Lab01

lab01_group29_inlab/adj.m

+function [b] = adj(k,l)                    %this function returns 1 if k th and l th cells are adjacent and zero otherwise
+row_k = floor(k/3);                        %computing the corresponding row and column corresponding to k th and l th cell
+row_l = floor(l/3);
+col_k = mod(k,3);
+col_l = mod(l,3);
+if((row_k-row_l)^2+(col_k-col_l)^2 <= 1)   
+b = 1;
+else
+b = 0;
+endif
+endfunction

lab01_group29_inlab/final_state_3.m

+function [B] = final_state_3(initial,action) % initial is the matrix to be updated after the operating action
+A = genmatrix();                      % generate a matrix (9*9) where (i,j)th entry is 1 if i th and j th cells are adjacent and 0 otherwise
+action = action';                     
+action = reshape(action,9,1);         
+action = (A*action);                  
+action = reshape(action,3,3);
+action = action';
+B = mod((initial + action),2);        %computing the effect of action by using the logic given at http://www.ueda.info.waseda.ac.jp/~n-kato/lightsout/
+return;
+endfunction 
\ No newline at end of file

lab01_group29_inlab/genmatrix.m

+function [A] = genmatrix()     %this function generates a matrix (9*9) where (i,j)th entry is 1 if i th and j th cells are adjacent and 0 otherwise
+A = zeros(9);
+for i=0:8
+for j=0:8
+if ( adj(i,j) == 1)
+A(i+1,j+1)= 1;
+endif
+end
+end
+%display(A)
+endfunction

lab01_group29_inlab/lookup_solve_3.m

+%A = floor(2*rand(3,3)); % make an abritrary input
+%disp('The matrix to be solved is:')
+%disp(A); % print it
+
+function [answer] = lookup_solve_3(A)  		% takes A as an arguement
+
+lookup_matrix = [3 7 4 6 5 1 2];		% decimal form of the given table
+
+answer = zeros(3);                              % the required final action matrix
+
+dec_equiv = 1;                                  % to enter the below loop for first time
+
+while dec_equiv > 0                             % checks decimal equivalent of the last row
+  for variable = [2 3]                          
+    action = zeros(3);                          % action matrix : 0 means no flipping of switch
+    action(variable,:) = A(variable-1,:);       % ith row of action matrix is updated with the (i-1)th row of input A
+    A = final_state_3(A,action);
+    answer = answer+action;                     % each action matrix is added to this final answer matrix
+  end
+  
+  dec_equiv = 4*A(3,1)+2*A(3,2)+A(3,3);         % calculates decimal equivalent of the last binary row. 
+  if(dec_equiv>0)
+    num = lookup_matrix(dec_equiv);	        % num is the corresponding decimal from the lookup_matrix 
+    C = zeros(3,1);                             
+    
+    for var = [3 2 1]                           % breaking num into binary form and storing it in the row vector C 
+      C(var) = mod(num,2);
+      num = floor(num/2);
+    end
+    
+    %change A/
+    action1 = zeros(3);				
+    action1(1,:) = C;				% action to be taken in the first row
+    A = final_state_3(A,action1);			% state of A after action1
+    answer = answer+action1;                    % updating the final answer matrix
+    answer = mod(answer,2);                   
+  endif
+endwhile
+
+answer = mod(answer,2);
+
+%disp('The answer is :')
+%disp(answer);
+return 
+endfunction

lab01_group29_inlab/readme.txt

+Group No - 29
+Group Name - puzzle
+
+(Ayush Raj , 150050042):100%
+(Umesh Kumar , 150050052):100%
+(Ankan Sardar , 150050064):100%
+
+Honor Code:
+
+ " We pledge on our honor that we have not given or received any unauthorised assistance for this assignment or any previous task!"
+
+
+citations:
+  https://en.wikibooks.org/wiki/Octave_Programming_Tutorial
+  http://www.ueda.info.waseda.ac.jp/~n-kato/lightsout/
+  http://www.gnu.org/software/octave/octave.pdf
+  http://www.dm.unibo.it/~lenci/teaching/14/maa/octavetut.pdf
+
+Reflection Essay:
+
+	We got to know an algorithmic solution to "lights Out" puzzle. We also learnt the implementations of various operations on matrices in octave environment and how to write functions in octave.