Commit ede95118 authored by AYUSH RAJ's avatar AYUSH RAJ


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;
b = 0;
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
\ No newline at end of file
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;
%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
dec_equiv = 4*A(3,1)+2*A(3,2)+A(3,3); % calculates decimal equivalent of the last binary row.
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);
%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);
answer = mod(answer,2);
%disp('The answer is :')
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!"
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.
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