Programming:Fast Square Root
Input: A = The number you want to find the square root of
Output: A = Square root of the number supplied in A
Call: CALL SqrtA
;this routine written 10 - 28 - 2003 ;ported from the 68k version (by Frank Yaul) by konrad meyer ;this routine is not as memory efficient as it could be; this is ;done to save cycles ;note: this routine is using the fastest available method ;and will come within 1 below the # ;note: this routine is inaccurate in #s below 4 SqrtA: LD (Asqr),A SRL A JR DataOver Asqr: .DB 0 Bsqr: .DB 0 Csqr: .DB 0 DataOver: LD (Bsqr),A LD B,A LD (Csqr),A iterate: LD A,(Asqr) LD B,(Bsqr) LD D,A LD E,B divideDbyEreturnA: RL D RLA SUB E JR nc,$+3 ADD A,E LD E,A LD A,D CPL LD B,(Bsqr) ADD A,B SRL A LD (Bsqr),A LD A,(Csqr) DEC A LD B,(Bsqr) CP B JR z,done LD (Csqr),B JR iterate done: LD A,(Bsqr) RET
Since there are only 15 possible results for the square root of an 8 bit number, it's really just a series of comparisons, optimised to use a binary comparison. This routine also leaves all other registers intact:
SqrtA: CP 8 * 8 JR NC,ge8 CP 4 * 4 JR NC,ge4 CP 2 * 2 JR NC,ge2 OR A RET Z LD A,1 RET ge2: CP 3 * 3 SBC A ADD 3 RET ge4: CP 6 * 6 JR NC,ge6 CP 5 * 5 SBC A ADD 5 RET ge6: CP 7 * 7 SBC A ADD 7 RET ge8: CP 12 * 12 JR NC,ge12 CP 10 * 10 JR NC,ge10 CP 9 * 9 SBC A ADD 9 RET ge10: CP 11 * 11 SBC A ADD 11 RET ge12: CP 14 * 14 JR NC,ge14 CP 13 * 13 SBC A ADD 13 RET ge14: CP 15 * 15 SBC A ADD 15 RET
The average time for this routine, including CALL and RET is 26.9us.