maybe you would have an idea on how to solve the following problem.
John decided to buy his son Johnny some mathematical toys. One of his most favorite toy is blocks of different colors. John has decided to buy blocks of C different colors. For each color he will buy googol (10^100) blocks. All blocks of same color are of same length. But blocks of different color may vary in length.
Jhonny has decided to use these blocks to make a large 1 x n block. He wonders how many ways he can do this. Two ways are considered different if there is a position where the color differs. The example shows a red block of size 5, blue block of size 3 and green block of size 3. It shows there are 12 ways of making a large block of length 11.
Each test case starts with an integer 1 ≤ C ≤ 100. Next line consists c integers. ith integer 1 ≤ leni ≤ 750 denotes length of ith color. Next line is positive integer N ≤ 10^15.
This problem should be solved in 20 seconds for T <= 25 test cases. The answer should be calculated MOD 100000007 (prime number).
It can be deduced to matrix exponentiation problem, which can be solved relatively efficiently in O(N^2.376*log(max(leni))) using Coppersmith-Winograd algorithm and fast exponentiation. But it seems that a more efficient algorithm is required, as Coppersmith-Winograd implies a large constant factor. Do you have any other ideas? It can possibly be a Number Theory or Divide and Conquer problem
Firstly note the number of blocks of each colour you have is a complete red herring, since 10^100 > N always. So the number of blocks of each colour is practically infinite.
Now notice that at each position, p (if there is a valid configuration, that leaves no spaces, etc.) There must block of a color, c. There are len[c] ways for this block to lie, so that it still lies over this position, p.
My idea is to try all possible colors and positions at a fixed position (N/2 since it halves the range), and then for each case, there are b cells before this fixed coloured block and a after this fixed colour block. So if we define a function ways(i) that returns the number of ways to tile i cells (with ways(0)=1). Then the number of ways to tile a number of cells with a fixed colour block at a position is ways(b)*ways(a). Adding up all possible configurations yields the answer for ways(i).
Now I chose the fixed position to be N/2 since that halves the range and you can halve a range at most ceil(log(N)) times. Now since you are moving a block about N/2 you will have to calculate from N/2-750 to N/2-750, where 750 is the max length a block can have. So you will have to calculate about 750*ceil(log(N)) (a bit more because of the variance) lengths to get the final answer.
So in order to get good performance you have to through in memoisation, since this inherently a recursive algorithm.
So using Python(since I was lazy and didn't want to write a big number class):
T = int(raw_input())
for case in xrange(T):
#read in the data
C = int(raw_input())
lengths = map(int, raw_input().split())
minlength = min(lengths)
n = int(raw_input())
#setup memoisation, note all lengths less than the minimum length are
#set to 0 as the algorithm needs this
memoise = {}
memoise[0] = 1
for length in xrange(1, minlength):
memoise[length] = 0
def solve(n):
global memoise
if n in memoise:
return memoise[n]
ans = 0
for i in xrange(C):
if lengths[i] > n:
continue
if lengths[i] == n:
ans += 1
ans %= 100000007
continue
for j in xrange(0, lengths[i]):
b = n/2-lengths[i]+j
a = n-(n/2+j)
if b < 0 or a < 0:
continue
ans += solve(b)*solve(a)
ans %= 100000007
memoise[n] = ans
return memoise[n]
solve(n)
print "Case %d: %d" % (case+1, memoise[n])
Note I haven't exhaustively tested this, but I'm quite sure it will meet the 20 second time limit, if you translated this algorithm to C++ or somesuch.
EDIT: Running a test with N = 10^15 and a block with length 750 I get that memoise contains about 60000 elements which means non-lookup bit of solve(n) is called about the same number of time.
A word of caution: In the case c=2, len1=1, len2=2, the answer will be the N'th Fibonacci number, and the Fibonacci numbers grow (approximately) exponentially with a growth factor of the golden ratio, phi ~ 1.61803399. For the
huge value N=10^15, the answer will be about phi^(10^15), an enormous number. The answer will have storage
requirements on the order of (ln(phi^(10^15))/ln(2)) / (8 * 2^40) ~ 79 terabytes. Since you can't even access 79
terabytes in 20 seconds, it's unlikely you can meet the speed requirements in this special case.
