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I tried sympy.factorint() but it returns a dictionary of prime factors with count.
I need all factors in a list in ascending order.
import sympy
sympy.factorint(567)
output
{3: 4, 7: 1}
I am looking for output like this
[1, 3, 7, 9, 21, 27, 63, 81, 189, 567]
Found solution I should use sympy.divisors() instead
import sympy
sympy.divisors(n)
I know it's probably a duplicate but I didn't manage to find it. but anyways, every time I try to run my code in vscode, it exits normally with code=0, but it doesn't display anything at all on output. how can I fix this?
this is the simple code that I'm trying to run
from datetime import datetime
odds = [ 1, 3 , 5 , 7, , 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, ]
rightThisMinute = datetime.today().minute
if rightThisMinute in odds:
print("this minute seems a little odd!")
else:
print("not an odd minute.")`
and this is the output.
https://stackoverflow.com/questions/62461708/exited-with-code-0-in-0-074-seconds-output-window-has-no-output-in-visual-stud#:~:text=The%20reason%20you%20have%20been,while%20running%20the%20python%20program.&text=Then%20open%20the%20vscode%20and,to%20get%20the%20desired%20results.
This question might be helpful. As the answers already point out, you might have multiple extensions running on VSCode.
Also try simply printing Hello World to check if it works.
It is a problem with your list. It should look like this:
odds = [ 1, 3 , 5 , 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35 ]
With the list above the code works fine for me. But please read about the modulo function. You really don't need a list of odd numbers.
if rightThisMinute % 2 == 0:
print("Even number")
else:
print("not an even number")
How can I combine the 2 lists below as in the example? len(list1+list2) = 5 but len(output)=2. Thank you for attention
List1 = [array([12, 10, 8, 5, 7, 3]),
array([24, 58, 49, 30,70,6])]
len(list1) = 2
List2 = [array([63, 15, 72, 23,89,3]),
array([27, 60, 47, 29,57,6]),
array([47, 28, 50, 35,21,8])]
len(list2) = 3
Example :
Output = [
[[12, 10, 8, 5, 7, 3], [24, 58, 49, 30,70,6]],
[[63, 15, 72, 23,89,3],[27, 60, 47, 29,57,6],[47, 28, 50, 35,21,8]]
]
len(output) = 2
Just use append command.
Output = []
Output.append(List1)
Output.append(List2)
The reason you got two is because there are two elements in the list(two lists).
If have a list you can use the extend method to add a list like this:
lis1 = [34,32,12]
lis2 = [56,21,54]
lis1.extend(lis2)
print(lis1) # [34, 32, 12, 56, 21, 54]
You need to append each variable in the list to the new list
output = []
for i in range(len(list1):
output.append(list1[i])
And than do it for the second list
If you’ll just do output.append(list1) than list 2
it will be the same as your first output the one with len = 2
how can i do this using simple code in python3
matrix = [[98, 19, 1, 46, 51, 33, 3, 33, 80, 40], [26, 88, 79, 10, 63, 76, 18, 49, 47, 44], [18, 53, 8, 96, 40, 53, 73, 8, 31, 43], [8, 40, 31, 98, 19, 39, 15, 9, 58, 32], [76, 45, 1, 5, 15, 14, 20, 88, 51, 48]
You can flatten your list via itertools.chain.from_iterable, then get the largest even and odd using %==0 for evens and %!=0 for odds:
import itertools
flat = list(itertools.chain.from_iterable(matrix))
even_max = max(i for i in flat if i%2==0)
odd_max = max(i for i in flat if i%2!=0)
>>> even_max
98
>>> odd_max
79
If you prefer to avoid itertools, you can flatten your 2d matrix by list comprehension:
flat = [v for i in matrix for v in i]
even_max = max(i for i in flat if i%2==0)
odd_max = max(i for i in flat if i%2!=0)
More options, with numpy (might be overkill for your matrix, but could be beneficial if your matrix were huge):
import numpy as np
m = np.array(matrix)
even_max = max(m[m%2==0])
odd_max = max(m[m%2!=0])
This Question is about challenge number 6 in set number 1 in the challenges of "the cryptopals crypto challenges".
The challenge is:
There's a file here. It's been base64'd after being encrypted with repeating-key XOR.
Decrypt it.
After that there's a description of steps to decrypt the file, There is total of 8 steps. You can find them in the site.
I have been trying to solve this challenge for a while and I am struggling with the final two steps. Even though I've solved challenge number 3, and it contains the solution for these steps.
Note: It is, of course, possible that there is a mistake in the first 6 steps but they seems to work well after looking at the print after every step.
My code:
Written in Python 3.6.
