A really nice problem : concrete origin, careful formulation, then a good walk in 3D geometry, analysis, integration and at the end a single formula. The critical value cuts the triangle in four pieces : my son said it was Zelda !
Raphael: I agree completely!
Perhaps you could have a precision requirement a bit more liberal in order to allow numerical approximation algorithms or other approaches as well? just my two cents, the problem looks great
Tip: the probability grows from the vertices of simplex (triforce/triangle) toward its center (which has the highest frequency).
This is not a solution, and it should be deleted. And If it is possible, then more test cases should be added.
I am pleased that you solved this problem, David. Congratulations! I didn't find any particularly easier way of solving it. The crucial step is showing that the probability density is proportional to your 1/y^3 for points within the corresponding "kite-shaped region". I used the Jacobian between two coordinate systems to show that. After dividing that region into two halves everything falls into place, though in my dotage I had to make heavy use of the Symbolic Toolbox to check for errors. (I hope this problem will serve as a warning to people who recommend this method of producing random numbers with a predetermined sum.) R. Stafford
It is inherent in the definition of P here that the density, dP/dA, must increase as P increases and therefore dA/dP must decrease. In your proposed solution you have dA/dP increasing as P increases. R. Stafford
Renaming a field in a structure array
Given an unsigned integer x, find the largest y by rearranging the bits in x
Four quadrant inverse tangent function.
Back to basics 8 - Matrix Diagonals
Sum the Infinite Series II
Subdivide the Segment
Sum the Infinite Series
Find the treasures in MATLAB Central and discover how the community can help you!
Choose a web site to get translated content where available and see local events and offers. Based on your location, we recommend that you select: .
You can also select a web site from the following list:
Select the China site (in Chinese or English) for best site performance. Other MathWorks country sites are not optimized for visits from your location.
Contact your local office