Your best hope occurs when C is not too large, and leni is large for all i. In such cases, the answer will
still grow exponentially with N, but the growth factor may be much smaller.
I recommend that you first construct the integer matrix M which will compute the (i+1,..., i+k)
terms in your sequence based on the (i, ..., i+k-1) terms. (only row k+1 of this matrix is interesting).
Compute the first k entries "by hand", then calculate M^(10^15) based on the repeated squaring
trick, and apply it to terms (0...k-1).
The (integer) entries of the matrix will grow exponentially, perhaps too fast to handle. If this is the case, do the
very same calculation, but modulo p, for several moderate-sized prime numbers p. This will allow you to obtain
your answer modulo p, for various p, without using a matrix of bigints. After using enough primes so that you know their product
is larger than your answer, you can use the so-called "Chinese remainder theorem" to recover
your answer from your mod-p answers.
I'd like to build on the earlier #JPvdMerwe solution with some improvements. In his answer, #JPvdMerwe uses a Dynamic Programming / memoisation approach, which I agree is the way to go on this problem. Dividing the problem recursively into two smaller problems and remembering previously computed results is quite efficient.
I'd like to suggest several improvements that would speed things up even further:
Instead of going over all the ways the block in the middle can be positioned, you only need to go over the first half, and multiply the solution by 2. This is because the second half of the cases are symmetrical. For odd-length blocks you would still need to take the centered position as a seperate case.
In general, iterative implementations can be several magnitudes faster than recursive ones. This is because a recursive implementation incurs bookkeeping overhead for each function call. It can be a challenge to convert a solution to its iterative cousin, but it is usually possible. The #JPvdMerwe solution can be made iterative by using a stack to store intermediate values.
Modulo operations are expensive, as are multiplications to a lesser extent. The number of multiplications and modulos can be decreased by approximately a factor C=100 by switching the color-loop with the position-loop. This allows you to add the return values of several calls to solve() before doing a multiplication and modulo.
A good way to test the performance of a solution is with a pathological case. The following could be especially daunting: length 10^15, C=100, prime block sizes.
Hope this helps.
In the above answer
ans += 1
ans %= 100000007
could be much faster without general modulo :
ans += 1
if ans == 100000007 then ans = 0
Please see TopCoder thread for a solution. No one was close enough to find the answer in this thread.
Related
So, in CLRS, there's this quote
A prime not too close to an exact power of 2 is often a good choice for m.
Several Questions...
I understand how a power of 2 will just be the lower order bits of your key...however, say you have keys from a universe of 1 to 1 million, with each key having an equal probability of being any number from universe (which I'm guessing is a common assumption about your universe if given no other data?) then wouldn't taking say the 4 lower order bits result in (2^4) lower order bit patterns that were pretty much equally likely for the keys from 1 to 1 million? How am I thinking about this incorrectly?
Why a prime number? So, if power of 2's aren't a good idea, why is a prime number a better choice as opposed to a composite number close to a power of 2 (Also why should it be close to a power of 2...lol)?
You are trying to find a hash table that works well for typical input data, and typical input data does things that you wouldn't expect from good random number generators. Very often you get formatted or semi-formatted strings which, when converted to numbers, end up as K, K+A, K+2A, K+3A,.... for some integers K and A. If K+xA and K+yA hash to the same number mod m, then (x-y)A must be 0 mod m. If m is prime, this can only happen if A = 0 mod m or if x = y mod m, so one time in m. But if m=pq and A happens to be divisible by p, then you get a collision every time x-y is divisible by q, which is more often since q < m.
I guess close to a power of 2 because it might be convenient for the memory management system to have blocks of memory of the resulting size - I really don't know. If you really care, and if you have the time, you could try different primes with some representative data and see which of them are best in practice.
I am looking to enumerate a random permutation of the numbers 1..N in fixed space. This means that I cannot store all numbers in a list. The reason for that is that N can be very large, more than available memory. I still want to be able to walk through such a permutation of numbers one at a time, visiting each number exactly once.
I know this can be done for certain N: Many random number generators cycle through their whole state space randomly, but entirely. A good random number generator with state size of 32 bit will emit a permutation of the numbers 0..(2^32)-1. Every number exactly once.
I want to get to pick N to be any number at all and not be constrained to powers of 2 for example. Is there an algorithm for this?
