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Old 2007-03-21, 12:53   #1
davieddy
 
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Default Rocky table

Since Mally has gone off in a huff, I shall put this teaser to you.

A four legged table typically rocks when placed on an unlevel floor.

However, with two simple assumptions, you can rectify the
problem by rotating the table through an angle <90 degrees.

How come? and what are the assumptions?

David
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Old 2007-03-21, 13:01   #2
R.D. Silverman
 
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Quote:
Originally Posted by davieddy View Post
Since Mally has gone off in a huff, I shall put this teaser to you.

A four legged table typically rocks when placed on an unlevel floor.

However, with two simple assumptions, you can rectify the
problem by rotating the table through an angle <90 degrees.

How come? and what are the assumptions?

David
I can do better: ONE simple assumption:

The height of the floor at each point is uniformly random. [the interval
does not matter]
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Old 2007-03-21, 13:11   #3
S485122
 
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Round table, the four legs are inside the table top.

You tilt the table, it rests on three points only and thus does not rock anymore.
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Old 2007-03-21, 13:24   #4
davieddy
 
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Neither of these were the answers I had in mind.
But at least I got some instantaneous response
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Old 2007-03-21, 13:29   #5
davieddy
 
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Quote:
Originally Posted by S485122 View Post
Round table, the four legs are inside the table top.

You tilt the table, it rests on three points only and thus does not rock anymore.
Until you try to cut the steak you're eating.
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Old 2007-03-21, 13:31   #6
davieddy
 
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Quote:
Originally Posted by R.D. Silverman View Post
I can do better: ONE simple assumption:

The height of the floor at each point is uniformly random. [the interval
does not matter]
I'm not sure I dined at that restaurant.
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Old 2007-03-21, 13:36   #7
hhh
 
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Quote:
Originally Posted by R.D. Silverman View Post
I can do better: ONE simple assumption:

The height of the floor at each point is uniformly random. [the interval
does not matter]
Is that you, Mr. Silverman?

Uniformly? random? interval? Can you explain what you mean please?

The only thing I can imagine you might be meaning is that the floor say of my office being discribed by the rectangle [0,3]x[0,5], for every point in that space, there is an identically and independently distributed (iid) (uniform) U[0,b] random variable, where b would be the end of the interval you are referring to.
That would be a hell of a floor, and it is not obvious (at least for me) why you should get the table fixed.

Yours, H.
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Old 2007-03-21, 14:08   #8
davieddy
 
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When I first told my good friend (now a professor at Princeton)
of my discovery, he went into deep thoght for a minute
then replied in excitement "Yes. And we can make it rigorous".
(That was about 35 years ago).

He didn't go into the details though.

David

Last fiddled with by davieddy on 2007-03-21 at 14:18
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Old 2007-03-21, 14:12   #9
R.D. Silverman
 
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Quote:
Originally Posted by hhh View Post
Is that you, Mr. Silverman?

Uniformly? random? interval? Can you explain what you mean please?

The only thing I can imagine you might be meaning is that the floor say of my office being discribed by the rectangle [0,3]x[0,5], for every point in that space, there is an identically and independently distributed (iid) (uniform) U[0,b] random variable, where b would be the end of the interval you are referring to.
That would be a hell of a floor, and it is not obvious (at least for me) why you should get the table fixed.

Yours, H.
The height of the floor is a uniform random variable taken over some interval
[a,b]. The values of a and b do not matter. Rotate the table using
one leg as a pivot. Any three legs must be co-planar. That there exists
*some* rotation where the 4th point is in the same plane follows from
e.g.: The Ham Sandwich Theorem, or Browder's Fixed Point Theorem
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Old 2007-03-21, 15:32   #10
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So, for the sake of understanding:

is it like I described, space $R^2$, for any $x\inR^2$ you take a U[0,1] r.v., all iid? This floor would be as I said highly discontinuous.

Or do you mean by "uniformly random" "arbitrary, but continuous". In this case, I can vaguely figure how it should work with what you cited (I think you meant Brouwer's fixed point theorem).
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Old 2007-03-21, 16:52   #11
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Quote:
Originally Posted by hhh View Post
So, for the sake of understanding:

is it like I described, space $R^2$, for any $x\inR^2$ you take a U[0,1] r.v., all iid? This floor would be as I said highly discontinuous.

Or do you mean by "uniformly random" "arbitrary, but continuous". In this case, I can vaguely figure how it should work with what you cited (I think you meant Brouwer's fixed point theorem).
Highly discontinuous? Indeed!! It is nowhere continuous.
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