Related
I saw this YouTube video online where this guy finds the largest prime factor of a number using a seemingly simple approach, but I don't quite understand the math of it. Here's the link https://m.youtube.com/watch?v=5kv9q7qgvlI
The code -
n=1234 # Some number whose largest factor I want to find
i=2 # It seems that he tries to start from the smallest prime number
while i**2<n: # I don't understand this part where he checks if the square of variable is less than the target number
while n%i==0: # I know this checks if n is divisible by i
n/=i
i+=1 # increments i's value
print(n)
I know that this code works, but why? I get the last two lines, but why is it necessary to check if the square of variable i is less than n?
If your number N is divisible by I (in other terms, I being a factor of N), and I is greater than sqrt(N), then there must be another factor of N, call it J = N/I, being smaller than sqrt(N).
If you exhaustively searched all potential factors up to I, you must have already found J, so you covered that factorization earlier.
We are looking for the largest prime factor, so the question remains whether the final N when terminating the loop is a prime number.
Whenever we encountered a factor of N, we divided N by this factor, so when considering a potential factor I, we can be sure that the current N won't have any factors below I any more.
Can the final N when terminating the loop be composed from more than one factor (i.e. not being prime)?
No.
If N isn't prime, it must be composed from K and L (N = K * L), and K and L must be both bigger than sqrt(N), otherwise we would have divided by K or L in an earlier step. But multiplying two numbers, each bigger than sqrt(N), will always be bigger than N, violating N = K * L.
So, assuming that the final N wasn't prime, runs into a contradiction, and thus we can be sure that the final N is a factor of the original N, and that this factor is indeed a prime number.
Caveats with the original code (thanks JohanC):
The original code checks for I being strictly less than SQRT(N) (while i**2<n). That will miss cases like N=9 where its square root 3 must be included in the iteration. So, this should better read while i**2<=n.
And the code risks some inaccuracies:
Using floating-point division (/= operator instead of //=) might give inexact results. This applies to Python, while in languages like Java, /= would be fine.
In Python, raising an integer to the power of 2 (while i**2<=n) seems to guarantee exact integer arithmetic, so that's ok in the Python context. In languages like Java, I'd recommend not to use the pow() function, as that typically uses floating-point arithmetic with the risk of inexact results, but to simply write i*i.
The problem is:
Given a range of numbers (x,y) , Find all the prime numbers(Count only) which are sum of the squares of two numbers, with the restriction that 0<=x<y<=2*(10^8)
According to Fermat's theorem :
Fermat's theorem on sums of two squares asserts that an odd prime number p can be
expressed as p = x^2 + y^2 with integer x and y if and only if p is congruent to
1 (mod4).
I have done something like this:
import math
def is_prime(n):
if n % 2 == 0 and n > 2:
return False
return all(n % i for i in range(3, int(math.sqrt(n)) + 1, 2))
a,b=map(int,raw_input().split())
count=0
for i in range(a,b+1):
if(is_prime(i) and (i-1)%4==0):
count+=1
print(count)
But this increases the time complexity and memory limit in some cases.
Here is my submission result:
Can anyone help me reduce the Time Complexity and Memory limit with better algorithm?
Problem Link(Not an ongoing contest FYI)
Do not check whether each number is prime. Precompute all the prime numbers in the range, using Sieve of Eratosthenes. This will greatly reduce the complexity.
Since you have maximum of 200M numbers and 256Mb memory limit and need at least 4 bytes per number, you need a little hack. Do not initialize the sieve with all numbers up to y, but only with numbers that are not divisible by 2, 3 and 5. That will reduce the initial size of the sieve enough to fit into the memory limit.
UPD As correctly pointed out by Will Ness in comments, sieve contains only flags, not numbers, thus it requires not more than 1 byte per element and you don't even need this precomputing hack.
You can reduce your memory usage by changing for i in range(a,b+1): to for i in xrange(a,b+1):, so that you are not generating an entire list in memory.
You can do the same thing inside the statement below, but you are right that it does not help with time.
return all(n % i for i in xrange(3, int(math.sqrt(n)) + 1, 2))
One time optimization that might not cost as much in terms of memory as the other answer is to use Fermat's Little Theorem. It may help you reject many candidates early.
More specifically, you could pick maybe 3 or 4 random values to test and if one of them rejects, then you can reject. Otherwise you can do the test you are currently doing.
First of all, although it will not change the order of your time-complexity, you can still narrow down the list of numbers that you are checking by a factor of 6, since you only need to check numbers that are either equal to 1 mod 12 or equal to 5 mod 12 (such as [1,5], [13,17], [25,29], [37,41], etc).
