I am trying to find the number of ways to construct an array such that consecutive positions contain different values.
Specifically, I need to construct an array with elements such that each element 1 between and k , all inclusive. I also want the first and last elements of the array to be 1 and x.
Complete problem statement:
Here is what I tried:
def countArray(n, k, x):
# Return the number of ways to fill in the array.
if x > k:
return 0
if x == 1:
return 0
def fact(n):
if n == 0:
return 1
fact_range = n+1
T = [1 for i in range(fact_range)]
for i in range(1,fact_range):
T[i] = i * T[i-1]
return T[fact_range-1]
ways = fact(k) / (fact(n-2)*fact(k-(n-2)))
return int(ways)
In short, I did K(C)N-2 to find the ways. How could I solve this?
It passes one of the base case with inputs as countArray(4,3,2) but fails for 16 other cases.
Let X(n) be the number of ways of constructing an array of length n, starting with 1 and ending in x (and not repeating any numbers). Let Y(n) be the number of ways of constructing an array of length n, starting with 1 and NOT ending in x (and not repeating any numbers).
Then there's these recurrence relations (for n>1)
X(n+1) = Y(n)
Y(n+1) = X(n)*(k-1) + Y(n)*(k-2)
In words: If you want an array of length n+1 ending in x, then you need an array of length n not ending in x. And if you want an array of length n+1 not ending in x, then you can either add any of the k-1 symbols to an array of length n ending in x, or you can take an array of length n not ending in x, and add any of the k-2 symbols that aren't x and don't repeat the last value.
For the base case, n=1, if x is 1 then X(1)=1, Y(1)=0 otherwise, X(1)=0, Y(1)=1
This gives you an O(n)-time method of computing the result.
def ways(n, k, x):
M = 10**9 + 7
wx = (x == 1)
wnx = (x != 1)
for _ in range(n-1):
wx, wnx = wnx, wx * (k-1) + wnx*(k-2)
wnx = wnx % M
return wx
print(ways(100, 5, 2))
In principle you can reduce this to O(log n) by expressing the recurrence relations as a matrix and computing the matrix power (mod M), but it's probably not necessary for the question.
[Additional working]
We have the recurrence relations:
X(n+1) = Y(n)
Y(n+1) = X(n)*(k-1) + Y(n)*(k-2)
Using the first, we can replace the Y(_) in the second with X(_+1) to reduce it down to a single variable. Then:
X(n+2) = X(n)*(k-1) + X(n+1)*(k-2)
Using standard techniques, we can solve this linear recurrence relation exactly.
In the case x!=1, we have:
X(n) = ((k-1)^(n-1) - (-1)^n) / k
And in the case x=1, we have:
X(n) = ((k-1)^(n-1) - (1-k)(-1)^n)/k
We can compute these mod M using Fermat's little theorem because M is prime. So 1/k = k^(M-2) mod M.
Thus we have (with a little bit of optimization) this short program that solves the problem and runs in O(log n) time:
def ways2(n, k, x):
S = -1 if n%2 else 1
return ((pow(k-1, n-1, M) + S) * pow(k, M-2, M) - S*(x==1)) % M
could you try this DP version: (it's passed all tests) (it's inspired by #PaulHankin and take DP approach - will run performance later to see what's diff for big matrix)
def countArray(n, k, x):
# Return the number of ways to fill in the array.
big_mod = 10 ** 9 + 7
dp = [[1], [1]]
if x == 1:
dp = [[1], [0]]
else:
dp = [[1], [1]]
for _ in range(n-2):
dp[0].append(dp[0][-1] * (k - 1) % big_mod)
dp[1].append((dp[0][-1] - dp[1][-1]) % big_mod)
return dp[1][-1]
Why does dividing by the larger factor pair result in slower execution?
My solution for https://codility.com/programmers/task/min_perimeter_rectangle/
from math import sqrt, floor
# This fails the performance tests
def solution_slow(n):
x = int(sqrt(n))
for i in xrange(x, n+1):
if n % i == 0:
return 2*(i + n / i))
# This passes the performance tests
def solution_fast(n):
x = int(sqrt(n))
for i in xrange(x, 0, -1):
if n % i == 0:
return 2*(i + n / i)
It's not division that slows it down; it's the number of iterations required.
Let L = xrange(0, x) (order doesn't matter here) and R = xrange(x, n+1). Every factor of n in L can be paired with exactly one factor of n in R. In general, x is much, much smaller than n/2, so L is much smaller than R. This means that there are far more elements of R that don't divide n than there are in L. In the case of a prime number, there are no factors, so the slow solution has to check every value of the much larger than instead of the much smaller set.
That's obvious. The first function loops many more times.
Note that sqrt(n) != n - sqrt(n)! in general sqrt(n) << n-sqrt(n) where << means much lesser than.
If n=1000 the first function is looping 969 times while the second one only 32.
