BuK/IIm20/Vorlesung 6 Grüning

Wiki-Quelltext dieser Seite.

<language>Python</language>
== Paar Liste Seite 18/19 ==

<code id="608c2edf0290c">
def check_solution(K, index_list, p = False):
  x = ""
  y = ""
  for index in index_list:
    index -= 1
    x += K[index][0]
    y += K[index][1]
  if p:
    print(x)
    print(y)
  return x == y
</code>
<run id="608c2edf0297a">
index_list = [2, 4, 3, 4, 4, 2, 1, 2, 4, 3, 4, 3, 4, 4, 3, 4, 4, 2, 1, 4, 4, 2, 1, 3, 4, 1, 1, 3, 4, 4, 4, 2, 1, 2, 1, 1, 1, 3, 4, 3, 4, 1, 2, 1, 4, 4, 2, 1, 4, 1, 1, 3, 4, 1, 1, 3, 1, 1, 3, 1, 2, 1, 4, 1, 1, 3]
K = [("001", "0"), ("01", "011"), ("01", "101"), ("10", "001")]
print(check_solution(K, index_list, True))
</run>

<code id="608c3c0157531">
from itertools import combinations, combinations_with_replacement, permutations, product
from math import log  

def find_solution(K):
  indexes = list(range(1, len(K) + 1))
  # len(indexes) ^ length
  stop_length = log(4**14, len(indexes))
  for length in range(1, int(stop_length + 1)):
    for combination in product(indexes, repeat = length):
      if check_solution(K, combination):
        return combination
    length += 1
</code>

<run id="608c3c01575bc">
K = [("1", "101"), ("10", "00"), ("011", "11")]
print(find_solution(K))
</run>

== TM als Akzeptor Seite 23 ==


== Verdopplungsmaschine Seite 25 ==
Band, Zustand, Lese/Schreibekopfposition relativ zum Start


$11, q0, 0$
$\$1, q1, 1$
$\$1\$, q1, 2$
$\$1\$\$, q2, 3$

$\$1\$1\$, q3, 4$

$\$1\$11, q4, 3$

\$1\$11, q4, 3

\$1\$11, q4, 2

\$1\$11, q5, 1

\$1\$11, q5, 0

\$1\$11, q0, 1

Runde 2

\$\$\$11, q1, 2

\$\$\$11, q2, 3

\$\$\$11, q2, 4

\$\$\$11\$, q2, 5

\$\$\$111\$, q3, 6

\$\$\$1111, q4, 5

\$\$\$1111, q4, 4

\$\$\$1111, q4, 3

\$\$\$1111, q4, 2

\$\$\$1111, q5, 1

\$\$\$1111, q0, 2

\$\$\$1111, q6, 3
$

== Eigenschaften der Rado-Funktion Seite 36 ==

'''1'''
$\Sigma(n) < \Sigma(n + 1)$