Here let be listed exercises for the readers of this wiki. You can allow yourself to as many helpers and resources as you find challenging: with each problem you should either find out you know the solution or learn something new while solving it.
2. Write a program in [C](c.md) that computes the value of [pi](pi.md) without using float/double and any libraries except for `stdio.h` and `stdint.h` -- you can only use built-in integer types and those from `stdint.h`. The program must compute pi as accurately as possible (at least 2 decimals) and write the value out as base 10 decimal.
3. Say we have an algorithm that finds all pairs of equal numbers in an array of numbers of length *N* and adds all of these (unordered) pairs to a set *S*. The algorithm is: `for i := 0 to N: for j := 0 to N: if numbers[i] == numbers[j]: add(S,set(i,j))`. How can we optimize the algorithm in terms of its execution speed (i.e. make it perform fewer operations)? How did the asymptotic time complexity ("big O") class change?
Both movements share very similar rules of licensing and technically free software and open-source are largely the same. However, free software is fundamentally aiming for the creation ethical software -- that which respects its user's freedom -- while open source is a later movement that tries to adapt free software for the business and abandons the pursuit of ethics.
In the given algorithm we compare all numbers twice. This can be avoided by not comparing a number to previous numbers in the array (because these have already been compared). Additionally we don't have to compare the same number to itself, a number will always be equal to itself:
```
for i := 0 to N:
add(S,i,i) // no need to compare
for i := 0 to N:
for j := i + 1 to N:
if numbers[i] == numbers[j]:
add(S,set(i,j))
```
While the first algorithm performs N^2 comparisons, the new one only needs N - 1 + N - 2 + N - 3 + ... ~= N * N / 2 = N^2 / 2 comparisons. Even though the new version is always twice as fast, its time complexity class remains the same, that is O(N^2).