Random Number Generator

Random Number Generator 2110301 Intro. to Discrete Structures Prabhas Chongstitvatana Prabhas Chongstitvatana

What is random number ? • Sequence of independent random numbers with a specified distribution such as uniform distribution (equally probable) • Tippett 1927 published a table of 40,000 random digits • Kendall and Babington-Smith 1939 built a machine to generate random number to produce a table of 100,000 random digits. Prabhas Chongstitvatana

Jon von Neumann 1946 suggested the production of random number using arithmetic operations of a computer, "middle square", square a previousrandom number and extract the middle digits, Example generate 10-digit numbers, was 5772156649, square 33317792380594909201the next number is 7923805949 Prabhas Chongstitvatana

The sequence is not random, but it appears to be. Sequences generated in a deterministic way are usually called pseudo-random sequences. "middle square" has proved to be a comparatively poor source of random numbers. If zero appear as a number of the sequence, it will continually perpetuate itself. Random numbers should not be generated with a method chosen at random. Some theory must be used. Prabhas Chongstitvatana

The use of random numbers 1. simulation 2. sampling 3. numerical analysis 4. computer programming 5. decision making randomness is an essential part of optimal strategies in the theory of games 6. recreation Prabhas Chongstitvatana

Linear Congruential Method xn+1= axn + c ( mod m ), 0 <= xn+1 < m m > 0, 2 <= a < m, 0 <= c < m, 0 <= x0 <m. TheoremThe terms of the sequence generated by the linear congruential method are given by x k= ak x0 + c(ak - 1) / ( a -1 ) (mod m ), 0 <= xk < m. Prabhas Chongstitvatana

Proof by mathematical induction. for k = 1 , the formula is obviously true, since x1= ax0 + c ( mod m), 0 <= x1 < m. Assume that the formula is valid for the k th term, so that x k= ak x0 + c(ak - 1) / ( a -1 ) (mod m ), 0 <= xk < m. Since x k+1= a xk + c (mod m ), 0 <= xk+1 < m. we have xk+1= a( ak x0 + c(ak -1)/(a-1)) + c = ak+1 x0 + c(a(ak -1)/(a-1) + 1 = ak+1 x0 + c(ak+1 -1)/(a-1) (mod m) Prabhas Chongstitvatana

which is the correct formula for the (k+1)th term. This demonstrates that the formula is correct for all positive integers k. The period length of a linear congruential pseudo-random number generator is the maximum length of the sequence obtained without repetition. Theorem The linear congruential generator produces a sequence of period length m if and only if (c,m) =1 , a = 1 (mod p) for all primes p dividing m, and a = 1(mod 4) if 4 | m Prabhas Chongstitvatana

For the proof see D. E. Knuth, “The art of computer programming” vol 2, “seminumerical algorithms”, 2nd ed Addison Wesley, 1981. pp. 9-20. A special case where c = 0 is called multiplicative congruential method. xn+1 = a xn ( mod m), 0 < xn+1 < m. or xn = an x0 (mod m), 0 < xn+1 < m. Prabhas Chongstitvatana

For many applications, the generator is used with the modulus m equal to the Mersenne prime M31 = 231 -1. When the modulus m is a prime, the maximum period length is m -1, and this is obtained when a is a primitive root of m. Find a primitive root of M 31 Prabhas Chongstitvatana

Number Theory “Elementary number theory and its applications”, 2ed, K. Rosen. Addison-Wesley, 1988. Definition Let m be a positive integer. If a and b are integers, we say that a is congruent to b modulo m if m | (a-b), denoted by a = b (mod m). Definition. The integers a and b are called relatively prime if a and b have greatest common divisor (a,b) = 1. Prabhas Chongstitvatana

Definition . Let n be a positive integer. The Euler phi-functionp(n) is defined to be the number of positive integers not exceeding n which are relatively prime to n. Prabhas Chongstitvatana

