Solutions to Prelim 1 Review Questions

NOTE: the original solutions posted to the newsgroup had some errors.

We strongly encourage you to solve the problems before looking at
the solutions below. It is hard, if not impossible, to learn
programming by only looking at examples developed by others.

In general, there are many solutions to any problem. We show only
some of these, which might not be the ones you came up with. As
long as your programs generate the correct result and satisfy the
style requirements, they are as good as the ones below.
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>> Problem 1:

Solution 1:

in = 'this is the string whose vowels should be replaced';
out = [];
% check each character in the string
    for c = in
          if c == 'a'
            out = [out '*'];
          elseif c == 'e'
            out = [out '*'];
          elseif c == 'i'
            out = [out '*'];
          elseif c == 'o'
            out = [out '*'];
          elseif c == 'u'
            out = [out '*'];
          else
            out = [out c];
          end
    end
% show the result
    out


Solution 2: 

in = 'this is the string whose vowels should be replaced';
out = [];
% check each character in the string
    for c = in
        if (c == 'a') | (c == 'e') | (c == 'i') | (c == 'o') | (c == 'u')
          out = [out '*'];
        else
          out = [out c];
        end
    end
% show the result
    out


Solution 3:

in = 'this is the string whose vowels should be replaced';
out = [];
% check each character in the string
    for c = in
      % use a trick to determine whether the current character
      % belongs to a set of characters (now, vowels).
          if sum('aeiou' == c) == 1
            out = [out '*'];
          else
            out = [out c];
          end
    end
% show the result
    out

Note: Even more compact solutions can be found. 

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>> Problem 2:

in = 'this is the string whose vowels should be replaced';
out = [];
% check each character in the string
    for c = in
      % use a trick to determine whether the current character
      % belongs to a set of characters (now, vowels).
          if sum('aeiou' == c) == 1
            out = [out c c];
          else
            out = [out c];
          end
    end
% show the result
    out

Note: All the ideas that solve problem 1 can be immediately adapted to
solve this one.

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>> Problem 3:

Solution 1:

in = [6 0 -3 -5 4 5 1 -9 0 3];
pos = []; % positive numbers from $in$
neg = []; % negative numbers from $in$
% check each element and place it into the appropriate array if not zero
    for x = in
      if x > 0
        pos = [pos x];
      elseif x < 0
        neg = [neg x];
      end
    end
% show results
    pos
    neg

Solution 2:

in = [6 0 -3 -5 4 5 1 -9 0 3];
pos = in(in > 0)
neg = in(in < 0)

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>> Problem 4:

Solution 1:

in = [1 3 2 5 7];
% take care of the first element, which is special
    out = (in(2) + in(end))/2;
% now deal with the 'regular' elements
    for i = 2:length(in)-1
      out = [out (in(i-1)+in(i+1))/2];
    end
% now take care of the last element, which is also special
    out = [out (in(end-1)+in(1))/2];
% show the result
out

Solution 2: no loop, only vector operations.

in = [1 3 2 5 7];
out = (in([end 1:end-1]) + in([2:end 1])) / 2

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>> Problem 5. 

% the array below holds the number of days in each month
    days = [31 28 31 30 31 30 31 31 30 31 30 31];
% loop through each month
    for month = 1:12
      % loop through each day of the month
          for day = 1:days(month)
             disp([ '[' num2str(month) '  ' num2str(day) ']' ]);
          end
    end

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>> Problem 6. Note that A is complementary to T, and G is
   complementary to C.

% input string
    in = 'taCaGCgATAcgtg';
% this will store the output string
    out = [];
% we convert the string to upper case to simplify comparisons
    for c = upper(in)
      if c == 'A'
        out = [out 'T'];
      elseif c == 'T'
        out = [out 'A'];
      elseif c == 'G'
        out = [out 'C'];
      else
        out = [out 'G'];
      end
    end
% show the resulting string
    out

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>> Problem 7.

% note: matlab pre-defines $pi$ = 3.14159...
N = 100; % number of terms to consider
d = 67; % angle in degrees
r = pi * d / 180; % angle in radians
cosine = 0; % approximation so far (avoid 'cos' because it is a function name)
% add all elements
    for i = 0 : N-1
      cosine = cosine + (-1)^i*r^(2*i)/factorial(2*i);
    end
% and this is the value we have computed
    cosine

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>> Problem 8.  

Solution 1:

N = 20;
x = [];
for i = 1:N
  x = [x floor(5*rand)-2];
end
x

Solution 2:

N = 20;
floor(5*(rand(1,N))-2)

==================================================


>> Problem 9.


Solution 1:

N = 100; % the number of steps we will simulate
x = 0; y = 0; % current coordinates of the drunkard
% simulate
    for i = 1:N
      % generate a random number in the range 1..4
      step = ceil(4*rand);
      if step == 1
        % step to the right
            x = x + 1;
      elseif step == 2
        % step to the left
            x = x - 1;
      elseif step == 3
        % step up
            y = y + 1;
      else
        % step down
            y = y - 1;
      end
    end
% compute and show distance reached ($sqrt$ is the square root function)
    sqrt(x*x + y*y)


Solution 2:

% note: matlab pre-defines $i$ = sqrt(-1)
N = 100; % the number of steps we will simulate
% simulate N random steps: position (x,y) is represented by x + i y
    pos = sum(i .^ floor(4*rand(1,N)));
% compute and show distance reached
    abs(pos)