Linear Interpolation Inversion for N even?

Under certain conditions the sampled function
x(1:N) can be reconstructed from the linearly
intepolated midpoints

xm = (x(1:end)+x(2:end))/2

For example, if one of the original points
(e.g. x(j)) is known, then

for i = j:N-1
x(i+1) = 2*xm(i)-x(i);
end
for i = j:-1:2
x(i-1) = 2*xm(i-1)-x(i);
end

If x(j) is arbitrarily initialized, the above
algorithm will yield errors that are constant
in magnitude and alternating in sign. Therefore,
if mean(x) = 0 and N is odd, the error is easily
removed by using sum(x).

y = zeros(N,1);
z = cos(pi*(0:N-1)'); % [+1,-1,...,-1]'

if rem(N,2) ~= 0 % N odd
for j=2:N
y(j) = 2*xm(j-1)-y(j-1);
end
s = sum(y)
x = y - s*z;
else
error('Reconstruction unknown for N even.')
end

The algorithm obviously doesn't work for N even
because sum(y) = 0 for arbitrary x(j).

Is there a way to recover x if no original points
are known and N is even?

Thanks in advance,

Greg