% Solution to Review question 1
% calculates the TAX to be paid by a citizen of Timbuktu

% defines income limits
limit1 = 30000;
limit2 = 60000; 

income = input('Please enter your income: ');

if (income < limit1)    % low income
    tax = 0.25*income;  % tax = 25% of income; 
elseif (income >= limit1) && (income < limit2)    % intermediate
                                                  % income
    tax = 0.25*limit1 + 0.30*(income - limit1);
elseif (income >= limit2) 
    tax =0.25*limit1 + 0.30*(limit2-limit1) + 0.40*(income -limit2);
end
 
fprintf('You should pay $%10.2f in tax.\n',tax);
fprintf('Thank you for living in Timbuktu\n'); 


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% Solution to review question 2
% ask for three numbers and print the maximum of the three 
% No arrays nor built in functions other than input

n1 = input('enter the  first number : ');
n2 = input('enter the second number : ');
n3 = input('enter the  third number : ');

if (n1 >= n2) && ( n1>= n3)   % n1 is the maximum 
    max = n1;
elseif (n2>= n1) && (n2>= n3) % n2 is the maximum 
    max = n2;
else                          % n3 must be the maximum 
    max = n3;
end
%prints out the maximum 
max


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Alternate solutions to review question 2

x = input('enter the  first number : ');
y = input('enter the second number : ');
z = input('enter the  third number : ');

if (x > y)
    if (x > z)
        fprintf('The maximum of numbers %d, %d, %d is %d',x,y,z,x);
    else
        fprintf('The maximum of numbers %d, %d, %d is %d',x,y,z,z);
    end
else
    if (y > z)
        fprintf('The maximum of numbers %d, %d, %d is %d',x,y,z,y);
    else
        fprintf('The maximum of numbers %d, %d, %d is %d',x,y,z,z);
    end
end


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% Solution to review question 3 
% Prints all integers between 1 and 1000
% that are divisible by 2 and 5 but not by 7 
%
% remember that the expression
%                 mod(n,d)==0
% evaluates to true if (n IS divisible by d) 
% otherwise it evaluates to false. 
% 
% Thus, the English sentence 
% "n is divisible by 2 and 5 but not by 7" 
% which is equivalent to the bad but simple sentence
% "(n is div. by 2) and (n is div. by 5) and (n is not div. by 7)" 
% can be translated into Matlab as 
% ( mod(n,2)==0 ) && (mod(n,5)==0 ) && ( mod(n,7)~=0 )
% the program is now simple to write

for n=1:1000 % make n take all integer values between and 1000
    if ( mod(n,2)==0 ) && (mod(n,5)==0 ) && ( mod(n,7)~=0 )
        % if n is div by 2 and 5 but not by 7 print it out
        fprintf('%d ',n); 
    end
    % the following block is just to get nice format 
    if (mod (n,50) ==0) 
        fprintf('\n'); % break the line once in a while
    end
end


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Solution to the review question 4
% The program simulates coin tossing. It prints "H" for heads 
% and "T" for tails. The simulation cointinues until the 
% first run of 6 heads occurs.  At the end the total number 
% of coin tosses in the simulation is printed.

run = 0;            % contains the number of heads in the current run
numtosses = 0;      % total number of toses
desiredlength = 6;  % length of run to stop 

while (run < desiredlength) 
    % Toss a coin
    toss = floor(rand*2); % produces 0 or 1 with equal probabilities 
    numtosses = numtosses +1 ; 
    if (toss ==0)   % this was tails
        fprintf('T'); 
        run= 0;     % the previous run of heads (if any) is over     
    else            % the toss was heads 
        fprintf('H'); 
        run= run+1; % the run of heads continues (or starts)
    end
    % The following block is for nice format
    if (mod(numtosses,50)==0) 
        fprintf('\n') % breaks the lines once in a while
    end
end

fprintf('\nTotal number of tosses %d\n', numtosses); 


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Solution to Review question 5
% Translate a time input in seconds 
% to the format hours:minutes:seconds

secinmin = 60; 
secinhour = 3600;

seconds = input('Introduce a time in seconds : ');
hours= floor(seconds/secinhour); 
                    % the integer number of hours
remainder = seconds - hours*secinhour; 
                    % the time (in secs) remaining after discounting the hours 
minutes= floor(remainder/secinmin);
sec = remainder - minutes*secinmin; 
                    % seconds remaining after discounting the minutes
fprintf('The time corresponds to %d:%d:%d\n',hours,minutes,sec);