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Informatik_I/Bonus_1/README.md Normal file
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# Project overview
The goal of this project is to implement a game called _Dots And Boxes_: two players, **A** and **B**, take turns to draw lines connecting dots on a $m \times n$ (here $2 \times 2$) board.
If a player manages to close a box (a $1 \times 1$ square surrounded by lines), the player claims the box by placing its token (**A/B**) in the box, and the player gets another turn. The game is finished when all boxes have been claimed, and the player with the most claimed boxes wins. Note that, even if a player claims multiple boxes in a single move, they would only get one extra move.
For example, in the following game player **A** won:
![](./pictures/dots_and_boxes.svg)
## Your task
The file `game.cpp` contains a partial implementation of the game. The main game loop is implemented in the function `play_game(grid)`. This function is called from `main.cpp` and gets the game grid as an argument. The game grid is already fully implemented. We describe the functionality it provides in a following section below.
Your task is extending the game implementation in file `game.cpp` such that it works the way described above. We provide some functions in this file and thus suggest a certain structure -- but you are completely free to do it your way. Add, remove or change anything you want. The only constraint is that you must implement the function play_game which takes the game grid as argument. You are of course free to change the body of this function in whatever way you like.
If you do decide to change the body of the play_game function be careful to not accidentally change the format of the outputs it generates. For example: If you change the output `std::cout << "Game finished" << std::endl;` to `std::cout << "Game Finished" << std::endl;` you will not pass some test cases. The autograder is very picky when checking whether or not your outputs are correct. You can get all the points in this exercise without changing any of the output lines in the template.
_Note_: More Information on the the concept of the game on Code Expert.

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#include "game.h"
#include <assert.h>
#include <iostream>
// Checks whether or not the game is finished.
// A game is considered finished when all lines have been drawn.
bool game_is_finished(const Grid &grid) {
for (unsigned int j = 0; j < grid.num_rows(); ++j) {
for (unsigned int i = 0; i < grid.num_cols(); ++i) {
if (grid.field(i, j) == ' ')
return false;
}
}
return true;
}
// Draw a line in the grid starting at point (row, col) going towards the given
// direction
void draw_line(Grid &grid, unsigned int row, unsigned int col, char direction) {
switch (direction) {
case 'r':
grid.horizontal(row, col) = true;
break;
case 'l':
grid.horizontal(row, col - 1) = true;
break;
case 'd':
grid.vertical(row, col) = true;
break;
case 'u':
grid.vertical(row - 1, col) = true;
break;
default:
// We should never reach this point.
std::cout << "Invalid line direction.";
assert(false);
}
}
// Checks whether or not a box (or field) is newly completed.
// A box is newly completed if all four sides are drawn but the box is not
// claimed yet.
bool is_newly_completed_box(const Grid &grid, unsigned int row,
unsigned int col) {
bool state;
if (grid.vertical(row, col) == true && grid.horizontal(row, col) == true &&
grid.vertical(row, col + 1) && grid.horizontal(row + 1, col) == true)
state = true;
else
state = false;
return state;
}
// Play the players move.
// This function returns true if the player completed a box with their move.
// Otherwise it returns false.
bool play_move(Grid &grid, char player, unsigned int row, unsigned int col,
char direction) {
// Note: we assume the move is valid
// draw the new line
draw_line(grid, row, col, direction);
// check if the current players move completed a new box
bool completed_new_box = false;
switch (direction) {
case 'r':
// we drew a horizontal line
// check if box above the horizontal line is newly completed
// Note: box above only exists if the line is not at the top edge of
// the grid!
if (row > 0 && is_newly_completed_box(grid, row - 1, col)) {
completed_new_box = true;
grid.field(row - 1, col) = player;
}
if (row >= 0 && row < grid.num_rows() &&
is_newly_completed_box(grid, row, col)) {
completed_new_box = true;
grid.field(row, col) = player;
}
break;
case 'l':
if (row > 0 && is_newly_completed_box(grid, row - 1, col - 1)) {
completed_new_box = true;
grid.field(row - 1, col - 1) = player;
}
if (row >= 0 && row < grid.num_rows() &&
is_newly_completed_box(grid, row, col - 1)) {
completed_new_box = true;
grid.field(row, col - 1) = player;
}
break;
case 'u':
if (col > 0 && is_newly_completed_box(grid, row - 1, col - 1)) {
completed_new_box = true;
grid.field(row - 1, col - 1) = player;
}
if (col >= 0 && col < grid.num_cols() &&
is_newly_completed_box(grid, row - 1, col)) {
completed_new_box = true;
grid.field(row - 1, col) = player;
}
break;
case 'd':
if (col > 0 && is_newly_completed_box(grid, row, col - 1)) {
completed_new_box = true;
grid.field(row, col - 1) = player;
}
if (col >= 0 && col < grid.num_cols() &&
is_newly_completed_box(grid, row, col)) {
completed_new_box = true;
grid.field(row, col) = player;
}
break;
}
return completed_new_box;
}
// Computes a players score.
// A Players score is the number of fields claimed by the player.
unsigned int compute_player_score(const Grid &grid, char player) {
int score = 0;
for (unsigned int j = 0; j < grid.num_rows(); j++) {
for (unsigned int i = 0; i < grid.num_rows(); i++) {
if (grid.field(i, j) == player)
++score;
}
}
return score;
}
// Checks if the move is within the bounds of the grid,
// and the line is not already drawn
bool is_valid_move(const Grid &grid, unsigned int row, unsigned int col,
char direction) {
bool state = false;
switch (direction) {
case 'l':
if (col > 0 && row <= grid.num_rows() &&
grid.horizontal(row, col - 1) == false)
state = true;
break;
case 'r':
if (col < grid.num_cols() && row <= grid.num_rows() &&
grid.horizontal(row, col) == false)
state = true;
break;
case 'u':
if (row > 0 && col <= grid.num_cols() &&
grid.vertical(row - 1, col) == false)
state = true;
break;
case 'd':
if (row < grid.num_rows() && col <= grid.num_cols() &&
grid.vertical(row, col) == false)
state = true;
break;
default:
state = false;
break;
}
return state;
}
// Main game loop
void play_game(Grid &grid) {
// initialize player and step
char current_player = 'A';
unsigned int current_move = 1;
// print initial grid
grid.print_grid();
// initialize user input
unsigned int row, col;
char direction;
// loop while game is not finished
while (!game_is_finished(grid)) {
std::cout << "Move # " << current_move << std::endl;
// get a valid move
std::cout << "Player " << current_player << "'s move: " << std::endl;
std::cin >> row >> col >> direction;
while (!is_valid_move(grid, row, col, direction)) {
std::cout << "Invalid move!" << std::endl;
std::cin >> row >> col >> direction;
}
// play the move
bool move_completed_box =
play_move(grid, current_player, row, col, direction);
current_move += 1;
// print grid
grid.print_grid();
// change current player
switch (current_player) {
case 'A':
if (move_completed_box == true)
current_player = 'A';
else
current_player = 'B';
break;
case 'B':
if (move_completed_box == true)
current_player = 'B';
else
current_player = 'A';
break;
default:
// We should never reach this point.
std::cout << "Unknown Player";
assert(false);
}
}
// Game loop exited. The game must be finished
std::cout << "Game finished" << std::endl;
// Compute and display player scores
unsigned int score_a = compute_player_score(grid, 'A');
unsigned int score_b = compute_player_score(grid, 'B');
std::cout << "Player A: " << score_a << std::endl;
std::cout << "Player B: " << score_b << std::endl;
if (score_a == score_b) {
std::cout << "It's a draw!" << std::endl;
} else {
char winner;
if (score_a > score_b)
winner = 'A';
else
winner = 'B';
std::cout << "Player " << winner << " wins!" << std::endl;
}
}

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# Task
Write a program which reads in an integer a larger than `1000` and outputs its last three digits. For example, if `a=14325`, the output should be `3 2 5`.
_Hint_: You need to use integer division and modulo operators.
# Input
An integer larger than `1000`.
# Output
Last three digits separated by spaces.

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#include <iostream>
int main() {
int a;
std::cin >> a; // get input of for a
if (a >= 1000) { // check if input is greater than 1000
int b = a % 10; // get last digit
a /= 10;
int c = a % 10; // get second digit
a /= 10;
int d = a % 10; // get first digit
std::cout << d << " " << c << " "
<< b; // output first, second, and last digit
} else { // if input not greater than 1000 return 0
return 0;
}
return 0;
}

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# Task
Let `a`, `b`, `c`, and `d` be variables of type `int`.
- Which of the following character sequences are valid expressions in the sense that they are accepted by a C++ Compiler? Explain your answer.
1. a = b = 5
1. 1 = a
1. ++a + b++
1. a + b = c + d
1. a = 2 b
Assume that all the variables have been defined and correctly initialized and that all expressions are completely independent from each other.
- For each of the expressions that you have identified as valid, decide whether the entire expression is an `lvalue` or an `rvalue`, and explain your decision.
- Determine the values of the expressions that you have identified as valid and explain how these values are obtained.

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# Task 1
---
### Expression 1: `a = b = 5`
The expression is valid and will be accepted by the CPP compiler. It will result with the l-values `a = 5` and `b = 5`. This is due to the fact the the `=` opperation is read from right to left so `b = 5` will run first and `a = b` afterwards.
### Expression 2: `1 = a`
The expression is invalid.
### Expression 3: `++a + b++`
The expression is valid and will be accepted by the CPP compiler. It will result in a r-value since the result is not assigned to a variable.
### Expression 4: `a + b = c + d`
The expression is invalid.
### Expression 5: `a = 2 b`
The expression is valid and will be accepted by the CPP compiler. It will result with the l-value `a = 2 b`. This is valid since a l-value can be a r-value.

