The answer gives the following code for computing floor(sqrt(x)) using just integers. Is it possible to use/modify it to return ceil(sqrt(x)) instead? Alternatively, what is the preferred way to calculate such value?
Edit: Thank you all so far and I apologise, I should have make it more explicit: I was hoping there is more "natural" way of doing this that using floor(sqrt(x)), possibly plus one. The floor version uses Newton's method to approach the root from above, I thought that maybe approaching it from below or similar would do the trick.
For example the answer even provides how to round to nearest integer: just input 4*x to the algorithm.
If x is an exact square, the ceiling and the floor of the square root are equal; otherwise, the ceiling is one more than the square root. So you could use (in Python),
result = floorsqrt(x)
if result * result != x:
result += 1
Modifying the code you linked to is not a good idea, since that code uses some properties of the Newton-Raphson method of calculating the square root. Much theory has been developed about that method, and the code uses that theory. The code I show is not as neat as modifying your linked code but it is safer and probably quicker than making a change in the code.
You can use this fact that:
floor(x) = (ceil(x) - 1) if x \not \in Z else ceil(x)
Hence, check if N is in the form 2^k, the code is the same, and if it is not, you can -1 the result of the current code.
Related
In traditional Simplex Algorithm notation, we have x at the current basis selection B as so:
xB = AB-1b - AB-1ANxN. How can I compute the AB-1AN term inside a separator in SCIP, or at least iterate over its columns?
I see three helpful methods: getLPColsData, getLPRowsData, getLPBasisInd. I'm just not sure exactly what data those methods represent, particularly the last one, with its negative row indexes. How do I use those to get the value I want?
Do those methods return the same data no matter what LP algorithm is used? Or do I need to account for dual vs primal? How does the use of the "revised" algorithm play into my calculation?
Update: I discovered the getLPBInvARow and getLPBInvRow. That seems to be much closer to what I'm after. I don't yet understand their results; they seem to include more/less dimensions than expected. I'm still looking for understanding at how to use them to get the rays away from the corner.
you are correct that getLPBInvRow or getLPBInvARow are the methods you want. getLPBInvARow directly returns you a of the simplex tableau, but it is not more efficient to use than getLPBInvRow and doing the multiplication yourself since the LP solver needs to also compute the actual tableau first.
I suggest you look into either sepa_gomory.c or sepa_gmi.c for examples of how to use these methods. How do they include less dimensions than expected? They both return sparse vectors.
This is a continuation of my questions:
Declaring a functional recursive sequence in Matlab
Is there a more efficient way of nesting logarithms?
Nesting a specific recursion in Pari-GP
But I'll keep this question self contained. I have made a coding project for myself; which is to program a working simple calculator for a tetration function I've constructed. This tetration function is holomorphic, and stated not to be Kneser's solution (as to all the jargon, ignore); long story short, I need to run the numbers; to win over the nay-sayers.
As to this, I have to use Pari-GP; as this is a fantastic language for handling large numbers and algebraic expressions. As we are dealing with tetration (think numbers of the order e^e^e^e^e^e); this language is, of the few that exist, the best for such affairs. It is the favourite when doing iterated exponential computations.
Now, the trouble I am facing is odd. It is not so much that my code doesn't work; it's that it's overflowing because it should over flow (think, we're getting inputs like e^e^e^e^e^e; and no computer can handle it properly). I'll post the first batch of code, before I dive deeper.
The following code works perfectly; and does everything I want. The trouble is with the next batch of code. This produces all the numbers I want.
\\This is the asymptotic solution to tetration. z is the variable, l is the multiplier, and n is the depth of recursion
\\Warning: z with large real part looks like tetration; and therefore overflows very fast. Additionally there are singularities which occur where l*(z-j) = (2k+1)*Pi*I.
\\j,k are integers
beta_function(z,l,n) =
{
my(out = 0);
for(i=0,n-1,
out = exp(out)/(exp(l*(n-i-z)) +1));
out;
}
\\This is the error between the asymptotic tetration and the tetration. This is pretty much good for 200 digit accuracy if you need.
\\modify the 0.000000001 to a bigger number to make this go faster and receive less precision. When graphing 0.0001 is enough
\\Warning: This will blow up at some points. This is part of the math; these functions have singularities/branch cuts.
tau(z,l,n)={
if(1/real(beta_function(z,l,n)) <= 0.000000001, //this is where we'll have problems; if I try to grab a taylor series with this condition we error out
-log(1+exp(-l*z)),
log(1 + tau(z+1,l,n)/beta_function(z+1,l,n)) - log(1+exp(-l*z))
)
}
\\This is the sum function. I occasionally modify it; to make better graphs, but the basis is this.
