A Game of Chance (Dana Whitestone, Book 2) by Jon Osborne

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.

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Additional resources for A Game of Chance (Dana Whitestone, Book 2)

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Sans partie else. La partie else omise est implicitement compl´et´ee par le compilateur, qui ajoute else (), autrement dit « sinon rien ». Ainsi, le code : if i >= 0 then begin ... end;; est compris par Caml comme si nous avions ´ecrit : if i >= 0 then begin ... end else ();; Cette compl´etion automatique vous explique pourquoi la phrase suivante est mal typ´ee : # if true then 1;; Entr´ ee interactive: >if true then 1;; > ^ Cette expression est de type int, mais est utilis´ ee avec le type unit.

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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.

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