---joke[Editors note: For the refs cited here, you will have to go to: {ga faq}.]
The correct spelling is "Gray"---not "gray", "Grey", or "grey"---
since Gray codes are named after the Frank Gray who patented their
use for shaft encoders in 1953 [1]. Gray codes actually have a
longer history, and the inquisitive reader may want to look up the
August, 1972, issue of Scientific American, which contains two
articles of interest: one on the origin of binary codes [2], and
another by Martin Gardner on some entertaining aspects of Gray
codes [3]. Other references containing descriptions of Gray codes
and more modern, non-GA, applications include the second edition of
Numerical Recipes [4], Horowitz and Hill [5], Kozen [6], and
Reingold [7].
A Gray code represents each number in the sequence of integers
{0...2^N-1} as a binary string of length N in an order such that
adjacent integers have Gray code representations that differ in only
one bit position. Marching through the integer sequence therefore
requires flipping just one bit at a time. Some call this defining
property of Gray codes the "adjacency property" [8].
Example (N=3): The binary coding of {0...7} is {000, 001, 010, 011,
100, 101, 110, 111}, while one Gray coding is {000, 001, 011, 010,
110, 111, 101, 100}. In essence, a Gray code takes a binary sequence
and shuffles it to form some new sequence with the adjacency
property. There exist, therefore, multiple Gray codings or any
given N. The example shown here belongs to a class of Gray codes
that goes by the fancy name "binary-reflected Gray codes". These are
the most commonly seen Gray codes, and one simple scheme for
generationg such a Gray code sequence says, "start with all bits zero
and successively flip the right-most bit that produces a new string."
Hollstien [9] investigated the use of GAs for optimizing functions of
two variables and claimed that a Gray code representation worked
slightly better than the binary representation. He attrributed this
difference to the adjacency property of Gray codes. Notice in the
above example that the step from three to four requires the flipping
of all the bits in the binary representation. In general, adjacent
integers in the binary representaion often lie many bit flips apart.
This fact makes it less likely that a MUTATION operator can effect
small changes for a binary-coded INDIVIDUAL.
A Gray code representation seems to improve a MUTATION operator's
chances of making incremental improvements, and a close examination
suggests why. In a binary-coded string of length N, a single
mutation in the most significant bit (MSB) alters the number by
2^(N-1). In a Gray-coded string, fewer mutations lead to a change
this large. The user of Gray codes does, however, pay a price for
this feature: those "fewer mutations" lead to much larger changes.
In the Gray code illustrated above, for example, a single mutation of
the left-most bit changes a zero to a seven and vice-versa, while the
largest change a single mutation can make to a corresponding binarycoded INDIVIDUAL is always four. One might still view this aspect of
Gray codes with some favor: most mutations will make only small
changes, while the occasional mutation that effects a truly big
change may initiate EXPLORATION of an entirely new region in the
space of CHROMOSOMEs.
The algorithm for converting between the binary-reflected Gray code
described above and the standard binary code turns out to be
surprisingly simple to state. First label the bits of a binary-coded
string B[i], where larger i's represent more significant bits, and
similarly label the corresponding Gray-coded string G[i]. We convert
one to the other as follows: Copy the most significant bit. Then
for each smaller i do either G[i] = XOR(B[i+1], B[i])---to convert
binary to Gray---or B[i] = XOR(B[i+1], G[i])---to convert Gray to
binary.
One may easily implement the above algorithm in C. Imagine you do
something like
typedef unsigned short ALLELE;
and then use type "allele" for each bit in your CHROMOSOME, then the
following two functions will convert between binary and Gray code
representations. You must pass them the address of the high-order
bits for each of the two strings as well as the length of each
string. (See the comment statements for examples.) NB: These
functions assume a chromosome arranged as shown in the following
illustration.
index: C[9] C[0]
*-----------------------------------------------------------*
Char C: | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 |
*-----------------------------------------------------------*
^^^^^ ^^^^^
high-order bit low-order bit
C CODE
/* Gray <==> binary conversion routines */
/* written by Dan T. Abell, 7 October 1993 */
/* please send any comments or suggestions */
/* to dabell@quark.umd.edu */
void gray_to_binary (Cg, Cb, n)
/* convert chromosome of length n from */
/* Gray code to binary representation */
allele *Cg,*Cb;
int n;
{
int j;
*Cb = *Cg; /* copy the high-order bit */
for (j = 0; j < n; j++) {
Cb--; Cg--; /* for the remaining bits */
*Cb= *(Cb+1)^*Cg; /* do the appropriate XOR */
}
}
void binary_to_gray(Cb, Cg, n)
/* convert chromosome of length n from */
/* binary to Gray code representation */
allele *Cb, *Cg;
int n;
{
int j;
*Cg = *Cb; /* copy the high-order bit */
for (j = 0; j < n; j++) {
Cg--; Cb--; /* for the remaining bits */
*Cg= *(Cb+1)^*Cb; /* do the appropriate XOR */
}
}
References
Hope this helps,Paul Holmes email: bsc0123@uk.ac.napier.dcs Computer Studies Department Napier University Edinburgh
Scotland
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via Fnet-EUnet id AA16477; Fri, 9 Dec 1994 05:23:28 +0100 (MET)
From: mgix@jpn.thomson-di.fr (Emmanuel Mogenet)
To: bru@kosice.upjs.sk
Cc: genetic-programming@cs.stanford.edu
Reply-To: mgix@jpn.thomson-di.fr
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Branislav Brutovsky writes:
>Hi everybody,>Would you send me the formula for construction decimal number from its >Gray counterpart, please. I would really appreciate it. >Thanks in advance.