I'd say the of iterations is the key which makes perfomance a little bit different between your functions as #Bakuriu already said. Also, xrange could be slightly more expensive than using a simple loop, for instance, take a look f3 will perform a little better than f1 & f2:
import timeit
from math import sqrt, floor
def f1(n):
x = int(sqrt(n))
for i in xrange(x, n + 1):
if n % i == 0:
return 2 * (i + n / i)
def f2(n):
x = int(sqrt(n))
for i in xrange(x, 0, -1):
if n % i == 0:
return 2 * (i + n / i)
def f3(n):
x = int(sqrt(n))
while True:
if n % x == 0:
return 2 * (x + n / x)
x -= 1
N = 30
K = 100000
print("Measuring {0} times f1({1})={2}".format(
K, N, timeit.timeit('f1(N)', setup='from __main__ import f1, N', number=K)))
print("Measuring {0} times f1({1})={2}".format(
K, N, timeit.timeit('f2(N)', setup='from __main__ import f2, N', number=K)))
print("Measuring {0} times f1({1})={2}".format(
K, N, timeit.timeit('f3(N)', setup='from __main__ import f3, N', number=K)))
# Measuring 100000 times f1(30)=0.0738177938151
# Measuring 100000 times f1(30)=0.0753000788315
# Measuring 100000 times f1(30)=0.0503645315841
# [Finished in 0.3s]
Next time, you got these type of questions, using a profiler is highly recommended :)
I've been trying to implement Dixon's factorization method in python, and I'm a bit confused. I know that you need to give some bound B and some number N and search for numbers between sqrtN and N whose squares are B-smooth, meaning all their factors are in the set of primes less than or equal to B. My question is, given N of a certain size, what determines B so that the algorithm will produce non-trivial factors of N? Here is a wikipedia article about the algorithm, and if it helps, here is my code for my implementation:
def factor(N, B):
def isBsmooth(n, b):
factors = []
for i in b:
while n % i == 0:
n = int(n / i)
if not i in factors:
factors.append(i)
if n == 1 and factors == b:
return True
return False
factor1 = 1
while factor1 == 1 or factor1 == N:
Bsmooth = []
BsmoothMod = []
for i in range(int(N ** 0.5), N):
if len(Bsmooth) < 2 and isBsmooth(i ** 2 % N, B):
Bsmooth.append(i)
BsmoothMod.append(i ** 2 % N)
gcd1 = (Bsmooth[0] * Bsmooth[1]) % N
gcd2 = int((BsmoothMod[0] * BsmoothMod[1]) ** 0.5)
factor1 = gcd(gcd1 - gcd2, N)
factor2 = int(N / factor1)
return (factor1, factor2)
Maybe someone could help clean my code up a bit, too? It seems very inefficient.
This article discusses the optimal size for B: https://web.archive.org/web/20160205002504/https://vmonaco.com/dixons-algorithm-and-the-quadratic-sieve/. Briefly, the optimal value is thought to be exp((logN loglogN)^(1/2)).
[ I wrote this for a different purpose, but you might find it interesting. ]
Given x2 ≡ y2 (mod n) with x ≠ ± y, about half the time gcd(x−y, n) is a factor of n. This congruence of squares, observed by Maurice Kraitchik in the 1920s, is the basis for several factoring methods. One of those methods, due to John Dixon, is important in theory because its sub-exponential run time can be proven, though it is too slow to be useful in practice.
Dixon's method begins by choosing a bound b ≈ e√(log n log log n) and identifying the factor base of all primes less than b that are quadratic residues of n (their jacobi symbol is 1).
function factorBase(n, b)
fb := [2]
for p in tail(primes(b))
if jacobi(n, p) == 1
append p to fb
return fb
Then repeatedly choose an integer r on the range 1 < r < n, calculate its square modulo n, and if the square is smooth over the factor base add it to a list of relations, stopping when there are more relations than factors in the factor base, plus a small reserve for those cases that fail. The idea is to identify a set of relations, using linear algebra, where the factor base primes combine to form a square. Then take the square root of the product of all the factor base primes in the relations, take the product of the related r, and calculate the gcd to identify the factor.
struct rel(x, ys)
function dixon(n, fb, count)
r, rels := floor(sqrt(n)), []
while count > 0
fs := smooth((r * r) % n, fb)
if fs is not null
append rel(r, fs) to rels
count := count - 1
r := r + 1
return rels
A number n is smooth if all its factors are in the factor base, which is determined by trial division; the smooth function returns a list of factors, which is null if n doesn't completely factor over the factor base.
function smooth(n, fb)
fs := []
for f in fb
while n % f == 0
append f to fs
n := n / f
if n == 1 return fs
return []
A factor is determined by submitting the accumulated relations to the linear algebra of the congruence of square solver.
