yˆ=(y<<a); y^=(y>>b); return (yˆ=(y<<c));

I looked at this in my last post, and had a brief look at the (a,b,c) triplets to use. However, this only yields a period length of the word size used. In my case, that's 2^16-1 -- not much. However, further on in the paper, section 3.1, a method is described for longer periods through additional state and promotion:

t=(xˆ(x<<a)); x=y; return y=(yˆ(y>>c))ˆ(tˆ(t>>b));

This is listed as 2^64-1 using 32-bit words, but there's no reason we can't use 16-bit words. This reduces the period to 2^32-1, and requires new triplets. I scanned (1,1,1) through till (15,15,15), here's the list:

(1,1,7) | (1,1,12) | (1,1,13) | (2,5,8) | (2,5,13) | (2,13,15) | (2,15,13) | (3,7,6) |

(5,3,1)* | (5,3,8) | (5,3,13)* | (5,7,4)* | (6,3,8)* | (7,1,6) | (7,1,15) | (7,2,1) |

(8,3,9)* | (9,14,5) | (11,8,5)* | (13,12,3) | (14,1,15) | (15,10,1) |

Values marked with an asterisk score well on the majority of diehard tests. It's probably not a good idea to use the other values, even though they do have a complete period. Notably, all values fail binary matrices tests for 31x31 and 32x32 - I think this is expected, and mentioned in Marsaglia2003.

So, the routine in C:

uint16_t rnd_xorshift_32() {

static uint16_t x=1,y=1;

uint16_t t=(x^(x<<5));

x=y;

return y=(y^(y>>1))^(t^(t>>3));

}

and in assembly:

; returns a 16-bit value >0, has a 2^32-1 period

random:

push cx

push dx

db 0b8h ; t=(x^(x shl a))

random_x: dw 1

mov dx,ax

mov cl,5

shl dx,cl

xor ax,dx

mov dx,ax ; y=(y^(y shr c))^(t^(t shr b))

mov cl,3 ; ^^^^^^^^^^^

shr dx,cl

xor ax,dx

push ax ; save t^(t shr b)

db 0b8h

random_y: dw 1

mov cs:[random_x],ax ; x=y

mov dx,ax ; y=(y^(y shr c))^(t^(t shr b))

shr dx,1 ; ^^^^^^^^^^^

xor ax,dx

pop dx

xor ax,dx

mov cs:[random_y],ax

pop dx

pop cx

ret

## 1 comment:

Which triplet passes the most tests? (5,3,1)?

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