>Branislav BrutovskyFind my personal brew attached.
Remarks:
a) Either your genetic code is seen has a collection of 32 bits chunks
over which the gray coding is applied as many times as they are chunks
b) Either your genetic code is seen has one long bit string
representing one single binary number over which gray coding is applied
once.
#include <stdio.h> #include <stdlib.h>
typedef int reg; typedef unsigned int uint32; typedef unsigned short uint16; typedef unsigned char uint8;
/*
static void BuildLookup()
{
for( reg i=0; i<2; ++i )
{
for( reg j=0; j<256; ++j )
{
uint16 last= (i<<8);
lookup[i][j]=j;
for( reg k=7; k>=0; --k )
{
last>>=1;
lookup[i][j] ^= last;
last= lookup[i][j]&(1<<k);
}
lookup[i][j] |= (last<<8);
}
}
} /*
void Gray( uint32 *gray, // (O) Gray Code uint32 *binary, // (I) Binary Codereg length, // (I) Length of the binary code (in 32 bit chunks) reg prop // (I) Propagate on the whole string of bits ?
)
{
uint32 last=0;
for( reg i=length-1; i>=0; --i )
{
gray[i]= binary[i] ^ (last|(binary[i]>>1));
last= prop ? binary[i]<<31 :0;
}
} /*
void Ungray( uint32 *binary, // (O) Binary Code uint32 *gray, // (I) Gray Codereg length, // (I) Length of the gray code (in 32 bit chunks) reg prop // (I) Propagate on the whole string of bits ?
)
{
if( !initDone )
{
initDone=1;
BuildLookup();
}
uint16 last=0;
for( reg i=length-1; i>=0; --i )
{
uint8 *s=(uint8*)(gray+i);
uint8 *d=(uint8*)(binary+i);
uint16 z;
z= lookup[last][ s[0] ];
d[0]= z&0xFF;
last= z>>8;
z= lookup[last][ s[1] ];
d[1]= z&0xFF;
last= z>>8;
z= lookup[last][ s[2] ];
d[2]= z&0xFF;
last= z>>8;
z= lookup[last][ s[3] ];
d[3]= z&0xFF;
last= prop ? z>>8 : 0;
}
}
static uint8 uniq[65535];
static char *bindump[]=
{
"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111",
"1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111",
};
/*
reg Hamming(
uint32 a,
uint32 b
)
{
reg res=0;
for( reg i=0; i<32; ++i,a>>=1,b>>=1 ) res+= ((a&0x1) != (b&0x1));
return res;
} /*
main()
{
reg i;
uint32 bin;
uint32 gray;
/* Visual Check
----------------*/
for( i=0; i<256; ++i )
{
bin=i;
Gray( &gray, &bin, 1, 0 );
Ungray( &bin, &gray, 1, 0 );
printf(
"%d: %s%s --> %s%s --> %s%s\n",
i,
bindump[ i>>4 ], bindump[ i&0xF ],
bindump[ gray>>4 ], bindump[ gray&0xF ],
bindump[ bin>>4 ], bindump[ bin&0xF ]
);
}
/* Check the first 64k
-----------------------*/
uint32 last=1;
for( i=0; i<65536; ++i )
{
bin=i;
Gray( &gray, &bin, 1, 0 );
Ungray( &bin, &gray, 1, 0 );
uniq[gray]++;
if( bin!=i || Hamming(last,gray)!=1 )
{
printf( "Failure at %d\n", i );
exit(1);
}
last=gray;
}
for( i=0; i<65536; ++i )
{
if( uniq[i]!=1 )
{
printf( "Mapping not 1 to 1\n" );
exit(1);
}
}
exit(0);
}
/*
// $Log: gcode.c++,v $
From: David_vun_Kannon_at_NY-RES@resgate.rch.gs.comI wrote some c code for this when I did GA work a couple of years ago.
However, I convinced myself that it wasn't really helpful to GA on
bitstrings. The basic argument for using Gray coded bits is that
single bit mutations in a Gray coded number will only change the value
by a little bit also, as opposed to the usual power of 2 code, where
changing one bit might change the value by as much as 2^(n-1) for n
its. Unfortunately, this arguement ignores the wraparound that occurs
between the first and last encoded values, which in Gray code are
always 000... and 100.... A mutation between these two strings is
still just a single bit, but it produces a value jump of 2^n. What
has happened in effect is that Gray code has saved up the possible
accumulated error and put it all in one place. put another way, I
think that total error over all one bit changes in any representation
is constant, all you can do is change the distribution of the error
weight over the set of changes. I didn't find the bimodal
distribution of error produced by Gray codes particularly more
compelling than the approximately Gaussian distribution of the binary
representation, and I couldn't come up with a representation that had
uniform distribution without resorting to a hash table. So I gave up
on Gray codes.