For example, consider the factorization of 143. Choose r = 17, so r2 ≡ 3 (mod 143). Then choose r = 19, so r2 ≡ 75 ≡ 3 · 52. Those two relations can be combined as (17 · 19)2 ≡ 32 · 52 ≡ 152 (mod 143), and the two factors are gcd(17·19 − 15, 143) = 11 and gcd(17·19 + 15, 143) = 13. This sometimes fails; for instance, the relation 212 ≡ 22 (mod 143) can be combined with the relation on 19, but the two factors produced, 1 and 143, are trivial.
Thanks for very interesting question!
In pure Python I implemented from scratch Dixon Factorization Algorithm in 3 different flavors:
Using simplest sieve. I'm creating u64 array with all numbers in range [N; N * 2), which signify z^2 value. This array hold result of multiplication of prime numbers. Then through sieving process I iterate all factor base prime numbers and do array[k] *= p in those k positions that are divisible by p. Finally when sieved array is ready I check both that a) array index k is a perfect square, b) and array[k] == k - N. Second b) condition means that all multiplied p primes give final number, this is only true if number is divisible only by factor-base primes, i.e. it is B-smooth. This is simplest and most slowest out of my 3 solutions.
Second solution uses SymPy library to factorize every z^2. I iterate all possible z and do sympy.factorint(z * z), this gives factorization of z^2. If this factorization contains only small primes, i.e. from factor base, then I collect such z and z^2 for later processing. This version of algorithm is also slow, but much faster than first one.
Third solution uses a kind of sieving used in Quadratic Sieve. This sieving process is fastest of all three algorithms. Basically what it does, it finds all roots of equation x^2 = N (mod p) for all primes in factor base, as I have just few primes root finding is done through simple loop through all variants, for bigger primes one can use Shanks Tonelli algorithm of finding root, which is really fast. Only around 50% of primes give a root solution at all, hence only half of primes are actually used in Quadratic Sieve. Roots of such equation can be used to generate lots of solutions at once, because root + k * p is also a valid solution for all k. Sieving is done through array[offset(root) :: p] += Log2(p). Here instead of multiplication of first algorithm I used adding a logarithm of prime. First it is a bit faster to add a number than to multiply. Secondly, what is more important is that it supports any size of number, e.g. even 256-bit. While multiplying is possible only till 64-bit number, because Numpy has no 128 or 256 bit integers support. After logartithms are added, I check which logarithms are equal to logarithm of original z^2 number, this numbers are final sieved numbers.
After all three algorithms above have sieved all z^2 then I do Linear Algebra stage through Gaussian Elemination algorithm. This stage is meant to find such combination of B-smooth z^2 numbers which after multiplication of their prime factors give final number with all EVEN prime powers.
Lets call a Relation a triple z, z^2, prime factors of z^2. Basically all relations are given to Gaussian Elemination stage, where even combinations are found.
Even powers of prime numbers give us equality a^2 = b^2 (mod N), from where we can get a factor by doing factor = GCD(a + b, N), here GCD is Greatest Common Divisor found through Euclidean Algorithm. This GCD sometimes gives trivial factors 1 and N, in this case other even combinations should be checked.
To be 100% sure to get even combinations I do Sieving stage till I find a bit more than amount of prime numbers amount of relations, actually around 105% of amount of prime numbers. This extra 5% of relations ensure us that we certainly will get dependent linear equations in Gaussian stage. All these dependent equation form even combinations.
Actually we need a bit more dependent equations, not just 1 more than amount of primes, but around 5%-10% more, only because some (50-60% of them as I can see experimentally) dependencies give only trivial factor 1 or N. Hence extra equations are needed.
Put a look at console output at the end of my post. This console output shows all the impressions from my program. There I run in parallel (multi-threaded) both 2nd (Sieve_B) and 3rd (Sieve_C) algorithms. 1st one (Sieve_A) is not run by my program because it is so slow that you'll wait forever for it to finish.
At the very end of source file you can tweak variable bits = 64 to some other size, like bits = 96. This is amount of bits in composite number N. This N is created as a product of just two random prime numbers of equal size. Such a composite consisting of two equal in size primes is usually called RSA Number.
Also find B = 1 << 10, this tells degree of B-smoothness, basically factor base consists of all possible primes < B. You may increase this B limit, this will give more frequent answers of sieved z^2 hence whole factoring becomes much faster. The only limitation of huge size of B is Linear Algebra stage (Gaussian Elemination), because with bigger factor base you have to solve more linear equations of bigger size. And my Gauss is done not in very optimal way, for example instead of keeping bits as np.uint8 you may keep bits as dense np.uint64, this will increase Linear Algebra speed by 8x times more.
You may also find variable M = 1 << 23, which tells how large is sieving array size, in other words it is block size that is processed at once. Bigger block is a bit faster, but not much. Bigger values of M will not give much difference because it only tells what size of tasks sieving process is split into, it doesn't influence any computation power. More than that bigger M will occupy more memory, so you can't increases it infinitely, only till you have enough memory.
Besides all mentioned above algorithms I also used Fermat Primality Test, also Sieve of Eratosthenes (for generating prime factor base).
