For one thing it looks like it can be done by inspection.

Big O Notation defines the behavior of a function as it approaches some value. Often, we say infinity, but generally "an arbitrarily large number" is sufficient.

It is clear that as n gets large the n^2 term is going to drown out the other terms. So we can say by inspection that O(f(n)) == O(n^2)

Or maybe you could try to show a bit more rigorously.

But this looks like overkill to me.

You might pick up a few hints on how engineers view this problem from here:

http://stackoverflow.com/questions/1513 … -fn-is-o2n

One of them uses a method similar to your professor.

]]>from zero, cause I don't know how to use arrays I'mm fully not understaing this topic in java]]>

A column of table correspond to field how can any one give me example or pic

]]>There are at least 9 different definitions of empirical quantiles.

So both numpy and Mathematica etc are correct, depending on what definition the textbook is using.

Is 100 percentile possible? I think theoretically no

I would think that 100 percentile would mean the value that 100 percent of the data would be less than. But some definitions obviously include equal to also. I have seen many cases of 100 percentile computed and urge you to look at this answer:

http://math.stackexchange.com/questions … tile#33502

whuber, is an expert on statistics and he seems to indicate 100 percentile is allowed.

]]>The depiction is not convincing.

?

]]>Generally I think it means a small part of a program. It means a bit more in M. Did you mean Modular Programming?

]]>A few months ago I was checking out the Puzzle Games at MathsIsFun and stumbled upon this Orientation Puzzle. At first I didn't quite get it, but after about one or two minutes I understood it all. So, after playing some levels I've had the idea of implementing an algorithm that could solve any level. So I did it (that was actually quite a while back) and decided to share it with ya: Here's the source code and the level files (C++).

It's a command line program. compile it and put the "Levels" folder in the same directory as the executable. Open a command prompt window on that directory. Every time you want the program to solve a level, just type the name of the executable on the prompt, press enter and type the name of a level file contained in the "Levels" folder, for example: Levels/1.txt. The program will then display all the steps needed to get into the exit.

EDIT: I don't seem to be able to post the download link to the compiled executable

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