# Re: An easy way to prove P != NP

*From*: Patricia Shanahan <pats@xxxxxxx>*Date*: Fri, 24 Nov 2006 18:51:28 GMT

Craig Feinstein wrote:

....

A person picks up a puppy in his hands and weighs himself and the puppy

together. Then he weighs himself and subtracts this weight from their

combined weight. This will be the weight of the puppy. This is clearly

the fastest way to weigh one puppy.

But is the fastest way to weigh 10 puppies to repeat this procedure 10

times? No, because the person does not have to weigh himself 10 times.

He only has to weigh himself once and weigh himself with another puppy

10 times.

By the Dynamic Programming Principle that I presented in my paper,

since he is solving each subproblem of getting the weight of each puppy

is the fastest way possible (by weighing himself with the puppy and

also weighing himself) without performing unnecessary work (he

doesn't weigh himself 10 times, which is unnecessary work; he only

weighs himself once), this is the fastest way to get the minimum weight

of the 10 puppies.

Suppose I add the information that nine of the puppies weigh the same,

and one is slightly lighter, not enough so to be visible given all the

squirming. The objective is still to find the weight of the lightest puppy.

How does your argument change?

Patricia

.

**References**:**An easy way to prove P != NP***From:*Craig Feinstein

**Re: An easy way to prove P != NP***From:*Craig Feinstein

**Re: An easy way to prove P != NP***From:*Patricia Shanahan

**Re: An easy way to prove P != NP***From:*Craig Feinstein

**Re: An easy way to prove P != NP***From:*Antti Virtanen

**Re: An easy way to prove P != NP***From:*Patricia Shanahan

**Re: An easy way to prove P != NP***From:*Craig Feinstein

**Re: An easy way to prove P != NP***From:*Patricia Shanahan

**Re: An easy way to prove P != NP***From:*Craig Feinstein

**Re: An easy way to prove P != NP***From:*Patricia Shanahan

**Re: An easy way to prove P != NP***From:*Craig Feinstein

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