The easiest way is probably to just create a full-range PRNG for a larger range than you care about, and when it generates a number larger than you want, just throw it away and get the next one.
Another possibility that's pretty much a variation of the same would be to use a linear feedback shift register (LFSR) to generate the numbers in the first place. This has a couple of advantages: first of all, an LFSR is probably a bit faster than most PRNGs. Second, it is (I believe) a bit easier to engineer an LFSR that produces numbers close to the range you want, and still be sure it cycles through the numbers in its range in (pseudo)random order, without any repetitions.
Without spending a lot of time on the details, the math behind LFSRs has been studied quite thoroughly. Producing one that runs through all the numbers in its range without repetition simply requires choosing a set of "taps" that correspond to an irreducible polynomial. If you don't want to search for that yourself, it's pretty easy to find tables of known ones for almost any reasonable size (e.g., doing a quick look, the wikipedia article lists them for size up to 19 bits).
If memory serves, there's at least one irreducible polynomial of ever possible bit size. That translates to the fact that in the worst case you can create a generator that has roughly twice the range you need, so on average you're throwing away (roughly) every other number you generate. Given the speed an LFSR, I'd guess you can do that and still maintain quite acceptable speed.
One way to do it would be
Find a prime p larger than N, preferably not much larger.
Find a primitive root of unity g modulo p, that is, a number 1 < g < p such that g^k ≡ 1 (mod p) if and only if k is a multiple of p-1.
Go through g^k (mod p) for k = 1, 2, ..., ignoring the values that are larger than N.
For every prime p, there are φ(p-1) primitive roots of unity, so it works. However, it may take a while to find one. Finding a suitable prime is much easier in general.
For finding a primitive root, I know nothing substantially better than trial and error, but one can increase the probability of a fast find by choosing the prime p appropriately.
Since the number of primitive roots is φ(p-1), if one randomly chooses r in the range from 1 to p-1, the expected number of tries until one finds a primitive root is (p-1)/φ(p-1), hence one should choose p so that φ(p-1) is relatively large, that means that p-1 must have few distinct prime divisors (and preferably only large ones, except for the factor 2).
Instead of randomly choosing, one can also try in sequence whether 2, 3, 5, 6, 7, 10, ... is a primitive root, of course skipping perfect powers (or not, they are in general quickly eliminated), that should not affect the number of tries needed greatly.
So it boils down to checking whether a number x is a primitive root modulo p. If p-1 = q^a * r^b * s^c * ... with distinct primes q, r, s, ..., x is a primitive root if and only if
x^((p-1)/q) % p != 1
x^((p-1)/r) % p != 1
x^((p-1)/s) % p != 1
...
thus one needs a decent modular exponentiation (exponentiation by repeated squaring lends itself well for that, reducing by the modulus on each step). And a good method to find the prime factor decomposition of p-1. Note, however, that even naive trial division would be only O(√p), while the generation of the permutation is Θ(p), so it's not paramount that the factorisation is optimal.
Another way to do this is with a block cipher; see this blog post for details.
The blog posts links to the paper Ciphers with Arbitrary Finite Domains which contains a bunch of solutions.
Consider the prime 3. To fully express all possible outputs, think of it this way...
bias + step mod prime
The bias is just an offset bias. step is an accumulator (if it's 1 for example, it would just be 0, 1, 2 in sequence, while 2 would result in 0, 2, 4) and prime is the prime number we want to generate the permutations against.
For example. A simple sequence of 0, 1, 2 would be...
0 + 0 mod 3 = 0
0 + 1 mod 3 = 1
0 + 2 mod 3 = 2
Modifying a couple of those variables for a second, we'll take bias of 1 and step of 2 (just for illustration)...
1 + 2 mod 3 = 0
1 + 4 mod 3 = 2
1 + 6 mod 3 = 1
You'll note that we produced an entirely different sequence. No number within the set repeats itself and all numbers are represented (it's bijective). Each unique combination of offset and bias will result in one of prime! possible permutations of the set. In the case of a prime of 3 you'll see that there are 6 different possible permuations:
0,1,2
0,2,1
1,0,2
1,2,0
2,0,1
2,1,0
If you do the math on the variables above you'll not that it results in the same information requirements...
1/3! = 1/6 = 1.66..
... vs...
1/3 (bias) * 1/2 (step) => 1/6 = 1.66..