Plus also implemented my own algorithm of filtering square numbers. For this I take some composite modulus that looks close to Primorial, like mod = 2 * 2 * 2 * 3 * 3 * 5 * 7 * 11 * 13. And inside boolean array I mark all numbers modulus mod that are squares. Later when any number K should be checked if it is square or not I get flag_array[K % mod] and if it is True then number is "Possibly" squares, while if it is False then number is "Definitely" not square. Thus this filter gives false positives sometimes but never false negatives. This filter checking stage filters out 95% of non-squares, remaining 5% of possibly squares can be double-checked through math.isqrt().
Please, click below on Try it online! link, to test run my program on online server of ReplIt. This will give you best impression, especially if you have no Python or no personal laptop. My code below can be just run straight away after only PIP-installing python -m pip numpy sympy.
Try it online!
import threading
def GenPrimes_SieveOfEratosthenes(end):
import numpy as np
composites = np.zeros((end,), dtype = np.uint8)
for p in range(2, len(composites)):
if composites[p]:
continue
if p * p >= end:
break
composites[p * p :: p] = 1
primes = []
for p in range(2, len(composites)):
if not composites[p]:
primes.append(p)
return np.array(primes, dtype = np.uint32)
def Print(*pargs, __state = (threading.RLock(),), **nargs):
with __state[0]:
print(*pargs, flush = True, **nargs)
def IsSquare(n, *, state = []):
if len(state) == 0:
import numpy as np
Print('Pre-computing squares filter...')
squares_filter = 2 * 2 * 2 * 3 * 3 * 5 * 7 * 11 * 13
squares = np.zeros((squares_filter,), dtype = np.uint8)
squares[(np.arange(0, squares_filter, dtype = np.uint64) ** 2) % squares_filter] = 1
state.extend([squares_filter, squares])
if not state[1][n % state[0]]:
return False, None
import math
root = math.isqrt(n)
return root ** 2 == n, root
def FactorRef(x):
import sympy
return dict(sorted(sympy.factorint(x).items()))
def CheckZ(z, N, primes):
z2 = pow(z, 2, N)
factors = FactorRef(z2)
assert all(p <= primes[-1] for p in factors), (primes[-1], factors, N, z, z2)
return z
def SieveSimple(N, primes):
import time, math, numpy as np
Print('Simple Sieve of B-smooth z^2...')
sieve_block = 1 << 21
rep0_time = 0
for iiblock, iblock in enumerate(range(N, N * 2, sieve_block)):
if time.time() - rep0_time >= 30:
Print(f'Block {iiblock:>3} (2^{math.log2(max(iblock - N, 1)):>5.2f})')
rep0_time = time.time()
iblock_end = iblock + sieve_block
sieve_arr = np.ones((sieve_block,), dtype = np.uint64)
iblock_modN = iblock % N
for p in primes:
mp = 1
while True:
if mp * p >= sieve_block:
break
mp *= p
off = (mp - iblock_modN % mp) % mp
sieve_arr[off :: mp] *= p
for i in range(1 if iblock == N else 0, sieve_block):
num = iblock + i
z2 = num - N
if sieve_arr[i] < z2:
continue
assert sieve_arr[i] == z2, (sieve_arr[i], round(math.log2(sieve_arr[i]), 3), z2)
is_square, z = IsSquare(num)
if not is_square:
continue
#Print('z', z, 'z^2', z2)
yield CheckZ(z, N, primes)
def SieveFactor(N, primes):
import math
Print('Factor Sieve of B-smooth z^2...')
for iz, z in enumerate(range(math.isqrt(N - 1) + 1, math.isqrt(N * 2 - 1) + 1)):
z2 = z ** 2 - N
assert 0 <= z2 and z2 < N, (z, z2)
factors = FactorRef(z2)
if any(p > primes[-1] for p in factors):
continue
#Print('iz', iz, 'z', z, 'z^2', z2, 'z^2 factors', factors)
yield CheckZ(z, N, primes)
def BinarySearch(begin, end, Test):
while begin + 1 < end:
mid = (begin + end - 1) >> 1
if Test(mid):
end = mid + 1
else:
begin = mid + 1
assert begin + 1 == end and Test(begin), (begin, end, Test(begin))
return begin
def ModSqrt(n, p):
n %= p
def Ret(x):
if pow(x, 2, p) != n:
return []
nx = (p - x) % p
if x == nx:
return [x]
elif x <= nx:
return [x, nx]
else:
return [nx, x]
#if p % 4 == 3 and sympy.isprime(p):
# return Ret(pow(n, (p + 1) // 4, p))
for i in range(p):
if pow(i, 2, p) == n:
return Ret(i)
return []
def SieveQuadratic(N, primes):
import math, numpy as np
# https://en.wikipedia.org/wiki/Quadratic_sieve
# https://www.rieselprime.de/ziki/Multiple_polynomial_quadratic_sieve
M = 1 << 23
def Log2I(x):
return int(round(math.log2(max(1, x)) * (1 << 24)))
def Log2IF(li):
return li / (1 << 24)
Print('Quadratic Sieve of B-smooth z^2...')
