Here is a link to recordings of invited and tutorial talks at the recent STACS 2015 conference. (Slides are available here.) These recordings include my tutorial talk on algorithmic game theory, all 3 hours of it, which may possibly be of use to writers of articles with titles like The Top Ten Mistakes you make When Giving a Technical Talk.
Possibly the commonest mistake (other than dressing badly) is to assume too much knowledge on the part of the audience. This mistake wastes the opportunity for a “big win” in which you present some fact or result that most people in the room don’t already know, but which should by rights be ingrained in their DNA. I got to do that via presenting Megiddo’s 1988 result that NP total search problems cannot be NP-complete unless NP=co-NP. To be clear, an NP total search problem is one where every instance has an easy-to-check solution, for example FACTORING, where the output is a prime factorization of an input number, or NASH, where the output is a Nash equilibrium of a given bimatrix game. This is the reason why we don’t bother to look for an NP-hardness result for NASH, and have to use an obscure-looking complexity class like PPAD.
It's possible to write a convincing proof sketch as follows. Consider what would it would mean if an NP total search problem (like FACTORING) was NP-complete. There should then be a poly-time reduction from (say) SAT to FACTORING. That is, an algorithm A that can solve SAT efficiently, provided that it has access to an efficient solver for FACTORING. Now consider what happens when pass a non-satisfiable formula Φ to A. What you end up concluding is that any run of A must constitute a (short) certificate that Φ is not satisfiable, implying NP=co-NP. A will process Φ, from time to time constructing instances of FACTORING that it can solve “for free”. Since every such instance of FACTORING must have a solution, A cannot fail for reasons of presenting the FACTORING oracle with an unsolvable instance, so must run its course. If it ends without finding a satisfying assignment, there isn’t one.
Possibly the commonest mistake (other than dressing badly) is to assume too much knowledge on the part of the audience. This mistake wastes the opportunity for a “big win” in which you present some fact or result that most people in the room don’t already know, but which should by rights be ingrained in their DNA. I got to do that via presenting Megiddo’s 1988 result that NP total search problems cannot be NP-complete unless NP=co-NP. To be clear, an NP total search problem is one where every instance has an easy-to-check solution, for example FACTORING, where the output is a prime factorization of an input number, or NASH, where the output is a Nash equilibrium of a given bimatrix game. This is the reason why we don’t bother to look for an NP-hardness result for NASH, and have to use an obscure-looking complexity class like PPAD.
It's possible to write a convincing proof sketch as follows. Consider what would it would mean if an NP total search problem (like FACTORING) was NP-complete. There should then be a poly-time reduction from (say) SAT to FACTORING. That is, an algorithm A that can solve SAT efficiently, provided that it has access to an efficient solver for FACTORING. Now consider what happens when pass a non-satisfiable formula Φ to A. What you end up concluding is that any run of A must constitute a (short) certificate that Φ is not satisfiable, implying NP=co-NP. A will process Φ, from time to time constructing instances of FACTORING that it can solve “for free”. Since every such instance of FACTORING must have a solution, A cannot fail for reasons of presenting the FACTORING oracle with an unsolvable instance, so must run its course. If it ends without finding a satisfying assignment, there isn’t one.
