def get_permutations(n):
"""
:param n:
:return: list of all permutations of n unique items
"""
uniques = list(SYMBOLS[:n])
return list(map(lambda tup: ''.join(tup), permute(uniques)))
def uniquely_permute(iterable, enforce_sort=False, r=None):
previous = tuple()
if enforce_sort: # potential waste of effort (default: False)
iterable = sorted(iterable)
for p in permute(iterable, r):
if p > previous:
previous = p
yield p
def chef(n,k,l):
l2=[s for s in range(1,k+1)]
l2=l2*n
l3=[]
for i in itertools.permute(l2,n):
l3.append(i)
l4=list(set(l3))
l4.sort()
gl.append(l4[l-1])
def run():
T = int(input())
for t in range(1, T + 1):
N, C = map(int, input().split(" "))
permutations = list(permute([i for i in range(1, N + 1)]))
# print(permutations)
for p in permutations:
i = reversort(list(p))
if reversort(list(p)) == C:
# print(i, C, p)
print(f"Case #{t}:", *p)
break
else:
print(f"Case #{t}: IMPOSSIBLE")
def permAlone(word):
if not word.isalpha(): return "Error: must contain only letters."
word = list(word.lower())
lenght = range(len(word) - 1)
if len(set(word)) == 1: return 0
ls = list(permute(word))
counter = 0
for sublist in ls:
for i in lenght:
if sublist[i] == sublist[i + 1]:
counter += 1
break
return len(ls) - counter
SessionLetter = ParticipantID[-1]
PartNumber = int(ParticipantID[-4:-1])
SessionNumber = ['A', 'B', 'C', 'D'].index(SessionLetter) + 1
# Between Subjects Design
if len(ParticipantID) == 7:
PartPrefix = ParticipantID[:3]
if PartPrefix not in ['ENG', 'CHN', 'THI', 'DOT']:
raise (ValueError(
'ID Lanuage Prefix must be either ENG, CHN, THI or DOT. Please change ID accordingly if you wish to proceed with a Between Subjects design.'
))
LanuageNumber = ['ENG', 'CHN', 'THI', 'DOT'].index(PartPrefix)
# Within Subjects Design
elif len(ParticipantID) == 4:
TaskSet = list(permute(range(4)))
# This is random order by which task set permutations may be presented, generated with below code.
#RandomisedOrder = range(len(TaskSet))
#random.shuffle(RandomisedOrder)
RandomisedOrder = [
20, 9, 13, 7, 15, 5, 8, 3, 17, 16, 19, 2, 11, 18, 22, 21, 1, 14, 6, 10,
23, 4, 12, 0
]
while len(RandomisedOrder) <= PartNumber:
RandomisedOrder += RandomisedOrder
TaskSet = TaskSet[RandomisedOrder[PartNumber]]
LanuageNumber = TaskSet[SessionNumber - 1]
PartPrefix = ['ENG', 'CHN', 'THI', 'DOT'][LanuageNumber]
ParticipantID = PartPrefix + ParticipantID
# Timing of Stim, Mask and Time Out - Changing will change experiment timing
#coding:utf-8 #Author: Alexander Hansson #Version: 2015-12-27 """ A permutation is an ordered arrangement of objects. For example, 3124 is one possible permutation of the digits 1, 2, 3 and 4. If all of the permutations are listed numerically or alphabetically, we call it lexicographic order. The lexicographic permutations of 0, 1 and 2 are: 012 021 102 120 201 210 What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9? """ from itertools import permutations as permute print "".join([str(x) for x in list(permute(range(10)))[10**6-1]]) #wtf hax
def permutate(self, samples): return [''.join([str(_) for _ in _]) for _ in list(permute(samples))]
def permutations(string):
return list(map(convertTuple,list(set(list(permute(string))))))
# Project euler problem 24
# Eric Jones
# 22 April 2014
from itertools import permutations as permute
if __name__ == '__main__':
a = [0,1,2,3,4,5,6,7,8,9]
z = permute(a) #z is an itertools object
z = [x for x in z]
z.sort()
print z[1000000-1]
def height_counts(n): return Counter(get_height(order) for order in permute(range(n)))