Notation: A single letter represents an 8 bit value
A single letter followed by a digit represents one bit in that 8 bit value.
Bits are numbered from right to left, so bit n has the value 2^n.

Start with the binary value b.
b=b7;b6;b5;b4;b3;b2;b1;b0

Compute its gray code, g = b ^ ( b >> 1 )
g=g7;g6;g5;g4;g3;g2;g1;g0
[0] g7=b7; g6=b7^b6; g5=b6^b5; g4=b5^b4; g3=b4^b3; g2=b3^b2; g1=b2^b1; g0=b1^b0

Now recover the original binary value from g in three steps
[1] x = g ^ (g >> 4)
x=x7;x6;x5;x4;x3;x2;x1;x0
x7=g7; x6=g6; x5=g5; x4=g4 x3=g3^g7 x2=g2^g6 x1=g1^g5; x0=g0^g4

[2] y = x ^ (x >> 2)
y=y7;y6;y5;y4;y3;y2;y1;y0
y7=x7; y6=x6; y5=x5^x7; y4=x4^x6; y3=x3^x5; y2=x2^x4; y1=x1^x3; y0=x0^x2

substitute eqn [1] into eqn [2]
[2a] y7=g7; y6=g6; y5=g5^g7; y4=g4^g6; y3=g3^g7^g5; y2=g2^g6^g4; y1=g1^g5^g3^g7; y0=g0^g4^g2^g6

[3] z = y ^ (y >> 1)
z7=y7; z6=y6^y7; z5=y5^y6; z4=y4^y5; z3=y3^y4; z2=y2^y3; z1=y1^y2; z0=y0^y1

Substitute eqn [2a] into eqn [3]
[3a] z7=g7; z6=g6^g7; z5=g5^g7^g6; z4=g4^g6^g5^g7; z3=g3^g7^g5^g4^g6; z2=g2^g6^g4^g3^g7^g5; z1=g1^g5^g3^g7^g2^g6^g4; z0=g0^g4^g2^g6^g1^g5^g3^g7

Substitute eqn [0] into eqn [3a]
[3b] z7=b7;
z6=b7^b6^b7;
z5=b6^b5^b7^b7^b6;
z4=b5^b4^b7^b6^b6^b5^b7;
z3=b4^b3^b7^b6^b5^b5^b4^b7^b6;
z2=b3^b2^b7^b6^b5^b4^b4^b3^b7^b6^b5;
z1=b2^b1^b6^b5^b4^b3^b7^b3^b2^b7^b6^b5^b4;
z0=b1^b0^b5^b4^b3^b2^b7^b6^b2^b1^b6^b5^b4^b3^b7

Rearrange the bits on the right
[3c] z7=b7;
z6=b6^b7^b7;
z5=b5^b6^b6^b7^b7;
z4=b4^b5^b5^b6^b6^b7^b7;
z3=b3^b4^b4^b5^b5^b6^b6^b7^b7;
z2=b2^b3^b3^b4^b4^b5^b5^b6^b6^b7^b7;
z1=b1^b3^b3^b2^b2^b4^b4^b5^b5^b6^b6^b7^b7;
z0=b0^b1^b1^b2^b2^b3^b3^b4^b4^b5^b5^b6^b6^b7^b7

Now, because x^x=0 and y^0=y, we can just remove each identical pair
[3c] z7=b7;
z6=b6;
z5=b5;
z4=b4;
z3=b3;
z2=b2;
z1=b1;
z0=b0;

And that’s how it works.