In order to not deal with web requests, and since it is not the purpose of this challenge. I just copied the content of the file to a string in the begging, You can do this as well before running the code.
import base64
# Encoding the file from base64 to binary
file = base64.b64decode("""HUIfTQsP...JwwRTWM=""")
print(file)
print()
# Step 1 - guess key size
KEYSIZE = 4
# Step 2 - find hamming distance - number of differing bits
def hamming2(s1, s2):
"""Calculate the Hamming distance between two bit strings"""
assert len(s1) == len(s2)
return sum(c1 != c2 for c1, c2 in zip(s1, s2))
def distance(a, b): # Hamming distance
calc = 0
for ca, cb in [(a[i], b[i]) for i in range(len(a))]:
bina = '{:08b}'.format(int(ca))
binb = '{:08b}'.format(int(cb))
calc += hamming2(bina, binb)
return calc
# Test step 2
print("distance: 'this is a test' and 'wokka wokka!!!' =", distance([ord(c) for c in "this is a test"], [ord(c) for c in "wokka wokka!!!"])) # 37 - Working
print()
# Step 3
key_sizes = []
# For each key size
for KEYSIZE in range(2, 41):
# take the first KEYSIZE worth of bytes, and the second KEYSIZE worth of bytes -
# file[0:KEYSIZE], file[KEYSIZE:2*KEYSIZE]
# and find the edit distance between them
# Normalize this result by dividing by KEYSIZE
key_sizes.append((distance(file[0:KEYSIZE], file[KEYSIZE:2*KEYSIZE]) / KEYSIZE, KEYSIZE))
key_sizes.sort(key=lambda a: a[0])
# Step 4
for val, key in key_sizes:
print(key, ":", val)
KEYSIZE = key_sizes[0][1]
print()
# Step 5 + 6
# Each line is a list of all the bytes in that index
splited_file = [[] for i in range(KEYSIZE)]
counter = 0
for char in file:
splited_file[counter].append(char)
counter += 1
counter %= KEYSIZE
for line in splited_file:
print(line)
print()
# Step 7
# Code from another level
# Gets a string and a single char
# Doing a single-byte XOR over it
def single_char_string(a, b):
final = ""
for c in a:
final += chr(c ^ b)
return final
# Going over all the bytes and listing the result arter the XOR by number of bytes
def find_single_byte(in_string):
helper_list = []
for num in range(256):
helper_list.append((single_char_string(in_string, num), num))
helper_list.sort(key=lambda a: a[0].count(' '), reverse=True)
return helper_list[0]
# Step 8
final_key = ""
key_list = []
for line in splited_file:
result = find_single_byte(line)
print(result)
final_key += chr(result[1])
key_list.append(result[1])
print(final_key)
print(key_list)
Output:
b'\x1dB\x1fM\x0b\x0f\x02\x1fO\x13N<\x1aie\x1fI...\x08VA;R\x1d\x06\x06TT\x0e\x10N\x05\x16I\x1e\x10\'\x0c\x11Mc'
distance: 'this is a test' and 'wokka wokka!!!' = 37
5 : 1.2
3 : 2.0
2 : 2.5
.
.
.
26 : 3.5
28 : 3.5357142857142856
9 : 3.5555555555555554
22 : 3.727272727272727
6 : 4.0
[29, 15, 78, 31, 19, 27, 0, 32, ... 17, 26, 78, 38, 28, 2, 1, 65, 6, 78, 16, 99]
[66, 2, 60, 73, 1, 1, 30, 3, 13, ... 26, 14, 0, 26, 79, 99, 8, 79, 11, 4, 82, 59, 84, 5, 39]
[31, 31, 19, 26, 79, 47, 17, 28, ... 71, 89, 12, 1, 16, 45, 78, 3, 120, 11, 42, 82, 84, 22, 12]
[77, 79, 105, 14, 7, 69, 73, 29, 101, ... 54, 70, 78, 55, 7, 79, 31, 88, 10, 69, 65, 8, 29, 14, 73, 17]
[11, 19, 101, 78, 78, 54, 100, 67, 82, ... 1, 76, 26, 1, 2, 73, 21, 72, 73, 49, 27, 86, 6, 16, 30, 77]
('=/n?3; \x00\x13&-,>1...r1:n\x06<"!a&n0C', 32)
('b"\x1ci!!>ts es(ogg ...5i<% tc:. :oC(o+$r\x1bt%\x07', 32)
('??:<+6!=ngm2i4\x0byD...&h9&2:-)sm.a)u\x06&=\x0ct&~n +=&*4X:<(3:o\x0f1<mE gy,!0\rn#X+\nrt6,', 32)
('moI.\'ei=Et\'\x1c:l ...6k=\x1b m~t*\x155\x1ei+=+ts/e*9$sgl0\'\x02\x16fn\x17\'o?x*ea(=.i1', 32)
('+3Enn\x16Dcr<$,)\x01...i5\x01,hi\x11;v&0>m', 32)
[32, 32, 32, 32, 32]
Notice that in the printing of the key as string you cannot see it but there is 5 chars in there.
It is not the correct answer since you can see that in the forth part - after the XOR, the results do not look like words... Probably a problem in the last two functions but I couldn't figure it out.
I've also tried some other lengths and It does not seems to be the problem.
So what I'm asking is not to fix my code, I want to solve this challenge by myself :). I would like you to tell me where I am wrong? why? and how should I continue?
Thank you for your help.
After a lot of thinking and checking the conclusion was that the problem is in step number 3. The result was not good enough since I looked only at the first two blocks.
I fixed the code so it will calculate the KEYSIZE according to all of the blocks.
The code of Step 3 now look like this:
# Step 3
key_sizes = []
# For each key size
for KEYSIZE in range(2, 41):
running_sum = []
for i in range(0, int(len(file) / KEYSIZE) - 1):
running_sum.append(distance(file[i * KEYSIZE:(i + 1) * KEYSIZE],
file[(i + 1) * KEYSIZE:(i + 2) * KEYSIZE]) / KEYSIZE)
key_sizes.append((sum(running_sum)/ len(running_sum), KEYSIZE))
key_sizes.sort(key=lambda a: a[0])
Thanks for any one who tried to help.