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I am trying to build list similar to this
Position = [1, 1, 1, 1, 1, 1, 1,
2, 2, 2, 2, 2, 2, 2,
3, 3, 3, 3, 3, 3, 3,
4, 4, 4, 4, 4, 4, 4,
5, 5, 5, 5, 5, 5, 5,
6, 6, 6, 6, 6, 6, 6,
7, 7, 7, 7, 7, 7, 7,
8, 8, 8, 8, 8, 8, 8,
9, 9, 9, 9, 9, 9, 9,
10, 10, 10, 10, 10, 10, 10,
11, 11, 11, 11, 11, 11, 11,
12, 12, 12, 12, 12, 12, 12,
13, 13, 13, 13, 13, 13, 13,
14, 14, 14, 14, 14, 14, 14,
15, 15, 15, 15, 15, 15, 15]
With the list comprehension like below (with one if statement), only the first 7 elements can be created. How can I add multiple if statements?
d = 1
position = [d if i<7 else d+1 for i in range (105)]
No need for even a single if; just use integer division
[1+i//7 for i in range (7*15)]
Use numpy for creating a list of some value with some repetition, append them together
import numpy as np
numbers_i_want = np.arange(1, 16, 1, dtype=int) #last number not included
arr = np.array([])
for n in numbers_i_want:
temp = np.full(
shape=10, # how many ints
fill_value=n #what int
)
arr= np.concatenate((arr,temp))
print(arr)
The simplest solution in the standard library is to chain multiple iterables together.
from itertools import chain, repeat
Position = list(chain.from_iterable(repeat(i, 7) for i in range(1, 16)))
or use two generators
Position = list(x for i in range(1, 16) for x in repeat(i, 7))
Or, since int values are immutable, list multiplication is a viable solution.
Position = list(chain.from_iterable([i]*7 for i in range(1, 16)))
Position = list(x for i in range(1, 16) for x in [i]*7)
[PYTHON] Hello i have to create a list which starts at 1 and ends at 20, and then convert this list into a nested list which should look like this: [[1,2,...,20],[20,19,...1]].
I did only this:
list = []
for i in range(1,21):
lista.append(i)
i += 1
which gives me a normal [1, ... 20] list,
but I don't know how to change it to a nested list with a reverse list.
You can use a list comprehension to simply do this (don't use list as variable name):
myList = [[x for x in range(1,21)],[y for y in range(21,0,-1)]]
Create a nested list to begin with:
l = [ [], [] ]
for i in range(20):
l[0].append(i+1) # fills the first inner list with 1...20
l[-1].append(20-i) # fills the last inner list with 20...1
print(l)
Output:
[[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20],
[20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1]]
my_list = [list(range(1, 21)), list(range(21, 0 ,-1))]
Reverse a list using slicing and concat both list.
arr = [1,2,3,4,5,6,7]
ans = [arr] + [arr[::-1]]
print(ans)
Output:
[[1, 2, 3, 4, 5, 6, 7], [7, 6, 5, 4, 3, 2, 1]]
Simple:
>>> x = [i for i in range(1, 21)]
>>> x
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]
>>> r = [x, x[::-1]]
>>> r
[[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20], [20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1]]
I am trying to implement a solution for outputting the sequence of moves for a 15-puzzle problem in Python. This is part of an optional assignment for a MOOC. The problem statement is given at this link.
I have a version of the program (given below) which performs valid transitions.
I am first identifying the neighbors of the empty cell (represented by 0) and putting them in a list. Then, I am randomly choosing one of the neighbors from the list to perform swaps with the empty cell. All the swaps are accumulated in a different list to record the sequence of moves to solve the puzzle. This is then outputted at the end of the program.