Since you only need to count the primes which are sum of squares of two numbers, the order doesn't matter. Therefore, you can change range(a,b+1) to range(1,b+1,12)+range(5,b+1,12).
Obviously, you can then remove the if n % 2 == 0 and n > 2 condition in function is_prime, and in addition, change the if is_prime(i) and (i-1)%4 == 0 condition to if is_prime(i).
And finally, you can check the primality of each number by dividing it only with numbers that are adjacent to multiples of 6 (such as [5,7], [11,13], [17,19], [23,25], etc).
So you can change this:
range(3,int(math.sqrt(n))+1,2)
To this:
range(5,math.sqrt(n))+1,6)+range(7,math.sqrt(n))+1,6)
And you might as well calculate math.sqrt(n))+1 beforehand.
To summarize all this, here is how you can improve the overall performance of your program:
import math
def is_prime(n):
max = int(math.sqrt(n))+1
return all(n % i for i in range(5,max,6)+range(7,max,6))
count = 0
b = int(raw_input())
for i in range(1,b+1,12)+range(5,b+1,12):
if is_prime(i):
count += 1
print count
Please note that 1 is typically not regarded as prime, so you might want to print count-1 instead. On the other hand, 2 is not equal to 1 mod 4, yet it is the sum of two squares, so you may leave it as is...
The following below is an algorithm that finds the prime factorization for a given number N. I'm wondering if there are any ways to make this faster using HUGE numbers. I'm talking like 20-35 digit numbers. I wanna try and get these to go as fast as possible. Any ideas?
import time
def prime_factors(n):
"""Returns all the prime factors of a positive integer"""
factors = []
divisor = 2
while n > 1:
while n % divisor == 0:
factors.append(divisor)
n /= divisor
divisor = divisor + 1
if divisor*divisor > n:
if n > 1:
factors.append(n)
break
return factors
#HUGE NUMBERS GO IN HERE!
start_time = time.time()
my_factors = prime_factors(15227063669158801)
end_time = time.time()
print my_factors
print "It took ", end_time-start_time, " seconds."
Your algorithm is trial division, which has time complexity O(sqrt(n)). You can improve your algorithm by using only 2 and the odd numbers as trial divisors, or even better by using only prime numbers as trial divisors, but the time complexity will remain O(sqrt(n)).
To go faster you need a better algorithm. Try this:
def factor(n, c):
f = lambda(x): (x*x+c) % n
t, h, d = 2, 2, 1
while d == 1:
t = f(t); h = f(f(h)); d = gcd(t-h, n)
if d == n:
return factor(n, c+1)
return d
To call it on your number, say
print factor(15227063669158801, 1)
That returns the (possibly composite) factor 2090327 virtually instantly. It uses an algorithm called the rho algorithm, invented by John Pollard in 1975. The rho algorithm has time complexity O(sqrt(sqrt(n))), so it's much faster than trial division.
There are many other algorithms for factoring integers. For numbers in the 20 to 35 digit range that interests you, the elliptic curve algorithm is well-suited. It should factor numbers of that size in no more than a few seconds. Another algorithm that is well-suited to such numbers, especially those that are semi-primes (have exactly two prime factors), is SQUFOF.
If you're interested in programming with prime numbers, I modestly recommend this essay on my blog. When you're finished with that, other entries on my blog talk about elliptic curve factorization, and SQUFOF, and various other even more-powerful methods of factoring ever-larger integers.
For example, list all prime factorization for a number 100.
Check 2 is one of factorizations or not. And then, 2 < 2*c <= 100 could be removed. Ex, 4, 6, 8, ... 98
Check 3 is one of factorizations or not. And then, 3 < 2*d <= 100 could be removed. Ex, 9, 12, ... 99
4 is removed from possible set.
Check 5, And then, 10, 15, 20, ..., 100 are removed.
6 is removed.
Check 7, ....
....
It seems like there is no check for divisors. Sorry if I am wrong but how do you know if divisor is prime or not? Your divisor variable is increasing by 1 after each loop so I assume it will generate a lot of composite numbers.
No optimizations to that algorithm will allow you to factor 35 digit numbers at least in the general case. The reason is that the number of primes up to 35 digits are too high to be listed in a reasonable amount of time let alone attempt to divide by each one. Even if one was inclined to try, the number of bits required to store them would be far too much as well. In this case you'll want to select a different algorithm from the list of general purpose factorization algorithms.
