BuK/IIm19/Ausarbeitungen Lüpke/Brainstorming VL03 Luepke

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<language>Python</language>
<language>Python</language>
== Entscheidbarkeit und Aufzählbarkeit ==

ÜA zu Beispiel 1 und 2: 
$A^* = \{a, b\}^*$ und $M \subset A^* = \{ w \space | \space w \in A^* \space\text{und}\space |w| = 3 \}$.<br>
$M = \{aaa,aba,aab,baa,bab,abb,bba,bbb\}$

<code id="5e8da562c99f4">
A = ['a', 'b']
M = ['aaa','aba','aab','baa','bab','abb','bba','bbb']

def inM(w):
    return w in M
</code>

<run id="5e8da562c9a7e">
print(inM('not'))
print(inM('a'))
print(inM('aba'))
</run>

$A^* = \{a, b\}^*$ und $M \subset A^* = \{ w = aw' \space | \space w' \in A^* \}$.

<code id="5e97538546e73">
import itertools
from itertools import permutations

def generate(f, x):
  return itertools.accumulate(itertools.repeat(x), lambda v,_:f(v))

def isWord(word): 
    if (word[0] != 'a'):
        prrint()
    for index, i in enumerate(generate(lambda x: 1+x, 0)):
        for aIndex, aPart in enumerate(generate(lambda x: x[0]+x, 'a')):
            for bIndex, bPart in enumerate(generate(lambda x: x[0]+x, 'b')):
                perms = ['a'+''.join(p) for p in permutations(aPart+bPart)]
                perms.append('a'+aPart)
                perms.append('a'+bPart)
                if checkWord in perms:
                    print("True")
                    return
                if (bIndex >= i):
                    break
            if (aIndex >= i):
                break
</code>

<run id="5e97538546f0b">
isWord('abbbabaa')
</run>