Restrictions are simple, bias must be within 0..P-1 and step must be within 1..P-1 (I have been functionally just been using 0..P-2 and adding 1 on arithmetic in my own work). Other than that, it works with all prime numbers no matter how large and will permutate all possible unique sets of them without the need for memory beyond a couple of integers (each technically requiring slightly less bits than the prime itself).
Note carefully that this generator is not meant to be used to generate sets that are not prime in number. It's entirely possible to do so, but not recommended for security sensitive purposes as it would introduce a timing attack.
That said, if you would like to use this method to generate a set sequence that is not a prime, you have two choices.
First (and the simplest/cheapest), pick the prime number just larger than the set size you're looking for and have your generator simply discard anything that doesn't belong. Once more, danger, this is a very bad idea if this is a security sensitive application.
Second (by far the most complicated and costly), you can recognize that all numbers are composed of prime numbers and create multiple generators that then produce a product for each element in the set. In other words, an n of 6 would involve all possible prime generators that could match 6 (in this case, 2 and 3), multiplied in sequence. This is both expensive (although mathematically more elegant) as well as also introducing a timing attack so it's even less recommended.
Lastly, if you need a generator for bias and or step... why don't you use another of the same family :). Suddenly you're extremely close to creating true simple-random-samples (which is not easy usually).
The fundamental weakness of LCGs (x=(x*m+c)%b style generators) is useful here.
If the generator is properly formed then x%f is also a repeating sequence of all values lower than f (provided f if a factor of b).
Since bis usually a power of 2 this means that you can take a 32-bit generator and reduce it to an n-bit generator by masking off the top bits and it will have the same full-range property.
This means that you can reduce the number of discard values to be fewer than N by choosing an appropriate mask.
Unfortunately LCG Is a poor generator for exactly the same reason as given above.
Also, this has exactly the same weakness as I noted in a comment on #JerryCoffin's answer. It will always produce the same sequence and the only thing the seed controls is where to start in that sequence.
Here's some SageMath code that should generate a random permutation the way Daniel Fischer suggested:
def random_safe_prime(lbound):
while True:
q = random_prime(lbound, lbound=lbound // 2)
p = 2 * q + 1
if is_prime(p):
return p, q
def random_permutation(n):
p, q = random_safe_prime(n + 2)
while True:
r = randint(2, p - 1)
if pow(r, 2, p) != 1 and pow(r, q, p) != 1:
i = 1
while True:
x = pow(r, i, p)
if x == 1:
return
if 0 <= x - 2 < n:
yield x - 2
i += 1
I'm pretty sure that this is the right site for this question, but feel free to move it to some other stackexchange site if it fits there better.
Suppose you have a sum of fractions a1/d1 + a2/d2 + … + an/dn. You want to compute a common numerator and denominator, i.e., rewrite it as p/q. We have the formula
p = a1*d2*…*dn + d1*a2*d3*…*dn + … + d1*d2*…d(n-1)*an
q = d1*d2*…*dn.
What is the most efficient way to compute these things, in particular, p? You can see that if you compute it naïvely, i.e., using the formula I gave above, you compute a lot of redundant things. For example, you will compute d1*d2 n-1 times.
My first thought was to iteratively compute d1*d2, d1*d2*d3, … and dn*d(n-1), dn*d(n-1)*d(n-2), … but even this is inefficient, because you will end up computing multiplications in the "middle" twice (e.g., if n is large enough, you will compute d3*d4 twice).
I'm sure this problem could be expressed somehow using maybe some graph theory or combinatorics, but I haven't studied enough of that stuff to have a good feel for it.
And one note: I don't care about cancelation, just the most efficient way to multiply things.
UPDATE:
I should have known that people on stackoverflow would be assuming that these were numbers, but I've been so used to my use case that I forgot to mention this.
We cannot just "divide" out an from each term. The use case here is a symbolic system. Actually, I am trying to fix a function called .as_numer_denom() in the SymPy computer algebra system which presently computes this the naïve way. See the corresponding SymPy issue.
Dividing out things has some problems, which I would like to avoid. First, there is no guarantee that things will cancel. This is because mathematically, (a*b)**n != a**n*b**n in general (if a and b are positive it holds, but e.g., if a == b ==-1 and n == 1/2, you get (a*b)**n == 1**(1/2) == 1 but (-1)**(1/2)*(-1)**(1/2) == I*I == -1). So I don't think it's a good idea to assume that dividing by an will cancel it in the expression (this may be actually be unfounded, I'd need to check what the code does).