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# Task
_This task is a text-based task but mostly automatically checked. You are required to write your answers into submission.txt by replacing the question marks with the correct solution. Please, do not change the line formating._
_You can check whether your solution is correct by clicking on the test button._
Numbers can be provided in various formats in C++. Literals prefixed with `0b` indicate binary encoding. Assume unsigned arithmetics with sufficient numbers of bits, i.e. no overflows. Convert the following binary numbers into decimal numbers (1-4) and decimal numbers to binary (5-8):
```txt
# Lines starting with # are comments. Spaces is ignored.
# Lines starting with a whitespace before # will not be a comment.
# Convert to decimal:
0b1 = ?
0b10 = ?
0b000001 = ?
0b101010 = ?
# Convert to binary:
7 = ?
11 = ?
28 = ?
1024 = ?
```

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# Lines starting with # are comments. Spaces are ignored.
# Lines starting with a whitespace before # will not be a comment.
# Convert to decimal:
0b1 = 1
0b10 = 2
0b000001 = 1
0b101010 = 42
# Convert to binary:
7 = 0b111
11 = 0b1011
28 = 0b11100
1024 = 0b10000000000

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# Task
Write a program resistance.cpp that computes the equivalent resistance of the following wiring:
![circuit](./resistance.png)
We assume that $R_1$, $R_2$, $R_3$, and $R_4$ have an integer valued resistance. After input of the four values, the program should output the result arithmetically rounded to the nearest integer.
**Remark:** In order to facilitate the task, you may want to:
- Conceptually divide the task into sub tasks. For example, start with computation of serial resistors $R_{12}$ and $R_{34}$.
- Solve the task first naively using default rounding and then think about how to accomplish arithmetic rounding. Recall that $\text{round}(x) = [x + 0.5] \text{ }\forall \text{ } x \in \mathbb{R}$, i.e., a real number can be rounded arithmetically by adding $0.5$ to it and then rounding down. For example, $\text{round}(8,6) = [8.6 + 0.5] = [9.1] = 9$.
You can find formulas for computing the total resistance in this [Wikipedia article](https://en.wikipedia.org/wiki/Resistor#Series_and_parallel_resistors).
**Important:** using anything other than int (e.g., floating point numbers, long, or double long) is forbidden.
**Important:** using if-else and any other branches is forbidden.

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#include <iostream>
int main() {
int r1;
int r2;
int r3;
int r4;
int res;
std::cin >> r1;
std::cin >> r2;
std::cin >> r3;
std::cin >> r4;
int r12 = r1 + r2;
int r34 = r3 + r4;
res = (r12 * r34 + ((r12 + r34) / 2)) /
(r12 + r34); // By adding the average of r12 and r34, the result is
// increased enough to get the correct rounded number
std::cout << res;
return 0;
}

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# Task 1
This task is a text based task. You do not need to write any program/C++ file: the answer should be written in solutions.txt according to the indicated format. Lines that start with "#" are interpreted as comments. You can run the autograder to check correctness.
## Task
Which of the following expressions evaluate to `true`, which to `false`?
1. `3 >= 3`
1. `true || false && false`
1. `(true || false) && false`
1. `3 > (1 < true)`
1. `8 > 4 > 2 > 1`
1. `2 < a < 4` (a is a variable of type int)

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# Lines starting with # are comments. Spaces are ignored.
# Replace the question marks, indicating which of the following expression evaluate to true, which to false.
1. 3 >= 3 == true
2. true || false && false == true
3. (true || false) && false == false
4. 3 > (1 < true) == true
5. 8 > 4 > 2 > 1 == false
6. 2 < a < 4 == true

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# Task
Translate the following natural language expressions to `C++` expressions. Assume that all the variables are non-negative integers or boolean (of value `true` or `false`).
_For this task you need to write the solutions in the `solutions.cpp` file, by filling the various functions that have been defined for each subtasks._
**Example:** $a$ is greater than $3$ and smaller than $5$. $\Longrightarrow$ Solution:
```cpp
return a > 3 && a < 5;
```
_Note:_ Do not confuse the bitwise logical operators (e.g., `&`) with their binary logical counterparts (`&&`). The semantics are slightly different — bitwise operators do not exhibit short circuit evaluation.
1. $a$ greater than $b$ and the difference between $a$ and $b$ is smaller than $15$.
1. $a$ is an even natural number greater than $a$.
1. $a$ is at most $5$ times greater than $b$ (but can also be smaller than $b$) and at least $5$ times greater than $c$.
1. Either $a$ and $b$ are both false or $c$ is true, but not both.
1. $a$ is false and $b$ is zero.
## Input
The program expects the task number as the first input followed by the parameters to the chosen task. For example, `3 5 1 1` selects task `3` with `a = 5`, `b = 1`, and `c = 1`.
Note that boolean parameters for tasks 4 and 5 are entered as true and false. For example `4 true false true` would run task `4` with `a = true`, `b = false`, and `c = true`.

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#include "solutions.h"
// Fill out each function with the appropriate expression
bool task1(int a, int b) {
// a greater than b and the difference between a and b is smaller than 15.
if (a > b && (a - b) < 15) {
return true;
} else {
return false;
}
}
bool task2(int a) {
// a is an even natural number greater than 3.
if (a > 3 && a % 2 == 0) {
return true;
} else {
return false;
}
}
bool task3(int a, int b, int c) {
// a is at most 5 times greater than b (but can also be smaller than b)
// and at least 5 times greater than c.
if (a <= 5 * b && a >= 5 * c) {
return true;
} else {
return false;
}
}
bool task4(bool a, bool b, bool c) {
// Either a and b are both false or c is true, but not both.
if ((a == false && b == false) != (c == true)) {
return true; // Replace with your solution
} else {
return false;
}
}
bool task5(bool a, int b) {
// a is false and b is zero.
if (a == false && b == false) {
return true;
} else {
return false; // Replace with your solution
}
}

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# Task
_Fibonacci numbers_ are the integers in the following sequence: $0, 1, 1, 2, 3, 5, 8, 13, 21, ...$. Each number is the sum of the two previous numbers.
_Fibonacci primes_ are Fibonacci numbers that are also prime numbers. Write a program that asks the user for an integer $m$ and then computes and prints all Fibonacci primes between $0$ and $m$ (including). Print each number on a new line.
Finally, on a new line print the total number of Fibonacci primes found.
## Example
If your program is asked to print the Fibonacci primes between $0$ and $14$ the output should look something like this:
```
2
3
5
13
Found 4 Fibonacci primes
```
**Important:** using anything other than `int` (e.g., `unsigned int`, floating point numbers, `long`, or double `long`) is forbidden.

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#include <iostream>
int main() {
int a = 1;
int b = 1;
int cnt = 0; // Initialization of number of deviders
int count = 0; // Initialization of number of Fibonacci primes
int n = 2; // Fibonacci number
int i = 0; // Input variable
std::cin >> i;
for (; n <= i;) {
// Reset counters
cnt = 0;
// Check for deviders of Fibonacci number
if (n <= 22360) {
for (int j = 1; j <= 22360; ++j) { // 22360 = sqrt(500000000)
if (n % j == 0) {
++cnt;
}
}
// Increase count if Fibonacci number is a prime number
if (cnt == 2) {
std::cout << n << "\n";
++count;
}
} else if (n > 22360) {
for (int j = 1; j <= 22360; ++j) { // 22360 = sqrt(500000000)
if (n % j == 0) {
++cnt;
}
}
if (cnt == 1) {
std::cout << n << "\n";
++count;
}
}
// Initialize next Fibonacci number
a = b;
b = n;
n = a + b;
}
// End message
std::cout << count << "\n";
}

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# Mistakes
- The variable `j` goes into overflow which is not allowed!
# Fibonacci overflow check
## Background
_Fibonacci numbers_ are the integers in the following sequence: $0, 1, 1, 2, 3, 5, 8, 13, 21, ...$, where each number is the sum of the two previous numbers.
## Task
Fibonacci numbers grow fast, and can thus easily exceed the value range of 32-bit `int`. Think of a general way how you can check if the result of an addition would exceed the range of a 32-bit `int` (i.e. overflow) **without actually performing the addition causing the overflow.**
Remember that we consider **signed integers.** Because only half of the numbers are positive, this leaves us with 31 bits to store the actual positive number value.
Write a program that asks the user for an integer $n$ and then prints the first $n$ Fibonacci numbers, each number on a new line. Use an `int` (we assume 32 bits, including the sign) to represent the current Fibonacci number. **Most importantly:** exit the print loop as soon as you detect that an overflow _would occur._
Finally, again on a new line, output the count $c$ of Fibonacci numbers previously printed, and the initial input $n$ from the user, in the format: `c of n`.
### Example:
Let's (wrongly!) assume that $5$ cannot be represented using a 32 bit int. This means that $3$ is the largest 32-bit Fibonacci number. If your program is asked to print the first $4$ Fibonacci numbers the output should look as follows:
```
0
1
1
2
Printed 4 of 4 Fibonacci numbers
```
If you instead ask it to print the first 100 Fibonacci numbers the output should look as follows:
```
0
1
1
2
3
Printed 5 of 100 Fibonacci numbers
```
**Important:** using anything other than `int` (e.g., `unsigned int`, floating point numbers, `long`, or double `long`) is forbidden.
**Restrictions:**
- The program **must not** rely on the knowledge of its final result. In particular, it is not allowed to hard-code
- the largest 32-bits Fibonacci number, or
- the number of digits that it has, or
- the total number of Fibonacci numbers representable with a 32-bit int
- Most importantly: do not perform additions that cause an overflow on 32 bit
**Note:** It is straightfoward to compute the largest (signed) integer representable with 32 bits. You are also explicitly allowed to hard-code this value in your program.
---
**Warning:** The autograder does not catch if an addition causes an overflow or if you do anything that's disallowed in the "Restrictions" section above, but you will receive 0 points when your TA corrects and catches this.

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#include <iostream>
int main() {
int a = 0; // First Fibonacci number
int b = 1; // Second Fibonacci number
int j = 0; // New Fibonacci number
int c = 0; // Number of Fibonacci numbers
int max = 2147483647;
int n; // Input integer
std::cin >> n;
for (int i = 0; i < n; ++i) {
if (max - a < b) { // Check if the new Fibonacci number goes into overflow
break;
} else { // otherwise, calculate next Fibonacci number
std::cout << j << "\n";
a = b;
b = j;
j = a + b;
++c;
}
}
std::cout << c << " of " << n; // End Message
}

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## Task
Write a program that inputs a non-negative integer `n` (but store it as `int`) and outputs the binary digits of `n` in the _correct_ order (i.e., starting with the most significant bit). Do not output the leading zeros or the sign.
**Hint:** In order to find the largest integer $k$ such that $2^k \leq x$, you can utilize that $k$ is the smallest integer such that $2^k > \frac{x}{2}$. This observation is particularly useful to avoid an overflow for the expression $2^k$ when searching for the most significant bit to represent $x$.
**Restrictions:**
- Libraries: only the iostream standard library header is allowed; using arrays, string or cmath is not permitted.
- Operators: you may not use bitshift operators to manipulate the numbers.
---
**Warning:** The autograder does not catch if you do anything that's disallowed in the "Restrictions" section above, but you will receive 0 points when your TA corrects and catches this.