plogs = {}
for p in primes:
plogs[int(p)] = Log2I(int(p))
qprimes = []
B = int(primes[-1]) + 1
for p in primes:
p = int(p)
res = []
mp = 1
while True:
if mp * p >= B:
break
mp *= p
roots = ModSqrt(N, mp)
if len(roots) == 0:
if mp == p:
break
continue
res.append((mp, tuple(roots)))
if len(res) > 0:
qprimes.append(res)
qprimes_lin = np.array([pinfo[0][0] for pinfo in qprimes], dtype = np.uint32)
yield qprimes_lin
Print('QSieve num primes', len(qprimes), f'({len(qprimes) * 100 / len(primes):.1f}%)')
x_begin0 = math.isqrt(N - 1) + 1
assert N <= x_begin0 ** 2
for iblock in range(1 << 30):
if (x_begin0 + (iblock + 1) * M) ** 2 - N >= N:
break
x_begin = x_begin0 + iblock * M
if iblock != 0:
Print('\n', end = '')
Print(f'Block {iblock} (2^{math.log2(max(1, x_begin ** 2 - N)):>6.2f})...')
a = np.zeros((M,), np.uint32)
for pinfo in qprimes:
p = pinfo[0][0]
plog = np.uint32(plogs[p])
for imp, (mp, roots) in enumerate(pinfo):
off_done = set()
for root in roots:
for off in range(mp):
if ((x_begin + off) ** 2 - N) % mp == 0 and off not in off_done:
break
else:
continue
a[off :: mp] += plog
off_done.add(off)
logs = np.log2(np.array((np.arange(M).astype(np.float64) + x_begin) ** 2 - N, dtype = np.float64))
logs2if = Log2IF(a.astype(np.float64))
logs_diff = np.abs(logs - logs2if)
for ix in range(M):
if logs_diff[ix] > 0.3:
continue
z = x_begin + ix
z2 = z * z - N
factors = FactorRef(z2)
assert all(p <= primes[-1] for p, c in factors.items())
#Print('iz', ix, 'z', z, 'z^2', z2, f'(2^{math.log2(max(1, z2)):>6.2f})', ', z^2 factors', factors)
yield CheckZ(z, N, primes)
def LinAlg(N, zs, primes):
import numpy as np
Print('Linear algebra...')
Print('Factoring...')
m = np.zeros((len(zs), len(primes) + len(zs)), dtype = np.uint8)
def SwapRows(i, j):
t = np.copy(m[i])
m[i][...] = m[j][...]
m[j][...] = t[...]
def MatToStr(m):
s = '\n'
for i in range(len(m)):
for j in range(len(m[i])):
s += str(m[i, j])
s += '\n'
return s[1:-1]
for iz, z in enumerate(zs):
z2 = z * z - N
fs = FactorRef(z2)
for p, c in fs.items():
i = np.searchsorted(primes, p, 'right') - 1
assert i >= 0 and i < len(primes) and primes[i] == p, (i, primes[i])
m[iz, i] = (int(m[iz, i]) + c) % 2
m[iz, len(primes) + iz] = 1
Print('Gaussian elemination...')
#Print(MatToStr(m)); Print()
one_col, one_rows = 0, 0
while True:
while True:
for i in range(one_rows, len(m)):
if m[i, one_col]:
break
else:
one_col += 1
if one_col >= len(primes):
break
continue
break
if one_col >= len(primes):
break
assert m[i, one_col]
assert np.all(m[i, :one_col] == 0)
for j in range(len(m)):
if i == j:
continue
if not m[j, one_col]:
continue
m[j][...] ^= m[i][...]
SwapRows(one_rows, i)
one_rows += 1
one_col += 1
assert np.all(m[one_rows:, :len(primes)] == 0)
zeros = m[one_rows:, len(primes):]
Print(f'Even combinations ({len(m) - one_rows}):')
Print(MatToStr(zeros))
return zeros
def ProcessResults(N, zs, la_zeros):
import math
Print('Computing final results...')
