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converted everything to orgmode and added solution to the README files
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Informatik_I/Exercise_2/task_1/README.md
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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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#+TITLE: Task 1: Expression Evaluation
#+AUTHOR: JirR02
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)
* Solutions
#+begin_src txt
# 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
#+end_src

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Informatik_I/Exercise_2/task_1/solutions.txt
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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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Informatik_I/Exercise_2/task_2/README.md
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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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#+TITLE: Task 2: From Natural Language to C++
#+AUTHOR: JirR02
* 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:*
#+begin_src cpp
return a > 3 && a < 5;
#+end_src
/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=.
* Solutions
#+begin_src cpp
#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
}
}
#+end_src
-----
Made by JirR02 in Switzerland 🇨🇭

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Informatik_I/Exercise_2/task_2/solutions.cpp
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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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Informatik_I/Exercise_2/task_3a/README.md
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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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#+TITLE: Task 3a: Fibonacci primes
* 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:
#+begin_src shell
2
3
5
13
Found 4 Fibonacci primes
#+end_src
*Important:* using anything other than =int= (e.g., =unsigned int=, floating point numbers, =long=, or double =long=) is forbidden.
* Solutions
#+begin_src cpp
#include <iostream>
int main() {
@@ -45,3 +72,8 @@ int main() {
// End message
std::cout << count << "\n";
}
#+end_src
-—---
Made by JirR02 in Switzerland 🇨🇭

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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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#+TITLE: Task 3b: Fibonacci overflow check
#+AUTHOR: JirR02
* 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:
#+begin_src shell
0
1
1
2
Printed 4 of 4 Fibonacci numbers
#+end_src
If you instead ask it to print the first 100 Fibonacci numbers the output should look as follows:
#+begin_src shell
0
1
1
2
3
Printed 5 of 100 Fibonacci numbers
#+end_src
*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.
* Mistakes
- The variable =j= goes into overflow which is not allowed!
* Solution
#+begin_src cpp
#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
}
#+end_src
—----
Made by JirR02 in Switzerland 🇨🇭

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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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Informatik_I/Exercise_2/task_4/README.md
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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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#+TITLE: Task 4: From decimal to binary representation
#+AUTHOR: JirR02
* 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.
* Solution
#+begin_src cpp
#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;
}
}
}
#+end_src

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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;
}
}
}