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Merge branch 'master' of https://whale.am28.uni-tuebingen.de/git/teaching/scientificComputing

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Jan Grewe 2019-11-11 19:52:41 +01:00
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programming/exercises/control_flow.tex
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@ -58,7 +58,7 @@ following pattern: ``variables\_datatypes\_\{lastname\}.m''
\question Indexing in vectors:
\begin{parts}
\part Define a vector \code{x} that contains values ranging from 101 to 200 in steps of 1.
\part Use a \code{for} loop to print each element of \code{x}. Use the running variable for the indexing.
\part Use a \code{for} loop to print each element of \code{x}. Use the running variable for indexing.
\begin{solution}
\code{for i = 1:length(x); disp(x(i)); end}
\end{solution}
@ -68,38 +68,161 @@ following pattern: ``variables\_datatypes\_\{lastname\}.m''
\end{solution}
\end{parts}
\question Create a vector \verb+x+ that contains 50 random numbers in the value range 0 - 10.
\question Create a vector \verb+x+ that contains 50 random numbers in the value range 0 --- 10.
\begin{parts}
\part Use a loop to calculate the arithmetic mean. The mean is defined as:
$\overline{x}=\frac{1}{n}\sum\limits_{i=0}^{n}x_i $.
\begin{solution}
\begin{verbatim}
x = rand(50, 1) .* 10;
total = 0;
for i = 1:length(x)
total = total + x(i);
end
average = total / length(x);
disp(average)
\end{verbatim}
\end{solution}
\part Use a loop to estimate the standard deviation:
$\sigma=\sqrt{\frac{1}{n}\sum\limits_{i=0}^{n}(x_i-\overline{x})^2}$).
\part Search the MATLAB help for functions that do the calculations for you :-).
$\sigma=\sqrt{\frac{1}{n}\sum\limits_{i=0}^{n}(x_i-\overline{x})^2}$.
\begin{solution}
\begin{verbatim}
x = rand(50, 1) .* 10;
total = 0;
for i = 1:length(x)
total = total + x(i);
end
average = total / length(x);
deviation = 0
for i = 1:length(x)
deviation = deviation + (x(i) - average)^2$;
end
standarddeviation = sqrt(deviation/length(x));
\end{verbatim}
\end{solution}
\part Search the MATLAB help for functions that do these calculations for you :-).
\begin{solution}
\verb+sqrt+ and \verb+mean+ do the job for you.
\end{solution}
\end{parts}
\question Implement a \code{while} loop
\question Implement a \code{while} loop that iterates ...
\begin{parts}
\part ... that iterates 100 times. Print out the current iteration number.
\part ... 100 times and displays the current iteration number.
\begin{solution}
\begin{verbatim}
max_iterations = 100;
iteration_count = 0;
while iteration_count < max_iterations
disp(iteration_count);
iteration_count = iteration_count + 1;
end
\end{verbatim}
\end{solution}
\part ... iterates endlessly. You can interrupt the execution with the command \code{Strg + C}.
\begin{solution}
\begin{verbatim}
while true
disp('i am still running');
end
\end{verbatim}
\end{solution}
\end{parts}
\question Use an endless \code{while} loop to individually display the elements of a vector that contains 10 elements. Stop the execution after all elements have been displayed on the command line.
\begin{solution}
\begin{verbatim}
x = rand(10, 1);
index = 0;
while true
disp(x(index);
if index >= length(x)
break;
end
index = index + 1;
end
\end{verbatim}
\end{solution}
\question Use the endless \verb+while+ loop to draw random numbers
(\verb+randn+) until you hit one that is larger than 1.33.
\begin{parts}
\part Count the number of required iterations.
\begin{solution}
\begin{verbatim}
count = 1;
threshold = 1.33;
while true
x = randn(1);
if x > threshold
break;
end
count = count + 1;
end
\end{verbatim}
\end{solution}
\part Use a \code{for} loop to run the previous test 1000 times. Store all counts and calculate the mean of it.
