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The problem is named after Flavius Josephus, a Jewish historian and leader who lived in the 1st century. According to Josephus's firsthand account of the siege of Yodfat, he and his 40 soldiers were trapped in a cave by Roman soldiers. They chose suicide over capture, and settled on a serial method of committing suicide by drawing lots.
A good example that highlights the pros and cons of using dynamic arrays vs. linked lists is by implementing a program that resolves the Josephus problem. The Josephus problem is an election method that works by having a group of people stand in a circle. Starting at a predetermined person, one may count around the circle n times.
NP-complete special cases include the edge dominating set problem, i.e., the dominating set problem in line graphs. NP-complete variants include the connected dominating set problem and the maximum leaf spanning tree problem. [3]: ND2 Feedback vertex set [2] [3]: GT7 Feedback arc set [2] [3]: GT8 Graph coloring [2] [3]: GT4
Near the end, the article says "Therefore, if we represent n as 2m + l, where 0 < = l < 2m, then f(n) = 2 * l + 1." I think the m's should be n's; there's no m in the problem. --ngb You are correct that there is no m in the statement of the problem.
In computer science, a Fibonacci heap is a data structure for priority queue operations, consisting of a collection of heap-ordered trees.It has a better amortized running time than many other priority queue data structures including the binary heap and binomial heap.
In the complementary case, a consumer might make the queue empty and wake up another consumer instead of a producer, and the consumer would go back to sleep. Using broadcast ensures that some thread of the right type will proceed as expected by the problem statement. Here is the variant using only one condition variable and broadcast:
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Using this method they obtain a structure that can maintain items with integer priorities in a range from 0 to a parameter C. The hot queue uses constant time per insertion or decrease-priority operation and amortized time O ( ( log C ) 1 / 3 ( log log C ) 1 / 2 ) {\displaystyle O((\log C)^{1/3}(\log \log C)^{1/2})} per extract-min ...