ABOUT Random Algorithm number
Multiplicative Congruential Method:
Basic Relationship
Xi+1 = a Xi (mod m), where a >= 0 and m >= 0
Most natural choice for m is one that equals to the capacity of a computer word.
m = 2b (binary machine), where b is the number of bits in the computer word.
m = 10d (decimal machine), where d is the number of digits in the computer word.
The max period(P) is:
For m a power of 2, say m = 2b, and c /= 0, the longest possible period is P = m = 2b , which is achieved provided that c is relatively prime to m (that is, the greatest common factor of c and m is 1), and a = 1 + 4k, where k is an integer.
For m a power of 2, say m = 2b, and c = 0, the longest possible period is P = m / 4 = 2b-2 , which is achieved provided that the seed X0 is odd and the multiplier, a, is given by a = 3 + 8k or a = 5 + 8k, for some k = 0, 1
For m a prime number and c = 0, the longest possible period is P = m - 1, which is achieved provided that the multiplier, a, has the property that the smallest integer k such that ak - 1 is divisible by m is k = m - 1,
(Example)
Using the multiplicative congruential method, find the period of the generator for a = 13, m = 26, and X0 = 1, 2, 3, and 4. The solution is given in next slide. When the seed is 1 and 3, the sequence has period 16. However, a period of length eight is achieved when the seed is 2 and a period of length four occurs when the seed is 4.
Period Determination Using Various seeds
i Xi Xi Xi Xi
0 1 2 3 4
1 13 26 39 52
2 41 18 59 36
3 21 42 63 20
4 17 34 51 4
5 29 58 23
6 57 50 43
7 37 10 47
8 33 2 35
9 45 7
10 9 27
11 53 31
12 49 19
13 61 55
14 25 11
15 5 15
16 1 3
Basic Relationship
Xi+1 = a Xi (mod m), where a >= 0 and m >= 0
Most natural choice for m is one that equals to the capacity of a computer word.
m = 2b (binary machine), where b is the number of bits in the computer word.
m = 10d (decimal machine), where d is the number of digits in the computer word.
The max period(P) is:
For m a power of 2, say m = 2b, and c /= 0, the longest possible period is P = m = 2b , which is achieved provided that c is relatively prime to m (that is, the greatest common factor of c and m is 1), and a = 1 + 4k, where k is an integer.
For m a power of 2, say m = 2b, and c = 0, the longest possible period is P = m / 4 = 2b-2 , which is achieved provided that the seed X0 is odd and the multiplier, a, is given by a = 3 + 8k or a = 5 + 8k, for some k = 0, 1
For m a prime number and c = 0, the longest possible period is P = m - 1, which is achieved provided that the multiplier, a, has the property that the smallest integer k such that ak - 1 is divisible by m is k = m - 1,
(Example)
Using the multiplicative congruential method, find the period of the generator for a = 13, m = 26, and X0 = 1, 2, 3, and 4. The solution is given in next slide. When the seed is 1 and 3, the sequence has period 16. However, a period of length eight is achieved when the seed is 2 and a period of length four occurs when the seed is 4.
Period Determination Using Various seeds
i Xi Xi Xi Xi
0 1 2 3 4
1 13 26 39 52
2 41 18 59 36
3 21 42 63 20
4 17 34 51 4
5 29 58 23
6 57 50 43
7 37 10 47
8 33 2 35
9 45 7
10 9 27
11 53 31
12 49 19
13 61 55
14 25 11
15 5 15
16 1 3
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