**Xorshift** is a simple, fast pseudorandom number generator developed by George Marsaglia. The generator combines three xorshift operations where a number is exclusive-ored with a shifted copy of itself:

```
/* 16-bit xorshift PRNG */
unsigned xs = 1;
unsigned xorshift( )
{
xs ^= xs << 7;
xs ^= xs >> 9;
xs ^= xs << 8;
return xs;
}
```

There are 60 shift triplets with the maximum period 2^{16}-1. Four triplets pass a series of lightweight randomness tests including randomly plotting various *n* × *n* matrices using the high bits, low bits, reversed bits, etc. These are: 6, 7, 13; 7, 9, 8; 7, 9, 13; 9, 7, 13.

7, 9, 8 is the most efficient when implemented in Z80, generating a number in 86 cycles. For comparison the example in C takes approx ~1200 cycles when compiled with HiSoft C v1.3.

```
; 16-bit xorshift pseudorandom number generator
; 20 bytes, 86 cycles (excluding ret)
; returns hl = pseudorandom number
; corrupts a
xrnd:
ld hl,1 ; seed must not be 0
ld a,h
rra
ld a,l
rra
xor h
ld h,a
ld a,l
rra
ld a,h
rra
xor l
ld l,a
xor h
ld h,a
ld (xrnd+1),hl
ret
```

Oh my gosh, my new favorite site! I tried implementing this a while back with no luck. I did however find that combining a simple 16-bit LFSR and a 16-bit LCG works well. It's not as fast (148cc), but it does pass CACert labs' testing. Not sure how to post code boxes, but:

ReplyDeleteprng16:

;collab with Runer112

;;Output:

;; HL is a pseudo-random int

;; A and BC are also, but much weaker and smaller cycles

;; Preserves DE

;;148cc, super fast

;;26 bytes

;;period length: 4,294,901,760

seed1=$+1

ld hl,9999

ld b,h

ld c,l

add hl,hl

add hl,hl

inc l

add hl,bc

ld (seed1),hl

seed2=$+1

ld hl,987

add hl,hl

sbc a,a

and 101101

xor l

ld l,a

ld (seed2),hl

add hl,bc

ret

Great stuff! I ported this to C64, 30 cycles without the RTS. I didn't need what is equivalent to the second lda a,l / rra because 6502 EOR does not touch carry:

ReplyDeleterng_zp_low = $02

rng_zp_high = $03

; seeding

LDA #1 ; seed, can be anything except 0

STA rng_zp_low

LDA #0

STA rng_zp_high

...

random

LDA rng_zp_high

LSR

LDA rng_zp_low

ROR

EOR rng_zp_high

STA rng_zp_high ; high part of x ^= x << 7 done

ROR ; A has now x >> 9 and high bit comes from low byte

EOR rng_zp_low

STA rng_zp_low ; x ^= x >> 9 and the low part of x ^= x << 7 done

EOR rng_zp_high

STA rng_zp_high ; x ^= x << 8 done

RTS

Thanks SO much for this. This is AWESOME!

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ReplyDeleteThe function is not correct, because the second XOR is not correctly implemented.

ReplyDeleteShifting the contents of HL >> 9, can never result in using any of the bits of the 8 bit register L, since they would have been completely shifted out after 8 right shifts.

So the second XOR snippet is incorrect.

Instead of this:

ld a,l

rra

ld a,h

rra

xor l

It should be:

ld a,h

rra

xor l

NOTE: Since carry is already reset, by the previous XOR, no extra instruction required to set Carry bit to zero before the RRA instruction.

And hence the complete (and shorter) function would become:

xrnd:

ld hl,1 ; seed must not be 0

ld a,h

rra

ld a,l

rra

xor h

ld h,a

ld a,h

rra

xor l

ld l,a

xor h

ld h,a

ld (xrnd+1),hl

ret

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