However, the random selection of numbers to make the swap with the empty cell is just going on forever. To avoid "infinite" (very long run) of loops, I have limited the number of swaps to 30 for now.
from random import randint
def find_idx_of_empty_cell(p):
for i in range(len(p)):
if p[i] == 0:
return i
def pick_random_neighbour_idx(neighbours_idx_list):
rand_i = randint(0, len(neighbours_idx_list)-1)
return neighbours_idx_list[rand_i]
def perform__neighbour_transposition(p, tar_idx, src_idx):
temp = p[tar_idx]
p[tar_idx] = p[src_idx]
p[src_idx] = temp
def solve_15_puzzle(p):
standard_perm = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,0]
neighbours_idx_list = []
moves_sequence = []
empty_cell_idx = find_idx_of_empty_cell(p)
previous_empty_cell_idx = empty_cell_idx
while (not(p == standard_perm) and len(moves_sequence) < 30):
if not (empty_cell_idx in [0,4,8,12]):
neighbours_idx_list.append(empty_cell_idx - 1)
if not (empty_cell_idx in [3,7,11,15]):
neighbours_idx_list.append(empty_cell_idx + 1)
if not (empty_cell_idx in [0,1,2,3]):
neighbours_idx_list.append(empty_cell_idx - 4)
if not (empty_cell_idx in [12,13,14,15]):
neighbours_idx_list.append(empty_cell_idx + 4)
if previous_empty_cell_idx in neighbours_idx_list:
neighbours_idx_list.remove(previous_empty_cell_idx)
chosen_neighbour_idx = pick_random_neighbour_idx(neighbours_idx_list)
moves_sequence.append(p[chosen_neighbour_idx])
perform__neighbour_transposition(p, empty_cell_idx, chosen_neighbour_idx)
previous_empty_cell_idx = empty_cell_idx
empty_cell_idx = chosen_neighbour_idx
neighbours_idx_list = []
if (p == standard_perm):
print("Solution: ", moves_sequence)
For the below invocation of the method, the expected output is [15, 14, 10, 13, 9, 10, 14, 15].
solve_15_puzzle([1, 2, 3, 4, 5, 6, 7, 8, 13, 9, 11, 12, 10, 14, 15, 0])
The 15-tiles problem is harder as it may seem at a first sight.
Computing the best (shortest) solution is a difficult problem and it has been proved than finding the optimal solution as N increases is NP-hard.
Finding a (non-optimal) solution is much easier. A very simple algorithm that can be made to work for example is:
Define a "distance" of the current position as the sum of the manhattan
distances of every tile from the position you want it to be
Start from the given position and make some random moves
If the distance after the moves improves or stays the same then keep the changes, otherwise undo them and return to the starting point.
This kind of algorithm could be described as a multi-step stochastic hill-climbing approach and is able to solve the 15 puzzle (just make sure to allow enough random moves to be able to escape a local minimum).
Python is probably not the best language to attack this problem, but if you use PyPy implementation you can get solutions in reasonable time.
My implementation finds a solution for a puzzle that has been mixed up with 1000 random moves in seconds, for example:
(1, 5, 43, [9, [4, 10, 14, 11, 15, 3, 8, 1, 13, None, 9, 7, 12, 2, 5, 6]])
(4, 17, 41, [9, [4, 10, 14, 11, 15, 3, 8, 1, 12, None, 6, 2, 5, 13, 9, 7]])
(7, 19, 39, [11, [4, 10, 14, 11, 15, 3, 1, 2, 12, 6, 8, None, 5, 13, 9, 7]])
(9, 54, 36, [5, [4, 14, 3, 11, 15, None, 10, 2, 12, 6, 1, 8, 5, 13, 9, 7]])
(11, 60, 34, [10, [4, 14, 3, 11, 15, 10, 1, 2, 12, 6, None, 8, 5, 13, 9, 7]])
