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In computer science and mathematics, the Josephus problem (or Josephus permutation) is a theoretical problem related to a certain counting-out game. Such games are used to pick out a person from a group, e.g. eeny, meeny, miny, moe. A drawing for the Josephus problem sequence for 500 people and skipping value of 6.
The problem for graphs is NP-complete if the edge lengths are assumed integers. The problem for points on the plane is NP-complete with the discretized Euclidean metric and rectilinear metric. The problem is known to be NP-hard with the (non-discretized) Euclidean metric. [3]: ND22, ND23
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The question is whether or not, for all problems for which an algorithm can verify a given solution quickly (that is, in polynomial time), an algorithm can also find that solution quickly. Since the former describes the class of problems termed NP, while the latter describes P, the question is equivalent to asking whether all problems in NP are ...
This article says, “Josephus had an accomplice; the problem was then to find the places of the two last remaining survivors (whose conspiracy would ensure their survival). It is alleged that he placed himself and the other man in the 31st and 16th place respectively (for k = 3 below).”
The activity selection problem is also known as the Interval scheduling maximization problem (ISMP), which is a special type of the more general Interval Scheduling problem. A classic application of this problem is in scheduling a room for multiple competing events, each having its own time requirements (start and end time), and many more arise ...
Uzquiano (2010) uses these techniques to provide a two question solution to the amended puzzle. [9] [10] Two question solutions to both the original and amended puzzle take advantage of the fact that some gods have an inability to answer certain questions. Neither True nor False can provide an answer to the following question.
Case 2.1 If the new backward(v) is not empty, then the components have not changed: there are other edges which connect v backwards. Process B halts (and process A is halted too). Case 2.2 If the new backward(v) is empty, then v is no longer connected to level i−1, and so its distance from the root is no longer i; it must be at least i+1.