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I have to find the set of integers that minimize this objective function:
The costraints are:
every x must be a non-negative integer
T, A and B are double known numbers.
I have been looking at the OR-Tools C++ library in order to solve this problem, specifically at the CP-SAT solver.
Is it the right tool from such problems?
If yes, would it be feasible to convert all the double to int in the objective function?
If not, what else do you suggest? (I'm also open to other open source C++ libraries)
It will fit in the CP-SAT solver. You will need to scale floating point coefficients to integers.
The objective function accepts floating point coefficients.
But (x1 + A1)^2 will propagate better if you keep it that way instead of A1^2 + 2 * A1 * x1 + x1^2. which fits into the linear with double coefficient limitation of CP-SAT, provided you use temporary variables sx1 = x1 * x1.
Then make sure to use at least 8 workers for that. (parameters num_search_workers:8).
Now, I believe there are least square solvers that are more suited for this.
I have simple code, which flags nodes with in region enclosed by cylinder. On implementing the code, the result is mild tilt of the cylinder observed case with 90 degrees
The actual issue:
The above algorithm is implemented in Fortran. The code checks for points in Cartesian grid if inside the cylinder. Following being the test case:
The cylinder makes an angle 90 degrees in the yz-plane with respect to y-axis. Therefore, the orientation vector $\vec{o}$ is (0, 1, 0).
Case 1:
Orientation vector is assigned directly with $\vec{o}=(0.0,1.0,0.0)$. This results in perfect cylinder with $\theta=90.$
Case 2:
Orientation vector is specified with intrinsic Fortran functions with double precision accuracy dsin and dcos with $\vec{o}=(0.0, \sin(\pi/2.0), \cos(\pi/2.0))$ with $\pi$ value assigned with more than 20 significant decimal points. The resulting cylinder results in mild tilt.
The highlighted region indicates the extra material due to tilt of cylinder with respect to Cartesian axes. I also tried architecture specific maximum precision "pi" value. This also results in same problem.
This shows like the actual angle made by cylinder is not 90 degrees. Can anyone suggest valid solution for this problem. I need to use the inbuilt trigonometric functions for arbitrary angles and looking for accurate cell flagging method.
Note: All operations are performed with double precision accuracy.
The actual function is below. rk is defined parameter with value 8
pure logical function in_particle(p,px,x)
type(md_particle_type),intent(in) :: p
real(kind=rk),intent(in) :: px(3),x(3)
real(kind=rk) :: r(3),rho(3),rop(2),ro2,rdiff,u
rop = particle_radii(p) ! (/R_orth,R_para/)
ro2 = rop(1)**2
rdiff = rop(2) - rop(1)
r = x-px
! Case 1:
! u = dot_product((/0.0_rk,-1.0_rk,0.0_rk/),r)
! rho = r-u*(/0.0_rk,-1.0_rk,0.0_rk/)
! Case 2:
u = dot_product((/0.0_rk,-dsin(pi/2.0_rk),dcos(pi/2.0_rk)/),r)
rho = r-u*(/0.0_rk,-dsin(pi/2.0_rk),dcos(pi/2.0_rk)/)
if((u.le.rdiff).and.(u.ge.-rdiff)) then
in_particle = dot_product(rho,rho) < ro2
else
in_particle = .false.
end if
end function in_particle
Note: The trigonometric operations are done inside the code to explain the problem better. However the original code reads the orientation in vector form from user. Then converts this information to quaternions for particle-particle collision operations. On converting quaternions back to orientation vector, this error is even more amplified. Even before the start of collision, the orientation of cylinder tends to be disoriented by 2 lattice cells.
cos(pi/2) is not necessarily going to give you exactly 0, no matter how exact you make the cos calculation, and no matter how many digits of pi you have, because:
pi, as an irrational number, will contain up to 1/2 ulp of error when represented as an FP number; and
sin and cos are not guaranteed by the IEEE-754 standard to be correctly rounded (or even implemented).
Now, sin(pi/2) is extremely likely to come out as 1 regardless of precision and FP architecture, simply because sin has such a low derivative around 1; with single-precision floats, it should come out to 1 if you're anywhere within about 3e-4 of the exact value of pi/2. The problematic call is the cos, which has lots of precision to play with around 0 and a derivative of about -1 in the neighborhood.
