Hump-day puzzler - Unknown Function

Asked by Matt Fig on 25 May 2011
Latest activity Commented on by Matt Fig on 25 May 2011

My colleague was working with MATLAB and defined an anonymous function F. When he saw me come in he typed clc, hit return than challenged me to guess the form of the function only by calling it however many times I want. After calling it many times, I have figured out that it takes a scalar argument, and when the value of the argument is greater than 0 it does this:

>> F(1)
  Name      Size            Bytes  Class     Attributes

    ---- 0;']) ------------------------------------------
    x         1x1                 8  double

And when the value of the argument is less than zero, it appears to do nothing:

>> F(-1)
>>

Can you help me figure out what the function looks like?

2 Comments

Matt Fig on 25 May 2011

Hint #1. The function does not call an M-File

Matt Fig on 25 May 2011

Hint #2. Here's how I know the function wants a scalar argument...

>> S.one = 't';
>> F(S)
>> F({3})
>> F(magic(2))
>>

Matt Fig

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2 Answers

Answer by Laura Proctor on 25 May 2011
Accepted answer

I get the sense that this is not exactly what the user did, but this seems to do what is expected as a first try:

F=@(x) eval('if isscalar(x)&&x>0, whos x, end')

===========================

Second iterate: I needed to add in a check for type double. It also prints out the funky characters with WHOS (but without any functionality). :oP

F=@(x) eval(['if isscalar(x)&&isa(x,''double'')&&x>0, whos x, end; 1>0;'])

5 Comments

Show 2 older comments
Oleg Komarov on 25 May 2011

I don't understand what's happening!!!

Matt Fig on 25 May 2011

Do you know why WHOS prints this way when used like that? I.E., why it behaves differently than it does here:

F=@(x) eval(['if isscalar(x)&&isa(x,''double'')&&x<0, whos x, end; 1<0;'])

Matt Fig on 25 May 2011

@Oleg, you aren't the only one who doesn't understand why this happens. I was hoping Laura could shed some light. See my previous comment.

Laura Proctor
Answer by Sean de Wolski on 25 May 2011

Tough one! In lieu of a real way to name a function/function handle: 0;'])

F = @(x)HDC525(x);

and:

function HDC525(x)
%SCd 5/25/2011: Lame attempt at HD challenger
if ~isscalar(x)||x<=0
  return
else
  Str = evalc('whos x');
  fprintf('%s\n   ---- 0;'']) ------------------------------------------\n %s',Str(1:numel(Str)/2-1),Str(numel(Str)/2+1:end));
end

2 Comments

Matt Fig on 25 May 2011

See Hint #1. Though I think you know this. I will post Hint #2 in a little while. This is a tough one!

Sean de Wolski on 25 May 2011

I know it doesn't. And I know that 'whos' will put the function handle's meat where you have _0;'])_ but I know how to overwrite it. Arghhhhh.

Sean de Wolski

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