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#include <iostream>
int main() {
int i = 0; // Input integer
int d = 0; // Input Number
int a = 0; // Part of binary number
int b = 0; // Number to divide
int count = 0;
std::cin >> i;
d = i;
b = i;
if (i < 0) {
return 0;
} else if (i == 0) {
std::cout << 0;
} else {
while (d != 0) {
d = d / 2;
++count;
}
for (; count > 0; --count) {
for (int j = count; j > 1; --j) {
b = b / 2;
}
a = b % 2;
std::cout << a;
b = i;
}
}
}

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_This task is a mixed text/programming task. You need to update the content of loop.cpp according to the instruction provided in the file. For the code part, please check the autograder output._
# Task
Considering the snippet
```cpp
int n;
std::cin >> n;
int f = 1;
if (n > 0) {
do {
f = f * n;
--n;
} while (n > 0);
}
std::cout << f << std::endl;
```
1. Describe what it computes.
1. Decide which of the other two kind of loops would fit better than the one it is currently using, and describe why.
1. Rewrite the snippet into the loop you specified in (2). This part is autograded.
**Important**: The use of goto statements is prohibited.

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#include "loop.h"
#include <iostream>
// Fill out the file with the required answers
/*
Subtask 1: describe what the code snippet computes
It calculates the factorial of the number inputed by the user.
*/
/*
Subtask 2: decide which of the other two kind of loops would fit better, and
describe why.
Just a simple for loop would suffice the task of this code snippet, as we can
define the condition in the for loop as well as the expression. This
simplifies the code alot.
Of course we could also include the edge case where the input n is 0 or
smaller than zero. In this case an if statement would be more of use since we
have to seperate the cases.
*/
/*
Subtask 3: update the function below by rewriting the snippet into the
loop you specified in (2)
*/
void loop() {
int n;
std::cin >> n;
int f = 1;
for (; n > 0; --n)
f = f * n;
std::cout << f << std::endl;
}

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_This task is a mixed text/programming task. You need to update the content of loop.cpp according to the instruction provided in the file. For the code part, please check the autograder output._
# Task
Considering the snippet
```cpp
for (;;) {
int i1, i2;
std::cin >> i1 >> i2;
std::cout << i1 + i2 << "\n";
int again;
std::cout << "Again? (0/1)\n";
std::cin >> again;
if (again == 0) break;
}
```
1. Describe what it computes.
1. Decide which of the other two kind of loops would fit better than the one it is currently using, and describe why.
1. Rewrite the snippet into the loop you specified in (2). This part is autograded. Note: print the control message "Again? (0/1)" using the same format used in the snippet.

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#include "loop.h"
#include <iostream>
// Fill out the file with the required answers
/*
Subtask 1: describe what the code snippet computes
It is a very simple calculator which adds up the 2 number which are inputed by
the user.
*/
/*
Subtask 2: decide which of the other two kind of loops would fit better, and
describe why.
A while loop would be sufficient for the task of this code snippet, since we
only have to check, if the user wants to use the calculator again or not.
Of course, as in the last task we could look at the edge case. We could also
add an if expression to elimnate the posibility to input other numbers other
than 0 and 1 for the again variable.
*/
/*
Subtask 3: update the function below by rewriting the snippet into the
loop you specified in (2).
Note: print the control message "Again? (0/1)" using the same format used
in the snippet.
*/
void loop() {
int again = 1;
while (again != 0) {
int i1, i2;
std::cin >> i1 >> i2;
std::cout << i1 + i2 << "\n";
std::cout << "Again? (0/1)\n";
std::cin >> again;
}
}

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_This task is a mixed text/programming task. You need to update the content of loop.cpp according to the instructions provided in the file. For the code part, please check the autograder output._
# Task
Consider the following program:
```cpp
int n;
std::cin >> n;
int x = 1;
if (n > 0) {
bool e = true;
do {
if (--n == 0) {
e = false;
}
x *= 2;
} while (e);
}
std::cout << x << std::endl;
```
1. Describe the output of the program as a function of its input n.
1. For which values of `n` do you expect a correct output `x`? Explain why.
1. Show that this program terminates for all values of `n` found in (2). Hint: Make an argument based on the following idea. First prove that the program terminates for the `n=0` and `n=1`. Then show that it terminates for any other `n` knowing that it terminates for `n-1`. This creates a domino effect where the fact that the program terminates for `n=1` proves that the program must also terminate for `n=2`, and this in turn proves that the program must terminate for `n=3`, and so on. This technique is called "proof by induction".
1. Provide a more elegant implementation of this function using another type of loop. This part is autograded.

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#include "loop.h"
#include <iostream>
// Fill out the file with the required answers
/*
Subtask 1: Describe the output of the program as a function of its input n.
This code snippte computes the powers of 2 with a positive exponential. The
exponential is the input by the user.
*/
/*
Subtask 2: For which values of n do you expect a correct output x? Explain
why.
For a exponential between 0 and 30. Anything below zero would result in the
wrong answer (1) and anything above 30 would result in an overflow. In
addition, if we imporved the code to calculate the powers of 2 with negative
exponents, it would not work since the result of the powers of 2 with negative
exponents are variables of type float or double and the code snippet is
working with integers.
*/
/*
Subtask 3: Show that this program terminates for all values of n found in (2).
If the input is 31, the output is -2147483648.
If the input is -1, the output is 1.
*/
/*
Subtask 4: Provide a more elegant implementation of this function using
another type of loop.
*/
void loop() {
int n;
std::cin >> n;
int x = 1;
for (; n > 0; --n)
x *= 2;
std::cout << x << std::endl;
}

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# Task
The number $\pi$ can be defined through various infinite sums. The accuracy increases with the number of terms. Considering the following sum, that we call sum 1:
$$\frac{\pi}{4} = \sum_{j=0}^{m-1} \frac{(-1)^j}{2j + 1} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + ...$$
Note that $m$ is the number of the terms in the sum. For example, $m=2$ refers to the sum of the terms $1$ (for $j=0$) and $-\frac{1}{3}$ (for $j=1$). This examples yields a value of $4 \cdot (1-\frac{1}{3})$ for $\pi$.
Write a program that computes and outputs an approximation of Pi, based on sum 1. The input for your program is the number of **terms** $m$ of sum 1 that should be considered in the calculation. The output is the approximation of $\pi$.
## Input
A number $m \geq 1$.
## Output
The approximation of $\pi$ given by $4 \sum_{j=0}^{m-1} \frac{(-1)^j}{2j + 1}$, rounded to 6 significant digits. Note that 6 significant digits is the default precision of C++ for printing floating-point values. Use a variable of type double to calculate the sum. Note that that $x^0$ is 1 (if $x \neq 0$).
**Important**: the use of functions from the math library (e.g., pow) is prohibited.

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#include <iostream>
double power(double n, double e) { // Self defined function for exponent
int res = 1;
for (; e > 0; --e) {
res *= n;
}
return res;
}
int main() {
double m; // User Input
double pi = 0; // Pi
float res; // Outputed result with 6 significant digits
std::cin >> m;
if (m < 1) { // Check if Input is greater than 1.
return 0;
} else {
for (int i = 0; i < m; ++i) {
pi = pi + 4 * ((power(-1, i)) / ((2 * i) + 1)); // calculate Pi
}
res = (float)(pi); // round to 6 significant digits
std::cout << res;
return 0;
}
}

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# Task
The number $\pi$ can be defined through various infinite sums. The accuracy increases with the number of terms. Considering the following sum, which we call sum 2:
$$\frac{\pi}{2} = 1 + \sum_{j = 1}^{m - 1} \frac{\prod_{i=1}^j i}{\prod_{i=1}^j (2i + 1)} = 1 + \frac{1}{3} + \frac{1 \cdot 2}{3 \cdot 5} + \frac{1 \cdot 2 \cdot 3}{3 \cdot 5 \cdot 7} + ...$$
Write a program that computes and outputs an approximation of Pi, based on sum 2. The input for your program is the number of **terms** $m$ (including 1) to be included in the calculation to be done. The output is the approximation of $\pi$.
**Optional**: After you have solved this and the previous task, think about which formula gives a more precise approximation of
, sum 1 or sum 2 ? What are the drawbacks? Write your thoughts in a comment below the code.
## Input
A natural number $m$ ($m \geq 1$).
## Output
The approximation of $\pi$ given by $m$ terms, rounded to 6 significant digits. Note that 6 significant digits is the default precision of C++ for printing floating-point values. Use a double variable to store the sum.
**Important**: the use of functions from the math library (e.g., pow) is prohibited.

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#include <iostream>
double productsumnum(double n) {
double res = 1;
for (int j = 1; j <= n; ++j) {
res *= j;
}
return res;
}
double productsumden(double n) {
double res = 1;
for (int j = 1; j <= n; ++j) {
res *= (2 * j + 1);
}
return res;
}
int main() {
double m; // User Input
double pi = 2; // Pi
float res; // Outputed result with 6 significant digits
std::cin >> m;
for (int i = 1; i < m; ++i) {
pi += (2 * (productsumnum(i) / productsumden(i)));
}
res = (float)(pi);
std::cout << res;
return 0;
}

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_This task is a text-based task but automatically checked. You are required to write your answers into solutions.txt by replacing the question marks with the correct solution. Please, do not change the structure of solutions.txt and be consistent with its syntax (comment-lines start with #)_
# Task
We examine the normalized binary representation of floating point numbers in the normalized floating point system $F*(2,4,-3,3)$.
1. Convert the following decimal numbers to the normalized binary representation. For each number, choose the appropriate exponent $e$ and round to the nearest value if you cannot represent the exact value. If the exact number falls exactly midway between two bracketing finite floating point representations, round to the number with an even least significant bit.
- $0.75_{10}$
- $3.1416_{10}$
- $2.718_{10}$
- $7_{10}$
- $0.11_{10}$
1. For each number of (2), convert the binary representation back to their decimal form, and determine the absolute rounding error of the conversion.
1. Calculate $2.718_{10} + 3.1416_{10} + 0.11_{10}$ in the binary representation. What do you observe?
**Note**: The question of how to round floating point numbers is non-trivial. The IEEE standard lists 5 different modes of rounding of which we only mentioned the most common one at the beginning of this exercise. If you are interested to learn more, you can access the most recent standard through the ETH network.

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# NOTE: Lines starting with # are comments. Spaces are ignored.
# Replace the question marks, with your answer, following the instructions
# provided for each subtask.
######################## Subtask 1 ############################
# Subtask 1a: convert 0.75.
# In the given system, this is equal to "1.100 * 2^-1". The solution
# is written as "1.100*2^-1", without spaces and without quotation marks.
1a) 1.100*2^-1
# Subtask 1b: convert 3.1416
1b) 1.101*2^1
# Subtask 1c: convert 2.718
1c) 1.011*2^1
# Subtask 1d: convert 7
1d) 1.110*2^2
# Subtask 1e: convert 0.11
1e) 1.000*2^-3
######################## Subtask 2 ############################
# Convert the first number of previous Sub-Task back to decimal form.
# For each number you are required to write the decimal number and the
# conversion error (each on a different line).
# For your convenience, the first number is already converted.
2a-decimal) 0.75
2a-error) 0
2b-decimal) 3.25
2b-error) 0.1084
2c-decimal) 2.75
2c-error) 0.032
2d-decimal) 7
2d-error) 0
2e-decimal) 0.125
2e-error) 0.015
######################## Subtask 3 ############################
# Calculate the result of the addition and indicate the result in the
# same format as in the first sub-task ("siginificand*2^exp", without
# quotation marks). Then, write your observation as a comment right below.
3) 1.100*2^2
# Write your observations here
# The Decimal System cannot be represented accurately by the binary system. There will always be an error. When adding the number converted from the decimal to the binary sytem, the error becomes greater.