Still, we're talking about extremely small values here. I think what's really potentiating the problem here is the in/out test you're doing, combined with ordinary FP rounding rules. I would guess, in fact, that if you were to bias your test points by, say, a quarter of the grid quantum, you'd see all straight verticals in your voxelization (though it might not be symmetrical around the minor axes).
Another option would be to actually discard some precision from your sin/cos calculation before doing the dot product, effectively quantizing your axes.
Short answer: Create a table of sin and cos of common angles (0, pi/6, pi/4, pi/3, pi/2, pi and their multiples) and compute only for uncommon angles. The reason being that errors with uncommon angles will be tolerated by most people while errors with common angles will likely not be tolerated.
Explanation:
Because floating point computation is not exact (that is its nature), you sometime need a little bit of compromise between the accuracy and the readability of the code.
One way of doing that is to avoid to compute something that is known exactly. To do that, you can check the value of the angle and do the actual computation only if it is not an obvious angle. For example angle 0, 90, 180 and 270 degrees have obvious values of sin and cos. More generally, the cos and sin of common angles (0, pi/6, pi/4, pi/3, pi/2, pi and their multiples) are known exactly (even if they are irrational numbers).
I am writing a program that solves this puzzle game: some numbers and a goal number is given, and you make the goal number using the n numbers and operators +, -, *, / and (). For example, given 2,3,5,7 and the goal number 10, the solutions are (2+3)*(7-5)=10, 3*5-(7-2)=10, and so on.
The catch is, if I implement it naively, I will get a bunch of identical solutions, like (2+3)*(7-5)=10 and (3+2)*(7-5)=10, and 3*5-(7-2)=10 and 5*3-(7-2)=10 and 3*5-7+2=10 and 3*5+2-7=10 and so on. So I'd like to detect those identical solutions and prune them.
I'm currently using randomly generated double numbers to detect identical solutions. What I'm doing is basically substituting those random numbers to the solution and check if there are any pairs of them that calculate to the same number. I have to perform the detection at every node of my search, so it has to be fast, and I use hashset for it now.
Now the problem is the error that comes with the calculation. Because even identical solutions do not calculate to the exactly same value, I currently round the calculated value to a precision when storing in the hashset. However this does not seem to work well enough, and gives different number of solutions every time to the same problem. Sometimes the random numbers are bad and prune some completely different solutions. Sometimes the calculated value lies on the edge of rounding function and it outputs two(or more) identical solutions. Is there a better way to do this?
EDIT:
By "identical" I mean two or more solutions(f(w,x,y,z,...) and g(w,x,y,z,...)) that calculate to the same number whatever the original number(w,x,y,z...) is. For more examples, 4/3*1/2 and 1*4/3/2 and (1/2)/(3/4) are identical, but 4/3/1/2 and 4/(3*1)/2 are not because if you change 1 to some other number they will not produce the same result.
It will be easier if you "canonicalize" the expressions before comparing them. One way would be to sort when an operation is commutative, so 3+2 becomes 2+3 whereas 2+3 remains as it was. Of course you will need to establish an ordering for parenthesized groups as well, like 3+(2*1)...does that become (1*2)+3 or 3+(1*2)? What the ordering is doesn't necessarily matter, so long as it is a total ordering.
Generate all possibilities of your expressions. Then..
When you create expressions, put them in a collection of parsed trees (this would also eliminate your parenthesis). Then "push down" any division and subtraction into the leaf nodes so that all the non-leaf nodes have * and +. Apply a sorting of the branches (e.g. regular string sort) and then compare the trees to see if they are identical.
I like the idea of using doubles. The problem is in the rounding. Why not use a container SORTED by the value obtained with one random set of double inputs. When you find the place you would insert in that container, you can look at the immediately preceding and following items. Use a different set of random doubles to recompute each for the more robust comparison. Then you can have a reasonable cutoff for "close enough to be equal" without arbitrary rounding.
If a pair of expressions are close enough for equal in both the main set of random numbers and the second set, the expressions are safely "same" and the newer one discarded. If close enough for equal in the main set but not the new set, you have a rare problem, that probably requires rekeying the entire container with a different random number set. If not close enough in either, then they are different.
For the larger n suggested by one of your recent comments, I think you would need the better performance that should be possible from a canonical by construction method (or maybe "almost" canonical by construction) rather than a primarily comparison based approach.
You don't want to construct an incredibly large number of expressions, then canonicalize and compare.
Define a doubly recursive function can(...) that takes as input:
A reference to a canonical expression tree.
