Message-ID: <3128@utah-cs.UUCP>
Date: Fri, 30-Nov-84 19:40:35 EST
Article-I.D.: utah-cs.3128
Posted: Fri Nov 30 19:40:35 1984
Date-Received: Sun, 2-Dec-84 06:05:59 EST
Organization: Univ of Utah CS Dept
Lines: 82
Xref: sdcsvax net.micro.6809:37 net.micro.cbm:928

[Let's give the line-eater terminal indeigestion!]

Here is my entry to the roman numeral contest (for the 6809, 48 bytes total):

*******************************************************************************
VALTBL	FCB	100,'C+$80
	FCB	90,'X+$80,'C+$80
	FCB	50,'L+$80
	FCB	40,'X+$80,'L+$80
	FCB	10,'X+$80
	FCB	9,'I+$80,'X+$80
	FCB	5,'V+$80
	FCB	4,'I+$80,'V+$80
	FCB	1,'I+$80
* The first byte after this table must have the MSB 0.  In this case my
* first instruction of the routine satisfies this.  If you don't like
* this type of trick, or want to put the code first, you'll have to add a
* zero byte.

* Enter with byte to be output as roman numerals in B register.  Destroys
* A, B, X, Y, and CCR
ROMAN	LEAX	 value) then after the characters are printed the same compare is
    performed again

T (total clock cycles) = 15 + 24*M + 9*C + (22+p)*P

I have also coded a 6502 version of the same algorithm (which I will provide on
request), which is 60 bytes long, primarily due to the availability of only one
accumulator, and the need for a four instruction sequence to transfer the X
register to the Y register (PHA, TXA, TAY, PLA).  If I remember correctly, the
65C02 has a TXY instruction which would save 6 bytes.

If anyone can provide a routine shorter than 48 bytes (code and data) for any
processor, I would very much like to see it.
-- 
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