Suppose you have a list of truth values with $2^k$ elements for any natural number $k$. If the first element of this list is denoted as $L(1)$, then we can come up with a new list by performing the following operation $Q(L)$.
$Q(L)$:
temp:=$L(2^k)$
$L(2^k):=L(2^k)\oplus L(1)$
$L(n):=L(n)\oplus L(n+1),1\leq n<2^k-1$
$L(2^k-1):=L(2^k-1)\oplus temp$
Show that for all $k$, repeated iterations of the operator $Q$ onto $L$ results in the list consisting of only false values.
Example:
Original: $L=\{true,false,true,true\}$
After 1 iteration: $\{true,true,false,false\}$
After 2 iterations: $\{false,true,false,true\}$
After 3 iterations: $\{true,true,true,true\}$
After 4 iterations: $\{false,false,false,false\}$
I originally wanted to use induction for this, but it seems this system will break near the edges because of the looping around effect.
It seems the following.
For the convenience we assume that the lists are written around a circle, so the last element of the circle precedes the first. Farther, if we have two such circular lists $L$ and $L’$ then we can define a new circular list $L’’=L\oplus L’$ as elementwise XOR of $L$ and $L’$, that is $L’’(n)=L(n)\oplus L’(n)$ for each index $n$. As $0$ we denote the zero circular list, that is such a list $Z$ such that $Z(n)$ is false for each index $n$. Clearly, that $L\oplus L=0$ for each circular list $L$. Moreover, if $L$ is a circular list then by $dL$ we denote its circular shift, that is $dL(n)=L(n+1)$, if $1\le n\le 2^k$ and $dL(2^k)=L(1)$. We can easily check the distributivity of $\oplus$ and $d$, that is $d(L\oplus L’)= dL\oplus dL’$ for each circular lists $L$ and $L’$.
Now we can describe the operation $Q(L)$ as $Q(L)=L\oplus dL$ for each list $L$. Then $$Q^2(L)=Q(L\oplus dL)= (L\oplus dL)\oplus d(L\oplus dL)=L\oplus dL\oplus dL\oplus ddL=L\oplus d^2 L,$$ where $A^l(L)$ denotes a result of $l$ repeated iterations of an operation $A$ to the circular list $L$. Repeating the iterations, we similarly obtain $Q^{2^l}(L)=L\oplus d^{2^l}L$ for each natural $l$. In particular, $$Q^{2^k}(L)=L\oplus d^{2^k}L= L\oplus L=0.$$