On my IT lessons i got this task:
Write function where user puts integer m (m>1) and non-negative n and
the function returns std::pow(n, 1.0/m). You cant use any library.
Which means, i have to use pow function... but i can't use it.
Since its not 2^2 or 2^7, i cant use simple for loop.
I've tried doing anything like adding, multiplication, nothing worked anywhere near.
Any ideas?
FYI:
It's not my homework, teacher just gave me this task to learn something.
std::pow(n, 1.0/m) is the m-th root of n. You can find an helpful algorithm of how to implement it on wikipedia:
Make an initial guess x0
Set x = 1.0/m * [ (n-1)*x + n/pow(x,m-1) ]1
Repeat step 2 until x is only changed a little or maxiter has been reached.
1) Obviously, you'll need to define pow(x,y) for integer y, which is trivial.
Related
So, I've got to write a recursive function which will count the number of "1"'s are in an any given number. For this part, I managed to create the function which converts a decimal to binary :
if dec = 0 then 0
else dec mod 2 + 10 *f(dec/2)
;;
but I've got no idea how to make the program check each digit and even count the wanted ones.
And I have to write a function which has to calculate the sum of the 1/(n!) series. Again, I tried for hours, but the best I could do was :
if n <= 1. then 1.
else 1./.(n*.e(n-.1.)) +. (e(n-.1.))
;;
which doesn't work the right way cause I guess the formula isn't right.
Can somebody help me? ;-; Please, I want to understand the way this works.
For the first question, you probably want to generate a list of bits in a first time rather than an integer. It should make further bit manipulation easier to handle.
For the second question, you should write a separate factorial function first and use it in your definition of e. Currently, you are mixing the sum computation and the factorial computation.
I need to print the number of coprime pairs (a,b), 0 < a <= b <= n and for n = 10^8 the program should run in less than 10 seconds. I have used this method : http://mathworld.wolfram.com/CarefreeCouple.html
But the program isn't as fast as I expected.
I have heard about an effient way of solving this problem by using something called 'Farey Sequence' but the code was written in PHP and I can only understand C.
So which method can help me solve the problem? thanks for the time.
Your stated interest is in co-prime pairs (a, b). The carefree couple adds an additional restriction that a is square-free. Therefore it is not the same problem, though some of the math is similar. As I understand your problem it is equivalent to summing the Euler totient function from 1 to n, the so-called Totient Summatory Function.
I do not know of any tricks that give one a simple closed form solution to come up with the answer. However, I think modifying a straightforward Sieve of Eratosthenes (SoT) should get you an answer in much less than 10 seconds in most programming languages.
Normally running the SoT simply yields a list of the primes <= n. Our goal will change, however, to computing the complete prime-power factorization of each integer between 1 and n inclusive. To do that we must store more than a single bit of information for each sieve entry, we must store a list. As we sieve a prime p through the array, we add (p, 1) to the list already stored at that entry. Then we sieve by p2 and change the (p,1) entries in each location we hit to (p,2), and so one for each power of p <= n and every p <= n. When it finishes, you
can compute the Totient function quickly for every value 0 <= x <= n and sum them up.
EDIT:
I see that there is already a much better set of answers to the question on math.stackexchange.com here. I'll leave this answer up for awhile until the disposition of the question is settled.
Introduction
Let’s assume that I need the Jacobian matrix for the following set of ODE:
dxdt[ 0 ] = -90.0 * x[0] - 50.0 * x[1];
dxdt[ 1 ] = x[0] + 3*x[1];
dxdt[ 2 ] = x[1] + 50*x[2];
In Matlab/Octave this would be pretty easy:
syms x0 x1 x2;
f = [-90.0*x0-50.0*x1, x0+3*x1, x1+50*x2]
v=[x0, x1, x2]
fp = jacobian(f,v)
This would results with following output matrix:
[-90 -50 0 ]
[ 1 3 0 ]
[ 0 1 50]
What I need
Now I want to reproduce the same results in C++. I can’t compute the Jacobian before and hard-code it, as it will depend for example on user inputs and time. So my question is: How to do this? Usually for mathematics operations, I use the Boost library, however in this case I can’t find any solution. There’s only short note about this in implicit systems, but the following code doesn’t work:
sys.second( x , jacobi , t )
It also requests the time (t), so it probably doesn’t generate an analytic form of solution. Do I misunderstand the documentations? Or should I use another function? I would prefer to stay within Boost, as I need the Jacobian as ublas::matrix and I want to avoid conversion.
EDIT:
More specific I will use Jacobian inside rosenbrock4 ODE solver. Example here - lines 47-52. I need automatic generation of this structure as the ODE set may be changed later and I want to avoid manually rewriting Jacobian ever time. Also some variables inside ODE definitions are not constant in time.
I know this is long after the fact, but I have recently been wanting to do the same thing and have come across many auto differentiation (AD) libraries that do this pretty well. I have mostly been using Eigen's AD because I am already using Eigen everywhere. Here's an example of how you can use Eigen's AD to get the jacobian like you asked.
There's also a long list of c++ AD libraries on autodiff.org.
Hope this helps someone!
The Jacobian is based on derivatives of the function. If the function f is only known at run-time (and there are no constraints such as linearity), you have to automatise the differentiation. If you want this to happen exactly (as opposed to a numerical estimation), you need to use symbolic computation. Look for example here and here for libraries supporting this.
Note that the Jacobian usually depends on the state and the time, so it’s impossible to represent it as a constant matrix (such as in your example), unless your problem is so boring that you can solve it analytically anyway.
