Algorithms



Hi Luc,
True. The most important thing he should do is limit the testing to sqrt(N).
To factorise a number N, you need to test up to sqrt(N), and there are approximately
sqrt(N)/log(sqrt(N)) primes to test. For example, to factorise 1,000,000 you only need to test the 168 primes less than 1000, or 16.8% of the numbers less than 1000.
Using the simple algorithm you test 100% of the numbers less than sqrt(N):
if you eliminate the multiples of 2 you are down to testing about 50%
then eliminating the multiples of 3 drops you to testing 33.3%
then eliminating the multiples of 5 drops you to testing 26.7%
and so on.
the first few are probably worthwhile as you point out, but you then rapidly get in to diminishing returns, and may easily get to the stage where it is not worth the effort to save the time of the line
if(InputCase%i==0)
I don't think multiple threads are worthwhile until you go past 32 bit arithmetic (even the simplest algorithm only has 2^16 tests to do).
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."
modified on Sunday, June 29, 2008 10:10 PM





cp9876 wrote: The most important thing he should do is limit the testing to sqrt(N).
I believe this is not correct, assume you have an x composed from 2p where p is a prime, in this case it is clear that in certain circumstance sqrt(x) can be less than p (eg: 15838 = 7919 * 2, where 7919 > sqrt(15838)), for primal testing the sqrt(n) bound is ok, however its not suitable for the upper bound in this problem.





Arash Partow wrote: I believe this is not correct, assume you have an x composed from 2p where p is a prime, in this case it is clear that in certain circumstance sqrt(x) can be less than p (eg: 15838 = 7919 * 2, where 7919 > sqrt(15838)), for primal testing the sqrt(n) bound is ok, however its not suitable for the upper bound in this problem.
He's studying for the ACM competition  I don't want to spoon feed him! He can limit his testing to sqrt(N) but he has to modify his algorithm slightly for when InputCase is not 1 at the end.
In the example you give, he only needs to test until floor(sqrt(15838))=125, he will have only found 2 as a factor and InputCase will be 7919 after the 124 tests. He can then conclude the the remaining value in InputCase is the remaining prime factor. This is much more efficient than testing the remaining 15,703 factors.
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."





cp9876 wrote: This is much more efficient than testing the remaining 15,703 factors.
true.





you're right, as long as one divides by the current number until the remainder is nonzero they will have taken out all of the multipleoffactors as well, no need to explicitly only test/divide by primes.
good catch!





hi
can anyone tell me how the vale is calculated using prob function of ms excel
it is used like this:
PROB
Returns the probability that values in a range are between two limits. If upper_limit is not supplied, returns the probability that values in x_range are equal to lower_limit.
Syntax
PROB(x_range,prob_range,lower_limit,upper_limit)
X_range is the range of numeric values of x with which there are associated probabilities.
Prob_range is a set of probabilities associated with values in x_range.
Lower_limit is the lower bound on the value for which you want a probability.
Upper_limit is the optional upper bound on the value for which you want a probability.
Remarks
If any value in prob_range = 0 or if any value in prob_range > 1, PROB returns the #NUM! error value.
If the sum of the values in prob_range ¹ 1, PROB returns the #NUM! error value.
If upper_limit is omitted, PROB returns the probability of being equal to lower_limit.
If x_range and prob_range contain a different number of data points, PROB returns the #N/A error value





Is it possible to get the rightmost NONZERO digit of a factorial without calculating for it?
Say 1000! = 2
Regards,
Ian
*NO, this is not a homework.
It is said that the most complex structures built by mankind are software systems. This is not generally appreciated because most people cannot see them. Maybe that's a good thing because if we saw them as buildings, we'd deem many of them unsafe.





I assume you mean without calculating the factorial but some simpler, faster calculation? You can rewrite a factorial as a polynomial in n but I'm not sure if that gives you an easy pattern for the last digit. And by definition, all factorials except 1! and 0! are even. Does this have a practical application or are you just curious?
If you don't have the data, you're just another a**hole with an opinion.





Nope. This is part of the question set my adviser gave me to prepare for a certain competition.
It is said that the most complex structures built by mankind are software systems. This is not generally appreciated because most people cannot see them. Maybe that's a good thing because if we saw them as buildings, we'd deem many of them unsafe.





Ian Uy wrote: It is said that the most complex structures built by mankind are software systems.
I've heard that bandied about but I think I'll disagree. They usually quote Vista at 50 million lines of code yet the latest generation of processor chips boast over 2 billion transistors. And you could argue that much of what gets counted in an operating system are simply utilities that are applications running on the base system so the core is smaller still.
If you don't have the data, you're just another a**hole with an opinion.







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