A reference to one subexpression of that tree.
A count N of inputs to be injected.
A set of flags for prohibiting some injections.
A leaf function to call.
If N is zero, can just calls the leaf function. If N is nonzero, can patches the subtree in every possible way that produces a canonical tree with N injected variables, and calls the leaf function for each and restores the tree, undoing each part of the patch as it is done with it, so we never need massive copying.
X is the subtree and K is a leaf representing variable N-1. First can would replace the subtree temporarily one at a time with subtrees representing some of (X)+K, (X)-K, (X)*K, (X)/K and K/(X) but both flags and some other rules would cause some of those to be skipped. For each not skipped, recursively call itself with the whole tree as both top and sub, with N-1, and with 0 flags.
Next drill into the two children of X and call recursively itself with that as the subtree, with N, and with appropriate flags.
The outer just calls can with a single node tree representing variable N-1 of the original N, and passing N-1.
In discussion, it is easier to name the inputs forward, so A is input N-1 and B is input N-2 etc.
When we drill into X and see it is Y+Z or Y-Z we don't want to add or subtract K from Y or Z because those are redundant with X+K or X-K. So we pass a flag that suppresses direct add or subtract.
Similarly, when we drill into X and see it is Y*Z or Y/Z we don't want to multiply or divide either Y or Z by K because that is redundant with multiplying or dividing X by K.
Some cases for further clarification:
(A/C)/B and A/(B*C) are easily non canonical because we prefer (A/B)/C and so when distributing C into (A/B) we forbid direct multiplying or dividing.
I think it takes just a bit more effort to allow C/(A*B) while rejecting C/(A/B) which was covered by (B/A)*C.
It is easier if negation is inherently non canonical, so level 1 is just A and does not include -A then if the whole expression yields negative the target value, we negate the whole expression. Otherwise we never visit the negative of a canonical expression:
Given X, we might visit (X)+K, (X)-K, (X)*K, (X)/K and K/(X) and we might drill down into the parts of X passing flags which suppress some of the above cases for the parts:
If X is a + or - suppress '+' or '-' in its direct parts. If X is a * or / suppress * or divide in its direct parts.
But if X is a / we also suppress K/(X) before drilling into X.
Since you are dealing with integers, I'd focus on getting an exact result.
Claim: Suppose there is some f(a_1, ..., a_n) = x where a_i and x are your integer input numbers and f(a_1, ..., a_n) represents any functions of your desired form. Then clearly f(a_i) - x = 0. I claim, we can construct a different function g with g(x, a_1, ..., a_n) = 0 for the exact same x and g only uses ()s, +, - and * (no division).
I'll prove that below. Consequently you could construct g evaluate g(x, a_1, ..., a_n) = 0 on integers only.
Example:
Suppose we have a_i = i for i = 1, ..., 4 and f(a_i) = a_4 / (a_2 - (a_3 / 1)) (which contains divisions so far). This is how I would like to simplify:
0 = a_4 / (a_2 - (a_3 / a_1) ) - x | * (a_2 - (a_3 / a_1) )
0 = a_4 - x * (a_2 - (a_3 / a_1) ) | * a_1
0 = a_4 * a_1 - x * (a_2 * a_1 - (a_3) )
In this form, you can verify your equality for some given integer x using integer operations only.
Proof:
There is some g(x, a_i) := f(a_i) - x which is equivalent to f. Consider any equivalent g with as few as possible division. Assume there is at least one (otherwise we are done). Assume within g we divide by h(x, a_i) (any of your functions, may contain divisions itself). Then (g*h)(x, a_i) := g(x, a_i) * h(x, a_i) has the same roots, as g has (multiplying by a root, ie. (x, a_i) where g(a_i) - x = 0, preserves all roots). But on the other hand, g*h is composed of one division fewer. A contradiction (g with minimum number of divisions), which is why g doesn't contain any division.
I've updated the example to visualize the strategy.
Update: This works well on rational input numbers (those represent a single division p/q). This should help you. Other input can't be provided by humans.
What are you doing to find / test f's? I'd guess some form of dynamic programming will be fast in practice.
I am trying to implement a root finding algorithm. I am using the hybrid Newton-Raphson algorithm found in numerical recipes that works pretty nicely. But I have a problem in bracketing the root.
While implementing the root finding algorithm I realised that in several cases my functions have 1 real root and all the other imaginary (several of them, usually 6 or 9). The only root I am interested is in the real one so the problem is not there. The thing is that the function approaches the root like a cubic function, touching with the point the y=0 axis...
