥ 0
Output: Their product
/
if y=0: return0
z = multiply(x, ?y/2?)
//每次遞歸調(diào)用,接收到返回值之后 向遞歸下一步執(zhí)行
if y is even:
  return 2z
else:
  return x + 2z
function divide(x,y)
Input: Two n-bit integers x and y, where y ≥ 1
Output: The quotient and remainder of x divided by y
/
if x = 0: return (q,r) = (0,0)
(q, r) = divide(?x/2?, y)
q=2·q, r=2·r
if x is odd: r=r+1
if r≥y: r=r?y, q=q+1
return (q,r)
1.2mod
  1. if x = qN + r with 0 ≤ r < N, then x modulo N is equal to r.

  2. x and y are congruent modulo N if they differ by a multiple of N , or in symbols:

x≡y (modN) ?? N divides (x?y).

  1. Substitution rule
    If x ≡ x′ (mod N) and y ≡ y′ (mod N), then:x+y≡x′+y′ (modN) and xy≡x′y′ (modN).

  2. Modular addition and multiplication:

  • addition: O(n),
    n = ?log N ? is the size of N ;(regard N as a binary number, n is the bits of this number, each bits need one operations)
    To add two numbers x and y modulo N, Since x and y are eachin the range 0 to N ?1, their sum is between 0 and 2(N ?1), The overall computation therefore consists of an addition, and possibly a subtraction

  • multiplication: O(n^2)
    using our quadratic-time division algorithm.Multiplication thus remains a quadratic operation.

  • Division: O(n^3)

  1. Modular exponentiation
    -Problem: compute x^y mod N for values of x, y, and N that are several hundred bits long
    -Sol1: x mod N →x^2 mod N →x^3 mod N →···→x^y mod N,
    -Sol2: x mod N →x^2 mod N →x^4 mod N →x8^ mod N →···→x2^?logy? mod N.
    a polynomial time algorithm:
function modexp(x, y, N)
Input: Two n-bit integers x and N, an integer exponent y
Output: x^y mod N
/
if y=0: return1
z = modexp(x, ?y/2?, N )
if y is even:
  return z^2 mod N
else:
  return x · z^2 mod N
  1. Euclid's Alg for Great Common Divisor

Euclid’s rule If x and y are positive integers with x ≥ y, then gcd(x, y) = gcd(x mod y, y).

Lemma If a ≥ b,then a mod b < a/2.

function Euclid(a,b)
Input: Two integers a and b with a≥b≥0
Output: gcd(a, b)
/
if b=0: return a
return Euclid(b, a mod b)

both arguments, a and b, If they are initially n-bit integers, then the base case will be reached within 2n recursive calls. And since each call involves a quadratic-time division, the total time is O(n3).

Lemma if d divides both a and b, and d = ax + by for some integers x and y(may be negative) , then necessarily d = gcd(a,b)

function extended-euclid(a,b)
Input: Two positive integers a and b with a ≥ b ≥ 0
Output: Integers x,y, d, such that d=gcd(a,b) and ax+by=d
/
if b = 0: return (1,0,a)
(x′, y′, d) = Extended-Euclid(b, a mod b)
return (y′, x′ ? ?a/b?y′, d)

模除法:gcd(a,N) = 1(即互質(zhì)) <==> 存在x,使得ax ≡ 1 (mod N) (可用反證法證明)
左推右:用extend-euclid algorithm可以得到x,y
右推左:如果ax+Ny=d(gcd<=d), 且d整除a, N(d <=gcd),那么d==gcd(a, N)

  1. x is the multiplicative inverse of a modulo N if ax ≡ 1 (mod N).

Modular division theorem For any a mod N, a has a multiplicative inverse modulo N if and only if it is relatively prime to N. When this inverse exists, it can be found in time O(n3)(where as usual n denotes the number of bits of N ) by running the extended Euclid algorithm.

1.3prime
function primality(N)
Input: Positive integer N
Output: yes/no
/
Pick a positive integer a < N at random 
if a^(N?1) ≡ 1 (mod N):
  return yes
else:
  return no

exercise都是clrs上的不附了

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