Second, I'd like to also apply a this algorithm to computing the sum of rational functions. In this case, the terms would automatically be multiplied together into a single polynomial, and "dividing" out each an would involve applying the polynomial division algorithm. You can see in this case, you really do want to compute the most efficient multiplication in the first place.
UPDATE 2:
I think my fears for cancelation of symbolic terms may be unfounded. SymPy does not cancel things like x**n*x**(m - n) automatically, but I think that any exponents that would combine through multiplication would also combine through division, so powers should be canceling.
There is an issue with constants automatically distributing across additions, like:
In [13]: 2*(x + y)*z*(S(1)/2)
Out[13]:
z⋅(2⋅x + 2⋅y)
─────────────
2
But this is first a bug and second could never be a problem (I think) because 1/2 would be split into 1 and 2 by the algorithm that gets the numerator and denominator of each term.
Nonetheless, I still want to know how to do this without "dividing out" di from each term, so that I can have an efficient algorithm for summing rational functions.
Instead of adding up n quotients in one go I would use pairwise addition of quotients.
If things cancel out in partial sums then the numbers or polynomials stay smaller, which makes computation faster.
You avoid the problem of computing the same product multiple times.
You could try to order the additions in a certain way, to make canceling more likely (maybe add quotients with small denominators first?), but I don't know if this would be worthwhile.
If you start from scratch this is simpler to implement, though I'm not sure it fits as a replacement of the problematic routine in SymPy.
Edit: To make it more explicit, I propose to compute a1/d1 + a2/d2 + … + an/dn as (…(a1/d1 + a2/d2) + … ) + an/dn.
Compute two new arrays:
The first contains partial multiples to the left: l[0] = 1, l[i] = l[i-1] * d[i]
The second contains partial multiples to the right: r[n-1] = 1, r[i] = d[i] * r[i+1]
In both cases, 1 is the multiplicative identity of whatever ring you are working in.
Then each of your terms on the top, t[i] = l[i-1] * a[i] * r[i+1]
This assumes multiplication is associative, but it need not be commutative.
As a first optimization, you don't actually have to create r as an array: you can do a first pass to calculate all the l values, and accumulate the r values during a second (backward) pass to calculate the summands. No need to actually store the r values since you use each one once, in order.
In your question you say that this computes d3*d4 twice, but it doesn't. It does multiply two different values by d4 (one a right-multiplication and the other a left-multiplication), but that's not exactly a repeated operation. Anyway, the total number of multiplications is about 4*n, vs. 2*n multiplications and n divisions for the other approach that doesn't work in non-commutative multiplication or non-field rings.
If you want to compute p in the above expression, one way to do this would be to multiply together all of the denominators (in O(n), where n is the number of fractions), letting this value be D. Then, iterate across all of the fractions and for each fraction with numerator ai and denominator di, compute ai * D / di. This last term is equal to the product of the numerator of the fraction and all of the denominators other than its own. Each of these terms can be computed in O(1) time (assuming you're using hardware multiplication, otherwise it might take longer), and you can sum them all up in O(n) time.
This gives an O(n)-time algorithm for computing the numerator and denominator of the new fraction.
It was also pointed out to me that you could manually sift out common denominators and combine those trivially without multiplication.
Lets say I have a large set of data .
Then I can divide it into two find mean of those two and calculate the mean of the last 2 values I get.
a) Is this the mean of the original big quantity ?
b) Can I do this sort of method for calculating standard deviation ??
a) only if the sets you divide into are always the same size, meaning that the original set size must be a power of 2.
For example, the mean of {6} is 6, and the mean of {3,6} is 4.5, but the mean of {3,6,6} is not 5.25, it's 5.
Certainly you could recursively divide into parts to calculate the sum, though, and divide by the total size at the end. Not sure if that does you any good.
b) no
For example, the s.d of {2} is 0, and the s.d. of {1} is 0, but the s.d of {1,2} is not 0.
