Nikolai Golovchenko says:
[The] debugged version: {ed: with correction from Vladyslav Borodavka Thank you!};*********************************************** ; Square 10 bits ; ; 6 Aug 2000 by Nikolai Golovchenko ; Based on the original version of John Payson ; ; Input: ; SrcH:SrcL ; Output: ; DstH:DstM:DstL ; ; Instructions: 68 ; Execution time(including return): 51+5*3+1=67 ; ; Description: ; The goal is calculation of Dst = Src^2 ; ; Src may be rewritten as: ; Src = SrcH*256 + SrcL ; Let's introduce variables a, ah, al, and b: ; a = ah*16 + al = SrcL ; b = SrcH ; Then ; Src^2 = (256*b+a)^2=65536*b^2+512*b*a+a^2 ; ; Sqr4 routine calculates squares for 4 bit data ; (b, al, and ah) using look-up table. ; ; Now we can find b^2 by Sqr4 and multiply it by 65536, ; and calculate 512*b*a. ; ; To calculate a^2 let's expand it in byte halves: ; a^2=(16*ah+al)^2=256*ah^2+32*ah*al+al^2 ; ; So, ; Src^2=65536*b^2+512*b*a+256*ah^2+32*ah*al+al^2 ; ; This is the algorithm of how to square a 10 bit number ; using Sqr4 look-up table. ; ;*********************************************** Sqr10 clrf DstH ;clear result clrf DstM clrf DstL clrc ;clear carry ;find 32*ah*al movf SrcL, w ;w = SrcL andlw 0x0F btfsc SrcL, 4 addwf DstM, f rrf DstM, f ;use carry after addition rrf DstL, f ;DstM:DstL=ah<0>*al*256/2 btfsc SrcL, 5 addwf DstM, f rrf DstM, f rrf DstL, f ;DstM:DstL=(ah<0>*al*256/2 + ah<1>*al*256)/2=64*al(ah<0>+2*ah<1>) btfsc SrcL, 6 addwf DstM, f rrf DstM, f rrf DstL, f ;DstM:DstL=32*al(ah<0>+2*ah<1>+4*ah<2>) btfsc SrcL, 7 addwf DstM, f ;DstM:DstL=32*al(ah<0>+2*ah<1>+4*ah<2>+8*ah<3>)=32*ah*al ;(maximum value = 0x1C20) ;Now add squared al and ah call Sqr4 addwf DstL, f ;add al^2 skpnc incf DstM, f ;propagate carry to DstM swapf SrcL, w andlw 0x0F ;w = ah call Sqr4 addwf DstM, f ;carry is reset ;At this point DstM:DstH contains a^2 ;Add 512*b*a to Dst ;512*b*a=512*b*a=512*(2*b<1>+b<0>)*<128*a<7>+a<0:6>)= ;=65536*b*a<7>+256*(2*a<0:6>*b<0>+2*a<0:6>*b<1>+2*a<0:6>*b<0>) movf SrcH, w ;Dst += 512*b*(128*a<7>) = 65536*b*a<7> btfsc SrcL, 7 addwf DstH, f ;carry is reset rlf SrcL, w ;w = a<0:6>*2 (a<7> is already used) clrc btfsc SrcH, 0 addwf DstM, f ;Dst += 512*b<0>*a<0:6> skpnc incf DstH, f clrc ;Dst += 512*b<1>*a<0:6> btfsc SrcH, 1 addwf DstM, f skpnc incf DstH, f clrc ;Dst += 512*b<1>*a<0:6> btfsc SrcH, 1 addwf DstM, f skpnc incf DstH, f ;Add 65536*b^2 movf SrcH, w call Sqr4 addwf DstH, f retlw 0 ; All done! Sqr4: ;Look-up table for 4 bit squares addwf PCL, f DT 0,1,4,9,16,25,36,49,64,81,100,121,144,169,196,225 ;***********************************************
The original version:
clrf DstH clrf DstM clrf DstL movf SrcL,w andlw $0F btfss Src,4 addwf DstM rrf DstM rrf DstL btfss Src,5 addwf DstM rrf DstM rrf DstL btfss Src,6 addwf DstM rrf DstM rrf DstL btfss Src,7 addwf DstM call Sqr4 addwf DstL swapf SrcL andlw $0F call Sqr4 addwf DstM ; At this point, 16-bit result is in DstM:DstH ; 25 words of code prior to this point (plus a ; 17-word table-lookup). Total execution time: ; 35 cycles up to this point. btfss SrcH,0 goto NoBit8 movf SrcL,w btfsc C incf DstH addwf DstM btfsc C incf DstH incf DstH ; Another 9 words for bit 8; 3 or 9 cycles to exec. NoBit8: btfss SrcH,1 goto NoBit9 movlw 4 btfss SrcH,0 movlw 8 addwf DstH rlf SrcL,w btfsc C incf DstH btfsc C incf DstH addwf DstM btfsc C incf DstH addwf DstM btfsc C incf DstH ; Another 17 words for bit 9; 3 or 17 cycles to execute ; Total worst-case time: 35+26 = 61 cycles. NoBit9: retlw 0 ; All done! Sqr4: addwf PC db 0,1,4,9,16,25,36,49,64,81,100,121,144,169,196,225
See also:
Questions:
| file: /Techref/microchip/math/sq/10b.htm, 5KB, , updated: 2010/3/30 07:47, local time: 2024/11/25 07:06,
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