1.txt

lab01_group24_final/backward_solve.m

+function R1 = backward_solve(A1)
+
+m = size(A1,1);
+
+n = size(A1,2);
+
+c = m*n;
+
+B = zeros(c,c);
+
+for i = 1:c
+
+	for j= 1:c
+
+		k1=(i-1-mod(i-1,n))/n;
+
+		k2=(1+mod(i-1,n));
+
+		l1=(j-1-mod(j-1,n))/n;
+
+		l2=(1+mod(j-1,n));
+
+		if(((abs(k1-l1)==0)&&(abs(k2-l2)==1))||((abs(k1-l1)==1)&&(abs(k2-l2)==0))) 
+
+			B(i,j) = 1;
+
+		end
+
+		if((k1==l1)&&(k2==l2))
+
+			B(i,j) = 1;
+
+		end
+
+	end
+
+end
+
+R = zeros(1,c);
+
+A = zeros(1,c);
+
+for i = 1:m
+
+	for j=1:n
+
+		A(1,(((i-1)*n)+j)) = A1(i,j); 
+
+	end
+
+end
+
+R = (A*(B^(-1))).*det(B);
+
+R = abs(R);
+
+R = int64(R);
+
+R = mod(R,2);
+
+R1 = zeros(m,n);
+
+for i = 1:m
+
+	for j=1:n
+
+		 R1(i,j) = R(1,(((i-1)*n)+j)); 
+
+	end
+
+end
+
+return 
+
+endfunction
+
+
+

lab01_group24_final/backward_solve_3.m

+function R1 = backward_solve_3(A1)
+
+L = [1 1 0; 1 1 1; 0 1 1 ];
+
+M = [1 0 0; 0 1 0; 0 0 1 ];
+
+N = zeros(3,3);
+
+B = [L M N; M L M; N M L ];
+
+A = [A1(1, :) A1(2, :) A1(3, :)];
+
+R = zeros(1,9);
+
+X = [0 0 1 1 0 1 0 1 1 ];
+
+%R = mod(A + X*B,2);
+
+%R = A/B;
+
+%R = (A + X*B);
+
+R = (A*(B^(-1))).*det(B);
+
+R = abs(R);
+
+R = int64(R);
+
+R = mod(R,2);
+
+R1 = zeros(3,3);
+
+R1 = [R(1,1:3); R(1,4:6); R(1,7:9)];
+
+return
+
+endfunction
+
+%disp(A)
+
+%disp(B)
\ No newline at end of file

lab01_group24_final/input_outlab_task_A1.txt

+180.0 100 0.05 5 3.5
\ No newline at end of file

lab01_group24_final/input_outlab_task_A2.txt

+0.0 0.0 76.63 27 0.027 0.0555 7 5 3.5 5
\ No newline at end of file

lab01_group24_final/main_task_A1.m

+clear; clc;
+
+file1 = fopen('input_outlab_task_A1.txt');
+
+A = fscanf(file1, '%f', [5,1]);
+
+fclose(file1);
+
+f = @ (a) (((A(2)-A(4))*tan(a)) + (A(4)*sin(a))/(sqrt((A(5))^2-((sin(a))^2))) - A(1));
+
+theta = fzero(f,[0,1.5]);
+
+theta = theta * (180/pi);
+
+file2 = fopen('output_outlab_task_A1.txt','w');
+
+fprintf(file2, '%f', theta);
+
+fclose(file2);
\ No newline at end of file

lab01_group24_final/main_task_A2.m

+clear; clc;
+
+file1 = fopen('input_outlab_task_A2.txt');
+
+A = fscanf(file1, '%f', [10,1]);
+
+fclose(file1);
+
+x1 = A(3) - A(1);
+
+y1 = A(4) - A(2);
+
+%secr = A(9)/((sqrt((A(9))^2-((sin(a))^2))));
+
+%a1 = (((sec(a))*(y1-A(8))) + ((A(9)/((sqrt((A(9))^2-((sin(a))^2)))))*(A(8))) + ((sec(a))*(A(5))*(A(10))))/(A(6)-(A(5)*(sec(a))));
+
+f = @ (a) ((A(5)*(A(10)+ ( ( ((sec(a))*(y1-A(8))) + (((A(9)/((sqrt((A(9))^2-((sin(a))^2)))))*(A(8)))*A(9)) + ((sec(a))*(A(5))*(A(10))) ) / (A(6)-(A(5)*(sec(a)))) ) )*tan(a)) + ((y1-A(8))*tan(a)) + ((A(8)*sin(a))/(sqrt((A(9))^2-((sin(a))^2)))) - x1) ;
+
+theta = fzero(f,[0.1,1.0]);
+
+A(10) = 3*A(10);
+
+a3 = (((sec(theta))*(y1-A(8))) + (((A(9)/((sqrt((A(9))^2-((sin(theta))^2)))))*(A(8)))*A(9)) + ((sec(theta))*(A(5))*(A(10))))/(A(6)-(A(5)*(sec(theta))));
+
+xf = x1;
+
+yf = y1 + (3*A(10)+a3)*A(5);
+
+theta = theta * (180/pi);
+
+%disp(theta)
+
+%disp(yf)
+
+file2 = fopen('output_outlab_task_A2.txt','w');
+
+fprintf(file2, '%f %f %f', theta, xf, yf);
+
+fclose(file2);
\ No newline at end of file

lab01_group24_final/output_outlab_task_A1.txt

+62.003645
\ No newline at end of file

lab01_group24_final/output_outlab_task_A2.txt

+41.070190 76.630000 93.409613
\ No newline at end of file

lab01_group24_final/readme.txt

+GROUP NUMBER - 24
+(150050071, Aahwan), (150050072, Sai Teja), (150050079, Rathod Dinesh).
+150050071- 100%
+150050072- 100%
+150050079- 100%
+Comments on the code :
+
+	-Task A2 had an important point. The consideration that laser travels with a velocity equal to velocity/refractiveIndex in the medium.
+
+	-There was a bit of an ambiguity in the task A2 description.Strictly considering the meaning, actually the second angle must be taken(because it is mentioned: to destroy the shield and it takes 2 hits to destroy it). But since the answer given is the angle for the first laser beam, the code for the same has been written.  
+
+	-An interesting pattern was observed for the action matrix for the general case of backward solve but that wasn't implemented because of the ease of generation using loops was more.
+
+Honour code :
+
+	We pledge on our honour that we have not given or received any unauthorized assistance on this assignment or any previous task.
+
+							-(150050071, Aahwan), (150050072, Sai Teja), (150050079, Rathod Dinesh).
+
+Citations :
+	
+	Octave learnt and understood from :
+
+	https://www.youtube.com/watch?v=gkVOSxmhtyk&index=17&list=PLHj4uM4X8Oso_M2yN6Fg0_Mp219z6f5-W
+		
+Reflection essay :
+
+	In the lights out problem found a pattern for the action matrix for different m*n matrices.
+
+	Solved the interesting problem of two spaceships.
+
+	Realised the point that laser travels with a velocity equal to velocity/refractive index in the medium!(that eureka moment :)) 
\ No newline at end of file