(12, 93, 33, [14, [4, 14, 11, 2, 15, 10, 3, 8, 12, 6, 1, 7, 5, 13, None, 9]])
(38, 123, 31, [11, [4, 14, 11, 2, 6, 10, 3, 8, 15, 12, 1, None, 5, 13, 9, 7]])
(40, 126, 30, [13, [15, 6, 4, 2, 12, 10, 11, 3, 5, 14, 1, 8, 13, None, 9, 7]])
(44, 172, 28, [10, [15, 4, 2, 3, 12, 6, 11, 8, 5, 10, None, 14, 13, 9, 1, 7]])
(48, 199, 23, [11, [15, 6, 4, 3, 5, 12, 2, 8, 13, 10, 11, None, 9, 1, 7, 14]])
(61, 232, 22, [0, [None, 15, 4, 3, 5, 6, 2, 8, 1, 12, 10, 14, 13, 9, 11, 7]])
(80, 276, 20, [10, [5, 15, 4, 3, 1, 6, 2, 8, 13, 10, None, 7, 9, 12, 14, 11]])
(105, 291, 19, [4, [9, 1, 2, 4, None, 6, 8, 7, 5, 15, 3, 11, 13, 12, 14, 10]])
(112, 313, 17, [9, [1, 6, 2, 4, 9, 8, 3, 7, 5, None, 14, 11, 13, 15, 12, 10]])
(113, 328, 16, [15, [1, 6, 2, 4, 9, 8, 3, 7, 5, 15, 11, 10, 13, 12, 14, None]])
(136, 359, 15, [4, [1, 6, 2, 4, None, 8, 3, 7, 9, 5, 11, 10, 13, 15, 12, 14]])
(141, 374, 12, [15, [1, 2, 3, 4, 8, 6, 7, 10, 9, 5, 12, 11, 13, 15, 14, None]])
(1311, 385, 11, [14, [1, 2, 3, 4, 8, 5, 7, 10, 9, 6, 11, 12, 13, 15, None, 14]])
(1329, 400, 10, [13, [1, 2, 3, 4, 6, 8, 7, 10, 9, 5, 11, 12, 13, None, 15, 14]])
(1602, 431, 9, [4, [1, 2, 3, 4, None, 6, 8, 7, 9, 5, 11, 10, 13, 15, 14, 12]])
(1707, 446, 8, [5, [1, 2, 3, 4, 6, None, 7, 8, 9, 5, 15, 12, 13, 10, 14, 11]])
(1711, 475, 7, [12, [1, 2, 3, 4, 6, 5, 7, 8, 9, 10, 15, 12, None, 13, 14, 11]])
(1747, 502, 6, [8, [1, 2, 3, 4, 6, 5, 7, 8, None, 9, 10, 12, 13, 14, 15, 11]])
(1824, 519, 5, [14, [1, 2, 3, 4, 9, 6, 7, 8, 5, 10, 15, 12, 13, 14, None, 11]])
(1871, 540, 4, [10, [1, 2, 3, 4, 9, 6, 7, 8, 5, 10, None, 12, 13, 14, 15, 11]])
(28203, 555, 3, [9, [1, 2, 3, 4, 5, 6, 7, 8, 9, None, 10, 12, 13, 14, 11, 15]])
(28399, 560, 2, [10, [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, None, 12, 13, 14, 11, 15]])
(28425, 581, 1, [11, [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, None, 13, 14, 15, 12]])
(28483, 582, 0, [15, [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, None]])
The last line means that after 24,483 experiments it found the target position after 582 moves. Note that 582 is for sure very far from optimal as it's known that no position in the classic version of the 15 puzzle requires more than 80 moves.
The number after the number of moves is the "manhattan distance", for example the fourth-last row is the position:
where the sum of manhattan distances from the solution is 3.
I have a problem checking multiple conditions at once.
import itertools
def repeats(input1, input2):
return [int(dz) for dz in input1 if int(dz) in input2]
n_combs = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25]
filters = [[[1, 2, 3, 6, 7, 8, 11, 12, 16, 17, 21, 22], [5]], [[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [7]], [[20, 21, 22, 23, 24], [2]]]
combinacoes = itertools.combinations(n_combs, 15)
for comb in combinacoes:
for filtro, maxx in filters:
if len(repeats(filtro, comb)) in maxx:
print(comb)
Basically, I need a combination to only print if:
contain 5 items from this list: [1, 2, 3, 6, 7, 8, 11, 12, 16, 17, 21, 22]
contain 7 items from this list: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
contain 2 items from this list: [20, 21, 22, 23, 24]
The above code does not do the 3 validations at the same time and that's what I need.