\begin{solution}
\begin{verbatim}
trials = 1000;
threshold = 1.33;
counts = zeros(trials, 1);
for i = 1:trials
while true
x = randn(1);
counts(i) = counts(i) + 1;
if x > threshold
break
end
end
end
\end{verbatim}
\end{solution}
\part Create a plot that shows for each of the 1000 trials, how often you had to draw.
\begin{solution}
\begin{verbatim}
plot(counts)
hold on
plot([0, trials], [mean(counts), mean(counts)])
xlabel('trials')
ylabel('number of samples')
\end{verbatim}
\end{solution}
\part Play around with the threshold. What happens?
\begin{solution}
When lowering the threshold, the average number of counts is reduced, when increasing it, the counts increase!
\end{solution}
\end{parts}
\question Create a vector \verb+x+ that contains 10 random numbers in the range 0:10.
\begin{parts}
\part Use a \code{for} loop to delete all those elements
(\code{x(index) = [];}) that are smaller than 5.
\part Delete all elements that are smaller than 5 and larger than 2.
\part Can you do the same without a loop?
(\code{x(index) = [];}) that are less than 5.
\begin{solution}
\begin{verbatim}
x = rand(10, 1) .* 10;
for i = length(x):-1:1
if x(i) < 5
x(i) = [];
end
end
\end{verbatim}
\end{solution}
\part Delete all elements that are less than 5 and greater than 2.
\begin{solution}
\begin{verbatim}
x = rand(10, 1) .* 10;
for i = length(x):-1:1
if x(i) < 5 & x(i) < 2
x(i) = [];
end
end
\end{verbatim}
\end{solution}
\part Can you do the same without a loop?
\begin{solution}
\verb+x((x < 5) & (x > 2))) = [];+
\end{solution}
\end{parts}
\question Test the random number generator! To do so count the
@ -107,6 +230,22 @@ following pattern: ``variables\_datatypes\_\{lastname\}.m''
[0.0, 0.2, 0.4, 0.6, 0.8, 1.0]. Store the results in a vector. Use a
loop to draw 1000 random numbers with \verb+rand()+ (see help). What
would be the expectation? What is the result?
\begin{solution}
We expect each catergory to contain the same amount of data points.
\begin{verbatim}
edges = [0.0, 0.2, 0.4, 0.6, 0.8, 1.0];
counts = zeros(length(edges)-1, 1);
trials = 1000;
for i = 1:trials
x = rand(1);
for j = 2:length(edges) % using a nested for loop we can avoid a complicated if elseif construct
if x > edges(j-1) & x < edges(j)
counts(j-1) = counts(j-1) + 1;
end
end
end
\end{verbatim}
\end{solution}
\question String parsing: It is quite common to use the names of datasets to encode the conditions under which they were recorded. To find out the required information we have to \emph{parse} the name.
\begin{parts}
@ -114,9 +253,32 @@ following pattern: ``variables\_datatypes\_\{lastname\}.m''
\verb+filename = '2015-10-12_100Hz_1.25V.dat'+. Obviously, the
underscore was used as a delimiter.
\part Use a \verb+for+ loop to find the positions of the underscores and store these in a vector.
\begin{solution}
\begin{verbatim}
positions = [];
for i = 1:length(filename)
if filename(i) == '_'
positions(end + 1) = i
end
end
\end{verbatim}
\end{solution}
\part Use a second loop to iterate over the previously filled `position' vector and use this information to
cut \verb+filename+ appropriately.
\part Print out the individual parts.
cut \verb+filename+ appropriately and display the individual parts.
\begin{solution}
Tricky thing here, we need to catch the corner cases!
\begin{verbatim}
for i = 1:length(positions)
if i == 1
disp(filename(1:positions(i)))
elseif i == length(positions)
disp(filename(positions(i-1):end))
else
disp(filename(positions(i-1):positions(i)))
end
end
\end{verbatim}
\end{solution}
\end{parts}
\end{questions}