查看完整版本: 世界上最难的逻辑题

世界上最难的逻辑题

[font=Times New Roman][size=4]世界最难逻辑题 [/size][/font]


[font=Times New Roman][size=4]The puzzle: Three gods A, B, and C are called, in some order, True, False, and　Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the　[/size][/font][font=Times New Roman][size=4]identities of A, B, and C by asking three yes-no questions; each question must be [/size][/font][font=Times New Roman][size=4]put to exactly one god. The gods understand English, but will answer all questions [/size][/font][font=Times New Roman][size=4]in their own language, in which the words for “yes” and “no” are “da” and “ja,” in [/size][/font][font=Times New Roman][size=4]some order. You do not know which word means which. [/size][/font]


[font=Times New Roman][size=4]有甲、乙、丙三个精灵，其中一个只说真话，另外一个只说假话。还有一个随机地决定何时说真话，何时说假话。你可以向这三个精灵发问三条是非题，而你的任务是从他们的答案找出谁说真话，谁说假话，谁是随机答话。这个难题困难的地方是这些精灵会以“Da”或“Ja”回答，但你并不知道它们的意思，只知道其中一个字代表“对”，另外一个字代表“错”。你应该问那三条问题呢？ [/size][/font]


[font=Times New Roman][size=4]《补充说明》 [/size][/font]


[font=Times New Roman][size=4]SOME YEARS AGO, THE LOGICIAN AND PUZZLE-MASTER [/size][/font][font=Times New Roman][size=4]Raymond Smullyan devised a logical puzzle that has no challengers I know [/size][/font][font=Times New Roman][size=4]of for the title of Hardest Logical Puzzle Ever. I’ll set out the puzzle here, give the solution, and then briefly discuss one of its more interesting aspects. [/size][/font]


[font=Times New Roman][size=4]（相对该文章）几年前，逻辑学家和难题制作大师Raymond Smullyan制作了一道就该文章作者所知无人挑战的逻辑难题，叫迄今最难的逻辑题。 [/size][/font]


[font=Times New Roman][size=4]The extra twist of not knowing which are the gods"" words for "yes" and "no" is due to the computer scientist John McCarthy. [/size][/font]


[font=Times New Roman][size=4]而有人还进一步加大该题的难度；计算机科学家 John McCarthy 将该题“扭”得更难：精灵的语言中哪个词表示“是”，哪个词表示“否”是未知的。 [/size][/font]


[font=Times New Roman][size=4]GEORGE BOOLOS IS [/size][/font][font=Times New Roman][size=4]Professor of Philosophy at [/size][/font][font=Times New Roman][size=4]MIT and one of the founders of [/size][/font][font=Times New Roman][size=4]the field of “Provability Logic.＂[/size][/font][font=Times New Roman][size=4]He is the author of The Logic of [/size][/font][font=Times New Roman][size=4]Provability (Cambridge, 1993) [/size][/font][font=Times New Roman][size=4]and, with Richard Jeffrey, [/size][/font][font=Times New Roman][size=4]Computability and Logic [/size][/font][font=Times New Roman][size=4](Cambridge, 1974). [/size][/font][font=Times New Roman][size=4]SPRING 1996 THE HARVARD REVIEW OF PHILOSOPHY [/size][/font]


[font=Times New Roman][size=4]George Boolos 是麻省理工学院哲学教授和“Provability Logic”的创始人之一，《The Logic of Provability》一书的作者。上述文章出现在1996年春的哈佛哲学评论.[/size][/font]


[font=Times New Roman][size=4]you can find the solution in chapter 29 of Boolos""s book Logic, Logic, Logic, published in 1998 by the Harvard University Press. [/size][/font]


[font=Times New Roman][size=4]文章的内容在Boolos的书《逻辑，逻辑，逻辑》的第29章。（1998年哈佛大学出版社出版）. [/size][/font]

[[i] 本帖最后由 genghis 于 2007-12-30 11:31 编辑 [/i]]

呵呵，有意思，初步分析，为了应对每一种da，ja的输出组合，提出的问题应该有普适性，既必须对所有可能的输出，有一个唯一的解释，毕竟提问机会只有一次嘛。这种问题做IC功能验证的兄弟应该比较在行哦。。。就是如何恰当的输入测试矢量，才能覆盖所有的可能性，并且最后将能够解释其输出。。。乱说一通 呵呵 ，再研究研究。。

ddddddddddddddddddddd

:victory: :victory: :victory:

赫赫，我去考考别人

呵呵，有意思

谢谢了，又学了一招

好玩！:lol :lol

:victory: 有意思！

很有意思

呵呵，有意思

的人太多让他的人当然

*** 作者被禁止或删除 内容自动屏蔽 ***

很有意思！！学习了

*** 作者被禁止或删除 内容自动屏蔽 ***

有点意思！

dddddddddddd

比较难。

的确很难

厉害啊