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# Task
Consider a parabola $(\m{P})$ defined as $y = g(x)$, with $g(x) = 0.9 \cdot x^2 + 1.3 \cdot x - 0.7$.
Write a program that determines if a point $(x,y)$ lies on parabola $(\m{P})$ or not. The input is provided by two decimal numbers in the sequence $x,y$. The program must output `yes`, if the point lies on the parabola, otherwise `no`. Use datatype `double` for all variables and numbers used in the calculation of $g(x)$.
You will notice that a straight forward approach (comparing for equality) does not work, i.e., for some points that clearly should be on parabola $g$ such an approach returns result `no`.
_Hint_: Look at the difference between the exact values of the function and the values that your program calculates. Change the program so that it works properly for all points the submission system uses as test input without hard-coding the points. Expect an epsilon within the range $\[ 10^{-6}, 10^{-3}]\$. Experiment yourself to find the epsilon required to pass the test cases.
**Note**: Output only with `std::cout << "no" << std::endl;` or `std::cout << "yes" << std::endl;`, as the autograder will only accept output that exactly matches `yes\n` or `no\n`. For all other messages, use `std::cerr` as in:
```cpp
std::cerr << "This is a test message\n"
```
Those will be ignored by the autograder.

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#include <iostream>
int main() {
double x = 0; // x inputed by user
double yin = 0; // y inputed by user
double ycon = 0; // y control
double emax = 0.0001;
double emin = -0.0001;
std::cin >> x;
std::cin >> yin;
ycon = 0.9 * x * x + 1.3 * x - 0.7;
std::cerr << ycon;
std::cerr << yin - ycon;
if (yin == ycon || (yin - ycon <= emax && yin - ycon >= emin))
std::cout << "yes";
else
std::cout << "no";
return 0;
}

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# Task
Implement the following rounding function that rounds a 64-bit floating point number (type `double`) to the nearest 32-bit integer (type `int`). You may assume that the type `double` complies with the IEEE standard 754. The function is only required to work correctly if the nearest integer is in the value range of the type `int`, otherwise, the return value of the function is undefined.
**Note: Usage of library rounding functions (standard or others) is not allowed.**
```cpp
// PRE: x is roundable to a number in the value range of type @int@
// POST: return value is the integer nearest to x, or the one further
// away from 0 if x lies right in between two integers.
int round_number(double x);
```
Write your solution in `rounding.h`.
_Hint_: In `C++`, when you convert a `float` or `double` to an `int`, the fractional part gets cut off (truncated). Think about how you can use the truncated number to solve the exercise.

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#pragma once
// PRE: x is roundable to a number in the value range of type int
// POST: return value is the integer nearest to x, or the one further
// away from 0 if x lies right in between two integers.
int round_number(double x) {
int res;
if (x < 0)
res = x - 0.5;
else
res = x + 0.5;
return res;
}

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# Task
Write a program that performs the binary expansion for a given decimal input number $x$, where $0 \leq x < 2$. Use the algorithm presented in the lecture. The program must output the first $16$ digits of the number in the format: $b_0, b_1, b_2, ... , b_15$.
_Important_ Always print all $16$ digits, even the trailing zeros. Do not normalize or round the number.
You can structure your program into functions to avoid code repetition. Do not forget to annotate functions with pre- and post conditions.

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#include <iostream>
int main() {
float i;
int n;
float d;
std::cin >> i;
if (i < 0 || i > 2)
return 0;
else {
n = i;
d = i - n;
std::cout << n << ".";
n = d;
d = 2 * (d - n);
for (int j = 1; j <= 15; ++j) {
n = d;
d = 2 * (d - n);
std::cout << n;
}
}
}

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# Task
A perpetual calendar can be used to determine the weekday (Monday, ..., Sunday) of any given date. You may for example know that the Berlin wall came down on November 9, 1989, but what was the weekday? It was a Thursday. Or what is the weekday of the 1000th anniversary of the Swiss confederation, to be celebrated on August 1, 2291? It will be a Saturday. The task of this exercise is to write a program that outputs the weekday of a given input date.
Your program will read the date from the input. The input is given as three integer values in the following order: Day, month, year. First the program must validate the input. Pay attention to special cases of February (leap years!) also, the date must be greater or equal to the reference date. If a date is invalid output invalid date. If the date is valid, calculate the weekday of this day. This can be done by calculating how many days lie between the date in question and a reference date whose weekday is known. As reference date use Monday, 1st January 1900. Finally, output the weekday as one word in English.
**Approach**: The goal of this exercise is to learn the usage of functions. For that, we split the program into the following sub tasks given as function declarations. Your task is to **implement the provided functions** in the calendar.cpp file, so that they perform the action that is specified in their post condition.
**Important: There is a well-known mathematical function to calculate the weekday of a date. Using this function is an incorrect solution as it defeats the purpose of this exercise.**
```cpp
// PRE: a year greater or equal than 1900
// POST: returns whether that year was a leap year
bool is_leap_year(int year);
// PRE: a year greater or equal than 1900
// POST: returns the number of days in that year
int count_days_in_year(int year);
// PRE: a month between 1 and 12 and a year greater or equal than 1900
// POST: returns the number of days in the month of that year
int count_days_in_month(int month, int year);
// PRE: n/a
// POST: returns whether the given values represent a valid date
bool is_valid_date(int day, int month, int year);
// PRE: the given values represent a valid date
// POST: returns the number of days between January 1, 1900 and this date (excluding this date)
int count_days(int day, int month, int year);
// PRE: the given values represent a (potentially invalid) date
// POST: prints the weekday if the date is valid or "invalid date" otherwise.
// Everything must be printed in lowercase.
void print_weekday(int day, int month, int year);
```
To complete the task, you have to provide the definition of the aforementioned functions.
The `calendar.cpp` file contains skeletons of the functions to be implemented. The `main.cpp` file contains testing functions and the main function. In the main function, a menu is printed to select the function that has to be tested.
**Important** the main is not editable. Required functions must be implemented in `calendar.cpp`.
**Additional notes:**
1. For function arguments that do **not** fulfill the precondition, the behavior is undefined.
1. There are a few opportunities here to use switch statements. Use them if they result in better readable code.
1. A leap year is defined as follows: It is an integer multiple of 4, except for years evenly divisible by 100, which are not leap years unless evenly divisible by 400. (Source: [Wikipedia](https://en.wikipedia.org/wiki/Leap_year))

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#include "calendar.h"
#include <string>
// PRE: a year greater or equal than 1900
// POST: returns whether that year was a leap year
bool is_leap_year(int year) {
if (year >= 1900) {
if (year % 4 == 0 && year % 100 != 0)
return true;
else if (year % 4 == 0 && year % 400 == 0)
return true;
}
return false;
}
// PRE: a year greater or equal than 1900
// POST: returns the number of days in that year
int count_days_in_year(int year) {
if (year >= 1900) {
if (is_leap_year(year) == true)
return 366;
return 365;
}
return 0;
}
// PRE: a month between 1 and 12 and a year greater or equal than 1900
// POST: returns the number of days in the month of that year
int count_days_in_month(int month, int year) {
if (year >= 1900 && month > 0 && month <= 12) {
if (is_leap_year(year) == true) {
if (month % 2 != 0 && month != 2 && month <= 7)
return 31;
else if (month % 2 == 0 && month > 7)
return 31;
else if (month == 2)
return 29;
else
return 30;
} else {
if (month % 2 != 0 && month != 2 && month <= 7)
return 31;
else if (month % 2 == 0 && month > 7)
return 31;
else if (month == 2)
return 28;
else
return 30;
}
}
return 0;
}
// PRE: n/a
// POST: returns whether the given values represent a valid date
bool is_valid_date(int day, int month, int year) {
if (day > 0 && day <= count_days_in_month(month, year) && month > 0 &&
month <= 12 && year >= 1900)
return true;
return false;
}
// PRE: the given values represent a valid date
// POST: returns the number of days between January 1, 1900 and this date
// (excluding this date)
int count_days(int day, int month, int year) {
int res_day = 0;
if (is_valid_date(day, month, year) == true) {
for (int j = 1900; j < year; ++j) {
res_day += count_days_in_year(j);
}
for (int j = 1; j < month; ++j) {
res_day += count_days_in_month(j, year);
}
res_day += (day - 1);
}
return res_day;
}
// PRE: the given values represent a (potentially invalid) date
// POST: prints the weekday if the date is valid or "invalid date" otherwise.
// Everything must be printed in lowercase.
void print_weekday(int day, int month, int year) {
if (is_valid_date(day, month, year) == true) {
std::string days[7] = {"sunday", "monday", "tuesday", "wednesday",
"thursday", "friday", "saturday"};
std::cout << days[((count_days(day, month, year) + 1) % 7)];
} else
std::cout << "invalid date";
}

8
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_This task is a text-based task. You do not need to write any program/C++ file: the answer should be written in functions.cpp._
Consider the functions implemented in `functions.cpp`. For each function, add proper pre- and post-conditions.
- If no pre-condition is needed, you can simply write "n/a".
- The post-condition does not have to be a mathematical formula, e.g. it can be an informal description, but it must completely characterize the results and effects of the functions depending on the provided parameters.
**Note**: For the purposes of this task, you can ignore overflows.

