By Jon Osborne
A felony mastermind is engaging in a dangerous video game of homicide at the streets of latest York. Following the principles of chess he strikes his sufferers round the urban, leaving his sinister calling playing cards at every one blood-soaked crime scene. yet because the physique count number mounts, the hot York police strength are not any closer to catching this sadistic killer. And specialist profiler, FBI specified Agent Dana Whitestone, is introduced in to help.
Still bruised from her come across with The Cleveland Slasher, Dana, and her accomplice Jeremy Brown, quickly turn into embroiled in a macabre online game of cat and mouse as they struggle to seek down this very smart yet completely ruthless assassin earlier than he claims his subsequent sufferer. after which they understand they could be up opposed to no longer one killer yet - and all hell breaks loose...
About the Author
Jon Osborne has been a newspaper reporter for a decade, so much lately for the Naples day-by-day information in Florida. he's a army veteran and lives in Bonita Springs, Florida.
Read Online or Download A Game of Chance (Dana Whitestone, Book 2) PDF
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Additional resources for A Game of Chance (Dana Whitestone, Book 2)
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Plus pr´ecis´ement, il faut n multiplications, n+1 appels r´ecursifs `a la fonction factorielle et n soustractions. Si l’on consid`ere que ces trois types d’op´erations ont des coˆ uts voisins, alors la complexit´e de factorielle est de l’ordre de 2n + (n + 1), c’est-` a-dire de l’ordre de 3n. On consid´erera donc que la fonction factorielle a une complexit´e qui augmente au mˆeme rythme que son argument, ce qu’on note O(n) et qu’on prononce « grand-o de n ». Plus pr´ecis´ement, O(n) signifie « un certain nombre de fois » n, plus des termes n´egligeables devant n quand n devient grand, comme par exemple une constante.
1: Comment r´esoudre le probl`eme des tours de Hanoi. 30 R´ecursivit´e Le programme Supposons que les tiges s’appellent A, B et C, que n soit le nombre de disques, tous pos´es au d´epart sur la tige A, et que nous devions les mettre sur la tige C. L’astuce consiste ` a se rendre compte que si nous savions comment r´esoudre le probl`eme pour n − 1 disques alors nous saurions le faire pour n, sans violer la r`egle. En effet, si l’on suppose les n − 1 disques d´ej` a pos´es sur la tige B, le dernier disque encore pos´e sur la tige A est le plus gros disque.