Write a function that estimates a d-dimensional integral to at least 1% relative precision.
Inputs:
Output:
1 1 1 / / / I = |dx_1 |dx_2 ...| dx_d fun([x_1,x_2,...,x_d]) / / / 0 0 0
Example:
fun = @(x) x(1)*x(2) d = 2
The result should be 0.25. An output I=0.2501 would be acceptable, because the relative deviation would be abs(0.25-0.2501)/0.25 which is smaller than 1%.
The functions in the test-suite are all positive and generally 'well behaved', i.e. not fluctuating too much. Some of the tests hav a relatively large d.
I'm confused by the 3rd test case. Can the integral inside an n-dimensional hypercube really be greater than 1?
In my comment, I mean an n-dimensional UNIT hypercube, which is what you integration limits impose.
Of course. It depends on the integrand. Even in 1d, if the integrand is e.g. 10x, the result will be 5.
I was confused. Thanks for clarifying.
I found it helpful to think about the problem as involving d+1 dimensions: the d dimensions of the input variables, and one more dimension for the (scalar) output variable. —DIV
Innovative, but a bit risky/lucky: roughly a 20–30% risk of failing the Test Suite, by my estimation [see Solutions 1572717–1572721, and Solution 1572713; cf. Solution 1572418].
Since the Monte-Carlo errors decrease proportionally to 1/sqrt(N), where N is the number of random numbers, you should always be able to choose a large enough N to nearly always pass the tests. This is only limited by the execution-time limit that cody enforces.
I agree that it is possible to choose N in such a way as to be practically guaranteed to pass the Test Suite. My point is that in this submission (and BTW also in Solution 188645) N has been chosen smaller than that, and so some luck was needed for this submission to pass — there was a significant risk that it could have failed.
just found this "cheat". don't hate! I've reported it to TMW
I adjusted the test suite, thanks for the hint...
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