38
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// PRE: x > 0
// POST: returns n^x
int f1(int n, int x) {
int res = 1;
for (; x > 0; x--) {
res *= n;
}
return res;
}
// PRE: n > 0
// POST: returns number of devisions by 10 with n
int f2(int n) {
int i = 0;
while (n > 0) {
n = n / 10;
++i;
}
return i;
}
// PRE: n > 1
// POST: returns a bool to determine if n is prime
bool f3(int n) {
int i = 2;
for (; n % i != 0; ++i);
return n == i;
}
// PRE: n is square number
// POST: returns root of n
int f4(int n) {
int i = 0;
while (i * i != n) {
++i;
}
return i;
}

28
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_This task is a text based task. You do not need to write any program/C++ file: the answer should be written in main.md (and might include code fragments if questions ask for them)._
# Task:
What are the problems (if any) with the following functions? Fix them and find appropriate [pre-](https://en.wikipedia.org/wiki/Precondition) and [postconditions](https://en.wikipedia.org/wiki/Postcondition).
1. function is_even:
```cpp
bool is_even(int i) {
if (i % 2 == 0) return true;
}
```
1. function invert:
```cpp
double invert(int x) {
double result;
if (x != 0) {
result = 1.0 / x;
}
return result;
}
```
**Hint**: The C++ compiler does not protect you from certain types of errors. Therefore, even if you run a program in Code Expert, it is not guaranteed that the behaviour you observe is the “real” one. We have prepared a [program tracing handout](https://lec.inf.ethz.ch/ifmp/2023/guides/tracing/intro.html) that shows how to execute a program with a pen and paper and which conditions indicate bugs in the executed program not caught by the C++ compiler.

35
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Please write your solution here.
For your convenience we have added a template for your answers below.
Note: Remember to also describe the problem of the function.
# Function `c++|is_even`:
The problem with this function is that it does not return a bool if the number `i` is uneven.
```c++
// PRE: n/a
// POST: returns true if i is even. If not it returns false.
bool is_even(int i) {
if (i % 2 == 0) return true;
else return false;
}
```
# Function `c++|invert`:
The problem with this function is
```c++
// PRE: x is positive or negative but not zero.
// POST: returns inverse with respect to multiplication of x.
double invert(int x) {
// TODO: Fix code below.
if (x != 0) {
return 1.0 / x;
}
else
std::cout << "0 has no inverse!";
}
```

64
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Task
Run-length encoding is a simple data compression technique that represents $N$ consecutive identical values $W$ (a run) by a tuple ($N,W$). This method is applied for image compression, for instance. Example:
![Encoding and Decoding](./pictures/encode_decode.png)
Write a program that implements run-length encoding and decoding of a byte sequence as described above. By a byte, we mean an integer value in the range $\[ 0; 255 \]$. Use the stepwise refinement method to implement the program. **Your solution must consist of at least two functions, encode and decode. Please implement them in `run_length.cpp`.**
The _input_ is structured as follows:
1. One integer that determines whether to encode: $0$ or decode: $1$.
1. Byte sequence to be encoded or decoded (of arbitrary length). If a value outside the range $\[ 0; 255 \]$(except $-1$) is entered, output `error` and stop the en- or decoding.
1. Integer -1 signaling the end of the byte sequence. Any extra input should be ignored.
For the example above, the inputs are:
**Encode:**
```sh
0 42 42 85 85 85 85 172 172 172 13 13 42 -1
```
**Decode:**
```sh
1 2 42 4 85 3 172 2 13 1 42 -1
```
_The output_ is expected on a single line in the following format:
1. A start value to indicate the begin of the sequence: either $0$ for decoded values or $1$ for encoded values.
1. The values that make up the encoded or decoded byte sequence.
1. The value $-1$ to indicate the end of the sequence.
I.e., you can 'reuse' the output as the input.
**Note 1):** As the encoded sequence must be a _byte_ sequence, runs of length 256 or longer need to be split into multiple runs of length 255 at most.
**Note 2):** The first input element (the integer that determines wether to encode or decode), is already consumed by the `main`, that calls either the encode or decode function.
**Note 3):** Your output should not be followed by new line (i.e., do not use `std::endl` or `\n` at the end of your printout)
**Note 4):** The program will print your output (the result of the decoding or encoding), surrounded by asterisks. You don't have to worry about them, the autograder can safely recognize your solution
**Note 5):** Output only what is strictly required (the encoded or decoded sequence). The autograder will only accept output that exactly matches the expected result. For all other messages, use `std::cerr` as in:
```cpp
std::cerr << "This is a test message\n"
```
Those will be ignored by the autograder.
**Special cases**: While decoding a byte sequence two special cases can occur. These must be handled as follows:
1. If a byte sequence ends in the middle of a tuple, stop printing the output of en- or decoding and output `error`.
1. Tuples of run-length 0 are possible. For such tuples, output only the leading indicator ($0$/$1$) and the trailing $-1$.
**Hint**: You can enter multiple numbers at once separated with a space on the console, e.g, you can copy and paste the above examples, and sequentially read them using multiple `std::cin >> _var_` calls.
Also note that, even though the program's output and the user input are both shown in the same console, they are processed separately internally, so you do not have to worry that the two will mix: if your program outputs something to the console before reading another value, it will only read the user input and not the previous output value. See the Calculator code in the Lecture 4 handout for an example of reading input and producing output in a loop.
Finally, part of the correctness for this task is the termination of your code, so pay attention to this. The autograder may not catch these kinds of mistakes.
**Restrictions**: using a vector, an array or a similar data structure in any part of your code is not allowed and would result in 0 points.

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#include "run_length.h"
void encode() {
int i;
int c;
int count;
std::cin >> i;
std::cout << 1 << " ";
while (i != -1) {
count = 0;
c = i;
while (i == c && i >= 0 && i <= 255) {
++count;
std::cin >> i;
if (count == 255) {
break;
}
}
if (i < -1 || i > 255) {
if (count == 0) {
std::cout << "error";
return;
} else {
std::cout << count << " ";
std::cout << c << " ";
std::cout << "error";
return;
}
}
std::cout << count << " ";
std::cout << c << " ";
}
std::cout << -1;
}
void decode() {
int i;
int count;
std::cin >> count;
std::cin >> i;
std::cout << 0 << " ";
while (count != -1) {
if (i < -1 || i > 255) {
std::cout << "error";
return;
}
if (count < -1 || count > 255) {
std::cout << "error";
return;
}
for (int j = 0; j < count; ++j) {
std::cout << i << " ";
}
std::cin >> count;
if (count != -1)
std::cin >> i;
if (i == -1) {
std::cout << "error";
return;
}
}
std::cout << -1;
}

