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Algorithms Design and its Applications. Algorithms with numbers. Back Ground. Number theory was once viewed as a beautiful but largely useless subject in pure mathematics.
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Algorithms Design and its Applications Algorithms with numbers
Back Ground • Number theory was once viewed as a beautiful but largely useless subject in pure mathematics. • Today number- theoretic algorithms are used widely , due in part to the invention of cryptographic schemes based on large prime numbers. • The feasibility of these schemes rests on our ability to find large primes easily, while their security rests on our inability to factor the product of large primes .
Size of Cost • In this lecture, a "large input" typically means an input containing "large integers" rather than an input containing "many integers". • We measure the size of an input in terms of the number of bits required to represent that input. • An algorithm with integer inputs a1, a2, ..., ak is a polynomial - time algorithm if it runs in time polynomial in lga1, lga2, ..., lgak, that is, polynomial in the lengths of its binary- encoded inputs.
Bases and logs • How many digits are needed to represent the number N>=0 in base b? • How much does the size of a number change when we change the base?
Addition • The sum of any three single-digit numbers (with base b>=2) is at most two digits long. • Given two binary number x and y, how long does our algorithm take to add them? O(n), n is the number of bits of x and y.
Multiplication • 13 * 11
Multiplication • 乘法的时间复杂度是多少呢？ • 对于长度为n的乘数来说，将产生n个中间结果，而对这些中间结果进行相加的次数是n-1次，从而时间复杂度为O(n(n-1))=O(n2)
Multiplication • another way to multiply
Multiplication • Al Khwarizmi乘法算法的时间复杂度？ • 由于乘数每次都被取半，对于二进制来说，取半意味着去掉最右边一位数，而乘数的长度为n，因此该算法在递归n次后终结。每次递归需要进行一次长度为n位的加法运算，其时间复杂度为O(n)。因此Al Khwarizmi乘法算法的时间复杂度为O(n2)。
Division • To divide an integer x by another integer y≠0 means to find a quotient q and a remainder r, where x = yq+r and r <y.
Modular Addition • x+y mod N • Since x and y are in the range 0 to N-1, their sum is between 0 and 2(N-1). If the sum exceeds N-1, subtract off N. • The overall computation consists of an addition, and possibly a subtraction, of numbers that never exceed 2N. the running time is O(n), where n = log N.
Modular multiplication • xy mod N • start with regular multiplication, then reduce the answer modulo N. The product can be as large as (N-1)2, at most 2n bits long since log(n-1)2 = 2log(N-1)≤2n. • The running time is O(n2).
Modular exponentiation • 在密码学中，常需计算xy mod N. 这个的x,y和N均为几百位长的整数。如何快速计算？ • 直接算xy，运算结果很大！即便x和y只有20位长，xy也要大概1千万位长。 • 为保证中间运算结果不要太大，每步运算都模N.
Modular exponentiation • Simple idea: repeatedly multiplying by x modulo N.problem: if y is 500 bits long, we need to perform y -1 ≈ 2500 multiplications!
Modular exponentiation • better idea: starting with x and squaring repeatedly modulo N, we getwe need to perform log y multiplications, ach takes O(log2N) to compute. • To determine xy mod N, we simply multiply together and appropriate subset of these powers, those corresponding to 1’s in the binary representation of y. • A polynomial-time algorithm is within reach!
sicily 1294. 高级机密 • 信息加密。 • 目前比较流行的编码规则称为RSA，是由美国麻省理工学院的三位教授发明的。这种编码规则是基于一种求密取模算法的：对于给出的三个正整数a,b,c，计算a的b次方除以c的余数。 • 题目要求：计算 ab mod c
sicily 1294. 高级机密问题分析 • 不好的算法： • 先求出a的b次方，再模c。但题目给出的a,b,c的范围比较大，要算出ab要用到高精度乘法，然后模c还要用到高精度除法； • 较好的算法： • 利用同余的性质，xy mod c = x * (y mod c) mod c
Euclid’s algorithm for greatest common divisor • Euclid’s ruleIfx and y are positive integer with x≥y, then gcd(x,y)= gcd(x mod y, y) • Proof.因为gcd(x,y)能整除x和y，因此整除x-y，即是x-y的因子，因此gcd(x,y)≤ gcd(x-y, y).而反过来推，同理可得gcd(x-y, y) ≤ gcd(x,y)。故gcd(x,y)=gcd(x-y, y)。由此显然可得结论。
Euclid’s algorithm for greatest common divisor • This means that after any two consecutive round, both a and b, are at the very least halved in value – the length of each decreases by at least one bit. If they are initially n-bits 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)
An extension of Euclid algorithm • 只要找到两个整数x和y，使得ax+by=d，且d是a和b的因子，则d就是a和b的最大公因子；如果d是a和b的最大公因子，则d一定可以表示为ax+by形式。 • 只要对欧几里得算法稍加扩展，即可找到所需的系数x和y。
An extension of Euclid algorithm • LemmaFor any positive integers a and b, the extended Euclid algorithm returns integers x, y, and d such that gcd(a,b) = d = ax+by • Proof. • 对b做归纳假设。当b=0, 验证可知算法正确。算法调用gcd(b,a mod b)来计算gcd(a,b)。由于a mod b< b， 由归纳假设知返回结果是正确的.
