Pomocný program k řešení úlohy do DIM
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#! /usr/bin/env python | |
""" | |
Zadání úlohy: | |
Kolik slov délky n lze sestavit v abecedě {A,B,C,D} za podmínky, že prvky A a B nesousedí | |
(ale A může sousedět s A, B může sousedět s B). | |
""" | |
import itertools | |
from math import sqrt | |
abeceda = ["A", "B", "C", "D"] | |
n = 10 | |
# zkontrolovat, jestli A a B nesousedí | |
def check(slovo): | |
for i in range(1, len(slovo)): | |
if slovo[i] == "A" and slovo[i-1] == "B": | |
return False | |
if slovo[i] == "B" and slovo[i-1] == "A": | |
return False | |
return True | |
# brute-force | |
def count_for_n(n): | |
counter = 0 | |
slova = itertools.product( *([abeceda] * n) ) | |
for slovo in slova: | |
if check(slovo): | |
counter += 1 | |
# print(slovo) | |
return counter | |
# napočítat členy rekurentní posloupnosti | |
def compute(n): | |
# počáteční podmínky | |
if n == 1: | |
compute.values.append(4) | |
elif n == 2: | |
compute.values.append(14) | |
else: | |
compute.values.append( 2*compute.values[n-2] + 3*compute.values[n-1] ) | |
return compute.values[n] | |
# inicializovat statické proměnné | |
compute.values = [0] # celkový počet možností | |
for i in range(1, n+1): # count from 1 to n inclusive | |
print("n =", i) | |
# spočítat metodou brute-force | |
counted = count_for_n(i) | |
print("počet slov =\t", counted) | |
# napočítat členy rekurentní posloupnosti | |
computed = compute(i) | |
print("rekurentně =\t", computed) | |
# vzorec pro n-tý člen | |
computed = round( (17+5*sqrt(17)) / 34 * ((3+sqrt(17))/2)**i + (17-5*sqrt(17)) / 34 * ((3-sqrt(17))/2)**i ) | |
print("n-tý člen =\t", computed) |
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