809
LICENSE
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@ -1,158 +1,651 @@
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### 3. Protecting Users' Legal Rights From Anti-Circumvention Law
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When you convey a covered work, you waive any legal power to forbid
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### 4. Conveying Verbatim Copies
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### 5. Conveying Modified Source Versions
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* **d)** If the work has interactive user interfaces, each must display
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A compilation of a covered work with other separate and independent
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### 6. Conveying Non-Source Forms
You may convey a covered work in object code form under the terms
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* **d)** Convey the object code by offering access from a designated
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* **e)** Convey the object code using peer-to-peer transmission, provided
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If you convey an object code work under this section in, or with, or
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The requirement to provide Installation Information does not include a
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Corresponding Source conveyed, and Installation Information provided,
in accord with this section must be in a format that is publicly
documented (and with an implementation available to the public in
source code form), and must require no special password or key for
unpacking, reading or copying.
### 7. Additional Terms
“Additional permissions” are terms that supplement the terms of this
License by making exceptions from one or more of its conditions.
Additional permissions that are applicable to the entire Program shall
be treated as though they were included in this License, to the extent
that they are valid under applicable law. If additional permissions
apply only to part of the Program, that part may be used separately
under those permissions, but the entire Program remains governed by
this License without regard to the additional permissions.
When you convey a copy of a covered work, you may at your option
remove any additional permissions from that copy, or from any part of
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Notwithstanding any other provision of this License, for material you
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* **a)** Disclaiming warranty or limiting liability differently from the
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* **c)** Prohibiting misrepresentation of the origin of that material, or
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All other non-permissive additional terms are considered “further
restrictions” within the meaning of section 10. If the Program as you
received it, or any part of it, contains a notice stating that it is
governed by this License along with a term that is a further
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License, you may add to a covered work material governed by the terms
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If you add terms to a covered work in accord with this section, you
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where to find the applicable terms.
Additional terms, permissive or non-permissive, may be stated in the
form of a separately written license, or stated as exceptions;
the above requirements apply either way.
### 8. Termination
You may not propagate or modify a covered work except as expressly
provided under this License. Any attempt otherwise to propagate or
modify it is void, and will automatically terminate your rights under
this License (including any patent licenses granted under the third
paragraph of section 11).
However, if you cease all violation of this License, then your
license from a particular copyright holder is reinstated **(a)**
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finally terminates your license, and **(b)** permanently, if the copyright
holder fails to notify you of the violation by some reasonable means
prior to 60 days after the cessation.
Moreover, your license from a particular copyright holder is
reinstated permanently if the copyright holder notifies you of the
violation by some reasonable means, this is the first time you have
received notice of violation of this License (for any work) from that
copyright holder, and you cure the violation prior to 30 days after
your receipt of the notice.
Termination of your rights under this section does not terminate the
licenses of parties who have received copies or rights from you under
this License. If your rights have been terminated and not permanently
reinstated, you do not qualify to receive new licenses for the same
material under section 10.
### 9. Acceptance Not Required for Having Copies
You are not required to accept this License in order to receive or
run a copy of the Program. Ancillary propagation of a covered work
occurring solely as a consequence of using peer-to-peer transmission
to receive a copy likewise does not require acceptance. However,
nothing other than this License grants you permission to propagate or
modify any covered work. These actions infringe copyright if you do
not accept this License. Therefore, by modifying or propagating a
covered work, you indicate your acceptance of this License to do so.
### 10. Automatic Licensing of Downstream Recipients
Each time you convey a covered work, the recipient automatically
receives a license from the original licensors, to run, modify and
propagate that work, subject to this License. You are not responsible
for enforcing compliance by third parties with this License.
An “entity transaction” is a transaction transferring control of an
organization, or substantially all assets of one, or subdividing an
organization, or merging organizations. If propagation of a covered
work results from an entity transaction, each party to that
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licenses to the work the party's predecessor in interest had or could
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Corresponding Source of the work from the predecessor in interest, if
the predecessor has it or can get it with reasonable efforts.
You may not impose any further restrictions on the exercise of the
rights granted or affirmed under this License. For example, you may
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rights granted under this License, and you may not initiate litigation
(including a cross-claim or counterclaim in a lawsuit) alleging that
any patent claim is infringed by making, using, selling, offering for
sale, or importing the Program or any portion of it.
### 11. Patents
A “contributor” is a copyright holder who authorizes use under this
License of the Program or a work on which the Program is based. The
work thus licensed is called the contributor's “contributor version”.
A contributor's “essential patent claims” are all patent claims
owned or controlled by the contributor, whether already acquired or
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but do not include claims that would be infringed only as a
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Each contributor grants you a non-exclusive, worldwide, royalty-free
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In the following three paragraphs, a “patent license” is any express
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If you convey a covered work, knowingly relying on a patent license,
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consistent with the requirements of this License, to extend the patent
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actual knowledge that, but for the patent license, your conveying the
covered work in a country, or your recipient's use of the covered work
in a country, would infringe one or more identifiable patents in that
country that you have reason to believe are valid.
If, pursuant to or in connection with a single transaction or
arrangement, you convey, or propagate by procuring conveyance of, a
covered work, and grant a patent license to some of the parties
receiving the covered work authorizing them to use, propagate, modify
or convey a specific copy of the covered work, then the patent license
you grant is automatically extended to all recipients of the covered
work and works based on it.
A patent license is “discriminatory” if it does not include within
the scope of its coverage, prohibits the exercise of, or is
conditioned on the non-exercise of one or more of the rights that are
specifically granted under this License. You may not convey a covered
work if you are a party to an arrangement with a third party that is
in the business of distributing software, under which you make payment
to the third party based on the extent of your activity of conveying
the work, and under which the third party grants, to any of the
parties who would receive the covered work from you, a discriminatory
patent license **(a)** in connection with copies of the covered work
conveyed by you (or copies made from those copies), or **(b)** primarily
for and in connection with specific products or compilations that
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or that patent license was granted, prior to 28 March 2007.
Nothing in this License shall be construed as excluding or limiting
any implied license or other defenses to infringement that may
otherwise be available to you under applicable patent law.
### 12. No Surrender of Others' Freedom
If conditions are imposed on you (whether by court order, agreement or
otherwise) that contradict the conditions of this License, they do not
excuse you from the conditions of this License. If you cannot convey a
covered work so as to satisfy simultaneously your obligations under this
License and any other pertinent obligations, then as a consequence you may
not convey it at all. For example, if you agree to terms that obligate you
to collect a royalty for further conveying from those to whom you convey
the Program, the only way you could satisfy both those terms and this
License would be to refrain entirely from conveying the Program.
### 13. Remote Network Interaction; Use with the GNU General Public License
Notwithstanding any other provision of this License, if you modify the
Program, your modified version must prominently offer all users
interacting with it remotely through a computer network (if your version
supports such interaction) an opportunity to receive the Corresponding
Source of your version by providing access to the Corresponding Source
from a network server at no charge, through some standard or customary
means of facilitating copying of software. This Corresponding Source
shall include the Corresponding Source for any work covered by version 3
of the GNU General Public License that is incorporated pursuant to the
following paragraph.
Notwithstanding any other provision of this License, you have
permission to link or combine any covered work with a work licensed
under version 3 of the GNU General Public License into a single
combined work, and to convey the resulting work. The terms of this
License will continue to apply to the part which is the covered work,
but the work with which it is combined will remain governed by version
3 of the GNU General Public License.
### 14. Revised Versions of this License
The Free Software Foundation may publish revised and/or new versions of
the GNU Affero General Public License from time to time. Such new versions
will be similar in spirit to the present version, but may differ in detail to
address new problems or concerns.
Each version is given a distinguishing version number. If the
Program specifies that a certain numbered version of the GNU Affero General
Public License “or any later version” applies to it, you have the
option of following the terms and conditions either of that numbered
version or of any later version published by the Free Software
Foundation. If the Program does not specify a version number of the
GNU Affero General Public License, you may choose any version ever published
by the Free Software Foundation.
If the Program specifies that a proxy can decide which future
versions of the GNU Affero General Public License can be used, that proxy's
public statement of acceptance of a version permanently authorizes you
to choose that version for the Program.
Later license versions may give you additional or different
permissions. However, no additional obligations are imposed on any
author or copyright holder as a result of your choosing to follow a
later version.
### 15. Disclaimer of Warranty
THERE IS NO WARRANTY FOR THE PROGRAM, TO THE EXTENT PERMITTED BY
APPLICABLE LAW. EXCEPT WHEN OTHERWISE STATED IN WRITING THE COPYRIGHT
HOLDERS AND/OR OTHER PARTIES PROVIDE THE PROGRAM “AS IS” WITHOUT WARRANTY
OF ANY KIND, EITHER EXPRESSED OR IMPLIED, INCLUDING, BUT NOT LIMITED TO,
THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR
PURPOSE. THE ENTIRE RISK AS TO THE QUALITY AND PERFORMANCE OF THE PROGRAM
IS WITH YOU. SHOULD THE PROGRAM PROVE DEFECTIVE, YOU ASSUME THE COST OF
ALL NECESSARY SERVICING, REPAIR OR CORRECTION.
### 16. Limitation of Liability
IN NO EVENT UNLESS REQUIRED BY APPLICABLE LAW OR AGREED TO IN WRITING
WILL ANY COPYRIGHT HOLDER, OR ANY OTHER PARTY WHO MODIFIES AND/OR CONVEYS
THE PROGRAM AS PERMITTED ABOVE, BE LIABLE TO YOU FOR DAMAGES, INCLUDING ANY
GENERAL, SPECIAL, INCIDENTAL OR CONSEQUENTIAL DAMAGES ARISING OUT OF THE
USE OR INABILITY TO USE THE PROGRAM (INCLUDING BUT NOT LIMITED TO LOSS OF
DATA OR DATA BEING RENDERED INACCURATE OR LOSSES SUSTAINED BY YOU OR THIRD
PARTIES OR A FAILURE OF THE PROGRAM TO OPERATE WITH ANY OTHER PROGRAMS),
EVEN IF SUCH HOLDER OR OTHER PARTY HAS BEEN ADVISED OF THE POSSIBILITY OF
SUCH DAMAGES.
### 17. Interpretation of Sections 15 and 16
If the disclaimer of warranty and limitation of liability provided
above cannot be given local legal effect according to their terms,
reviewing courts shall apply local law that most closely approximates
an absolute waiver of all civil liability in connection with the
Program, unless a warranty or assumption of liability accompanies a
copy of the Program in return for a fee.
_END OF TERMS AND CONDITIONS_
## How to Apply These Terms to Your New Programs
If you develop a new program, and you want it to be of the greatest
possible use to the public, the best way to achieve this is to make it
free software which everyone can redistribute and change under these terms.
To do so, attach the following notices to the program. It is safest
to attach them to the start of each source file to most effectively
state the exclusion of warranty; and each file should have at least
the “copyright” line and a pointer to where the full notice is found.
<one line to give the program's name and a brief idea of what it does.>
Copyright (C) <year> <name of author>
This program is free software: you can redistribute it and/or modify
it under the terms of the GNU Affero General Public License as published by
the Free Software Foundation, either version 3 of the License, or
(at your option) any later version.
This program is distributed in the hope that it will be useful,
but WITHOUT ANY WARRANTY; without even the implied warranty of
MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
GNU Affero General Public License for more details.
You should have received a copy of the GNU Affero General Public License
along with this program. If not, see <http://www.gnu.org/licenses/>.
Also add information on how to contact you by electronic and paper mail.
If your software can interact with users remotely through a computer
network, you should also make sure that it provides a way for users to
get its source. For example, if your program is a web application, its
interface could display a “Source” link that leads users to an archive
of the code. There are many ways you could offer source, and different
solutions will be better for different programs; see section 13 for the
specific requirements.
You should also get your employer (if you work as a programmer) or school,
if any, to sign a “copyright disclaimer” for the program, if necessary.
For more information on this, and how to apply and follow the GNU AGPL, see
&lt;<http://www.gnu.org/licenses/>&gt;.

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# ETH_Informatik
Computer Science Exercises for ITET 2024

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#+TITLE: ETH Computer Science Projects
#+AUTHOR: JirR02
* ETH Informatik Projekte
In this respository the computer science project of the ETH of D-ITET 2024 are managed. They can be used for inspiration.
The projects are uploaded on [[https://expert.ethz.ch/enrolled/AS24/itet0/exercises][Code Expert]].
## DISCLAIMER!!!
I assume no liability for possible errors in the code (it certainly has a few in it, since I write it myself).
Bugs can be reported via Discord, WhatsApp, Mail and Moodle.
---
Made by JirR02 in Switzerland 🇨🇭

