Cuestion 9 (Aritmetica, codigo Gray)

Solucion:

$B_{6}$	$B_{5}$	$B_{4}$	$B_{3}$	$B_{2}$	$B_{1}$
1	1	1	0	0	1

$\begin{displaymath}G_{i_{max}}=B_{i_{max}}\end{displaymath}$

$\begin{displaymath}G_i=B_{i+1}\oplus B_{i}\end{displaymath}$

$\begin{displaymath}B_{6}=G_{6}=1\end{displaymath}$

$\begin{displaymath}G_5=B_{6}\oplus B_{5}=1\oplus 1=0\end{displaymath}$

$\begin{displaymath}G_4=B_{5}\oplus B_{4}=1\oplus 1=0\end{displaymath}$

$\begin{displaymath}G_3=B_{4}\oplus B_{3}=1\oplus 0=1\end{displaymath}$

$\begin{displaymath}G_2=B_{3}\oplus B_{2}=0\oplus 0=0\end{displaymath}$

$\begin{displaymath}G_1=B_{2}\oplus B_{1}=0\oplus 1=1\end{displaymath}$

Con lo que la transformacion de binario a codigo de Gray sera 100101.

Vamos a comprobarlo con el Mathematica:

conversionBinarioGray[a_] := {a1 := IntegerDigits[a,2,7];
  l1 := Length[a1]; b2 = a1; a4 = a1[[1]];
  Do[b2[[i]] = BitXor[a1[[i]], a4]; a4 = a1[[i]], {i, 2, l1}];
  Print[b2]}

conversionBinarioGray[2^^111001]

2011-07-31