User ehud friedgut - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T23:33:17Z http://mathoverflow.net/feeds/user/2229 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/7493/a-riddle-about-zeros-ones-and-minus-ones A riddle about zeros, ones and minus-ones Ehud Friedgut 2009-12-01T20:00:07Z 2011-01-26T01:46:22Z <p>I was asked this years ago, but I don't remember by whom, and have never managed to solve it. Consider the $2^n \times n$ matrix of all vectors in {-1,1}$^n$. Someone comes and maliciously replaces some of the entries by zeros. Show that there still remains a non-empty subset of rows that add up to the all zero vector. </p> http://mathoverflow.net/questions/7470/is-there-a-neat-formula-for-the-volume-of-a-tetrahedron-on-the-surface-of-s3 Is there a neat formula for the volume of a tetrahedron on the surface of $S^3$? Ehud Friedgut 2009-12-01T18:13:10Z 2009-12-02T01:41:31Z <p>There is a nice formula for the area of a triangle on the surface of the 2-dimensional sphere; If the triangle is the intersection of three half spheres, and has angles $\alpha$, $\beta$ and $\gamma$, and we normalize the area of the whole sphere to be $4\pi$ then the area of the triangle is $$\alpha + \beta + \gamma - \pi.$$ The proof is a cute application of inclusion-exclusion of three sets, and involves the fact that the area we want to calculate appears on both sides of the equation, but with opposite signs.</p> <p>However, when trying to copy the proof to the three dimensional sphere the parity goes the wrong way and you get 0=0.</p> <p>Is there a simple formula for the volume of the intersection of four half-spheres of $S^3$ in terms of the 6 angles between the four bounding hyperplanes?</p> http://mathoverflow.net/questions/7470/is-there-a-neat-formula-for-the-volume-of-a-tetrahedron-on-the-surface-of-s3/7486#7486 Answer by Ehud Friedgut for Is there a neat formula for the volume of a tetrahedron on the surface of $S^3$? Ehud Friedgut 2009-12-01T19:30:03Z 2009-12-01T19:30:03Z <p>Nice answer, Greg. I looked at the linked paper and was sufficiently intimidated. I just want to point out, again, that for those (like me) who have a phobia of differential geometry, and hence don't want to use (generalized)Gauss-Bonnet, it is easy to see, using inclusion-exclusion, that the formula in even dimensions is a neat linear combination of the formulas in lower dimensions.</p> http://mathoverflow.net/questions/7155/famous-mathematical-quotes/7172#7172 Comment by Ehud Friedgut Ehud Friedgut 2009-12-02T05:49:53Z 2009-12-02T05:49:53Z My version of this quote, after working my way backwards through a series of papers, each relying on the previous ones: If I can't see a darn thing it's because I stand on the shoulders of giants... http://mathoverflow.net/questions/7493/a-riddle-about-zeros-ones-and-minus-ones/7518#7518 Comment by Ehud Friedgut Ehud Friedgut 2009-12-02T05:43:40Z 2009-12-02T05:43:40Z Great, thanks Alon. I've been haunted by this silly question for a long time. BTW, I believe it used to be used as a question in job interviews for Hi-Tech companies in Israel. I am sure every single candidate started by trying induction. The nice thing about your solution is that it is almost dimension-free...