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# Tutorium 07 - 01.12.2023
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## Execrise 06
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- Korrektur wieder am Samstag
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### Problematik ChatGPT und Plagiate
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- ChatGPT ist ein tolles Tool, warum?
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- Manchmal liefert es andere Lösungen zu Problemen
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- Grundverständnis bei neuen Problemen
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- integriert in die IDE (z.B. Github Copilot):
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- schneller Code schreiben
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<details>
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<summary>Wie viele Zeilen Code schreibt ein Entwickler durchschnittlich am Tag?</summary>
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10 bis 50 Codezeilen
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</details>
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- Leichtsinnsfehler ausbessern
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- Kurz: Es nimmt einen repetetive Arbeit ab
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#### Die Problematik?
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- Ein EidP soll das Grundverständnis von Programmieren vermittelt werden
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- Denkweise
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- Konzepte in der theoretischen Informatik
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- Konzepte in Programmiersprachen
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- Übung
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- Um ChatGPT sinnvoll zu nutzen müsst ihr diese Grundverständnis bereits besitzen
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- Auch Studierende mit Vorwissen profitieren davon die Übung sinnvoll zu bearbeiten
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- Wenn Ihr für die Aufgaben ChatGPT verwendet, dann habt ihr nicht genug Vorwissen
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<details>
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<summary>Studienleistung WS2022</summary>
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<img src="./img/ws2022-studienleistung.png" width=833 height=auto>
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</details>
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<details>
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<summary>Notenverteilung WS2022</summary>
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<img src="./img/ws2022-notenverteilung.png" width=833 height=auto>
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</details>
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#### Also, macht eure Aufgaben selber!
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----
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## Wichtiges/Hilfreiches für Exercise-07
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### Rekursion
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- Rekursion in Python sind Funktionen die sich selbst aufrufen
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```python
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def fac(n: int) -> int:
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if n <= 1: # Abbruchbedingung, kein Rekursiver Aufruf mehr!
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return 1
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return n * fac(n - 1) # Rekursiver Aufruf
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```
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- Eine Rekursion braucht eine **Abbruchbedingung**
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- primitive Rekursionen können auch einfach iterative gelöst werden
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```python
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def fac2(n: int) -> int:
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fac = 1
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for i in range(1, n + 1):
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fac *= i
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return fac
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```
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- Eine Rekursion kann mehrere Rekursionspfade haben! (Kaskadenförmige Rekursion), welche auch primitiv berechenbar sind!
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```python
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def fib(n: int) -> int:
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if n in {0, 1}: # Abbruchbedingung
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return n
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return fib(n - 1) + fib(n - 2) # mehrere Rekursionsaufrufe
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```
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- Wie Funktioniert das?
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- Es wird ein Rekursionsbaum aufgebaut
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- Wenn dieser Fertig ist wird berechnet
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- Z.b. `fac`:
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```
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fac(5)
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5 * fac(4)
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5 * 4 * fac(3)
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5 * 4 * 3 * fac(2)
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5 * 4 * 3 * 2 * fac(1)
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5 * 4 * 3 * 2 * 1
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120
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```
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```
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fib(4)
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fib(3) + fib(2)
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(fib(2) + fib(1)) + (fib(0) + fib(1))
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((fib(0) + fib(1)) + fib(1)) + (fib(0) + fib(1))
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((0 + 1) + 1) + (0 + 1)
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3
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```
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- Gibt es Rekursionen die nicht iterative berechenbar sind?
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- $\mu$-Rekursionen oder partiell Rekursionen
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- erste partiell rekursive Funktion von Wilhelm Ackermann 1926, die "Ackermannfunktion"
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$\alpha(0, m) = m + 1$
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$\alpha(n, 0) = \alpha(n - 1, 1)$
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$\alpha(n, m) = \alpha(n, \alpha(n, m - 1))$
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```python
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def ack(n: int, m: int) -> int:
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match (n, m):
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case (0, _):
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return m + 1
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case (_, 0):
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return ack(n - 1, 1)
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case _:
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return ack(n - 1, ack(n, m - 1))
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```
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#### Tipp:
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Man kann alles rekursiv Aufbauen mit Operatoren (`+, -, *, /, %, //, &&, and, ...`), also auch Listen oder Strings
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```python
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def all_fac(max: int) -> list[(int, int)]:
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if max == 0: # Abbruchbedingung
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return [(0, 1)]
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return [(max, fac(max))] + all_fac(max - 1) # Rekursion
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def all_fac_str(min: int, max: int) -> str:
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if min >= max: # Abbruchbedingung
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return f"{fac(min)}"
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return f"{fac(min)} " + all_fac_str(min + 1, max) # Rekursion
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def fib_str(n: int) -> str:
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if n in {0, 1}: # Abbruchbedingung
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return str(n)
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return f"({fib_str(n - 1)} + {fib_str(n - 2)})" # Rekursion
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```
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### Rekursion in Bäumen
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- Drei möglichkeiten einen Baum *abzulaufen*
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- **Pre-Order**: Knoten, links, rechts
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```python
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def preorder[T](tree: BTree[T]):
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match tree:
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case Node(value, left, right):
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print(value)
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preorder(left)
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preorder(right)
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case _:
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return
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```
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- **Post-Order**: links, rechts, Knoten
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```python
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def postorder[T](tree: BTree[T]):
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match tree:
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case Node(value, left, right):
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postorder(left)
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postorder(right)
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print(value)
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case _:
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return
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```
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- **In-Order**: links, Knoten, rechts
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```python
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def inorder[T](tree: BTree[T]):
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match tree:
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case Node(value, left, right):
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inorder(left)
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print(value)
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inorder(right)
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case _:
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return
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```
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BIN
Tutorium/tut07/img/ws2022-notenverteilung.png
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Tutorium/tut07/img/ws2022-notenverteilung.png