Abl(z,l,n) = {
beta_function(z,l,n) + tau(z,l,n)
}
Plugging this in, you get the following expressions:
Abl(1,log(2),100)
realprecision = 28 significant digits (20 digits displayed)
%109 = 0.15201551563214167060
exp(Abl(0,log(2),100))
%110 = 0.15201551563214167060
Abl(1+I,2+0.5*I,100)
%111 = 0.28416643148885326261 + 0.80115283113944703984*I
exp(Abl(0+I,2+0.5*I,100))
%112 = 0.28416643148885326261 + 0.80115283113944703984*I
And so on and so forth; where Abl(z,l,n) = exp(Abl(z-1,l,n)). There's no problem with this code. Absolutely none at all; we can set this to 200 precision and it'll still produce correct results. The graphs behave exactly as the math says they should behave. The problem is, in my construction of tetration (the one we actually want); we have to sort of paste together the solutions of Abl(z,l,n) across the value l. Now, you don't have to worry about any of that at all; but, mathematically, this is what we're doing.
This is the second batch of code; which is designed to "paste together" all these Abl(z,l,n) into one function.
//This is the modified asymptotic solution to the Tetration equation.
beta(z,n) = {
beta_function(z,1/sqrt(1+z),n);
}
//This is the Tetration function.
Tet(z,n) ={
if(1/abs(beta_function(z,1/sqrt(1+z),n)) <= 0.00000001,//Again, we see here this if statement; and we can't have this.
beta_function(z,1/sqrt(1+z),n),
log(Tet(z+1,n))
)
}
This code works perfectly for real-values; and for complex values. Some sample values,
Tet(1+I,100)
%113 = 0.12572857262453957030 - 0.96147559586703141524*I
exp(Tet(0+I,100))
%114 = 0.12572857262453957030 - 0.96147559586703141524*I
Tet(0.5,100)
%115 = -0.64593666417664607364
exp(Tet(0.5,100))
%116 = 0.52417133958039107545
Tet(1.5,100)
%117 = 0.52417133958039107545
We can also effectively graph this object on the real-line. Which just looks like the following,
ploth(X=0,4,Tet(X,100))
Now, you may be asking; What's the problem then?
If you try and plot this function in the complex plane, it's doomed to fail. The nested logarithms produce too many singularities near the real line. For imaginary arguments away from the real-line, there's no problem. And I've produced some nice graphs; but the closer you get to the real line; the more it misbehaves and just short circuits. You may be thinking; well then, the math is wrong! But, no, the reason this is happening is because Kneser's tetration is the only tetration that is stable about the principal branch of the logarithm. Since this tetration IS NOT Kneser's tetration, it's inherently unstable about the principal branch of the logarithm. Of course, Pari just chooses the principal branch. So when I do log(log(log(log(log(beta(z+5,100)))))); the math already says this will diverge. But on the real line; it's perfectly adequate. And for values of z with an imaginary argument away from zero, we're fine too.
So, how I want to solve this, is to grab the Taylor series at Tet(1+z,100); which Pari-GP is perfect for. The trouble?
Tet(1+z,100)
*** at top-level: Tet(1+z,100)
*** ^------------
*** in function Tet: ...unction(z,1/sqrt(1+z),n))<=0.00000001,beta_fun
*** ^---------------------
*** _<=_: forbidden comparison t_SER , t_REAL.
The numerical comparison I've done doesn't translate to a comparison between t_SER and t_REAL.
So, my question, at long last: what is an effective strategy at getting the Taylor series of Tet(1+z,100) using only real inputs. The complex inputs near z=0 are erroneous; the real values are not. And if my math is right; we can take the derivatives along the real-line and get the right result. Then, we can construct a Tet_taylor(z,n) which is just the Taylor Series expansion. Which; will most definitely have no errors when trying to graph.
Any help, questions, comments, suggestions--anything, is greatly appreciated! I really need some outside eyes on this.
Thanks so much if you got to the bottom of this post. This one is bugging me.
Regards, James
EDIT:
I should add that a Tet(z+c,100) for some number c is the actual tetration function we want. There is a shifting constant I haven't talked about yet. Nonetheless; this is spurious to the question, and is more a mathematical point.
This is definitely not an answer - I have absolutely no clue what you are trying to do. However, I see no harm in offering suggestions. PARI has a built in type for power series (essentially Taylor series) - and is very good at working with them (many operations are supported). I was originally going to offer some suggestions on how to get a Taylor series out of a recursive definition using your functions as an example - but in this case, I'm thinking that you are trying to expand around a singularity which might be doomed to failure. (On your plot it seems as x->0, the result goes to -infinity???)
