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seen Dec 6 '13 at 12:24

Apr
17
awarded  Yearling
Apr
17
awarded  Yearling
Feb
27
awarded  Nice Answer
Dec
14
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.
Dec
6
answered pigeonhole problem understanding a step
Dec
6
revised Maximize $x_1x_2+x_2x_3+\cdots+x_nx_1$
deleted 6 characters in body
Dec
6
revised Maximize $x_1x_2+x_2x_3+\cdots+x_nx_1$
added 1 characters in body
Dec
6
revised Maximize $x_1x_2+x_2x_3+\cdots+x_nx_1$
deleted 79 characters in body
Dec
6
answered Maximize $x_1x_2+x_2x_3+\cdots+x_nx_1$
Dec
3
answered Given a set of digits, what is the biggest number we can make using exponentiation - numberphile noodle quiz
Dec
3
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.
Dec
1
answered Shortest path between two points that stays away from its rotated images
Nov
30
revised Men on a boat problem
added 243 characters in body
Nov
30
revised Men on a boat problem
added 60 characters in body
Nov
30
comment Men on a boat problem
And from there a closed formula can be easily found...
Nov
30
answered Men on a boat problem
Nov
30
answered Symmetry of Solution to Classical 3-Dimensional Isoperimetric Problem
Nov
30
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.
Nov
23
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).
Nov
22
comment Finding minimal cost edge cover for a bipartie graph
Take a complete bipartite graph with $n$ vertex to the left and $n$ to the right, and with each edge with the same cost. A minimum spanning tree require $2n-1$ edge, but taking a maximum matching satisfy the question for only $n$ edge.