Mathematical procedures.

Greetings, All! So, procedure number 1: quick division 16/16. To the one who can do it speed up, it will be possible to erect a monument during his lifetime (at his expense). Acceleration by replacing jr with jp is not considered, because winning/losing in speed depends on the size of the arguments. At the entrance: de-divisible bc divisor Output: de-quotient hl-residue divide ld a,b ;change cpl ; ld b,a ;sign ld a,c ; cpl;y ldc,a ; inc bc ;divisor t=30 ld hl,0 ;reset the new dividend to zero ld a,d ;move first rla ;high byte of dividend ;t=18 rl l ;┐ add hl,bc ;│ jr c,$+4 ;│8 times sbc hl,bc ;│ rla ;┘ t=8*45=360 ld d,a ;st. result byte ld a,e ;now move rla ;low byte t=12 adc hl,hl ;┐ add hl,bc ;│ jr c,$+4 ;│8 times sbc hl,bc ;│ rla ;┘ t=8*52=416 ld e,a ;ml. result byte ;hl-remainder of division ;t=4 Total: 30+18+360+12+416+4=840 cycles. If the remainder is not needed, then the end of the process is fools should look like this: adc hl,hl ;┐ add hl,bc ;│ jr c,$+4 ;│7 times sbc hl,bc ;│ rla ;┘ t=7*52=364 adc hl,hl ; add hl,bc ;rla ; ld e,a ; t=34 Total: 30+18+360+12+364+34=818 cycles. P.S. If you divide by zero, then, as you would expect, the quotient will be equal to zero, and the remainder will be equal to the dividend, i.e. in this regard, the procedure there are no glitches. ;************************* Procedure number 2: taking the square root of a register pair. Here here we really need to erect a monument to someone who will speed up the procedure at least for 1 beat. Acceleration by replacing jr with jp is not considered, because gain/loss in speed depends on the size of the argument. At the entrance: hl-number from which to extract square root at the output: a-square root of hl sqrhl xor a ;zeroed a ;-) ld de,64 ;this is required by the algorithm ; t=14 sla h ;take the two leftmost adc a,a ;argument bit sla h; adc a,a ; jr z,$+4 ;if they are not equal dec a ;zero, then increase inc d ;result ; t=39 sla h ;┐follow adc a,a ;│extraction algorithm sla h ;│square root adc a,a ;│"column", only sla d ;│ ld c,d ;│3 times sli c ;│ cp c ;│ jr c,$+4 ;│ subc ;│ inc d ;┘ t=63*3=189 sla l ;┐do something similaradc a,a ;│operation with sla l ;│low byte adc a,a ;│argument sla d ;│ ld c,d ;│2 times sli c ;│ cp c ;│ jr c,$+4 ;│ subc ;│ inc d ;┘ t=63*2=126 ld h,a ; or a ;cleared the carry flag ; t=8 sbc hl,de ;now we lack jr nc,$+3 ;one register d for add hl,de ;execute the last ccf ;two cycles, will have to rl d ;use more add hl,hl ;slow operations add hl,hl ;with register pairs ; t=67 sbc hl,de ;and last time ccf ;perform calculations, not ld a,d ;restoring hl adc a,a ;in a-result ; t=27 Total: 14+39+189+126+8+67+27=470 cycles. Procedure number 3: extracting the square root of a three-byte number. Speeding up by replacing jr with jp is not considered, because winning/losing in speed depends on the value of the argument. At the entrance: dhl=d*65536+hl - the number itself Output: hl - the same square root sqrdhl xor a ;beginning of procedure ld bc,64 ;almost the same ;previous ; t=14 sla d ; adc a,a ; sla d ; adc a,a ; t=39 jr z,$+4 ; dec a ;inc b ; leaf d ;┐ adc a,a ;│ leaf d ;│ adc a,a ;│ page b ;│ ld e,b ;│3 times slip and ;│ cp and ;│ jr c,$+4 ;│ sub and ;│ inc b ;┘ t=3*63=189 leaf h ;┐ adc a,a ;│ leaf h ;│ adc a,a ;│ page b ;│ ld e,b ;│2 times slip and ;│ cp and ;│ jr c,$+4 ;│ sub and ;│ inc b ;┘ t=63*2=126 ld d,l ; ld l,h ; ld h,a ;change registers or a ; t=16 sbc hl,bc;┐ jr nc,$+3;│ add hl,bc;│ ccf ;│2 times rl b ;│ add hl,hl;│ add hl,hl;┘ t=67*2=134 ld l,h ;there is a suspicion that ld h,0 ;this block and previous ld c,b ;you can speed up a bit ld b,h ; rl h ; ld a,d ; t=31 rla ;┐yes and this unit is mine adc hl,hl;│dislike. rla ;│very likely that adc hl,hl;│it can be fine sla c ;│accelerate. But how? rl b ;│ ld d,b ;│ ld and c ;│3 times slip and ;│ rl d ;│ sbc hl,of;│ jr nc,$+4;│ add hl,of;│ Dec c ;│ inc c ;┘ t=3*119=357 (!) ;horror! Think about how;speed up! rla; this block is similar adc hl,hl;previous with rla ;small difference adc hl,hl;at the end. It's done ld d,b ;for the sake of promotion ld e,c ;calculation speed sla e; rl d ; sli e; rl d ; sbc hl,de; ccf ; t=97 ld h,b ; ld l,c ; adc hl,hl;in hl - result ; t=23 Total: 14+39+189+126+16+134+31+357+97+23=1026. ;******************* Well, and the procedure for calculating the address in attributes using coordinates in familiar places. You can't do it faster if you don't use tables. At the entrance: d-y e-x Output: hl-address in attributes ld a,d rrca rrca rrca ld h,a and #e0 or e ld l,a xor e xor h or #58 ld h,a t=54 clock cycles. I wish you health, happiness and creative Uzbeks. Aleksey Malov aka VIVID/Brainwave.

I`m happy to meet you again, Aleksey! last time (04 Apr 00 01:03:19) you talk about something AM> So, procedure number 1: quick division 16/16. To those who can AM> speed it up, it will be possible to erect a monument during his lifetime (at his expense). AM> divide ld a,b ;change AM> cpl ; AM> ld b,a ;sign AM> ld a,c ; AM> cpl ;у AM> ld c,a ; AM> inc bc ;divisor t=30 immediately caught my eye xor a sub c ldc,a sbc a,b sub c ld b,a 30 -> 24. I'll watch the rest when I have time. Don't forget that they exist and other division algorithms... catch my respects, Alexey