factors = []
for i in range(len(la_zeros)):
zero = la_zeros[i]
assert len(zero) == len(zs)
cz = []
for j in range(len(zero)):
if not zero[j]:
continue
z = zs[j]
z2 = z * z - N
cz.append((z, z2, FactorRef(z2)))
a = 1
for z, z2, fs in cz:
a = (a * z) % N
cnts = {}
for z, z2, fs in cz:
for p, c in fs.items():
cnts[p] = cnts.get(p, 0) + c
cnts = dict(sorted(cnts.items()))
b = 1
for p, c in cnts.items():
assert c % 2 == 0, (p, c, cnts)
b = (b * pow(p, c // 2, N)) % N
factor = math.gcd(a + b, N)
Print('a', str(a).rjust(len(str(N))), ' b', str(b).rjust(len(str(N))), ' factor', factor if factor != N else 'N')
if factor != 1 and factor != N:
factors.append(factor)
return factors
def SieveCollectResults(N, its):
import time, threading, queue, traceback, math
K = len(its)
qs = [queue.Queue() for i in range(K)]
last_dot, finish = False, False
def Get(it, ty, need, compul):
nonlocal last_dot, finish
try:
cnt = 0
for iz, z in enumerate(it):
if finish:
break
if iz < 4:
z2 = z * z - N
Print(('\n' if last_dot else '') + 'Sieve_' + ('C', 'B', 'A')[K - 1 - ty], ' iz', iz,
'z', z, 'z^2', z2, f'(2^{math.log2(max(1, z2)):>6.2f})', ', z^2 factors', FactorRef(z2))
last_dot = False
else:
Print(('.', 'b', 'a')[K - 1 - ty], end = '')
last_dot = True
qs[ty].put(z)
cnt += 1
if cnt >= need:
break
except:
Print(traceback.format_exc())
thr = []
for ty, (it, need, compul) in enumerate(its):
thr.append(threading.Thread(target = Get, args = (it, ty, need, compul), daemon = True))
thr[-1].start()
for ithr, t in enumerate(thr):
if its[ithr][2]:
t.join()
finish = True
if last_dot:
Print()
zs = [[] for i in range(K)]
for iq, q in enumerate(qs):
while not qs[iq].empty():
zs[iq].append(qs[iq].get())
return zs
def DixonFactor(N):
import time, math, numpy as np, sys
B = 1 << 10
primes = GenPrimes_SieveOfEratosthenes(B)
Print('Num primes', len(primes), 'last prime', primes[-1])
IsSquare(0)
it = SieveQuadratic(N, primes)
qprimes = next(it)
zs = SieveCollectResults(N, [
#(SieveSimple(N, primes), 3, False),
(SieveFactor(N, primes), 3, False),
(it, round(len(qprimes) * 1.06 + 0.5), True),
])[-1]
la_zeros = LinAlg(N, zs, qprimes)
fs = ProcessResults(N, zs, la_zeros)
if len(fs) > 0:
Print('Factored, factors', sorted(set(fs)))
else:
Print('Failed to factor! Try running program again...')
def IsPrime_Fermat(n, *, ntrials = 32):
import random
if n <= 16:
return n in (2, 3, 5, 7, 11, 13)
for i in range(ntrials):
if pow(random.randint(2, n - 2), n - 1, n) != 1:
return False
return True
def GenRandom(bits):
import random
return random.randrange(1 << (bits - 1), 1 << bits)
def RandPrime(bits):
while True:
n = GenRandom(bits) | 1
if IsPrime_Fermat(n):
return n
def Main():
import math
bits = 64
N = RandPrime(bits // 2) * RandPrime((bits + 1) // 2)
Print('N to factor', N, f'(2^{math.log2(N):>5.1f})')
DixonFactor(N)
if __name__ == '__main__':
Main()
Console output:
N to factor 10086068308526249063 (2^ 63.1)
Num primes 172 last prime 1021
Pre-computing squares filter...
Quadratic Sieve of B-smooth z^2...
Factor Sieve of B-smooth z^2...
QSieve num primes 78 (45.3%)
Block 0 (2^ 32.14)...
Sieve_C iz 0 z 3175858067 z^2 6153202727426 (2^ 42.48) , z^2 factors {2: 1, 29: 2, 67: 1, 191: 1, 487: 1, 587: 1}
Sieve_C iz 1 z 3175859246 z^2 13641877439453 (2^ 43.63) , z^2 factors {31: 1, 61: 1, 167: 1, 179: 1, 373: 1, 647: 1}
Sieve_C iz 2 z 3175863276 z^2 39239319203113 (2^ 45.16) , z^2 factors {31: 1, 109: 1, 163: 1, 277: 1, 311: 1, 827: 1}
Sieve_C iz 3 z 3175867115 z^2 63623612174162 (2^ 45.85) , z^2 factors {2: 1, 29: 1, 41: 1, 47: 1, 61: 1, 127: 1, 197: 1, 373: 1}
.........................................................................
Sieve_B iz 0 z 3175858067 z^2 6153202727426 (2^ 42.48) , z^2 factors {2: 1, 29: 2, 67: 1, 191: 1, 487: 1, 587: 1}
......
Linear algebra...
Factoring...
Gaussian elemination...