Euler’s theorem. If m is a positive integer and a is an integer with (a,m) = 1, then a p(m)= 1 (mod m). Definition. Let a and m be relatively prime positive integers. Then the least positive integer x such that a x= 1 (mod m) is called the order of a modulo m denoted by ord m a. Prabhas Chongstitvatana

Example Find the order of 2 modulo 7. 2 1 = 2 (mod 7), 2 2 = 4 (mod 7), 2 3 = 1 (mod 7). Therefore ord 7 2 = 3. Definition . If r and n are relatively prime integers with n > 0 and if ord nr = p(n) , then r is called a primitive root modulo n. Prabhas Chongstitvatana

Theorem 1 If ord ma = t and if u is a positive integer, then ord m (a u ) = t / (t,u) Corollary 1 Let r be a primitive root modulo m where m is an integer m > 1. Then r u is a primitive root modulo m if and only if (u,p(m) ) = 1. Proof By Theorem 1 we know that ord m r u = ord m r / (u, ord m r ) = p(m) / (u, p(m) ). Consequently, ord m r u = p(m), and r u is a primitive root modulo m, if and only if (u, p(m)) = 1. Prabhas Chongstitvatana

To find a primitive root of M31 that can be used with the good results, we first demonstrate that 7 is a primitive root of M31 Theorem The integer 7 is a primitive root of M31 = 231 - 1. Proof To show that 7 is a primitive root of M31, it is sufficient to show that 7(M31 - 1)/q/= 1 ( mod M31) for all prime divisors q of M31 -1. With this information we can conclude that ordM31 7 = M31 -1. Prabhas Chongstitvatana

To find factorization of M31 -1, we note that M31 - 1 = 231 -2 = 2(230 - 1) = 2(215 - 1)(215 + 1) = 2(25-1)(210+25+1)(25+1)(210-25+1) = 2.32.7.11.31.151.331 if we show that 7(M31 -1 )/q/=1 (mod M31) for q = 2,3,7,11,31,151,331, then we know that 7 is a primitive root of M31 = 2147483647. Prabhas Chongstitvatana

Since 7(M31 -1)/2= 2147483546 /= 1 (mod M31) 7(M31 -1)/3= 1513477735 /= 1 (mod M31) 7(M31 -1)/7= 120536285 /= 1 (mod M31) 7(M31 -1)/11 = 1969212174 /= 1 (mod M31) 7(M31 -1)/31= 512 /= 1 (mod M31) 7(M31 -1)/151= 535044134 /= 1 (mod M31) 7(M31 -1)/331= 1761885083 /= 1 (mod M31) we see that 7 is a primitive root of M31. Prabhas Chongstitvatana

In practice we do not want to use the primitive root 7 as the generator, since the first few integers generated are small. We find a larger primitive root using Corollary 1. We take a power of 7 where the exponent is relatively prime to M31 - 1. For instance, since (5,M31 - 1) = 1, Corollary 1 tells us that 75 = 16807 is also a primitive root. Since (13,M31 -1 ) = 1, another possibility is to use 713= 252246292 (mod M31) as the multiplier. Prabhas Chongstitvatana

Choice of modulus Let w be the computer’s word size, or 2 e on an e-bit binary computer. Use m = w + - 1. Why not m = w ? When m = w the right-hand digits of x n are much less random than the left-hand digits. Prabhas Chongstitvatana

If d is a divisor of m, and if y n = x n mod d we can easily show that y n+1 = (a y n + c ) mod d for x n+1 = a x n + c – qm for some integer q, and taking both sides mod d cause the quantity qm to drop out when d is a factor of m. This shows that the low-order form a congruential sequence that has a period of length d or less. Prabhas Chongstitvatana

Other generators Linear congruential method can be generalized to, say, a quadratic congruential method x n+1 = ( d x 2n + a x n + c ) mod m additive number generator (Mitchell and Moore 1958) x n = (x n-24 + x n-55) mod m , n >= 55 m is even, x 0 . . . x 54 not all even. The least significant bits “x n mod 2” have a period of length 255 – 1. Therefore the generator must have a period at least this long. Prabhas Chongstitvatana

Random Number Generator