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-   -   Question on puzzle from lecture #5 (http://book.caltech.edu/bookforum/showthread.php?t=375)

 jcatanz 04-18-2012 05:32 PM

Question on puzzle from lecture #5

In considering the possible dichotomies of 3 points (subject to the restriction of a breakpoint at k=2) you said that

o o o
o o x
o x o

are allowed. Here, o and x can represent -1 and +1.

Then you said

o x x is not allowed,

x o o is allowed,

x o x,

x x o, and

x x x are not allowed.

So that for 3 points, half of the possible dichotomies are eliminated.

If there is a breakpoint at k = 2, all we can say is that not all of the 2^2 = 4 possible dichotomies {+1,-1}, {+1,+1} {-1,+1} {-1,-1} for two points are allowed.

How did you determine which dichotomies are not allowed?

:clueless:

 kkkkk 04-18-2012 08:02 PM

Re: Question on puzzle from lecture #5

I believe there are other possible solutions. So long that for any 2 of the 3 columns, if there are less than 2^N = 4 unique patterns, the break point of 2 is satisifed.

 jcatanz 04-19-2012 07:49 AM

Re: Question on puzzle from lecture #5

Quote:
 Originally Posted by kkkkk (Post 1440) I believe there are other possible solutions. So long that for any 2 of the 3 columns, if there are less than 2^N = 4 unique patterns, the break point of 2 is satisifed.
I agree.

But in the lecture Prof. Abu-Mostafa seemed to prompt the class, asking if each successive dichotomy (composed from counting to 8 in binary) were allowed or not allowed, as if the answer were unique.

 yaser 04-19-2012 12:27 PM

Re: Question on puzzle from lecture #5

Quote:
 Originally Posted by jcatanz (Post 1457) I agree. But in the lecture Prof. Abu-Mostafa seemed to prompt the class, asking if each successive dichotomy (composed from counting to 8 in binary) were allowed or not allowed, as if the answer were unique.
The unique is the numerical values of the maximum, namely 4. The identity of the 4 patterns is not unique.

 jcatanz 04-19-2012 02:38 PM

Re: Question on puzzle from lecture #5

OK, that's what I thought. Thanks, Prof. Abu-Mostafa.
--Joe

 vikram360 01-28-2013 11:18 AM

Re: Question on puzzle from lecture #5

Quote:
 Originally Posted by yaser (Post 1468) The unique is the numerical values of the maximum, namely 4. The identity of the 4 patterns is not unique.
So, ( I'm asking just to reinforce my understanding of this)
What Prof. Abu-Mostafa was saying was, given
o o o
o o x
o x o

are allowed, the following i.e

o x x
x o x
x x o
x x x

are not allowed ?

 vikram360 01-28-2013 11:44 AM

Re: Question on puzzle from lecture #5

Ok, I got it now. I hadn't watched the Q&A after the lectures (until now) so for anybody who's still having a problem, here's what I've understood.

* You're given 3 points and the break point k (=2 in this case).
* Start enumerating the possibilities and for convenience, lets do this in a binary sequence.
* Note that you don't know which 4th pattern of any two points is not allowed. You just know that one of the four is not allowed (because of the breakpoint).
* (In the lecture), black dots and white dots can be +1 or -1 ( or -1 and +1 - it doesn't matter). All that matters is that not all four combinations of 2 points are allowed. (In my explanation I use o and x)
* So let's start enumerating the possibilities
Possibility 1 (all white)
x1 x2 x3
0 0 0 (allowed because irrespective of which 2 points you take, you've seen just one pattern).
0 0 x (allowed because if you take x1 and x2, you've seen one pattern so far (00) and for x1 and x2 (or x2 and x3) you've seen two patterns (00 and 0x)
0 x 0 (allowed because for x1 and x2, you've seen two patterns so far (00 and 0x) and for x2 and x3 you've seen three patterns so far(00, 0x and x0) (still not 4!))
0 x x (BAM! If you take points x2 and x3, you've seen 4 patterns thus far i.e (00, 0x, x0 and xx) and this contradicts the behavior that is indicated by the breakpoint =2).

Similarly , all other combinations (in increasing binary sequence - if continued from the sequence above) except x 0 0 are 'illegal'.

 John Bang 01-28-2013 11:49 AM

Re: Question on puzzle from lecture #5

Quote:
 Originally Posted by vikram360 (Post 9036) So, ( I'm asking just to reinforce my understanding of this) What Prof. Abu-Mostafa was saying was, given o o o o o x o x o are allowed, the following i.e o x x x o x x x o x x x are not allowed ?
vikram360,

Assuming you meant to include [x o o] in the "allowed" group, then yes.

 yaser 01-28-2013 11:50 AM

Re: Question on puzzle from lecture #5

Quote:
 Originally Posted by vikram360 (Post 9037) Ok, I got it now. I hadn't watched the Q&A after the lectures (until now)
One of the advantages of the recorded Q&A session is that it was done live after the actual lecture, so if there was a part that was unclear during the lecture, the chances are there was a question about it and it was explained further in that session.

 Elroch 04-17-2013 09:40 AM

Re: Question on puzzle from lecture #5

Quote:
 Originally Posted by vikram360 (Post 9036) So, ( I'm asking just to reinforce my understanding of this) What Prof. Abu-Mostafa was saying was, given o o o o o x o x o are allowed, the following i.e o x x x o x x x o x x x are not allowed ?
I think John Bang's inference that you meant to include x o o in the "allowed list" is correct. But, as an alternative, your

o x x
x o x
x x o
x x x

would be allowed on their own but, similarly, it would not be possible to add any other.

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