Once you've calculated the mean of the whole set, you can recursively divide to calculate the sum square deviation from the mean, and as with the mean calculation, divide by the total size and take square root at the end. [Edit: in fact all you need to calculate s.d is the sumsquare, the sum, and the count. Forgot about that. So you don't have to calculate the mean first]
It is incorrect, but if you can express the mean and standard deviation of a set from the means, standard deviations, and size of the sets which that set is divided into.
Specifically, if m_x, s_x and n_x are the means, standard deviations, and sizes of x, and X is partitioned into many x's, then
n_X = sum_x(n_x)
m_X = sum_x(n_x m_x)/n_X
s_X^2 = (sum_x(n_x(s_x^2 + m_x^2)) - m_X)/n_X
assuming the standard deviation is of the form sum(x - mean(x))/n; if it is the sample unbiased estimator, just adjust the weights accordingly.
Sure you can. No need for equal sets, power of two. Pseudo code:
N1,mean1,s1;
N2,mean2,s2;
N12,mean12,s12;
N12 = N1+N2;
mean12 = ((mean1*N1) + (mean2*N2)) / N12;
s12 = sqrt( (s1*s1*N1 + s2*s2*N2) / N12 + N1*N2/(N12*N12)*(s1-s2)*(s1-s2) );
http://en.wikipedia.org/wiki/Weighted_mean
http://en.wikipedia.org/wiki/Standard_deviation#Combining_standard_deviations
On (a) - it's only precisely correct if you precisely divided the set into two. If there were an odd number of items, for instance, there is a slight weighting toward the smaller "half". The larger the set, the less significant the problem. However, the problem recurs for the smaller sets as you subdivide. You get very large error when dividing a set of three items into a single item and a pair - each item in the pair is only half as significant to the final result as the single item.
I don't see the gain, though. You still do as many additions. You even end up doing more divisions. More importantly, you access memory in a non-sequential order, leading to poor cache performance.
The usual approach for a mean and standard deviation is to first calculate the sum of all items, and the sum of the squares - both in the same loop. Old calculators used to handle this with running totals, also keeping count of the number of items as they went. At the end, those three values (n, sum-of-x and sum-of-x-squared) are all you need - the rest is just substitution into the standard formulae for the mean and standard deviation.
EDIT
If you're dead set on using recursion for this, look up "tail recursion". Mathematically, tail recursion and iteration are equivalent - different representations of the same thing. In implementation terms tail recursion might cause a stack overflow where iteration would work, but (1) some languages guarantee this will not happen (e.g. Scheme, Haskell), and (2) many compilers will handle this as an optimisation anyway (e.g. GCC for C or C++).
I need to find out the prime factors of over 300 billion. I have a function that is adding to the list of them...very slowly! It has been running for about an hour now and i think its got a fair distance to go still. Am i doing it completly wrong or is this expected?
Edit: Im trying to find the largest prime factor of the number 600851475143.
Edit:
Result:
{
List<Int64> ListOfPrimeFactors = new List<Int64>();
Int64 Number = 600851475143;
Int64 DividingNumber = 2;
while (DividingNumber < Number / DividingNumber)
{
if (Number % DividingNumber == 0)
{
ListOfPrimeFactors.Add(DividingNumber);
Number = Number/DividingNumber;
}
else
DividingNumber++;
}
ListOfPrimeFactors.Add(Number);
listBox1.DataSource = ListOfPrimeFactors;
}
}
Are you remembering to divide the number that you're factorizing by each factor as you find them?
Say, for example, you find that 2 is a factor. You can add that to your list of factors, but then you divide the number that you're trying to factorise by that value.
Now you're only searching for the factors of 150 billion. Each time around you should start from the factor you just found. So if 2 was a factor, test 2 again. If the next factor you find is 3, there's no point testing from 2 again.
And so on...
Finding prime factors is difficult using brute force, which sounds like the technique you are using.
Here are a few tips to speed it up somewhat:
Start low, not high
Don't bother testing each potential factor to see whether it is prime--just test LIKELY prime numbers (odd numbers that end in 1,3,7 or 9)
Don't bother testing even numbers (all divisible by 2), or odds that end in 5 (all divisible by 5). Of course, don't actually skip 2 and 5!!
When you find a prime factor, make sure to divide it out--don't continue to use your massive original number. See my example below.
If you find a factor, make sure to test it AGAIN to see if it is in there multiple times. Your number could be 2x2x3x7x7x7x31 or something like that.
