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//Eg-15.5
//pg-619
clear
clc
close()
y = zeros(5,5);
y(1:5,1) = 10;
y(1:5,5) = 10;
y(1,5:4) = 0;
//y(1,2:4) = 10;
for(i = 3:5)
for(j = 2:4)
y(i,j) = y(i-1,j+1) + y(i-1,j-1) - y(i-2,j);
end
end
printf('The table of the y values with different rows as different timelines :\n')
Y = y(2:5,:)
disp(Y)
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