In particular if I compute:
log(beta(z+1, 100))
log(log(beta(z+2, 100)))
log(log(log(beta(z+3, 100))))
log(log(log(log(beta(z+4, 100)))))
...
The different series are not converging to anything. Even the constant term of the series is getting smaller with each iteration, so I am not entirely sure there is even a Taylor series expansion about x = 0.
Questions/suggestions:
Should you be expanding about a different point? (say where the curve
crosses the x-axis).
Does the Taylor series satisfy some recursive relation? For example: A(z) = log(A(z+1)). [This doesn't work, but perhaps there is another way to write it].
I suspect my answer is unlikely to be satisfactory - but then again your question is more mathematical than a practical programming problem.
So I've successfully answered my question. I haven't programmed in so long; I'm kind of shoddy. But I figured it out after enough coffee. I created 3 new functions, which allow me to grab the Taylor series.
\\This function attempts to find the number of iterations we need.
Tet_GRAB_k(A,n) ={
my(k=0);
while( 1/real(beta(A+k,n)) >= 0.0001, k++);
return(k);
}
\\This function will run and produce the same results as Tet; but it's slower; but it let's us estimate Taylor coefficients.
\\You have to guess which k to use for whatever accuracy before overflowing; which is what the last function is good for.
Tet_taylor(z,n,k) = {
my(val = beta(z+k,n));
for(i=1,k,val = log(val));
return(val);
}
\\This function produces an array of all the coefficients about a value A.
TAYLOR_SERIES(A,n) = {
my(ser = vector(40,i,0));
for(i=1,40, ser[i] = polcoeff(Tet_taylor(A+z,n,Tet_GRAB_k(A,n)),i-1,z));
return(ser);
}
After running the numbers, I'm confident this works. The Taylor series is converging; albeit rather slowly and slightly less accurately than desired; but this will have to do.
Thanks to anyone who read this. I'm just answering this question for completeness.
I have a function f(x,y) whose outcome is random (I take mean from 20 random numbers depending on x and y). I see no way to modify this function to make it symbolic.
And when I run
x,y = var('x,y')
d = plot_vector_field((f(x),x), (x,0,1), (y,0,1))
it says it can't cast symbolic expression to real or rationa number. In fact it stops when I write:
a=matrix(RR,1,N)
a[0]=x
What is the way to change this variable to real numbers in the beginning, compute f(x) and draw a vector field? Or just draw a lot of arrows with slope (f(x),x)?
I can create something sort of like yours, though with no errors. At least it doesn't do what you want.
def f(m,n):
return m*randint(100,200)-n*randint(100,200)
var('x,y')
plot_vector_field((f(x,y),f(y,x)),(x,0,1),(y,0,1))
The reason is because Python functions immediately evaluate - in this case, f(x,y) was 161*x - 114*y, though that will change with each invocation.
My suspicion is that your problem is similar, the immediate evaluation of the Python function once and for all. Instead, try lambda functions. They are annoying but very useful in this case.
var('x,y')
plot_vector_field((lambda x,y: f(x,y), lambda x,y: f(y,x)),(x,0,1),(y,0,1))
Wow, I now I have to find an excuse to show off this picture, cool stuff. I hope your error ends up being very similar.
I have a function that takes a floating point number and returns a floating point number. It can be assumed that if you were to graph the output of this function it would be 'n' shaped, ie. there would be a single maximum point, and no other points on the function with a zero slope. We also know that input value that yields this maximum output will lie between two known points, perhaps 0.0 and 1.0.
I need to efficiently find the input value that yields the maximum output value to some degree of approximation, without doing an exhaustive search.
I'm looking for something similar to Newton's Method which finds the roots of a function, but since my function is opaque I can't get its derivative.
I would like to down-thumb all the other answers so far, for various reasons, but I won't.
An excellent and efficient method for minimizing (or maximizing) smooth functions when derivatives are not available is parabolic interpolation. It is common to write the algorithm so it temporarily switches to the golden-section search (Brent's minimizer) when parabolic interpolation does not progress as fast as golden-section would.
I wrote such an algorithm in C++. Any offers?
UPDATE: There is a C version of the Brent minimizer in GSL. The archives are here: ftp://ftp.club.cc.cmu.edu/gnu/gsl/ Note that it will be covered by some flavor of GNU "copyleft."
