Mike Paterson Overhang Academic Sponsors’ Day, MSRI, March 2, 2012 Peter Winkler Uri Zwick Yuval...
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Transcript of Mike Paterson Overhang Academic Sponsors’ Day, MSRI, March 2, 2012 Peter Winkler Uri Zwick Yuval...
![Page 1: Mike Paterson Overhang Academic Sponsors’ Day, MSRI, March 2, 2012 Peter Winkler Uri Zwick Yuval Peres Mikkel Thorup.](https://reader036.fdocuments.in/reader036/viewer/2022062515/56649cee5503460f949bc0d3/html5/thumbnails/1.jpg)
Mike Paterson
Overhang
Academic Sponsors’ Day, MSRI, March 2, 2012
Peter Winkler
Uri Zwick
Yuval Peres
Mikkel Thorup
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A Crow Problem:
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How long does it take to drive off the crow?
-n 0 1-1-2 2 n
Simple random walk: about n2 throws.
One way to see that: consider the probability distribution of crow’s location; its variance
goes up by 1 after each throw.
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A new problem, brought to MSRI in spring ’05
by Zwick: the crow comes back…
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…at night!
Now what---your first stone will hit the crow and dislodge him, but after that
you’re increasingly unsure where he is.
You can certainly get him off the wallin order n3 throws, and you certainly
still need at least n2 . Which is the truth?
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Theorem: Order n3 throws are necessary.
Proof: Uses two different potential functions, each for the wrong problem.
An unusual case of two wrongs making a right.
-n 0 n
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The overhang problem
How far off the edge of the table can we reach by stacking n identical blocks of length 1?
“Real-life” 3D version
Idealized 2D version
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Back in time with the overhang problem…
John F. Hall, Fun with Stacking Blocks, Am. J. Physics December 2005.
Martin Gardner - Scientific American’s “Mathematical Games” column, 1969.
J.G. Coffin – Problem 3009, American Mathematical Monthly, 1923.
George M. Minchin, A Treatise on Statics with Applications to Physics, 6th ed. (Clarendon, Oxford, 1907), Vol. 1, p. 341.
William Walton, A Collection of Problems in Illustration of the Principles of Theoretical Mechanics 2nd ed. (Deighton, Bell, Cambridge, 1855), p. 183.
J.B. Phear, Elementary Mechanics (MacMillan, Cambridge, 1850), pp. 140–141.
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The classical solution
“Harmonic Stack”
Using n bricks we can get an overhang
of
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Is the classical solution optimal?
Apparently not. How can we improve the construction?
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Inverted pyramids?
Claimed to be stable in Mad About Physics, by Chris Jargodzki and Franklin Potter,
but…
![Page 12: Mike Paterson Overhang Academic Sponsors’ Day, MSRI, March 2, 2012 Peter Winkler Uri Zwick Yuval Peres Mikkel Thorup.](https://reader036.fdocuments.in/reader036/viewer/2022062515/56649cee5503460f949bc0d3/html5/thumbnails/12.jpg)
They are unbalanced, when the number of layers exceeds 2.
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Diamonds?
The 4-diamond is balanced…
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But the 5-diamond is …
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not.
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What really happens?
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What really happens!
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Why is this unbalanced?
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… and this balanced?
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Equilibrium
F1 + F2 + F3 = F4 + F5
x1 F1+ x2 F2+ x3 F3 = x4 F4+ x5 F5
Force equation
Moment equation
F1
F5F4
F3
F2
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Forces between bricks
Assumption: No friction.
All forces are vertical.
Equivalent sets of forces
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Balanced Stacks
Definition: A stack of bricks is balanced iff there is an admissible set of forces under which each brick is in equilibrium.
1 1
3
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How can we tell if a stack is balanced?
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Checking for balance
F1F2 F3 F4 F5 F6
F7F8 F9 F10
F11 F12
F13F14 F15 F16
F17 F18
Equivalent to the feasibilityof a set of linear
inequalities:
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Stability and Collapse
A feasible solution of the primal system gives a set of balancing
forces.
A feasible solution of the dual system describes an infinitesimal
motion that decreases the potential energy.
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Small optimal stacks
Overhang = 1.16789Bricks = 4
Overhang = 1.30455Bricks = 5
Overhang = 1.4367
Bricks = 6Overhang = 1.53005
Bricks = 7
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Small optimal stacks
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Small optimal stacks
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Small optimal stacks
Overhang = 2.14384Bricks = 16
Overhang = 2.1909Bricks = 17
Overhang = 2.23457Bricks = 18
Overhang = 2.27713Bricks = 19
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Support and counterweight bricks
Support
set
Counter-weights
These examples are “spinal”: support stack has only one brick per level, so overhang increases
with height.Spinal stacks can achieve overhang S(n) ~ log n.