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Tutorium/tut07/img/ws2022-studienleistung.png
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Tutorium/tut07/img/ws2022-studienleistung.png
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After Width: | Height: | Size: 10 KiB |
57
Tutorium/tut07/src/solutions-exercise-06/mastermind.py
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Tutorium/tut07/src/solutions-exercise-06/mastermind.py
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import random
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def remove_char(string: str, c: str) -> str:
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for i, s in enumerate(string):
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if s == c:
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return string[:i] + string[i + 1:]
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return string
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def perfect_chars(inp: str, sol: str) -> str:
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chars = ""
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for c, s in zip(inp, sol):
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if c == s:
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chars = chars + c
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return chars
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def correct_chars(inp: str, sol: str) -> str:
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res = ""
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for c in inp:
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if c in sol:
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res += c
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sol = remove_char(sol, c)
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return res
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def compare(inp: str, sol: str) -> tuple[int, int]:
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perfect = len(perfect_chars(inp, sol))
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correct = len(correct_chars(inp, sol))
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return perfect, correct - perfect
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def compare_alt(inp: str, sol: str) -> tuple[int, int]:
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perfect = perfect_chars(inp, sol)
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for c in perfect:
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sol = remove_char(sol, c)
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inp = remove_char(inp, c)
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correct = correct_chars(inp, sol)
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return len(perfect), len(correct)
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def game(length: int, symbols: str):
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solution = "".join(random.choices(symbols, k=length))
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print("Länge:", length, "Zeichen:", symbols)
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result = (0, 0)
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while result != (length, 0):
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c = input()
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if len(c) != length:
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print("Try again!")
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continue
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result = compare(c, solution)
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print("Antwort:", result[0] * "X" + result[1] * "-")
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if __name__ == '__main__':
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game(5, "ABCDE")
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Tutorium/tut07/src/solutions-exercise-06/typevars.py
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Tutorium/tut07/src/solutions-exercise-06/typevars.py
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from typing import Optional
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def head[T](xs: list[T]) -> Optional[T]:
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if not xs:
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return None
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return xs[0]
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def tail[T](xs: list[T]) -> Optional[list[T]]:
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if not xs:
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return None
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return xs[1:]
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def concat[T](xss: list[list[T]]) -> list[T]:
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outer = list()
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for xs in xss:
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outer += xs
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return outer
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def zip[T, U](xs: list[T], ys: list[U]) -> list[tuple[T, U]]:
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out = list()
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for i, x in enumerate(xs):
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if i >= len(ys):
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return out
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else:
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out += [(x, ys[i])]
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return out
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def assoc[T, U, V](t: tuple[tuple[T, U], V]) -> tuple[T, tuple[U, V]]:
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(x, y), z = t
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return (x, (y, z))
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Tutorium/tut07/src/trees.py
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Tutorium/tut07/src/trees.py
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@dataclass
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class Node[T]:
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value: T
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left: Optional['Node[T]'] = None
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right: Optional['Node[T]'] = None
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type BTree[T] = Node[T] | None
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def preorder[T](tree: BTree[T]):
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match tree:
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case Node(value, left, right):
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print(value)
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preorder(left)
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preorder(right)
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case _:
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return
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def postorder[T](tree: BTree[T]):
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match tree:
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case Node(value, left, right):
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postorder(left)
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postorder(right)
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print(value)
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case _:
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return
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def inorder[T](tree: BTree[T]):
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match tree:
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case Node(value, left, right):
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inorder(left)
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print(value)
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inorder(right)
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case _:
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return
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Tutorium/tut07/src/tutorium_07.py
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Tutorium/tut07/src/tutorium_07.py
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def fibonacci(n: int) -> int:
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if n in {0, 1}: # Abbruchbedingung
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return n
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return fibonacci(n - 1) + fibonacci(n - 2) # mehrere Rekursionsaufrufe
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def fac(n: int) -> int:
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if n <= 0: # Abbruchbedingung, kein Rekursiver Aufruf mehr!
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return 1
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return n * fac(n - 1) # Rekursiver Aufruf
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def fac2(n: int) -> int:
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num = 1
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for i in range(1, n + 1):
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num *= i
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return num
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def ack(n: int, m: int) -> int:
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match (n, m):
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case (0, _):
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return m + 1
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case (_, 0):
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return ack(n - 1, 1)
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case _:
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return ack(n - 1, ack(n, m - 1))
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def all_fac(max: int) -> list[(int, int)]:
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if max == 0: # Abbruchbedingung
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return [(0, 1)]
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return [(max, fac(max))] + all_fac(max - 1) # Rekursion
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def all_fac_str(min: int, max: int) -> str:
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if min >= max: # Abbruchbedingung
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return f"{fac(min)}"
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return f"{fac(min)} " + all_fac_str(min + 1, max) # Rekursion
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def fib_str(n: int) -> str:
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if n in {0, 1}:
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return str(n)
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return f"({fib_str(n - 1)} + {fib_str(n - 2)})"
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if __name__ == "__main__":
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assert [fibonacci(n) for n in range(15)] == [0, 1, 1, 2,
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3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377]
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assert [fac(n) for n in range(11)] == [1, 1, 2, 6, 24,
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120, 720, 5040, 40320, 362880, 3628800]
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assert [fac(n) for n in range(10)] == [fac2(n) for n in range(10)]
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assert list(reversed(all_fac(10))) == [(n, fac(n)) for n in range(11)]
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assert all_fac_str(0, 10) == "1 1 2 6 24 120 720 5040 40320 362880 3628800"
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