Newton-Rapson method needs some brackets of different sign and all the bracketing methods I found don't work for this specific case.
What can I do? It is pretty important to find that root in my program...
EDIT: more problems: sometimes due to reaaaaaally small numerical errors, say a variation of 1e-6 in some value the "cubic" function does NOT have that real root, it is just imaginary with a neglectable imaginary part... (checked with matlab)
EDIT 2: Much more information about the problem.
Ok, I need root finding algorithm.
Info I have:
The root I need to find is between [0-1] , if there are more roots outside that part I am not interested in them.
The root is real, there may be imaginary roots, but I don't want them.
Probably all the rest of the roots will be imaginary
The root may be double in that point, but I think that actually doesn't mater in numerical analysis problems
I need to use the root finding algorithm several times during the overall calculations, but the function will always be a polynomial
In one of the particular cases of the root finding, my polynomial will be similar to a quadratic function that touches Y=0 with the point. Example of a real case:
The coefficient may not be 100% precise and that really slight imprecision may make the function not to touch the Y=0 axis.
I cannot solve for this specific case because in other cases it may be that the polynomial is pretty normal and doesn't make any "strange" thing.
The method I am actually using is NewtonRaphson hybrid, where if the derivative is really small it makes a bisection instead of NewRaph (found in numerical recipes).
Matlab's answer to the function on the image:
roots:
0.853553390593276 + 0.353553390593278i
0.853553390593276 - 0.353553390593278i
0.146446609406726 + 0.353553390593273i
0.146446609406726 - 0.353553390593273i
0.499999999999996 + 0.000000040142134i
0.499999999999996 - 0.000000040142134i
The function is a real example I prepared where I know that the answer I want is 0.5
Note:
I still haven't check completely some of the answers I you people have give me (Thank you!), I am just trying to give al the information I already have to complete the question.
Assuming you have a one-dimensional polynomial problem (which I assume from the imaginary solutions) you can use Sturm sequences to bracket all real roots. See Sturm's theorem.
Welcome to the wonderful world of numerical methods. Watch your hairline; it might start receding as you pull your hair out in frustration.
First off, with numerical root finding, you are toast if you can't bracket the problem. Newton Raphson is nice for polishing off a solution once you get close, and it only works if the derivative near the root is well away from zero. You always need to have some slower technique at hand as a backup because Newton Raphson can send you off to never-never land (i.e., somewhere well outside the bracket). If your function is not a polynomial, the first thing to try is Brent's method. If your function is a polynomial, try Laguerre's method or Jenkins-Traub.
BTW, it sounds like you have a pathological problem. You shouldn't expect particularly good performance. Pathological problems are, well, pathological.
Addendum
If you are having problems with things that appear to be roots, but aren't, you need to take care how you evaluate your function. If you do have a polynomial, form each term of the polynomial, sort by absolute value, and add smallest to largest. This produces better accuracy most of the time, but fails if you have large terms whose sum is nearly zero. If that's the case, you might want to add those canceling terms separately, add the rest smallest to largest, and then compute a grand total -- and your still kinda screwed. That big addition that nearly cancels loses a lot of precision. There's no escape other than extended precision arithmetic.
Ander, thanks for responding to my question (about the interval); sorry for the delay in following up - I have very busy work. Also - before I found the additional information you've provided - I had in mind to explain quite a few things how to handle this and was contemplating how to present that. However, I now believe your case is not too difficult and we can get at it without too much additional stuff, since you apparently have an explicit polynomial expression (coefficients to the various powers).
Let's start with a simple case, to pinpoint the approach.
Step 1.
If you have a 2nd degree polynomial, its derivative is first order and has a simple zero (which you can find by bracketing or simply by explicitly solving the equation). (Yes, I know there's a closed formula for the roots of a 2nd degree polynomial also, but for the sake of the current argument, let us forget that).
The zero's of the 2nd degree polynomial are then located one at the left side and one at the right side of the zero of the derivative. So, if you also have the interval where the roots of the original function (the 2nd degree polynomial) are to be found, you now have two intervals - left and right of the derivative-zero, each with one zero.
It is important to realize that the original function is MONOTONIC on each subinterval (decreasing on one of them, increasing on the other). Therefore, simply by checking the function values at the ends of the (sub)interval you can determine whether or not they actually bracket a zero. If not, there's a multiple zero (double, in this case) exactly at the zero of the derivative IF the function is zero there (otherwise, it is a double imaginary root of which you've now found the real part).