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100 bricks example
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But are spinal stacks optimal?
No! When # bricks reaches 20 . . .
Support set is not spinal.
Overhang = 2.32014, slightly exceeding S(20).
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Optimal weight 100 construction
Overhang = 4.20801
Bricks = 47Weight = 100
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Brick-wall constructions
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Brick-wall constructions
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“Parabolic” construction
5-stack
Number of bricks: Overhang:Stable!
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Thus: n bricks can achieve an overhang of order n1/3 ...
an exponential improvement over the order log n overhang
of spinal stacks.
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Mayan from 900 BC---no keystone
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Yes! Argument is based on the idea that laying bricks is like stoning crows.
Each additional brick…
spreads forces the same way that throwing a stone (at night) spreads the crow’s probability
bar.
The Upper BoundIs order n1/3 best possible??
![Page 40: Mike Paterson Overhang Academic Sponsors’ Day, MSRI, March 2, 2012 Peter Winkler Uri Zwick Yuval Peres Mikkel Thorup.](https://reader036.fdocuments.in/reader036/viewer/2022062515/56649cee5503460f949bc0d3/html5/thumbnails/40.jpg)
In particular, a stack of only n bricks cannot overhang by more than 6n1/3 brick
lengths. The parabolic construction gives overhang (3/16)1/3 n1/3 ~ .572357121 n1/3, so we have the order right but the constant is off by an
order of magnitude.
Simulations suggest that the constant can be improved by adjusting the shape
of the brick wall construction…
A generalized version of the “stoning crows” analysis shows that it takes order n3 bricks to
get the stack to lean out by n .
![Page 41: Mike Paterson Overhang Academic Sponsors’ Day, MSRI, March 2, 2012 Peter Winkler Uri Zwick Yuval Peres Mikkel Thorup.](https://reader036.fdocuments.in/reader036/viewer/2022062515/56649cee5503460f949bc0d3/html5/thumbnails/41.jpg)
“Vases”
Weight = 1151.76
Bricks = 1043
Overhang = 10
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“Vases”
Weight = 115467.
Bricks = 112421
Overhang = 50
![Page 43: Mike Paterson Overhang Academic Sponsors’ Day, MSRI, March 2, 2012 Peter Winkler Uri Zwick Yuval Peres Mikkel Thorup.](https://reader036.fdocuments.in/reader036/viewer/2022062515/56649cee5503460f949bc0d3/html5/thumbnails/43.jpg)
“Oil lamps”
Weight = 1112.84
Bricks = 921
Overhang = 10
giving overhang of about 1.02 n1/3 .
![Page 44: Mike Paterson Overhang Academic Sponsors’ Day, MSRI, March 2, 2012 Peter Winkler Uri Zwick Yuval Peres Mikkel Thorup.](https://reader036.fdocuments.in/reader036/viewer/2022062515/56649cee5503460f949bc0d3/html5/thumbnails/44.jpg)
How about using the third dimension?Our upper bound proof makes no use of the fact
that bricks cannot overlap in space! Hence, the 6n1/3 bound applies even in 3D, as long as there are no non-vertical forces.
However, the constant can be improved in space by skintling,
Effectively increasing the brick length to (1+w)1/2 .
![Page 45: Mike Paterson Overhang Academic Sponsors’ Day, MSRI, March 2, 2012 Peter Winkler Uri Zwick Yuval Peres Mikkel Thorup.](https://reader036.fdocuments.in/reader036/viewer/2022062515/56649cee5503460f949bc0d3/html5/thumbnails/45.jpg)
Open problems● What is the correct constant in the maximum
overhang, in the rectilinear case? In the general 3-dimensional case?
● What is the asymptotic shape of “vases”?● What is the asymptotic shape of “oil lamps”?● What is the gap between brick-wall
constructionsand general constructions?
● Can the proof be extended to cover non-vertical forces (if, indeed, they are possible for 3D bricks)?
● How much friction is needed to change the 1/3 exponent for overhang?
![Page 46: Mike Paterson Overhang Academic Sponsors’ Day, MSRI, March 2, 2012 Peter Winkler Uri Zwick Yuval Peres Mikkel Thorup.](https://reader036.fdocuments.in/reader036/viewer/2022062515/56649cee5503460f949bc0d3/html5/thumbnails/46.jpg)
Thank you for your attention. Happy stacking…