Even combinations (7):
01000000000000000000000000000000000000000000000000001100000000000000000000000000000
11010100000010000100100000010011100000000001001001001001011001000000110001010000000
11001011000101111100011111001011010011000111101000001001011000001111100101001110000
11010010010000110110101100110101000100001100010011100011101000100010011011001001000
00010110111010000010000010000111010001010010111001000011011011101110110001001100100
00000010111000110010100110001111010101001000011010110011101000110001101101100100010
10010001111111101100011110111110110100000110111011010001010001100000010100000100001
Computing final results...
a 9990591196683978238 b 9990591196683978238 factor 1
a 936902490212600845 b 3051457985176300292 factor 3960321451
a 1072293684177681642 b 8576178744296269655 factor 2546780213
a 1578121372922149955 b 1578121372922149955 factor 1
a 2036768191033218175 b 8049300117493030888 factor N
a 1489997751586754228 b 2231890938565281666 factor 3960321451
a 9673227070299809069 b 3412883990935144956 factor 3960321451
Factored, factors [2546780213, 3960321451]
What is the fastest way to swap two digits in a number in Python? I am given the numbers as strings, so it'd be nice if I could have something as fast as
string[j] = string[j] ^ string[j+1]
string[j+1] = string[j] ^ string[j+1]
string[j] = string[j] ^ string[j+1]
Everything I've seen has been much more expensive than it would be in C, and involves making a list and then converting the list back or some variant thereof.
This is faster than you might think, at least faster than Jon Clements' current answer in my timing test:
i, j = (i, j) if i < j else (j, i) # make sure i < j
s = s[:i] + s[j] + s[i+1:j] + s[i] + s[j+1:]
Here's my test bed should you want to compare any other answers you get:
import timeit
import types
N = 10000
R = 3
SUFFIX = '_test'
SUFFIX_LEN = len(SUFFIX)
def setup():
import random
global s, i, j
s = 'abcdefghijklmnopqrstuvwxyz'
i = random.randrange(len(s))
while True:
j = random.randrange(len(s))
if i != j: break
def swapchars_martineau(s, i, j):
i, j = (i, j) if i < j else (j, i) # make sure i < j
return s[:i] + s[j] + s[i+1:j] + s[i] + s[j+1:]
def swapchars_martineau_test():
global s, i, j
swapchars_martineau(s, i, j)
def swapchars_clements(text, fst, snd):
ba = bytearray(text)
ba[fst], ba[snd] = ba[snd], ba[fst]
return str(ba)
def swapchars_clements_test():
global s, i, j
swapchars_clements(s, i, j)
# find all the functions named *SUFFIX in the global namespace
funcs = tuple(value for id,value in globals().items()
if id.endswith(SUFFIX) and type(value) is types.FunctionType)
# run the timing tests and collect results
timings = [(f.func_name[:-SUFFIX_LEN],
min(timeit.repeat(f, setup=setup, repeat=R, number=N))
) for f in funcs]
timings.sort(key=lambda x: x[1]) # sort by speed
fastest = timings[0][1] # time fastest one took to run
longest = max(len(t[0]) for t in timings) # len of longest func name (w/o suffix)
print 'fastest to slowest *_test() function timings:\n' \
' {:,d} chars, {:,d} timeit calls, best of {:d}\n'.format(len(s), N, R)
def times_slower(speed, fastest):
return speed/fastest - 1.0
for i in timings:
print "{0:>{width}}{suffix}() : {1:.4f} ({2:.2f} times slower)".format(
i[0], i[1], times_slower(i[1], fastest), width=longest, suffix=SUFFIX)
Addendum:
For the special case of swapping digit characters in a positive decimal number given as a string, the following also works and is a tiny bit faster than the general version at the top of my answer.
The somewhat involved conversion back to a string at the end with the format() method is to deal with cases where a zero got moved to the front of the string. I present it mainly as a curiosity, since it's fairly incomprehensible unless you grasp what it does mathematically. It also doesn't handle negative numbers.
n = int(s)
len_s = len(s)
ord_0 = ord('0')
di = ord(s[i])-ord_0
dj = ord(s[j])-ord_0
pi = 10**(len_s-(i+1))
pj = 10**(len_s-(j+1))
s = '{:0{width}d}'.format(n + (dj-di)*pi + (di-dj)*pj, width=len_s)
It has to be of a mutable type of some sort, the best I can think of is (can't make any claims as to performance though):
def swapchar(text, fst, snd):
ba = bytearray(text)
ba[fst], ba[snd] = ba[snd], ba[fst]
return ba
>>> swapchar('thequickbrownfox', 3, 7)
bytearray(b'thekuicqbrownfox')
You can still utilise the result as a str/list - or explicitly convert it to a str if needs be.
>>> int1 = 2
>>> int2 = 3
>>> eval(str(int1)+str(int2))
23
I know you've already accepted an answer, so I won't bother coding it in Python, but here's how you could do it in JavaScript which also has immutable strings:
function swapchar(string, j)
{
return string.replace(RegExp("(.{" + j + "})(.)(.)"), "$1$3$2");
}
Obviously if j isn't in an appropriate range then it just returns the original string.
