Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

[EDITED] (Changed the question)

Hello, now I'm making tables to review it, but it's slow... Xor with product in Z/(2^n) are a ring?

I have a "linear" system like this (we know $a,b,c,d,e,f,g,h$, we want to know $x_1,x_2,x_3,x_4$):

$ \begin{array}{ccccccccl} a\cdot x_1 & \oplus & b\cdot x_2 & \oplus & 0 & \oplus & 0 & = e \\ 0 & \oplus & 0 & \oplus & a\cdot x_3 & \oplus & b\cdot x_4 & = f \\ c\cdot x_1 & \oplus & d\cdot x_2 & \oplus & 0 & \oplus & 0 & = g \\ 0 & \oplus & 0 & \oplus & c\cdot x_3 & \oplus & d\cdot x_4 & = h \\ \end{array} $

We can transform the system to two independent equations system, but even if we are working only with 2 unknowns I can't find the way to work with XOR and the typical addition together.

Thanks in advance for any idea :) .

share|improve this question
    
XOR does not distribute over the regular sum of $\mathbb{Z}/2^n\mathbb{Z}$: for example, with $n=3$, $3$XOR$(1+1) = 3$XOR$2 = 1$, but $3$XOR$1 + 3$XOR$1 = 2+2=4$. Nor does the sum distribute over XOR: $1+(3$XOR$1) = 1+2 = 3$, but $(1+3)$XOR$(1+1) = 4$XOR$2 = 6$. –  Arturo Magidin Mar 24 '11 at 19:41
    
Sorry, in any case,¿there's a way to prove that this system is (not) resoluble ? –  castarco Mar 24 '11 at 19:43
2  
@castarco: En ingles, se dicen "unknowns" (=incognitas). Y no se usan los signos de interrogacion o admiracion izquierdos. –  Arturo Magidin Mar 24 '11 at 19:45
    
@Arturo gracias, thanks. –  castarco Mar 24 '11 at 19:50
    
@castarco: Is $\oplus$ meant to be XOR? –  Arturo Magidin Mar 24 '11 at 20:03

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.