[b;zeros(2,numel(b))]

It is not clear the overlap values. I assumed the following: k33+k11 k33+k12 K33+k13 K33+k23 ..., Until the last is k33+k33 The other values is just copy from the given location. Then, you might use only for loop for this. Check in the following: k19 = reshape(k(1:3,1:3),1,9); % Reshape for [k11 k12 k12 k21 k22 k23 k31 k32 k33] j = 1; for i=4:2:21 k(i,i-1) = k(2,1); k(i-1,i) = k(2,1); k(i+1,i-1) = k(3,1); k(i-1,i+1) = k(3,1); k(i+1,i) = k(3,2); k(i,i+1) = k(3,2); k(i,i) = k(2,2); k(i+1,i+1) = k(3,3) + k19(j); % Overlap goes here j = j + 1; end

Here is one approach using simple loops and indexing. You will have to do some fine-tuning to get it to suit your requirements. A = [0,0,0;0,2,1;0,1,0]; % Input 3x3 matrix. % M = [... % base cube '~~~______~'; '~~/ /|'; '~/ / |'; '/_____/ |'; '| | /'; '| | /~'; '|_____|/~~']; % Sa = size(A); Sm = size(M); Xm = M~='~'; Xr = cell(Sa); Xc = cell(Sa); % Generate indices: for ii = 1:Sa(1) % back -> front. for jj = 1:Sa(2) % left -> right. for kk = 1:A(ii,jj) % bottom -> top. R = ii*3 - kk*3; C = jj*6 - ii*3 - kk; % try without kk. Xr{ii,jj} = [Xr{ii,jj};R+1-Sm(1):R]; Xc{ii,jj} = [Xc{ii,jj};C+1-Sm(2):C]; end end end % Normalize indices: Xr = vertcat(Xr{:}); Xc = vertcat(Xc{:}); Xr = 1+Xr-min(Xr(:)); Xc = 1+Xc-min(Xc(:)); % Place cube at required locations: Z = repmat(' ',max(Xr(:)),max(Xc(:))); for nn = 1:sum(A(:)) tmp = Z(Xr(nn,:),Xc(nn,:)); tmp(Xm) = M(Xm); % keep char outside the cube. Z(Xr(nn,:),Xc(nn,:)) = tmp; end disp(Z) Displays this in the command window: ______ / /| / / | /_____/ |______ | | / /| | | / / | |______/_____/ | / /| | / / / | | / /_____/ |_____|/ | | / | | / |_____|/ Tested with the first example input matrix: >> A = [0,0,0;0,1,1;0,0,0] A = 0 0 0 0 1 1 0 0 0 Giving ____________ / / /| / / / | /_____/_____/ | | | | / | | | / |_____|_____|/

Hi, for the lower right 3x3 matrix use: A = k(4:end,4:end) for the marked area (4×3): A = k(3:end,4:end) An upper left 2×2 would be: A = k(1:2,1:2) Best regards Stephan