减法求最大公约数 于大整数而言，取模运算（其中用到除法）是非常昂贵的开销，将成为整个算法的瓶颈。有没有办法能够不用取模运算呢？ 如果一个数能够同时整除x和y，则必能同时整除x-y和y；而能够同时整x-y和y的数也必能同时整除x和y，即x和y的公约数与x-y和y的公约数是相同的，其最大公约数也是相同的，即f（x, y）= f（x-y, y），那么就可以不再需要进行大整数的取模运算，而转换成简单得多的大整数的减法。 实例：f（42, 30）=f（30, 12）=f（12, 18）= f（18, 12）= f（12, 6）= f（6, 6）= f（6, 0）= 6 不足之处。最大的瓶颈就是迭代的次数比之前的算法多了不少，如果遇到（10 000 000 000 000, 1）
求最大公约数算法三 算法一(欧几里得算法)的问题在于计算复杂的大整数除法运算，而算法二虽然将大整数的除法运算转换成了减法运算，降低了计算的复杂度，但它的问题在于减法的迭代次数太多，如果遇到（10 000 000 000 000, 1）的情况就很糟糕。 能否结合算法一和算法二从而使其成为一个最佳的算法呢？
求最大公约数算法三 记x和y的最大公约数为f（x, y）。 若x, y均为偶数，f（x, y）= 2 * f（x/2, y/2）= 2 * f（x>>1, y>>1） 若x为偶数，y为奇数，f（x, y）= f（x/2, y）= f（x>>1, y） 若x为奇数，y为偶数，f（x, y）= f（x, y/2）= f（x, y>>1） 若x, y均为奇数，f（x, y）= f（x, x- y）， 那么在f（x, y）= f（x, x- y）之后，（x- y）是一个偶数，下一步一定会有除以2的操作。 最坏情况下的时间复杂度是O（log2（max（x, y））。
求最大公约数算法三 示例： f（42, 30）= f（1010102, 111102） = 2 * f（101012, 11112） = 2 * f（11112, 1102） = 2 * f（11112, 112） = 2 * f（11002, 112） = 2 * f（112, 112） = 2 * f（02, 112） = 2 * 112 = 6
同余 • 同余 • 设m是正整数，a,b是整数，如果m|(a-b)，则称a和b关于模m同余，记作a≡b(mod m)或者说，如果a,b除以m的余数相等，则称a和b关于模m同余 • 同余的性质 • a≡a(mod m) • 如果a≡b(mod m)，则b≡a(mod m) • 如果a≡b(mod m)且b≡c(mod m)， a≡c(mod m) • 如果a≡b(mod m)且c≡d(mod m)，则a±c≡b± d(mod m)， ac≡bd(mod m)
同余 • 同余的性质(cont.) • 如果a≡b(mod m)，则an≡bn(mod m)，n∈N • 如果ac≡bc(mod m)，则a≡b(mod (m/gcd(c,m)) • 如果a≡b(mod m)且d|m，则a≡b(mod d) • 如果a≡b(mod m)，则ad≡bd(mod m) • 如果a≡b(mod mi)，i=1,2,…,n，l=lcm(m1,m2,…,mn)，则a≡b(mod l) • 如果p为素数，则ap ≡ a(mod p)；如果gcd(a,p)=1，则ap-1 ≡ 1(mod p)
代码（筛法求素数） for (inti = 2; i <= (int) floor(sqrt(MAX)); ++i) { if (prime[i]) { int j = i * 2; while (j <= MAX) { prime[j] = false; j += i; } } } }
Fermat’s little theorem • Proof.
An algorithm for testing primality, with low error probability
Carmichael numbers • 561 = 3*11*17, not a prime. • fool the Fermat test, because a560 ≡1 (mod 561) for all values of a relatively prime to 561. • Rabin and Miller algorithm.