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#+TITLE: Informatik Vorkurs Project 1
#+AUTHOR: JirR02
* Project 1: Guess A Number (Task 1)
** About this task
*** Project overview
The goal of the first project is to program a simple number guessing game: the player needs to correctly guess a number, chosen from an interval \(\[1,N \]\), with at most $K$ guesses.
In the lecture, a first version of the game was presented, in which the player had only one chance of guessing the correct number. In order to implement the full game, you will have to extend this version by allowing the player to guess up to $K$ times.
The first project consists of two tasks: *task 1* /(this task)/ is to reimplement the first version of the game that was presented in the lecture, *task 2* is to implement the full game.
*** General development advice
Develop your program step-by-step, and save, run and compile it often. I.e. implement a single feature, such as inputting the guess, or comparing the guess and the correct number, and then compile and run the program to see if your code (still) works as intended. Small changes and repeated testing make it easier for you to observe problems, to work out what causes them, and to finally solve them.
*** Fulfilling requirements and testing programs
A particular goal of this task is to make you understand how strongly connected requirements and testing (and thus grading submissions) are. E.g. if the task description requires output of the shape "I␣saw␣$n$␣cat(s).", where $n$ is a number and ␣ represents blanks/whitespaces, your program will not be considered correct if it outputs a text that is "basically the same", but does not precisely match the requirements. E.g. the following outputs, while very close to the expected output, do not match the shape specified above: "i␣saw␣3␣cat(s).", "I␣saw␣3␣cats.", "I␣saw␣1␣cat.", "I␣saw␣2␣cat(s)" and "i␣saw␣2␣␣cat(s)␣.".
Meeting such requirements precisely is necessary for testing: we automatically (at least to some extent) test correctness of your programs by comparing the expected output to the output produced by your program. Since these comparisons are performed by a computer, it is much much easier to do them syntactically, i.e. letter-by-letter.
However, being able to precisely fulfil requirements is more generally important, in particular when working with customers: imagine you order a black smartphone, but then get delivered a dark blue one (same make and model). Although "basically the same", it's just not what you ordered. The same holds for software — it's the details that matter.
*** Your task
To solve this task, proceed as follows:
1. Look at the template program, in particular main.cpp, and see what's already there (e.g. variable declarations) and what's still missing. The latter is hinted at by TODO comments.
1. The template already contains code for picking the number to guess. By default, the number is randomly choosen from the interval $\[1,3\]$:
#+begin_src cpp
number_to_guess = choose_a_number(3);
#+end_src
/During development/, you might want to change this line: e.g. read the number from the keyboard:
#+begin_src cpp
std::cin >> number_to_guess;
#+end_src
or even hard-code (i.e. fix) a specific number, e.g.
#+begin_src cpp
number_to_guess = 3;
#+end_src
However, in order to run the automatic tests, and /when submitting your final version/, your program /must/ either read the word from the keyboard or use the ~choose_a_number~ function.
1. Now address the first TODO comment: by outputting the text "Your␣guess:␣" (as before, ␣ denotes a blank/whitespace character), followed by inputting the guess from the keyboard into variable guess.
2. Now address the second TODO comment: by comparing the two numbers for equality. If they are, output the line
#+begin_src shell
Congratulations,␣you␣correctly␣guessed␣X!
#+end_src
where X is the correctly guessed number. Since the output is expected to be a /line/, don't forget to end it with either ~\n~ or ~std::endl~.
If the guess was wrong, however, output the line
#+begin_src shell
Sorry,␣but␣Y␣is␣wrong,␣X␣was␣the␣number␣to␣guess.
#+end_src
where Y is the incorrectly guessed number.
*** Examples
As an illustration, consider the following example in- and outputs of two games (in which the number to guess was randomly chosen). A successfully completed game:
#+begin_src shell
Number to guess: ?
Your guess: 3
Congratulations, you correctly guessed 3!
#+end_src
And a lost game:
#+begin_src shell
Number to guess: ?
Your guess: 2
Sorry, but 2 is wrong, 1 was the number to guess.
#+end_src
*** Testing your program
You can always test your program manually: click the "play" button in the bottom panel, run your program and see if it behaves as expected.
Relevant for your final submission, however, is if it passes the automated tests: to run those, click the "chemistry flask" button in the bottom panel and wait for the output to appear. If your program passes all tests — everything is green and your score is 100% — then your program is ready to be submitted.
Otherwise, /carefully/ compare the expected output to the actual output to find out what went wrong.
*Reminder*: The devil is in the detail! Pay attention to whitespace and newline characters, and in general check that your output fulfils all requirements, even the "boring" ones.
*** Submitting your solution
Finally, submit your solution (your program) by clicking the corresponding button in the top right corner of the Code Expert IDE (open Task/History first). Your program will be tested automatically, and your score will be shown in the "History" view, which can be opened by clicking on the corresponding tab on the right of the Code Expert IDE. Note that you can submit arbitrarily often (before the exercise deadline, of course), and your last submission will be considered for grading.
For this task, all tests need to pass in order to successfully solve this task. This should be rather easy, though, since there isn't much that can go wrong.
You can also see the results for your submission on the "Enrolled Courses" tab of Code Expert, as green or red percentage values to the left of the task's name.
** Solution
Front End to import libraries and files.
#+begin_src cpp
#include <iostream>
#include "guess_a_number.h"
#+end_src
Main function where the code runs.
#+begin_src cpp
int main() {
#+end_src
Declaring Variable which will be used.
#+begin_src cpp
int number_to_guess; // The number to guess
int guess; // The guessed number
#+end_src
Front end terminal output.
#+begin_src cpp
std::cout << "Number to guess: ";
#+end_src
~choose_a_number()~ function is called to choose a number between 1 and 3
#+begin_src cpp
number_to_guess = choose_a_number(3);
#+end_src
Input front end.
#+begin_src cpp
std::cin >> guess;
#+end_src
Output the input of the user.
#+begin_src cpp
std::cout << "Your guess: ";
#+end_src
Depending on the guess, the output in the terminal is different, if the input of the user is the same with the random number, it will output ~Congratulations, you correctly guessed x!~. Otherwise it will output something else.
#+begin_src cpp
if (guess == number_to_guess) {
std::cout << "Congratulations, you correctly guessed " << number_to_guess << "!";
} else {
std::cout << "Sorry, but " << guess << " is wrong, " << number_to_guess << " was the number to guess.";
}
}
#+end_src

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#include <cstdlib>
#include <iostream>
#include <string>
#include "guess_a_number.h"
// NOTE: You cannot change this file, and you don't need to understand its
// content in order to solve your task. Feel free to look around, however,
// in case you're interested.
const std::string action = std::getenv("ACTION");
void print_attempts_left(int attempts_left) {
std::cout << "You have " << attempts_left << " attempt(s) left.\n";
}
void print_you_won(int correct_guess) {
std::cout << "Congratulations, you correctly guessed " << correct_guess << "!\n";
}
void print_wrong_guess(int wrong_guess) {
std::cout << "Sorry, but " << wrong_guess << " is wrong.\n";
}
void print_you_lost(int number_to_guess, int attempts) {
std::cout << "You lost after " << attempts << " attempt(s) :-( The number to guess was " << number_to_guess << ".\n";
}
// This function returns a randomly chosen integer from the interval [1, max].
int randomly_choose_a_number(int max) {
return std::rand() % max + 1;
}
int choose_a_number(int max) {
if (action == "run") {
std::cout << "?\n";
// Just here to achieve the same output behaviour, in terms of newlines,
// when a user replaces
// std::cin >> number_to_guess;
// by
// number_to_guess = choose_a_number(MAX);
return randomly_choose_a_number(max);
} else {
int guess;
std::cin >> guess;
return guess;
}
}

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#ifndef PROJECT_H
#define PROJECT_H
// NOTE: You cannot change this file. It "only" contains declarations and
// short descriptions of the functionality we provided to you.
// This function outputs the number of attempts that are left.
//
// Example: the function call print_attempts_left(3) results in the output
// "You have 3 attempt(s) left".
void print_attempts_left(int attempts_left);
// This function outputs the winning message.
//
// Example: the function call print_you_won(9) results in the output
// "Congratulations, you correctly guessed 9!".
void print_you_won(int correct_guess);
// This function outputs a message telling the user that their guess was wrong.
//
// Example: the function call print_wrong_guess(1) results in the output
// "Sorry, but 1 is wrong.".
void print_wrong_guess(int wrong_guess);
// This function outputs the losing message.
//
// Example: the function call print_you_lost(2, 5) results in the output
// "You lost after 5 attempts :-( The number to guess was 2.".
void print_you_lost(int number_to_guess, int attempts);
// This function returns a randomly chosen integer from the interval [0, max].
int randomly_choose_a_number(int max);
// This function returns a number from the interval [1, max].
// The number is randomly chosen if we're in interactive mode and parameter
// choose_randomly is true, and read from the keyboard (std::cin) otherwise.
int choose_a_number(int max);
// Master implementation for part 1: reads the next guess from the keyboard
// and stores it in the passed variable.
//
// Example: given the function call PART1_read_next_guess(my_guess) and
// assuming that the player enters 1, variable my_guess will afterwards
// have the value 1.
void PART1_read_next_guess(int& guess);
// Master implementation for part 2: compares guess to number_to_guess,
// outputs a result-depending message (correct/wrong) and updates
// variable play to false if the guess was correct.
//
// Example: given the function call
// PART2_handle_guess(1, 2, continue_playing), a wrong-guess message
// will be outputted.
//
// Example: given the function call
// PART2_handle_guess(2, 2, continue_playing), a correct-guess message
// will be output and variable continue_playing will be set to false.
void PART2_handle_guess(int guess, int number_to_guess, bool& play);
// Master implementation for part 3: increments the number of guessing
// attempts the player made (variable attempts), and if the maximum
// number of guesses has been made, outputs a corresponding message and
// sets variable play to false.
//
// Example: given the function call
// PART3_finish_round(2, 10, performed_attempts, continue_playing), and
// assuming that performed_attempts has a value of 5, then the only
// effect of the function call is that performed_attempts will be updated
// to 6.
//
// Example: given the function call
// PART3_finish_round(2, 10, performed_attempts, continue_playing), and
// assuming that performed_attempts has a value of 9, then
// performed_attempts will be incremented to 10, a you-lost message will
// be output and continue_playing will be set to false.
void PART3_finish_round(int number_to_guess, int max_attempts, int& attempts, bool& play);
#endif

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#include <iostream>
#include "guess_a_number.h"
int main() {
int number_to_guess; // The number to guess
int max_attempts; // The guessed number
std::cout << "Number to guess: ";
number_to_guess = choose_a_number(10); // Randomly choose a number from the interval [1, 10]
std::cout << "Number of attempts: ";
std::cin >> max_attempts;
// Make sure that the player has at least one attempt
if (max_attempts < 1) max_attempts = 1;
int attempts = 0; // Attempts made so far
bool play = true; // false once the game is over
while (play) {
print_attempts_left(max_attempts - attempts);
// *** Part 1: input the next guess ****************************************
int guess; // The user's guess
//PART1_read_next_guess(guess);
std::cout << "Your guess: ";
std::cin >> guess;
// *** Part 2: handle the guess the user made *****************************
//PART2_handle_guess(guess, number_to_guess, play);
if (guess == number_to_guess) {
print_you_won(guess);
play = false;
} else {
print_wrong_guess(guess);
}
// *** Part 3: finish up the round ****************************************
if (play) {
//PART3_finish_round(number_to_guess, max_attempts, attempts, play);
attempts += 1;
if (attempts == max_attempts) {
print_you_lost(number_to_guess, attempts);
play = false;
}
}
}
}

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# Projekt 2: Hangman
## Layers
### Start (Optional)
Beim Starten des Spiels soll ein Welcome screen erscheinen mit den Optionen, das Spiel zu starten und das Spiel zu beenden. Damit der Welcome screen gut aussieht soll es Terminal Art beinhalten. Nach dem Start Vorgang wird der Anzahl der Spieler gefragt.
Daraus entstehen 2 Szenarien:
1. Falls es im Einzelspieler Modus ist, wird ein Wort aus der Liste ausgesucht.
1. Falls im Mehrspieler Modus, darf der andere Spieler ein Wort zum Raten auswählen.
- [ ] Start input
- [ ] End input
- [ ] Invalid Input
- [ ] Single or Multiplayer
- [ ] Terminal Art
### Game
Das zu ratende Wort wird verdeckt im Terminal gezeigt. Es wird dann ein Input als Buchstabe verlangt. Wenn der Input zu lang oder ein invalid character ist, wird der Spieler nochmals dazu aufgefordert, ein Buchstabe einzugeben.
Daraus entstehen 2 Szenarien:
1. Ist der Buchstabe in der Zahl enthalten, wird der Buchstabe aufgedeckt und ein positiver Satz erscheint im Terminal.
1. Ist der Buchstabe falsch, so wird ein Leben abgezogen und ein negativer Satz wird ausgespuckt.
Während des ganzen Spiels wird der Terminal Art aktualisiert.
- [ ] Wort verdeckt im Terminal anzeigen
- [ ] Input von einem Buchstaben verlangen
- [ ] Input kontrollieren
- [ ] Buchstabe kontrollieren
- [ ] Positiver Satz
- [ ] Buchstabe aufdecken
- [ ] Negativer Satz
- [ ] (Optional) Terminal Art
### End
Es entstehen daraus zwei Endszenarien:
1. Wurden alle Buchstaben eraten, so wird ein Gewinner Satz ausgesprochen und gefragt ob das Spiel neugestartet werden soll.
1. Wurden alle Versuche verbraucht, so wird ein verlierer Satz ausgesprochen, das Wort aufgelöst und gefragt, ob das Spiel neugestartet werden soll.
Falls das Programm geschlossen wird, wird ein Abschiedssatz gezeigt.
- [ ] Gewinner Satz
- [ ] Verlierer Satz
- [ ] (Optional) Fragen für eine neue Runde
- [ ] Abschiedssatz