Stop when you reach >= sqrt(remaining large number)
Edit: A simple example:
You are finding the factors of 275.
Test 275 for divisibility by 2. Does 275/2 = int(275/2)? No. Failed.
Test 275 for divisibility by 3. Failed.
Skip 4!
Test 275 for divisibility by 5. YES! 275/5 = 55. So your NEW test number is now 55.
Test 55 for divisibility by 5. YES! 55/5 = 11. So your NEW test number is now 11.
BUT 5 > sqrt (11), so 11 is prime, and you can stop!
So 275 = 5 * 5 * 11
Make more sense?
Factoring big numbers is a hard problem. So hard, in fact, that we rely on it to keep RSA secure. But take a look at the wikipedia page for some pointers to algorithms that can help. But for a number that small, it really shouldn't be taking that long, unless you are re-doing work over and over again that you don't have to somewhere.
For the brute-force solution, remember that you can do some mini-optimizations:
Check 2 specially, then only check odd numbers.
You only ever need to check up to the square-root of the number (if you find no factors by then, then the number is prime).
Once you find a factor, don't use the original number to find the next factor, divide it by the found factor, and search the new smaller number.
When you find a factor, divide it through as many times as you can. After that, you never need to check that number, or any smaller numbers again.
If you do all the above, each new factor you find will be prime, since any smaller factors have already been removed.
Here is an XSLT solution!
This XSLT transformation takes 0.109 sec.
<xsl:stylesheet version="2.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xs="http://www.w3.org/2001/XMLSchema"
xmlns:saxon="http://saxon.sf.net/"
xmlns:f="http://fxsl.sf.net/"
exclude-result-prefixes="xs saxon f"
>
<xsl:import href="../f/func-Primes.xsl"/>
<xsl:output method="text"/>
<xsl:template name="initial" match="/*">
<xsl:sequence select="f:maxPrimeFactor(600851475143)"/>
</xsl:template>
<xsl:function name="f:maxPrimeFactor" as="xs:integer">
<xsl:param name="pNum" as="xs:integer"/>
<xsl:sequence select=
"if(f:isPrime($pNum))
then $pNum
else
for $vEnd in xs:integer(floor(f:sqrt($pNum, 0.1E0))),
$vDiv1 in (2 to $vEnd)[$pNum mod . = 0][1],
$vDiv2 in $pNum idiv $vDiv1
return
max((f:maxPrimeFactor($vDiv1),f:maxPrimeFactor($vDiv2)))
"/>
</xsl:function>
</xsl:stylesheet>
This transformation produces the correct result (the maximum prime factor of 600851475143) in just 0.109 sec.:
6857
The transformation uses the f:sqrt() and f:isPrime() defined in FXSL 2.0 -- a library for functional programming in XSLT. FXSL is itself written entirely in XSLT.
f:isPrime() uses Fermat's little theorem so that it is efficient to determine primeality.
One last thing nobody has mentioned, perhaps because it seems obvious. Every time you find a factor and divide it out, keep trying the factor until it fails.
64 only has one prime factor, 2. You will find that out pretty trivially if you keep dividing out the 2 until you can't anymore.
$ time factor 300000000000 > /dev/null
real 0m0.027s
user 0m0.000s
sys 0m0.001s
You're doing something wrong if it's taking an hour. You might even have an infinite loop somewhere - make sure you're not using 32-bit ints.
The key to understanding why the square root is important, consider that each factor of n below the square root of n has a corresponding factor above it. To see this, consider that if x is factor of n, then x/n = m which means that x/m = n, hence m is also a factor.
I wouldn't expect it to take very long at all - that's not a particularly large number.
Could you give us an example number which is causing your code difficulties?
Here's one site where you can get answers: Factoris - Online factorization service. It can do really big numbers, but it also can factorize algebraic expressions.
The fastest algorithms are sieve algorithms, and are based on arcane areas of discrete mathematics (over my head at least), complicated to implement and test.
