SX Microcontroller Math Method

32 bit Square Root binary restoring method

Andy David says:

Here's my 32 bit routine as written for the 17c43 taken from a mail I sent Scott just after I wrote it, hence the comments about the implementations I used.

Looks a lot like Scott's original 16-bit sqrt. As the root is going to be a 16 bit number the last subtract is awkward, so the 24-bit sqrt method wasn't appropriate. I did actually write this myself rather than automatically converting Scott's code to 32 bit. I DID, however, consciously and unashamedly steal two parts - how to carry out the final 17-bit subtraction and how to count iterations - the extra 'counting' bit in the mask was quite a devious idea. This one took a little longer to write than the 24-bit, probably because it needs to iterate more times...

Standard disclaimer applies

;=========================================================================
; brSQRT32
;
; Calculates the square root of a thirtytwo bit number using the
; binary restoring method.
;
; Result in ACCaHI:ACCaLO
; Mask in ACCbHI:ACCbLO
; Input in ACCcHI:ACCcLO:ACCdHI:ACCdLO
;
; Takes between 392 and 439 cycles (incl. call and return).
; Uses 58 words ROM, 8 bytes RAM including 4 holding the input.
;
;-------------------------------------------------------------------------

brSQRT32:
	movlw 0x40 ; Initial value for Result is...
	movwf ACCaHI ; ... 01000000 00000000
	clrf ACCaLO,f ;

	movlw 0xC0 ; Initial value for mask is...
	movwf ACCbHI ; ... 11000000 00000000
	clrf ACCbLO,f ; (second '1' is loop counter).

Sub_Cmp:movfp ACCaLO,WREG ; Compare root-so-far with current
	subwf ACCcLO,f ; ... remainder.
	movfp ACCaHI,WREG ;
	subwfb ACCcHI,f ;
	btfss ALUSTA,C ;
	goto brstr ; (result is -ve, need to restore).

In1:	movfp ACCbLO,WREG ; set the current bit in the result.
	iorwf ACCaLO,f ;
	movfp ACCbHI,WREG ;
	iorwf ACCaHI,f ;

ShftUp:	rlcf ACCdLO,f ;
	rlcf ACCdHI,f ;
	rlcf ACCcLO,f ;
	rlcf ACCcHI,f ;

	rrcf ACCbHI,f ; Shift mask right for next bit, whilst
	rrcf ACCbLO,f ; ... shifting IN MSB from remainder.
	btfsc ACCbHI,7 ; If MSB is set, unconditionally set the
	goto USet1 ; ... next bit.

	movfp ACCbLO,WREG ; Append '01' to root-so-far
	xorwf ACCaLO,f ;
	movfp ACCbHI,WREG ;
	xorwf ACCaHI,f ;

	btfss ALUSTA,C ; If second '1' in mask is shifted out,
	goto Sub_Cmp ; ... then that was the last normal iteration.

	movfp ACCaLO,WREG ; Last bit Generation.
	subwf ACCcLO,f ; ... The final subtract is 17-bit (15-bit root
	movfp ACCaHI,WREG ; ... plus '01'). Subtract 16-bits: if result
	subwfb ACCcHI,f ; ... generates a carry, last bit is 0.
	btfss ALUSTA,C ;
	return

	movlw 1 ; If result is 0 AND msb of is '0', result bit
	btfsc ALUSTA,Z ; ... is 0, otherwise '1'.
	btfsc ACCdHI,7 ;
	xorwf ACCaLO,f ;
	return

USet1:	btfsc ALUSTA,C ; If mask has shifted out, leave. final bit
	return ; ... has been set by iorwf at in1.
	bcf ACCbHI,7 ; clear bit shifted in from input.

	movfp ACCbLO,WREG ; Append '01' to root-so-far
	xorwf ACCaLO,f ;
	movfp ACCbHI,WREG ;
	xorwf ACCaHI,f ;

	movfp ACCaLO,WREG ; This subtraction is guaranteed not to
	subwf ACCcLO,f ; ... cause a borrow, so subtract and
	movfp ACCaHI,WREG ; ... jump back to insert a '1' in the
	subwfb ACCcHI,f ; ... root.
	goto In1 ;

brstr:	movfp ACCaLO,WREG ; A subtract above at Sub_Cmp was -ve, so
	addwf ACCcLO,f ; ... restore the remainder by adding.
	movfp ACCaHI,WREG ; The current bit of the root is zero.
	addwfc ACCcHI,f ;
	goto ShftUp ;

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