In case the zero of the derivative lies OUTSIDE the total interval, you will have at most one root inside your interval and you need to check only that particular (sub)interval.
Step 2.
Consider now a 3rd order polynomial.
Its derivative is 2nd order.
The derivative of THAT 2nd order polynomial is again 1st order and you proceed as before to get two subintervals to find the roots of the derivative of the original function. These two roots give you THREE (at most) intervals where you will find the 3 roots of the original (3rd order) function.
And also here, you will have intervals (3) where the original function is monotonic (alternatingly increasing/decreasing), making the analysis per subinterval quite easy.
Again, zeros may coincide (2 or even all 3) and may in addition turn out to be complex-valued (i.e. having non-zero imaginary parts). The analysis of the cases is straightforward: check function values at the borders of the intervals to assess whether not there's a sign-change (function is monotonic on each subinterval) and/or whether the function is zero at one of the subinterval-borders.
Step 4.
Generalize this with the known polynomial. Let's say - your example - it is 6th order:
a) construct the 5th derivative (i.e. reducing the original to a 1st order polynomial). Compute it's zero (it is at precisely 0.5 in your example). In this case you're already done, but suppose you don't realize that. So you have now 2 intervals 0..0.5 and 0.5..1
b) construct the 4th derivative. Inspect its values at the subinterval-boundaries (0, 0.5, 1)
For each subinterval determine if it has a real zero inside. If so, you re-partition your original interval in 3 subintervals, using the two found zeros (you forget about the zero of the 5th derivative). If they coincide (at the previous cut, 0.5) you stick with that 0.5 (don't care whether you've found a true double zero of your 4th derivative there or a "double imaginary") and still have only 2 intervals, but for the sake of the argument let's say you now have 3.
c) construct the 3rd derivative and do likewise as before. You will then have 4 (at most) intervals.
d) And so on. After having processed the 2nd derivative in this fashion you have 5 (at most) intervals, and after processing the 1st derivative you have 6 intervals (or less...) and knowing the function is monotonic on each subinterval, you'll quickly determine in each of them if there's a real root, as always using the know monotonicity of the function in each of the final subintervals.
Adding a note on numerical accuracy at evaluating a function:
A first (probably sufficient, in this case) method to reduce noise is NOT to evaluate your function in the way suggested by the original form (i.e. a6 x*6 + a5 x*5 +..), but to rewrite it as:
a0 + x*(a1 + x*(a2 + x*(a3 + x*(a4 + x*(a5 + x*a6)))))
So, in evaluating you proceed:
tmp = a6
tmp = x*tmp + a5
tmp = x*tmp + a4
etcetera.
In case this little rewriting is not sufficient for numerical stability, you should rewrite your polynomial in (for instance) a chebyshev-polynomial expansion and evaluate that one with its recurrence relations. Both (getting the expansion and applying the recurrence relations for evaluation) are rather simple. I can explain, if you need help, but I guess it won't be necessary here.
In all cases, you HAVE to allow for some inaccuracy, i.e. accept that a computation will, generally speaking, NEVER give you the mathematically exact function value. So the assessment whether the function is presumably zero at some point must include some "tolerance", there's no way around this, unfortunately; the best you can aim for is to minimize the noise.
Well, if your function touches zero but never crosses it, you seem to be looking for a minimum (or a maximum). In which case, you're better off telling computer to do exactly that --- either find the root of a derivative (if you can calculate it analytically), or use a minimization routine. Then check that the function value at the minimum is 'close enough' to zero.
Just to reiterate what was already said by other people:
don't start with Newton-Raphson method; it's almost always better to start with Brent or even a straightforward bisection (provided you can bracket the root).
An instability where 'small numerical errors' of the order of 1e-6 have bad effects is worth investigating. Immediate suspects: mixing floats and doubles, loss of precision somewhere etc.
EDIT: So, depending on some parameters, your function has either a zero crossing, or a minimum with zero value, is this correct? In this case, what I'd do is this: use a simple and robust bracketing strategy (e.g. start from [-1, 1], multiply the endpoints by 1.1, check the signs, keep multiplying, something like this). If that succeeds, there's a zero crossing, use a root finding routine. If bracketing fails, use minimization.
