%% Temple university CIS 2033
% A modern Introduction to Probability and Statistics
% Lab 1 assignment 
% URL: http://astro.temple.edu/~tuf28053/CIS2033_Spring2015/lab_assignments/lab_assignment_1.pdf
%
% Author: Djordje Gligorijevic
% email: gligorijevic@temple.edu

%% Problem 1 solutions

% create a matrix A with 5 rows and 3 columns, filled with random variables 
% between 0 and 10
A = rand(5,3);

% create an identity matrix B with the size of 3 times 3
B = eye(3);

% get the submatrix C of A by extracting the elements of the last three rows
C = A(end-2:end, :);

% perform element-wise multiplication between matrix B and matrix C. 
% The result is denoted as the matrix D
D = B .* C; 

% concatenate A and D to form matrix E, whose first 5 rows are from A and 
% the last 3 rows are from D
E = vertcat(A, D);

% plot a histogram of each column of matrix E, what can you tell about 
% the distribution of values in columns? (Please label the axes and add a 
% title. You should also specify other properties such as the line width, 
% the font size and the color. Please save the figure as .png)
figure;
hist(E);
xlabel('Spread of Values');
ylabel('Contained Values');
title('Plot of E');
set(gca, 'LineWidth', 2);
set(gca, 'FontSize', 12);
set(gca, 'YColor', [0.2 0.4 0.6]);
set(gca, 'XColor', [0.1 0.3 0.5]);

% write a function to calculate intersection between two sets 
% A = 1 : 5 : 200 and B = 1 : 3 : 190.
A2 = 1:5:200;
B2 = 1:3:190;
results = Intersection(A2,B2);

%% Problem 2 solution

% Before taking a look at potential homweork solutions please refer to this
% blog post on this no three coincident birthdays problem which was main
% inspiration of this lab assignment problem: 

web('http://drmoron.org/3-birthday-problem/');

% The problem solved here is slighlty different as it measures probability
% that three people have coincident birthday.
% Statement of the author of the blog:
% 
% "So what is the probability for, given a group of people, that three have
% the same birthday. The answer is… I don’t know. The trick I used for a 
% pair of people doesn’t work for 3, and the counting just gets too 
% complicated for my little brain.
%
% Fortunately, I have another trick up my sleeve. Rather than figure out 
% the exact solution, I can write a simulator program to estimate the 
% probability. Here’s how it works. The simulator sets up an empty (virtual)
% room. It then lets in people one at a time. Each person coming in has a 
% random birthday. The simulation continues doing this until someone comes 
% in with a birthday that completes a triple.
%
% That, of course, does not give us any probabilities. If we run the 
% simulation again, we will (probably) get a different answer. But, if we 
% run the simulation enough times, we can estimate the probabilities as the
% percentage of times the simulated group had 3 people with the same 
% birthday. There is a very rigorous branch of computational science to 
% determine the number of samples needed and the uncertainty remaining. 
% But screw all that; I just ran the simulation for an excessive 100,000 
% times."

maxBirthdays = 6;
numDays = 365;
numTrials = 1000;
maxGroupSize = 100;

probabilities = NCoincidentBirthdaysProbability( ...
    maxBirthdays, numDays, numTrials, maxGroupSize);

colors = {'r','b','k','m','c'};

figure;
hold on;
x = 1:maxGroupSize;
% Note that making an inverse probability plot simple requires putting 
% 1 - probabilities(i,:)
for i = 1:maxBirthdays-1
   plot(x, 1-probabilities(i,:), colors{i}, 'lineWidth', 2); 
end
title('N coincident birthday problem');
legend('2 b-days','3 b-days','4 b-days','5 b-days','6 b-days');
hold off;

%% Problem 3 solution

% For a given probability mass function please calculate and plot in MATLAB
% proper distribution function.

figure;
hold on;
title('Distribution Function from Given PMF', 'FontSize', 16);
xlabel('a', 'FontSize', 14);
ylabel('F(a)', 'FontSize', 14);
a = [-1, 0, 1, 2, 3];
F = [1/24, 13/24, 19/24, 23/24, 1];
stairs(a, F);
hold off;