I'm having a trouble with my math:
Assume that we have a function: F(x,y) = P; And my question is: what would be the most efficient way of counting up suitable (x,y) plots for this function ? It means that I don't need the coordinates themself, but I need a number of them. P is in a range: [0 ; 10^14]. "x" and "y" are integers. Is it solved using bruteforce or there are some advanced tricks(math / programming language(C,C++)) to solve this fast enough ?
To be more concrete, the function is: x*y - ((x+y)/2) + 1.
x*y - ((x+y)/2) + 1 == P is equivalent to (2x-1)(2y-1) == (4P-3).
So, you're basically looking for the number of factorizations of 4P-3. How to factor a number in C or C++ is probably a different question, but each factorization yields a solution to the original equation. [Edit: in fact two solutions, since if A*B == C then of course (-A)*(-B) == C also].
As far as the programming languages C and C++ are concerned, just make sure you use a type that's big enough to contain 4 * 10^14. int won't do, so try long long.
You have a two-parameter function and want to solve it for a given constant.
This is a pretty big field in mathematics, and there are probably dozens of algorithms of solving your equation. One key idea that many use is the fact that if you find a point where F<P and then a point F>P, then somewhere between these two points, F must equal P.
One of the most basic algorithms for finding a root (or zero, which you of course can convert to by taking F'=F-P) is Newton's method. I suggest you start with that and read your way up to more advanced algorithms. This is a farily large field of study, so happy reading!
Wikipedia has a list of root-finding algorithms that you can use as a starting place.
I understand this is a classic programming problem and therefore I want to be clear I'm not looking for code as a solution, but would appreciate a push in the right direction. I'm learning C++ and as part of the learning process I'm attempting some programming problems. I'm attempting to write a program which deals with numbers up to factorial of 1billion. Obviously these are going to be enormous numbers and way too big to be dealing with using normal arithmetic operations. Any indication as to what direction I should go in trying to solve this type of problem would be appreciated.
I'd rather try to solve this without using additional libraries if possible
Thanks
PS - the problem is here http://www.codechef.com/problems/FCTRL
Here's the method I used to solve the problem, this was achieved by reading the comments below:
Solution -- The number 5 is a prime factor of any number ending in zero. Therefore, dividing the factorial number by 5, recursively, and adding the quotients, you get the number of trailing zeros in the factorial result
E.G. - Number of trailing zeros in 126! = 31
126/5 = 25 remainder 1
25/5 = 5 remainder 0
5/5 = 1 remainder 0
25 + 5 + 1 = 31
This works for any value, just keep dividing until the quotient is less
than 5
Skimmed this question, not sure if I really got it right but here's a deductive guess:
First question - how do you get a zero on the end of the number? By multiplying by 10.
How do you multiply by 10? either by multiplying by either a 10 or by 2 x 5...
So, for X! how many 10s and 2x5s do you have...?
(luckily 2 & 5 are prime numbers)
edit: Here's another hint - I don't think you need to do any multiplication. Let me know if you need another hint.
Hint: you may not need to calculate N! in order to find the number of zeros at the end of N!
To solve this question, as Chris Johnson said you have to look at number of 0's.
The factors of 10 will be 1,2,5,10 itself. So, you can go through each of the numbers of N! and write them in terms of 2^x * 5^y * 10^z. Discard other factors of the numbers.
Now the answer will be greaterof(x,y)+z.
One interesting thing I learn from this question is, its always better to store factorial of a number in terms of prime factors for easy comparisons.
To actually x^y, there is an easy method used in RSA algorithm, which don't remember. I will try to update the post if I find one.
This isn't a good answer to your question as you've modified it a bit from what I originally read. But I will leave it here anyway to demonstrate the impracticality of actually trying to do the calculations by main brute force.
One billion factorial is going to be out of reach of any bignum library. Such numbers will require more space to represent than almost anybody has in RAM. You are going to have to start paging the numbers in from storage as you work on them. There are ways to do this. The guy who recently calculated π out to 2700 billion places used such a library
Do not use the naive method. If you need to calculate the factorial, use a fast algorithm: http://www.luschny.de/math/factorial/FastFactorialFunctions.htm
I think that you should come up with a way to solve the problem in pseudo code before you begin to think about C++ or any other language for that matter. The nature of the question as some have pointed out is more of an algorithm problem than a C++ problem. Those who suggest searching for some obscure library are pointing you in the direction of a slippery slope, because learning to program is learning how to think, right? Find a good algorithm analysis text and it will serve you well. In our department we teach from the CLRS text.
You need a "big number" package - either one you use or one you write yourself.
I'd recommend doing some research into "large number algorithms". You'll want to implement the C++ equivalent of Java's BigDecimal.
Another way to look at it is using the gamma function. You don't need to multiply all those values to get the right answer.
To start you off, you should store the number in some sort of array like a std::vector (a digit for each position in the array) and you need to find a certain algorithm that will calculate a factorial (maybe in some sort of specialized class). ;)
//SIMPLE FUNCTION TO COMPUTE THE FACTORIAL OF A NUMBER
//THIS ONLY WORKS UPTO N = 65
//CAN YOU SUGGEST HOW WE CAN IMPROVE IT TO COMPUTE FACTORIAL OF 400 PLEASE?
#include <iostream>
#include <cmath>
using namespace std;
int factorial(int x); //function to compute factorial described below
int main()
{
int N; //= 150; //you can also get this as user input using cin.
cout<<"Enter intenger\n";
cin>>N;
factorial(N);
return 0;
}//end of main
int factorial(int x) //function to compute the factorial
{
int i, n;
long long unsigned results = 1;
for (i = 1; i<=x; i++)
{
results = results * i;
}
cout<<"Factorial of "<<x<<" is "<<results<<endl;
return results;
}