The simplest algorithm for factoring is probably (as others have said) the Sieve of Eratosthenes. Things to remember about using this to factor a number N:
general idea: you're checking an increasing sequence of possible integer factors x to see if they evenly divide your candidate number N (in C/Java/Javascript check whether N % x == 0) in which case N is not prime.
you just need to go up to sqrt(N), but don't actually calculate sqrt(N): loop as long as your test factor x passes the test x*x<N
if you have the memory to save a bunch of previous primes, use only those as the test factors (and don't save them if prime P fails the test P*P > N_max since you'll never use them again
Even if you don't save the previous primes, for possible factors x just check 2 and all the odd numbers. Yes, it will take longer, but not that much longer for reasonable sized numbers. The prime-counting function and its approximations can tell you what fraction of numbers are prime; this fraction decreases very slowly. Even for 264 = approx 1.8x1019, roughly one out of every 43 numbers is prime (= one out of every 21.5 odd numbers is prime). For factors of numbers less than 264, those factors x are less than 232 where about one out of every 20 numbers is prime = one out of every 10 odd numbers is prime. So you'll have to test 10 times as many numbers, but the loop should be a bit faster and you don't have to mess around with storing all those primes.
There are also some older and simpler sieve algorithms that a little bit more complex but still fairly understandable. See Dixon's, Shanks' and Fermat's factoring algorithms. I read an article about one of these once, can't remember which one, but they're all fairly straightforward and use algebraic properties of the differences of squares.
If you're just testing whether a number N is prime, and you don't actually care about the factors themselves, use a probabilistic primality test. Miller-Rabin is the most standard one, I think.
I spent some time on this since it just sucked me in. I won't paste the code here just yet. Instead see this factors.py gist if you're curious.
Mind you, I didn't know anything about factoring (still don't) before reading this question. It's just a Python implementation of BradC's answer above.
On my MacBook it takes 0.002 secs to factor the number mentioned in the question (600851475143).
There must obviously be much, much faster ways of doing this. My program takes 19 secs to compute the factors of 6008514751431331. But the Factoris service just spits out the answer in no-time.
The specific number is 300425737571? It trivially factors into 131 * 151 * 673 * 22567.
I don't see what all the fuss is...
Here's some Haskell goodness for you guys :)
primeFactors n = factor n primes
where factor n (p:ps) | p*p > n = [n]
| n `mod` p /= 0 = factor n ps
| otherwise = p : factor (n `div` p) (p:ps)
primes = 2 : filter ((==1) . length . primeFactors) [3,5..]
Took it about .5 seconds to find them, so I'd call that a success.
You could use the sieve of Eratosthenes to find the primes and see if your number is divisible by those you find.
You only need to check it's remainder mod(n) where n is a prime <= sqrt(N) where N is the number you are trying to factor. It really shouldn't take over an hour, even on a really slow computer or a TI-85.
Your algorithm must be FUBAR. This only takes about 0.1s on my 1.6 GHz netbook in Python. Python isn't known for its blazing speed. It does, however, have arbitrary precision integers...
import math
import operator
def factor(n):
"""Given the number n, to factor yield a it's prime factors.
factor(1) yields one result: 1. Negative n is not supported."""
M = math.sqrt(n) # no factors larger than M
p = 2 # candidate factor to test
while p <= M: # keep looking until pointless
d, m = divmod(n, p)
if m == 0:
yield p # p is a prime factor
n = d # divide n accordingly
M = math.sqrt(n) # and adjust M
else:
p += 1 # p didn't pan out, try the next candidate
yield n # whatever's left in n is a prime factor
def test_factor(n):
f = factor(n)
n2 = reduce(operator.mul, f)
assert n2 == n
def example():
n = 600851475143
f = list(factor(n))
assert reduce(operator.mul, f) == n
print n, "=", "*".join(str(p) for p in f)
example()
# output:
# 600851475143 = 71*839*1471*6857
(This code seems to work in defiance of the fact that I don't know enough about number theory to fill a thimble.)
Just to expand/improve slightly on the "only test odd numbers that don't end in 5" suggestions...
All primes greater than 3 are either one more or one less than a multiple of 6 (6x + 1 or 6x - 1 for integer values of x).
It shouldn't take that long, even with a relatively naive brute force. For that specific number, I can factor it in my head in about one second.
You say you don't want solutions(?), but here's your "subtle" hint. The only prime factors of the number are the lowest three primes.
Semi-prime numbers of that size are used for encryption, so I am curious as to what you exactly want to use them for.
That aside, there currently are not good ways to find the prime factorization of large numbers in a relatively small amount of time.