Using Newton-Raphson is an act of desperation. You are much better off finding the continued fraction that represents your function and calculating that. A CF will converge much faster and will produce the real root(s). Also, because the CF produces a ratio of two integers you have tight control over numeric precision and don't have to worry about accumulation of rounding errors and other similar hair-pulling-out problems.
To find the real roots of any polynomial function refer to "A Continued Fraction Algorithm for Approximating All Real Polynomial Roots" by David Rosen (1978).
------------ ADDENDUM 1 --- 11 OCT-----------------
Ok, you are solving a sextic. You have several options. The simplest is to use a Taylor approximation (say to the 3rd degree) in conjunction with Halley's method. This is much superior to Newton because it has cubic convergence and you can detect imaginary solutions. The disadvantage is that you will have rounding problems which may result in an incorrect answer.
The ideal option is to find the continued fraction that represents the monic root, because this CF will be computable as an integer ratio of any desired precision, thus elminating the problem of rounding.
One approach to computing this CF is via the Jacobi-Perron algorithm. See the paper Hendy and Jeans: http://www.ams.org/mcom/1981-36-154/S0025-5718-1981-0606514-X/S0025-5718-1981-0606514-X.pdf. This paper shows the exact algorithm for computing cubic and quartic roots via CF approximation.
Note that if the sextic is reducible then it can converted into a quartic and quadratic: http://elib.mi.sanu.ac.rs/files/journals/tm/21/tm1124.pdf. The quartic is then solvable by the algorithm in the Hendy paper.
The general solution to generate a CF for a sextic can be done via the Rogers-Ramunajan CF. See the following paper for the method: http://arxiv.org/pdf/1111.6023v2. This will generate the CF for any sextic.
As in your case, you are interested in the real factorization of a real polynomial. One may see that all complex roots come in conjugate pairs which correspond to a real quadratic factor. By finding this real quadratic and completing the square to get the form (x-r)^2 + s you will be able to see the "real" even order root r with an "error" given by s. If s > 0 is too large, you may discard it as probably being complex. If s < 0 is also large, then you have two faraway real roots given by x = r ± √(-s). If s is very small then you might suspect r is a real double root and keep it.
Finding such a quadratic factor may be done using Bairstow's method, which actually applies a two-dimensional Newton method. This gives x^2 + ux + v and r = -u/2; s = v - r^2.
So we have some function like (pow(e,(-a*x)))/(sqrt(x)) where a, e are const floats. we have some float eps=pow (10,(-4)). We need to find out starting from which x integral of that function from that x to infinety is less than eps? We can not use functions for special default integration function just standart math like operators. point is to achive max evaluetion speed.
If you perform the u-substitution u=sqrt(x), your integral will become 2 * integral e^(-au^2) du. With one more substitution you can reduce it to a standard normal. Once you have it in standard normal form, this reduces to calculating erf(x). The substitutions can be done abstractly for any a, and the results hardcoded for simplicity and speed.
To calculate this integral you need calculate Error function. If you use gcc you can find erf(...) function in math.h, but it doesn't take params to get exact precise. But you can evaluate Error function's value by youself just using Taylor's series. With given eps it possible to calc the exact number of terms of the series.
Hmm, no one seems to understand the question. The question is: given some function f, find the smallest x such that Integral _ x ^ +inf f(x) < eps. That's the question. So basically we try x = 0, then x = 0.1 then x = 0.2 ... until the integral, for all intents and purposes, vanishes.
For example, given the bell curve for IQ of programmers on SO, at what IQ is the cumulative intelligence of programmers with higher IQ vanishingly small? If we pick x = 100 we know at least half the programmers will have a higher IQ than 100, if we pick 120, how many are left? What about 200? If we have 10,000 programmers here and eps = 1/10000 we're basically asking what IQ the top 0.01% of SO contributors have.
The question is: what is the most efficient way to find this number, given that nothing is known about f other than that is decreases fast enough that its the integral from x to infinity approaches zero as x approaches infinity?
The general answer is: you must start with a guess of some kind. If the result is too big, double your guess, and keep going until you satisfy the requirement. Then, go back to the last value you had (which didn't) and do a binary chop to find the smallest x satisfying the requirement.
To make a good guess is hard. One way is to use a Chebychev approximation of the function, integrate it analytically, solve the problem with the resulting polynomial, and use the solution as your starting guess. The assumption is that all functions look like polynomials off sufficiently high order in any given range.