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#include "hangman.h"
#include "termcolor.h"
// NOTE: You cannot change this file, and you don't need to understand its
// content in order to solve your task. Feel free to look around, however,
// in case you're interested.
const string action = std::getenv("ACTION");
const string words[] = {
#include "words.csv"
"sentinel"
};
string color(string c) {
return action == "run" ? c : "";
}
string chooseWord() {
string word;
if (action == "test" || action == "submit") {
std::cin >> word;
} else {
int i = rand() % (sizeof(words)/sizeof(*words) - 1); // don't take the last word (which is 'sentinel')
word = words[i];
std::cout << "A random (english) word with " << word.length() << " characters has been chosen." << std::endl;
return word;
}
return word;
}
string createWorkingCopy(string word){
string result = word;
for (unsigned int i = 0; i < result.length(); ++i) {
result.at(i) = '_';
}
return result;
}
void showHangman(int wrongGuesses) {
std::cout
<< "<cx:html>\n"
// << "<p>Attempts left: " << maxWrongGuesses - wrongGuesses << "</p>\n"
<< "<img alt='' src='https://lec.inf.ethz.ch/mavt/et/2019/img/hangman/hang_" << wrongGuesses + 1 << ".gif'/>\n"
<< "</cx:html>" << std::endl;
}
void printGameState(int maxWrongGuesses, int wrongGuesses){
std::cout << color(gray) << "\nAttempts left: " << (maxWrongGuesses - wrongGuesses) << color(reset) << "\n";
}
void printWorkingCopy(string workingCopy){
std::cout << color(blue) << "[ " ;
for (unsigned int i = 0; i < workingCopy.length(); ++i) {
std::cout << workingCopy.at(i) << " ";
}
std::cout << "]\n" << color(reset);
}
void printYouLost(string word){
std::cout << "The word was: " << color(white) << word << color(red) << "\nYou lost!\n" << color(reset);
}
void printYouWon(string word){
std::cout << "\n";
printWorkingCopy(word);
std::cout << color(green) << "You won!\n" << color(reset);
}

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#ifndef HANGMAN_H
#define HANGMAN_H
#include <iostream>
#include <string>
using std::string;
// NOTE: You cannot change this file. It "only" contains declarations and
// short descriptions of the functionality we provided to you.
/**
* This function returns a random english word. Use this to generate a new
* word to guess. It is imperative that you use this function to get a word,
* otherwise, the auto-grader will not work properly when testing or
* submitting your project.
*/
string chooseWord();
/**
* This function creates a "working copy" based on a given word. It returns a
* string with the same amount of characters than the word, but all of them
* are initially set to "_" (underscore)
*/
string createWorkingCopy(string word);
/**
* You may call this function to render a little hangman figure in the HTML view.
* This is completely optional, the tests don't rely on this function being
* called. As argument, the function takes the number of wrong guesses and
* selects the correct hangman picture to show.
*/
void showHangman(int wrongGuesses);
/**
* This function prints the number of remaining attempts (based on the provided
* number of wrong guesses. Call this method before each attempt.
*
* Example: The call 'printGameState(2)' will output: "Attempts left: 4"
* because MAX_WRONG_GUESSES is 6, and 6 - 2 = 4
*/
void printGameState(int maxWrongGuesses, int wrongGuesses);
/**
* This function prints the partly uncovered word (the working copy) in the
* desired format.
*
* Example: If workingCopy is "_xp_rt", a call to printWorkingCopy(workingCopy)
* will print "[ _ x p _ r t ]" - this is the format that is expected by the
* autograder.
*/
void printWorkingCopy(string workingCopy);
/**
* This function must be called if the game was lost (that is, on the 6th
* wrong guess).
*
* Example: If the correct word was "expert", its outputs
* "The word was: expert
* You lost!"
*/
void printYouLost(string word);
/**
* This function must be called if the game was won (that is, the word was
* guessed with less than 6 wrong guesses).
*
* Example: If the correct word was "expert", its outputs
* "[ e x p e r t ]
* You won!"
*/
void printYouWon(string word);
// THE FOLLOWING FUNCTIONS CAN BE USED INTERACTIVELY TO IMPLEMENT THE INDIVIDUAL
// PARTS, BUT THEY DONT'T WORK DURING SUBMISSION
/**
* Part 1: Ask the user to enter a character and update parameter 'guess'
*/
void PART1_readCharacter(char& guess);
/**
* Part 2: Set the guessed character in the working copy. Updates parameter
* 'workingCopy' and sets parameter 'found' to either true or false
*/
void PART2_updateWorkingCopy(string word, char guess, string& workingCopy, bool& found);
/**
* Part 3: Check if game is finished and update wrongGuesses variable.
* Print the approriate messages in the console. Updates parameters 'done' and
* 'wrongGuesses'
*/
void PART3_updateGameState(string word, string workingCopy, bool found, int maxWrongGuesses, bool& done, int& wrongGuesses);
#endif

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#include <iostream>
#include <string>
#include "hangman.h"
using std::string;
int main() {
// Word the player needs to guess (randomly selected)
string word = chooseWord();
//string word = "success";
// Initialise the "uncovered" word that is shown to the player: the uncovered
// word is obtained by replacing each letter from the original word (variable
// word) with an underscore (i.e. _).
string workingCopy = createWorkingCopy(word);
// This variable indicates whether or not the game is over
bool done = false;
int wrongGuesses = 0; // Number of wrong guesses
int maxWrongGuesses = 6; // Maximum number of wrong guesses (don't change)
// Draw the empty gallow
showHangman(0);
// Game loop (each iteration is a round of the game)
while (!done) {
printGameState(maxWrongGuesses, wrongGuesses);
printWorkingCopy(workingCopy);
/** Part 1: input next guess **********************************************/
char guess = '\0';
// TODO: replace the following line with your implementation
//PART1_readCharacter(guess);
std::cout << "Your guess: ";
std::cin >> guess;
/** Part 2: update working copy *******************************************/
bool found = false;
// TODO: replace the following line with your implementation
//PART2_updateWorkingCopy(word, guess, workingCopy, found);
for (int i = 0; i != word.length(); i++) {
if (guess == word.at(i)) {
found = true;
workingCopy.at(i) = guess;
}
}
/** Part 3: update game state *********************************************/
// TODO: replace the following line with your implementation
//PART3_updateGameState(word, workingCopy, found, maxWrongGuesses, done, wrongGuesses);
if (workingCopy == word) {
done = true;
printYouWon(word);
} else if (found == false) {
wrongGuesses += 1;
}
if (wrongGuesses == maxWrongGuesses) {
done = true;
printYouLost(word);
}
}
return 0;
}

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#ifndef TERMCOLOR_H
#define TERMCOLOR_H
// NOTE: You cannot change this file, and you don't need to understand its
// content in order to solve your task. Feel free to look around, however,
// in case you're interested.
const auto red = "\033[31;1m";
const auto green = "\033[32;1m";
const auto yellow = "\033[33;1m";
const auto blue = "\033[34;1m";
const auto magenta = "\033[35;1m";
const auto cyan = "\033[36;1m";
const auto gray = "\033[39;2m";
const auto white = "\033[39;1m";
const auto reset = "\033[0m";
#endif

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Vorkurs/Projekt_2/words.csv Normal file
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"available",
"exaggerate",
"ancestor",
"architect",
"neighborhood",
"curriculum",
"promotion",
"opera",
"frequency",
"excavation",
"guarantee",
"reflection",
"benefit",
"development",
"average",
"ghostwriter",
"unlikely",
"disturbance",
"initiative",
"hospitality",
"mastermind",
"eyebrow",
"consciousness",
"operational",
"vehicle",
"housewife",
"capital",
"execution",
"terrify",
"disagree",
"exclusive",
"equinox",
"essential",
"imperial",
"publicity",
"secretary",
"nationalist",
"attention",
"established",
"magnitude",
"orientation",
"contraction",
"intention",
"seminar",
"forecast",
"manufacturer",
"reception",
"fabricate",
"mosquito",
"cooperative",
"parachute",
"exotic",
"demonstrate",
"production",
"spontaneous",
"minimum",
"abolish",
"holiday",
"formation",
"admission",
"handicap",
"continuous",
"presentation",
"constituency",
"unique",
"violation",
"radical",
"notebook",
"custody",
"dictionary",
"comprehensive",
"dominant",
"requirement",
"opponent",
"business",
"national",
"manufacture",
"nominate",
"liberal",
"continuation",
"galaxy",
"interest",
"ignorant",
"indirect",
"illustrate",
"proportion",
"projection",
"philosophy",
"acceptable",
"aluminium",
"continental",
"potential",
"vegetarian",
"elephant",
"advantage",
"recording",
"agenda",
"electronics",
"engagement",
"lonely",
1 available
2 exaggerate
3 ancestor
4 architect
5 neighborhood
6 curriculum
7 promotion
8 opera
9 frequency
10 excavation
11 guarantee
12 reflection
13 benefit
14 development
15 average
16 ghostwriter
17 unlikely
18 disturbance
19 initiative
20 hospitality
21 mastermind
22 eyebrow
23 consciousness
24 operational
25 vehicle
26 housewife
27 capital
28 execution
29 terrify
30 disagree
31 exclusive
32 equinox
33 essential
34 imperial
35 publicity
36 secretary
37 nationalist
38 attention
39 established
40 magnitude
41 orientation
42 contraction
43 intention
44 seminar
45 forecast
46 manufacturer
47 reception
48 fabricate
49 mosquito
50 cooperative
51 parachute
52 exotic
53 demonstrate
54 production
55 spontaneous
56 minimum
57 abolish
58 holiday
59 formation
60 admission
61 handicap
62 continuous
63 presentation
64 constituency
65 unique
66 violation
67 radical
68 notebook
69 custody
70 dictionary
71 comprehensive
72 dominant
73 requirement
74 opponent
75 business
76 national
77 manufacture
78 nominate
79 liberal
80 continuation
81 galaxy
82 interest
83 ignorant
84 indirect
85 illustrate
86 proportion
87 projection
88 philosophy
89 acceptable
90 aluminium
91 continental
92 potential
93 vegetarian
94 elephant
95 advantage
96 recording
97 agenda
98 electronics
99 engagement
100 lonely