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80 lines
3.4 KiB
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80 lines
3.4 KiB
Markdown
# Exercises
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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.
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1. What's the difference between [free software](free_software.md) and [open source](open_source.md)?
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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.
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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?
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## Solutions
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A solution to each problem should be listed here -- keep in mind there may exist other solutions that those listed here.
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**solution 1**:
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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.
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**solution 2**:
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```
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#include <stdio.h>
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#include <stdint.h>
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#define DECIMALS 10000
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int main(void)
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{
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int64_t previousError = 10000000000;
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uint64_t previousPi = 0;
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for (uint64_t gridSize = 2; gridSize < 200000; gridSize *= 2)
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{
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/* We'll sample a grid of points and count those that fall inside the
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circle with radius of gridSize. Thanks to the 8-symmtery of a circle we
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only sample the 1/8th of the plane. */
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uint64_t inCircle = 0;
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for (int y = 0; y < gridSize; ++y)
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for (int x = y; x <= gridSize; ++x)
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if ((x * x + y * y) / gridSize <= gridSize) // if distance is < radius
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inCircle++; // count the point
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// compute pi from the formula for circle area (area = 2 * pi * r):
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uint64_t pi = (inCircle * 8 * DECIMALS) / (gridSize * gridSize);
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int64_t error = pi - previousPi;
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if (error < 0)
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error *= -1;
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if (error > previousError) // error got bigger due to overflows, stop
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{
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puts("that's it");
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break;
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}
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previousError = error;
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previousPi = pi;
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printf("%d.%d\n",pi / DECIMALS,pi % DECIMALS);
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}
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}
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```
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**solution 3**:
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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:
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```
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for i := 0 to N:
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add(S,i,i) // no need to compare
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for i := 0 to N:
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for j := i + 1 to N:
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if numbers[i] == numbers[j]:
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add(S,set(i,j))
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```
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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). |