I think your loop is working properly only that the last element in filters does not meet the conditions you have specified. Edit it to include smaller numbers like I have done below. Also, I have reduced the number of elements in n_combs so that my answer becomes easy to understand. This is because looping through the more than 3 million entries in 25 combination 15 is not a very good idea for demonstration purposes. Try to run the edited version below and you will see what I am saying.
import itertools
def repeats(input1, input2):
return [int(dz) for dz in input1 if int(dz) in input2]
n_combs = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]
filters = [[[1, 2, 3, 6, 7, 8, 11, 12, 16, 17, 21, 22], [5]], [[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [7]], [[2, 4, 7,9], [2]]]
combinacoes = itertools.combinations(n_combs, 15)
for comb in combinacoes:
for filtro, maxx in filters:
if len(repeats(filtro, comb)) in maxx:
print(comb, maxx)
Let me know if this solves your problem.
I assume you want to apply an arbitrary number of tests to some value.
Since in Python functions are first class citizens, you can have a list of validator functions:
def iterable_contains_1(value):
return 1 in value
def iterable_contains_2(value):
return 2 in value
validators = [iterable_contains_1, iterable_contains_2]
Then you can call all your validations:
for item in ([1, 2, 3], [2, 3, 4], [1, 3, 4], [3, 4, 5]):
if all(validator(item) for validator in validators):
print('do something with', item)
This will print only do something with [1, 2, 3] as it is the only list passing both tests.
[edit]
I think you are looking for set.
def validator1(iterable):
return len(
set(iterable).intersection(
[1, 2, 3, 6, 7, 8, 11, 12, 16, 17, 21, 22]
)) >= 5
def validator2(iterable):
return len(
set(iterable).intersection(
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
)) >= 7
def validator3(iterable):
return len(
set(iterable).intersection(
[20, 21, 22, 23, 24]
)) >= 2
I found a resolution:
import itertools
def repeats(input1, input2):
return [int(dz) for dz in input1 if int(dz) in input2]
n_combs = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25]
filters = [[[1, 2, 3, 6, 7, 8, 11, 12, 16, 17, 21, 22], [5]], [[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [7]], [[20, 21, 22, 23, 24], [2]]]
combinacoes = itertools.combinations(n_combs, 15)
for comb in combinacoes:
if all([len(repeats(filtro, comb)) in qtd for filtro, qtd in filters]):
print(comb)
Remembering that the values used in this post were just examples, to simplify and better understand logic.
Now the algorithm is able to validate a combination on all filters at the same time, and allow one condition to be satisfied for all.
I am trying to implement differential cryptanalysis on an sbox for a DES algorithm, and I'm receiving this error.
This is my code so far:
s1=[14, 4, 13, 1, 2, 15, 11, 8, 3, 10, 6, 12, 5, 9, 0, 7,
0, 15, 7, 4, 14, 2, 13, 1, 10, 6, 12, 11, 9, 5, 3, 8,
4, 1, 14, 8, 13, 6, 2, 11, 15, 12, 9, 7, 3, 10, 5, 0,
15, 12, 8, 2, 4, 9, 1, 7, 5, 11, 3, 14, 10, 0, 6, 13]
b=[]
for i in range(0, 63):
j=ZZ(i).binary().zfill(6)
b += [s1[ZZ('0b'+j[0]+j[5])][ZZ('0b'+j[1]+j[2]+j[3]+j[4])]]
s1=mq.SBox(b)
ddt=s1.difference_distribution_matrix()
for i in range(63):
print i,ddt[i]
The error is on
b += [s1[ZZ('0b'+j[0]+j[5])][ZZ('0b'+j[1]+j[2]+j[3]+j[4])]]
s1 is a list of a single dimension.
s1[ZZ('0b'+j[0]+j[5])][ZZ('0b'+j[1]+j[2]+j[3]+j[4])] is trying to index into the return of an index.
s1[ZZ('0b'+j[0]+j[5])] pulls back the integer value in s1 is position [ZZ('0b'+j[0]+j[5])].
Then you are trying to index into the integer at position [ZZ('0b'+j[1]+j[2]+j[3]+j[4])].
That is my best guess without knowing what is happening in ZZ().