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 Apr17 awarded Yearling Apr17 awarded Yearling Apr17 awarded Yearling Feb27 awarded Nice Answer Dec14 comment How to find perpendicular vector to another vector? There are many possible notation, I choose to use the same notation of the question, but other choice are good as well. $i$,$j$,$k$ refers to vectors $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$, so it is basically the same thing after you do vector-scalar multiplication. Dec6 answered pigeonhole problem understanding a step Dec6 revised Maximize $x_1x_2+x_2x_3+\cdots+x_nx_1$ deleted 6 characters in body Dec6 revised Maximize $x_1x_2+x_2x_3+\cdots+x_nx_1$ added 1 characters in body Dec6 revised Maximize $x_1x_2+x_2x_3+\cdots+x_nx_1$ deleted 79 characters in body Dec6 answered Maximize $x_1x_2+x_2x_3+\cdots+x_nx_1$ Dec3 answered Given a set of digits, what is the biggest number we can make using exponentiation - numberphile noodle quiz Dec3 comment Game theory problem: Poker with bluffing Yes, should be solvable even in this case, but you must remove strategy with the same payoff. For example "RC,RC,RC" dominates "RC,RC,RD" (and the opposite too), because you can always play the one of them instead of the other and get a greater or equal pay-off. Note that you can have some rounding errors in the code that lead to a fail of equality check, I suggest multiplying each payoff by 9, so you can work with integer. Dec1 answered Shortest path between two points that stays away from its rotated images Nov30 revised Men on a boat problem added 243 characters in body Nov30 revised Men on a boat problem added 60 characters in body Nov30 comment Men on a boat problem And from there a closed formula can be easily found... Nov30 answered Men on a boat problem Nov30 answered Symmetry of Solution to Classical 3-Dimensional Isoperimetric Problem Nov30 comment Solve Leonardo's purse problem: three merchants find a purse in road You meant $d=2x$, not $4x$, otherwise the first equation do not holds. Nov23 comment A game played on a rectangle Even by even is always a win for the second player (he can always play the move central symmetric to the last one). And similar to this odd by odd is always a win for the first player (play the central position, and mantain central symmetry).