However, if all the prime factors are small enough (say below 10^12 or so), then you could use a segmented Sieve of Eratosthenes, or simply find a list of primes up to some practical number (say 10^12 or so) online and use that instead of trying to calculate the primes and hope the list is large enough.
num = input ()
fact = 0
while fact != num:
fact = fact + 1
rem = num % fact
if rem == 0:
print fact
You only need to go to the square root of the input number to get all the factors (not as far as half the number, as suggested elsewhere). For example, 24 has factors 1, 2, 3, 4, 6, 8, 12, 24. sqrt(24) is approx 4.9. Check 1 and also get 24, check 2 and also get 12, check 3 and also get 8, check 4 and also get 6. Since 5 > 4.9, no need to check it. (Yes, I know 24 isn't the best example as all whole numbers less than sqrt(24) are factors of 24.)
factors = set()
for i in xrange(math.floor(math.sqrt(x))+1):
if x % i == 0:
factors.add(i)
factors.add(x/i)
print factors
There are some really complicated ways to do better for large numbers, but this should get you a decent runtime improvement. Depending on your application, caching could also save you a lot of time.
Use for loops, for starters. Then, let Python increment for you, and get rid of the unnecessary rem variable. This code does exactly the same as your code, except in a "Pythonic" way.
num = input()
for x in xrange(1, num):
if (num % x) == 0:
print fact
xrange(x, y) returns a generator for all integers from x up to, but not including y.
So that prints out all the factors of a number? The first obvious optimisation is that you could quit when fact*2 is greater than num. Anything greater than half of num can't be a factor. That's half the work thrown out instantly.
The second is that you'd be better searching for the prime factorisation and deriving all the possible factors from that. There are a bunch of really smart algorithms for that sort of thing.
Once you get halfway there (once fact>num/2), your not going to discover any new numbers as the numbers above num/2 can be discovered by calculating num/fact for each one (this can also be used to easily print each number with its pair).
The following code should cust the time down by a few seconds on every calculation and cut it in half where num is odd. Hopefully you can follow it, if not, ask.
I'll add more if I think of something later.
def even(num):
'''Even numbers can be divided by odd numbers, so test them all'''
fact=0
while fact<num/2:
fact+=1
rem=num % fact
if rem == 0:
print '%s and %s'%(fact,num/fact)
def odd(num):
'''Odd numbers can't be divided by even numbers, so why try?'''
fact=-1
while fact<num/2:
fact+=2
rem=num % fact
if rem == 0:
print '%s and %s'%(fact,num/fact)
while True:
num=input(':')
if str(num)[-1] in '13579':
odd(num)
else:
even(num)
Research integer factorization methods.
Unfortunately in Python, the divmod operation is implemented as a built-in function. Despite hardware integer division often producing the quotient and the remainder simultaneously, no non-assembly language that I'm aware of has implemented a /% or //% basic operator.
So: the following is a better brute-force algorithm if you count machine operations. It gets all factors in O(sqrt(N)) time without having to calculate sqrt(N) -- look, Mum, no floating point!
# even number
fact = 0
while 1:
fact += 1
fact2, rem = divmod(num, fact)
if not rem:
yield fact
yield fact2
if fact >= fact2 - 1:
# fact >= math.sqrt(num)
break
Yes. Use a quantum computer
Shor's algorithm, named after mathematician Peter Shor, is a quantum
algorithm (an algorithm which runs on a quantum computer) for integer
factorization formulated in 1994. Informally it solves the following
problem: Given an integer N, find its prime factors.
On a quantum computer, to factor an integer N, Shor's algorithm runs
in polynomial time (the time taken is polynomial in log N, which is
the size of the input). Specifically it takes time O((log N)3),
demonstrating that the integer factorization problem can be
efficiently solved on a quantum computer and is thus in the complexity
class BQP. This is exponentially faster than the most efficient known
classical factoring algorithm, the general number field sieve, which
works in sub-exponential time — about O(e1.9 (log N)1/3 (log log
N)2/3). The efficiency of Shor's algorithm is due to the efficiency
of the quantum Fourier transform, and modular exponentiation by
squarings.
I am trying to solve a problem involving printing the product of all divisors of a given number. The number of test cases is a number 1 <= t <= 300000 , and the number itself can range from 1 <= n <= 500000
I wrote the following code, but it always exceeds the time limit of 2 seconds. Are there any ways to speed up the code ?
from math import sqrt
def divisorsProduct(n):
ProductOfDivisors=1
for i in range(2,int(round(sqrt(n)))+1):
if n%i==0:
ProductOfDivisors*=i
if n/i != i:
ProductOfDivisors*=(n/i)
if ProductOfDivisors <= 9999:
print ProductOfDivisors
else:
result = str(ProductOfDivisors)
print result[len(result)-4:]
T = int(raw_input())
for i in range(1,T+1):
num = int(raw_input())
divisorsProduct(num)
Thank You.
