[SOLVED] Distributed arithmetic: 2BAAT hardware with a single LUT

N = 16;
ids = dec2bin(([0:N-1]));

k0 = fi(0.72, true, N, N-1);
k1 = fi(-0.3, true, N, N-1);
k2 = fi(0.95, true, N, N-1);
k3 = fi(0.11, true, N, N-1);

multiplicand = @(k) bin2dec(k);
mem = zeros(N, 1);

for i=1:N
    mask = ids(i, :);
    k = k0 * multiplicand(mask(1)) + k1 * multiplicand(mask(2)) + ...
    k2 * multiplicand(mask(3)) + k3 * multiplicand(mask(4));
    mem(i, 1) = k;
end

x0 = fi(0.5, true, N, N-1);
x1 = fi(0.25, true, N, N-1);
x2 = fi(0.5, true, N, N-1);
x3 = fi(0.25, true, N, N-1);

y = x0*k0 + x1*k1 + x2 * k2 + x3 * k3;

acc = 0;

for i=N:-1:1
    bid1 = bin(x0);
    id1 = bid1(i);
    bid1 = bin(x1);
    id2 = bid1(i);
    bid1 = bin(x2);
    id3 = bid1(i);
    bid1 = bin(x3);
    id4 = bid1(i);
    id = bin2dec([id1 id2 id3 id4]) + 1;
    if i > 1
        acc = (acc + mem(id))/2;
    else
        if i == 0
            acc = acc - mem(id);
        end
    end
end
    disp(['acc = ' num2str(acc)])
    disp(['y = ' num2str(y)])

clc;
clear;
n = 4;
N = 2^(n*2);
ids = dec2bin(([0:N-1]));

k0 = fi(0.72, true, N, N-1);

k1 = fi(-0.3, true, N, N-1);

k2 = fi(0.95, true, N, N-1);

k3 = fi(0.11, true, N, N-1);

multiplicand = @(k) bin2dec(k);
mem = zeros(N, 1);

for i=1:N
    mask = ids(i, :);
    k = ( (k0 * multiplicand(mask(1))) + (k0 * multiplicand(mask(2))) * 0.5 ) + ...
        ( (k1 * multiplicand(mask(3))) + (k1 * multiplicand(mask(4))) * 0.5 ) + ...
        ( (k2 * multiplicand(mask(5))) + (k2 * multiplicand(mask(6))) * 0.5 ) + ...
        ( (k3 * multiplicand(mask(7))) + (k3 * multiplicand(mask(8))) * 0.5 );
    mem(i, 1) = k;
end

mem_q15 = fi(mem, true, 16, 15);
%%
s = 0;
% x0 - 0.100 (0.5)
% x1 - 0.010 (0.25)
% x2 - 0.100
% x3 - 0.010

adr1 = bin2dec('00010001') + 1;
adr2 = bin2dec('10001000') + 1;

v1 = mem(adr1);
v2 = mem(adr2);
s = (s + v1) / 2;
s = s - v2;

disp(s)