Exercise 7
Added Exercise 7 to Repository
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Informatik_I/Exercise_7/Task_1/README.org
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Informatik_I/Exercise_7/Task_1/README.org
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#+title: Task 1: Recursive function analysis
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#+author: JirR02
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/This task is a text based task. You do not need to write any program or C++ file: the answer should be written in main.md (and might include code fragments if questions ask for them)./
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* Task
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:PROPERTIES:
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:CUSTOM_ID: task
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:END:
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For each of the following recursive functions:
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1.
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#+begin_src cpp
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bool f(const int n) {
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if (n == 0) return false;
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return !f(n - 1);
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}
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#+end_src
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2. [@2]
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#+begin_src cpp
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void g(const int n) {
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if (n == 0) {
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std::cout << "*";
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return;
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}
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g(n - 1);
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g(n - 1);
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}
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#+end_src
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1) Formulate pre- and post conditions.
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2) Show that the function terminates. *Hint*: No proof expected, proceed
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similar as with the lecture example of the factorial function.
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3) Determine the number of functions calls as mathematical function of
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parameter n. *Note*: include the first non-recursive function call.
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* Solution
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:PROPERTIES:
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:CUSTOM_ID: solution
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:END:
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#+begin_src markdown
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1. i) Pre- and post conditions
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```c++
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// PRE: n is a positive integer
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// POST: returns true if n is even and false if n is odd.
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```
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ii) The function =plain|f= terminates because it has a termination
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condition. It will terminate once n reaches 0. If the number is
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negative, it will be in an infinite loop.
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iii) $\text{Calls}_{f}(n) = n$
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2. i) Pre- and post conditions
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```cpp
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// PRE: n is a positive integer
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// POST: print 2^n *
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```
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ii) The function =plain|g= terminates because it has a termination condition. It will terminate once n reaches 0. If the number is negative, it will be in an infinite loop.
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iii) $\text{Calls}_{g}(n) = 2^n$
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#+end_src
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