Given an integer n and two (zero-started) indexes i and j of digits to swap, this can be done using powers of ten to locate the digits, division and modulo operations to extract them, and subtraction and addition to perform the swap.
def swapDigits(n, i, j):
# These powers of 10 encode the locations i and j in n.
power_i = 10 ** i
power_j = 10 ** j
# Retrieve digits [i] and [j] from n.
digit_i = (n // power_i) % 10
digit_j = (n // power_j) % 10
# Remove digits [i] and [j] from n.
n -= digit_i * power_i
n -= digit_j * power_j
# Insert digit [i] in position [j] and vice versa.
n += digit_i * power_j
n += digit_j * power_i
return n
For example:
>>> swapDigits(9876543210, 4, 0)
9876503214
>>> swapDigits(9876543210, 7, 2)
9826543710
I found an implementation of the thomas algorithm or TDMA in MATLAB.
function x = TDMAsolver(a,b,c,d)
%a, b, c are the column vectors for the compressed tridiagonal matrix, d is the right vector
n = length(b); % n is the number of rows
% Modify the first-row coefficients
c(1) = c(1) / b(1); % Division by zero risk.
d(1) = d(1) / b(1); % Division by zero would imply a singular matrix.
for i = 2:n-1
temp = b(i) - a(i) * c(i-1);
c(i) = c(i) / temp;
d(i) = (d(i) - a(i) * d(i-1))/temp;
end
d(n) = (d(n) - a(n) * d(n-1))/( b(n) - a(n) * c(n-1));
% Now back substitute.
x(n) = d(n);
for i = n-1:-1:1
x(i) = d(i) - c(i) * x(i + 1);
end
end
I need it in python using numpy arrays, here my first attempt at the algorithm in python.
import numpy
aa = (0.,8.,9.,3.,4.)
bb = (4.,5.,9.,4.,7.)
cc = (9.,4.,5.,7.,0.)
dd = (8.,4.,5.,9.,6.)
ary = numpy.array
a = ary(aa)
b = ary(bb)
c = ary(cc)
d = ary(dd)
n = len(b)## n is the number of rows
## Modify the first-row coefficients
c[0] = c[0]/ b[0] ## risk of Division by zero.
d[0] = d[0]/ b[0]
for i in range(1,n,1):
temp = b[i] - a[i] * c[i-1]
c[i] = c[i]/temp
d[i] = (d[i] - a[i] * d[i-1])/temp
d[-1] = (d[-1] - a[-1] * d[-2])/( b[-1] - a[-1] * c[-2])
## Now back substitute.
x = numpy.zeros(5)
x[-1] = d[-1]
for i in range(-2, -n-1, -1):
x[i] = d[i] - c[i] * x[i + 1]
They give different results, so what am I doing wrong?
I made this since none of the online implementations for python actually work. I've tested it against built-in matrix inversion and the results match.
Here a = Lower Diag, b = Main Diag, c = Upper Diag, d = solution vector
import numpy as np
def TDMA(a,b,c,d):
n = len(d)
w= np.zeros(n-1,float)
g= np.zeros(n, float)
p = np.zeros(n,float)
w[0] = c[0]/b[0]
g[0] = d[0]/b[0]
for i in range(1,n-1):
w[i] = c[i]/(b[i] - a[i-1]*w[i-1])
for i in range(1,n):
g[i] = (d[i] - a[i-1]*g[i-1])/(b[i] - a[i-1]*w[i-1])
p[n-1] = g[n-1]
for i in range(n-1,0,-1):
p[i-1] = g[i-1] - w[i-1]*p[i]
return p
For an easy performance boost for large matrices, use numba! This code outperforms np.linalg.inv() in my tests:
import numpy as np
from numba import jit
#jit
def TDMA(a,b,c,d):
n = len(d)
w= np.zeros(n-1,float)
g= np.zeros(n, float)
p = np.zeros(n,float)
w[0] = c[0]/b[0]
g[0] = d[0]/b[0]
for i in range(1,n-1):
w[i] = c[i]/(b[i] - a[i-1]*w[i-1])
for i in range(1,n):
g[i] = (d[i] - a[i-1]*g[i-1])/(b[i] - a[i-1]*w[i-1])
p[n-1] = g[n-1]
for i in range(n-1,0,-1):
p[i-1] = g[i-1] - w[i-1]*p[i]
return p
There's at least one difference between the two:
for i in range(1,n,1):
in Python iterates from index 1 to the last index n-1, while
for i = 2:n-1
iterates from index 1 (zero-based) to the last-1 index, since Matlab has one-based indexing.
In your loop, the Matlab version iterates over the second through second-to last elements. To do the same in Python, you want:
for i in range(1,n-1):
(As noted in voithos's comment, this is because the range function excludes the last index, so you need to correct for this in addition to the change to 0 indexing).
Writing somthing like this in python is going to be really slow. You would be much better off using LAPACK to do the numerical heavy lifting and use python for everything around it. LAPACK is compiled so it will run much faster than python it is also much more higly optimised than it is feasible for most of us to match.
SciPY provides low level wrappers for LAPACK so that you can call it from python very simply, the one you are looking for can be found here:
https://docs.scipy.org/doc/scipy/reference/generated/scipy.linalg.lapack.dgtsv.html#scipy.linalg.lapack.dgtsv