You need to clarify by what you mean by "product of divisors." The code posted in the question doesn't work for any definition yet. This sounds like a homework question. If it is, then perhaps your instructor was expecting you to think outside the code to meet the time goals.
If you mean the product of unique prime divisors, e.g., 72 gives 2*3 = 6, then having a list of primes is the way to go. Just run through the list up to the square root of the number, multiplying present primes into the result. There are not that many, so you could even hard code them into your program.
If you mean the product of all the divisors, prime or not, then it is helpful to think of what the divisors are. You can make serious speed gains over the brute force method suggested in the other answers and yours. I suspect this is what your instructor intended.
If the divisors are ordered in a list, then they occur in pairs that multiply to n -- 1 and n, 2 and n/2, etc. -- except for the case where n is a perfect square, where the square root is a divisor that is not paired with any other.
So the result will be n to the power of half the number of divisors, (regardless of whether or not n is a square).
To compute this, find the prime factorization using your list of primes. That is, find the power of 2 that divides n, then the power of 3, etc. To do this, take out all the 2s, then the 3s, etc.
The number you are taking the factors out of will be getting smaller, so you can do the square root test on the smaller intermediate numbers to see if you need to continue up the list of primes. To gain some speed, test p*p <= m, rather than p <= sqrt(m)
Once you have the prime factorization, it is easy to find the number of divisors. For example, suppose the factorization is 2^i * 3^j * 7^k. Then, since each divisor uses the same prime factors, with exponents less than or equal to those in n including the possibility of 0, the number of divisors is (i+1)(j+1)(k+1).
E.g., 72 = 2^3 * 3^2, so the number of divisors is 4*3 = 12, and their product is 72^6 = 139,314,069,504.
By using math, the algorithm can become much better than O(n). But it is hard to estimate your speed gains ahead of time because of the relatively small size of the n in the input.
You could eliminate the if statement in the loop by only looping to less than the square root, and check for square root integer-ness outside the loop.
It is a rather strange question you pose. I have a hard time imagine a use for it, other than it possibly being an assignment in a course. My first thought was to pre-compute a list of primes and only test against those, but I assume you are quite deliberately counting non-prime factors? I.e., if the number has factors 2 and 3, you are also counting 6.
If you do use a table of pre-computed primes, you would then have to also subsequently include all possible combinations of primes in your result, which gets more complex.
C is really a great language for that sort of thing, because even suboptimal algorithms run really fast.
Okay, I think this is close to the optimal algorithm. It produces the product_of_divisors for each number in range(500000).
import math
def number_of_divisors(maxval=500001):
""" Example: the number of divisors of 12 is 6: 1, 2, 3, 4, 6, 12.
Given a prime factoring of n, the number of divisors of n is the
product of each factor's multiplicity plus one (mpo in my variables).
This function works like the Sieve of Eratosthenes, but marks each
composite n with the multiplicity (plus one) of each prime factor. """
numdivs = [1] * maxval # multiplicative identity
currmpo = [0] * maxval
# standard logic for 2 < p < sqrt(maxval)
for p in range(2, int(math.sqrt(maxval))):
if numdivs[p] == 1: # if p is prime
for exp in range(2,50): # assume maxval < 2^50
pexp = p ** exp
if pexp > maxval:
break
exppo = exp + 1
for comp in range(pexp, maxval, pexp):
currmpo[comp] = exppo
for comp in range(p, maxval, p):
thismpo = currmpo[comp] or 2
numdivs[comp] *= thismpo
currmpo[comp] = 0 # reset currmpo array in place
# abbreviated logic for p > sqrt(maxval)
for p in range(int(math.sqrt(maxval)), maxval):
if numdivs[p] == 1: # if p is prime
for comp in range(p, maxval, p):
numdivs[comp] *= 2
return numdivs
# this initialization times at 7s on my machine
NUMDIV = number_of_divisors()
def product_of_divisors(n):
if NUMDIV[n] % 2 == 0:
# each pair of divisors has product equal to n, for example
# 1*12 * 2*6 * 3*4 = 12**3
return n ** (NUMDIV[n] / 2)
else:
# perfect squares have their square root as an unmatched divisor
return n ** (NUMDIV[n] / 2) * int(math.sqrt(n))
# this loop times at 13s on my machine
for n in range(500000):
a = product_of_divisors(n)
On my very slow machine, it takes 7s to compute the numberofdivisors for each number, then 13s to compute the productofdivisors for each. Of course it can be sped up by translating it into C. (#someone with a fast machine: how long does it take on your machine?)