As I write this, the latest-and-greatest appears to be gsl-1.14.tar.gz. The minimizer is located in the file gsl-1.14/min/brent.c. It appears to have termination criteria similar to what I implemented. I have not studied how it decides to switch to golden section, but for the OP, that is probably moot.
UPDATE 2: I googled up a public domain java version, translated from FORTRAN. I cannot vouch for its quality. http://www1.fpl.fs.fed.us/Fmin.java I notice that the hard-coded machine efficiency ("machine precision" in the comments) is 1/2 the value for a typical PC today. Change the value of eps to 2.22045e-16.
Edit 2: The method described in Jive Dadson is a better way to go about this. I'm leaving my answer up since it's easier to implement, if speed isn't too much of an issue.
Use a form of binary search, combined with numeric derivative approximations.
Given the interval [a, b], let x = (a + b) /2
Let epsilon be something very small.
Is (f(x + epsilon) - f(x)) positive? If yes, the function is still growing at x, so you recursively search the interval [x, b]
Otherwise, search the interval [a, x].
There might be a problem if the max lies between x and x + epsilon, but you might give this a try.
Edit: The advantage to this approach is that it exploits the known properties of the function in question. That is, I assumed by "n"-shaped, you meant, increasing-max-decreasing. Here's some Python code I wrote to test the algorithm:
def f(x):
return -x * (x - 1.0)
def findMax(function, a, b, maxSlope):
x = (a + b) / 2.0
e = 0.0001
slope = (function(x + e) - function(x)) / e
if abs(slope) < maxSlope:
return x
if slope > 0:
return findMax(function, x, b, maxSlope)
else:
return findMax(function, a, x, maxSlope)
Typing findMax(f, 0, 3, 0.01) should return 0.504, as desired.
For optimizing a concave function, which is the type of function you are talking about, without evaluating the derivative I would use the secant method.
Given the two initial values x[0]=0.0 and x[1]=1.0 I would proceed to compute the next approximations as:
def next_x(x, xprev):
return x - f(x) * (x - xprev) / (f(x) - f(xprev))
and thus compute x[2], x[3], ... until the change in x becomes small enough.
Edit: As Jive explains, this solution is for root finding which is not the question posed. For optimization the proper solution is the Brent minimizer as explained in his answer.
The Levenberg-Marquardt algorithm is a Newton's method like optimizer. It has a C/C++ implementation levmar that doesn't require you to define the derivative function. Instead it will evaluate the objective function in the current neighborhood to move to the maximum.
BTW: this website appears to be updated since I last visited it, hope it's even the same one I remembered. Apparently it now also support other languages.
Given that it's only a function of a single variable and has one extremum in the interval, you don't really need Newton's method. Some sort of line search algorithm should suffice. This wikipedia article is actually not a bad starting point, if short on details. Note in particular that you could just use the method described under "direct search", starting with the end points of your interval as your two points.
I'm not sure if you'd consider that an "exhaustive search", but it should actually be pretty fast I think for this sort of function (that is, a continuous, smooth function with only one local extremum in the given interval).
You could reduce it to a simple linear fit on the delta's, finding the place where it crosses the x axis. Linear fit can be done very quickly.
Or just take 3 points (left/top/right) and fix the parabola.
It depends mostly on the nature of the underlying relation between x and y, I think.
edit this is in case you have an array of values like the question's title states. When you have a function take Newton-Raphson.
I'm trying to write a program that will help someone study for the GRE math. As many of you may know, fractions are a big part of the test, and calculators aren't allowed. Basically what I want to do is generate four random numbers (say, 1-50) and either +-/* them and then accept an answer in fraction format. The random number thing is easy. The problem is, how can I 1) accept a fractional answer and 2) ensure that the answer is reduced all the way?
I am writing in ASP.NET (or jQuery, if that will suffice). I was pretty much wondering if there's some library or something that handles this kind of thing...
Thanks!
have a look at
http://www.geekpedia.com/code73_Get-the-greatest-common-divisor.html
http://javascript.internet.com/math-related/gcd-lcm-calculator.html
Since fractions are essentially divisions you can check to see if the answer is partially correct by performing the division on the fraction entries that you're given.
[pseudocode]
if (answer.contains("/"))
int a = answer.substring(1,answer.instanceof("/"))
int b = answer.substring(answer.instanceof("/"))
if (a/b == expectedAnswer)
if (gcd(a,b) == 1)
GOOD!
else
Not sufficiently reduced
else
WRONG!
To find out whether it's reduced all the way, create a GCD function which should evaluate to the value of the denominator that the user supplied as an answer.
Learn Python and try fractions module.