C++ Pseudo-RSA solving for d (decryption key) quickly with large numbers

C++ Pseudo-RSA solving for d (decryption key) quickly with large numbers

By : Peter Chung
Date : November 17 2020, 11:52 AM
this one helps. the problem you are running into is: your integers are too small and will overflow when the values get larger ...
fixed size integers for things like RSA ... no good idea unless you happen to have integers with a few thousand bits length... instead of normal ints or even uint64_t, try an arbitrary precision integer library like GMP ...
code :

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Comparing large numbers of files in Bash quickly

Comparing large numbers of files in Bash quickly

By : user3751755
Date : March 29 2020, 07:55 AM
I hope this helps . I downloaded many files (~10,000) from a website, most of which are a bunch of useless html that all say the same thing. However, there are some files in this haystack that have useful information (and are thus fairly different files) and I need a quick way to separate those from the rest. I know I can go through all of the files one by one and use cmp to compare to a template and see if they are the same, and the delete them. However, this is rather slow. Is there a faster way to do this? I don't mind if I only have a 99% recovery rate. , This one lists the unique files in the tree passed as the argument:
code :
declare -A uniques
while IFS= read -r file; do
    [[ ! "${uniques[${file%% *}]}" ]] && uniques[${file%% *}]="${file##* }"
done< <(find "$1" -type f -exec md5sum -b "{}" \;)

for file in ${uniques[@]}; do
    echo "$file"
declare -a files uniques

while IFS= read -r -d $'\0' file; do
done< <(find "$1" -type f -print0)

uniques=( ${files[@]} )
for file in "${files[@]}"; do
    for unique in "${!uniques[@]}"; do
        [[ "$file" != "${uniques[$unique]}" ]] && cmp -s "$file" "${uniques[$unique]}" && && unset -v uniques[$unique]

for unique in "${uniques[@]}"; do
    echo "$unique"
i was solving a sequence of large numbers, but python shell stop responding

i was solving a sequence of large numbers, but python shell stop responding

By : Abhijit
Date : March 29 2020, 07:55 AM
around this issue You should use a modular exponentiation method. Python builtin pow does that for you:
code :
def f(x,e,m):
    X = x
    E = e
    Y = 1
    while E > 0:
        if E % 2 == 0:
            X = (X * X) % m
            E = E/2
            Y = (X * Y) % m
            E = E - 1
    return Y
>>> n=1000000000000000009
>>> m=1000000007
>>> n*pow(2,n-1, m) % m
Solving linear equation with large numbers

Solving linear equation with large numbers

By : Mr.L
Date : March 29 2020, 07:55 AM
Hope this helps You want to find x such that x = a (mod b) and x = c (mod d). For then, n1 = (x - a) / b and n2 = (x - c) / d.
If b and d are coprime, then the existence of x is guaranteed by the Chinese Remainder Theorem -- and a solution can be found using the Extended Euclidean Algorithm.
code :
def egcd(a, b):
    if a == 0:
        return (b, 0, 1)
        g, y, x = egcd(b % a, a)
        return (g, x - (b // a) * y, y)

def modinv(a, m):
    return egcd(a, m)[1] % m

def solve(a, b, c, d):
    gcd = egcd(b, d)[0]
    if gcd != 1:
        if a % gcd != c % gcd:
            raise ValueError('no solution')
        a, c = a - a % gcd, c - c % gcd
        a //= gcd
        b //= gcd
        c //= gcd
        d //= gcd
    x = a * d * modinv(d, b) + c * b * modinv(b, d)
    return (x - a) // b, (x - c) // d
import sys

import random

digit = '0123456789'

def rn(k):
    return int(''.join(random.choice(digit) for _ in xrange(k)), 10)

k = 1000

for _ in xrange(1000):
    a, b, c, d, = rn(k), rn(k), rn(k), rn(k)
    print a, b, c, d
        n1, n2 = solve(a, b, c, d)
    except ValueError, exn:
        print 'no solution'
    if a + b * n1 != c + d * n2:
        raise AssertionError('failed!')
    print 'found solution:', n1, n2
How to quickly generating a list of all pairs from a large set of numbers?

How to quickly generating a list of all pairs from a large set of numbers?

By : 于培华
Date : March 29 2020, 07:55 AM
I wish this help you You can use itertools.combinations but that will probably also take a little while like so:
code :
import itertools as it

n = 131072
pairs = it.combinations(range(n), 2)
>>> pairs
<itertools.combinations at 0x7fb939a72a48>
pairs = list(it.combinations(range(n), 2)
import numpy as np

pairs = np.transpose(np.triu_indices(n, 1))
quickly adding large numbers of mesh primitives in blender

quickly adding large numbers of mesh primitives in blender

By : George2
Date : March 29 2020, 07:55 AM
will be helpful for those in need I'm trying to add tens of thousands of mesh primitives to a scene in blender using its Python interface. I've been using something to the effect of:
code :
import bpy
from mathutils import Vector;

n = "cube";
orig_cube = bpy.context.active_object;

for i in range(10000):
    m = orig_cube.data.copy();
    o = bpy.data.objects.new(n, m);
    o.location = Vector((i, i, i));

import bpy;
import bmesh;
from mathutils import Vector;

orig_cube = bpy.context.active_object;

o = bpy.context.active_object;
me = o.data;
bm = bmesh.new();
for i in range(10000):
    bm.verts.new().co=Vector((i, i, i));
o.dupli_type = 'VERTS';
orig_cube.parent = o;
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