Four Squares Reborn

gAr · 2014-03-01 18:51:27

Hi,

How do you get the first 100 squares with your favorite language?

In J, it's simple:

(1+i.100)^2

Range[100]^2 in Mm?

bobbym · 2014-03-01 18:59:37

Hi;

gAr · 2014-03-01 19:05:25

Functional programming is very powerful once we get a hang of it.

bobbym · 2014-03-01 19:09:59

I agree, I am hoping I can convince Agnishom of that.

gAr · 2014-03-01 19:12:16

Anybody will, once they realize how compact the code can get!

bobbym · 2014-03-01 19:18:58

Yes, he is a very bright guy and luckily he does not have lots of years of programming in procedural languages to make it difficult.

gAr · 2014-03-01 19:21:45

That's right.

bobbym · 2014-03-01 19:33:30

Did you see the new problem in Bafflers?

gAr · 2014-03-02 00:16:05

Yes, I was trying a way to use only the formulas for sector and square, could not proceed that way.

bobbym · 2014-03-02 00:19:46

Oh okay, you will get it.

Agnishom · 2014-03-02 01:08:13

bobbym wrote:

How do you get the first 100 squares with your favorite language?

Python Code:

for i in xrange(1,101): print i**2

Functional Way (Python):

map(lambda x: x**2,xrange(1,101))

Maxima Code:

for a: 1 thru 100 do display(a^2);

Functional Way (Maxima):

makelist(k^2,k,1,100);

Mathematica Code:

Function[x,x^2] /@ Range[1,100]

or

Map[Function[x,x^2],Range[1,100]]

bobbym · 2014-03-02 01:14:01

Simpler is Range[100]^2.

Agnishom · 2014-03-02 01:21:47

But that is a matrix operation of which I know nothing.

he does not have lots of years of programming in procedural languages to make it difficult.

I've got atleast 7 years of Procedural experience.

bobbym · 2014-03-02 01:27:30

Nope, that is not a matrix operation. You are thinking it is like m x m but it is not.

Range[100] generates a list from1 to 100. Range[100]^2 means square every element of the list. This is why loops are unnecessary. In functional languages commands work on entire lists as if they were single objects. This is more readable and saves the programmer the details of indexing an array.

Agnishom · 2014-03-02 01:30:24

Range[100]^2 means Range[100] multiplied with Range[100]. Which is multiplying the matrix [1,2,3,...100] with the matrix [1,2,3,...100]. That is how it works.

bobbym · 2014-03-02 01:35:11

It is not a matrix, it is a list. Let me illustrate.

What do you think Range[10]^(1/2) does?

Agnishom · 2014-03-02 01:36:07

If you are thinking that InsertSomeFunction[Range[100]] is the same as InsertSomeFunction /@ Range[100] then you are wrong.

Proof. Assume the contrary.

See this counterexample: there is a difference between Speak[Range[100]] is not the same as Speak /@ Range[100]

It works in the other case because it is a matrix operation.

What do you think Range[10]^(1/2) does?

It multiplies the matrix {1,2,3,4,5,6,7,8,9,10} with the constant 1/2

Last edited by Agnishom (2014-03-02 01:37:32)

bobbym · 2014-03-02 01:37:14

It is not a matrix, it is a list. Let me illustrate.

What do you think Range[10]^(1/2) does?

Agnishom · 2014-03-02 01:38:41

What do you think Range[10]^(1/2) does?

It multiplies the matrix {1,2,3,4,5,6,7,8,9,10} with the constant 1/2

A matrix and a list is the same thing

Last edited by Agnishom (2014-03-02 01:40:22)

bobbym · 2014-03-02 01:43:33

That is incorrect.

Range[10]^(1/2)

Range[10]^n

See how it was mapped to each element. Single brackets are lists.

Sin[Range[10]]

Agnishom · 2014-03-02 01:45:42

That is incorrect.

Sorry, I meant that it raises the matrix {1,2,3,4,5,6,7,8,9,10} to the power 0.5

Then why isn't Speak[Range[100]] and Speak /@ Range[100] the same?

bobbym · 2014-03-02 01:49:56

Speak does that because it treats the argument as a string. Just reading what it sees. Because for most functions have the attribute called listable. That means that they map themselves over lists.

{1,2,3,4,5,6,7,8,9,10} is not a matrix, it is a list. In M, matrices are lists inside of lists. {{1,2,3},{4,5,6},{7,8,9}} that is a matrix. You see, in M everything is a list just like lisp.

Agnishom · 2014-03-02 01:53:53

Is it okay to say that for any list L, f[L] does the same thing as f /@ L?

bobbym · 2014-03-02 01:57:16

Yes, in all cases that I have seen that will be true. The \@ means map. Just like a mapping in mathematics. It maps the function f onto all the elements of the list.

Please do not get confused here. In M, the list is used to represent vectors, matrices, sets, and arrays. It depends on the context and what commands you are using afterward whether you are speaking of a list or a vector.

Agnishom · 2014-03-02 02:03:58

Okay, can we get back to explaining that code?

Math Is Fun Forum

#26 2014-03-01 18:51:27

Re: Four Squares Reborn

#27 2014-03-01 18:59:37

Re: Four Squares Reborn

#28 2014-03-01 19:05:25

Re: Four Squares Reborn

#29 2014-03-01 19:09:59

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#30 2014-03-01 19:12:16

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#31 2014-03-01 19:18:58

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#32 2014-03-01 19:21:45

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#33 2014-03-01 19:33:30

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#34 2014-03-02 00:16:05

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#35 2014-03-02 00:19:46

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#36 2014-03-02 01:08:13

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#37 2014-03-02 01:14:01

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#38 2014-03-02 01:21:47

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#39 2014-03-02 01:27:30

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#40 2014-03-02 01:30:24

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#41 2014-03-02 01:35:11

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#42 2014-03-02 01:36:07

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#43 2014-03-02 01:37:14

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#44 2014-03-02 01:38:41

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#45 2014-03-02 01:43:33

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