A Conjecture for the Packing Density of 2413

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A Conjecture for the Packing Density of 2413

• Walter Stromquist Swarthmore College
• (joint work with Cathleen Battiste Presutti,
  Ohio University at Lancaster)
• PP2007
• St. Andrews, Scotland
• June 12, 2007

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Outline

• 1. About packing densities
• 2. Packing rates for measures
• a. There is an optimal measure for every
  pattern
• b. Packing rate Packing density
• 3. An optimal measure for 2413
• the best we can do without recursion
• 4. Amazing things happen with recursion bubbles
• 5. Conjecture for 2413 0.10472422757673209041
  

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About packing densities

• A pattern is a permutation ? in Sm.
• Let ?n ? Sn. An occurrence of ? in ?n is an
  m-element
• subsequence of ?n that has the same order
  type as ?.
• reduces to ?.
• Define
• The packing density of ? in ?n is
• Clearly,

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About packing densities

• Were concerned with permutations ?n?Sn that
  maximize the
• packing density ?( ?, ?n ). So, define
• Any permutation ?n that achieves this maximum
  (for a given
• size n) is called an optimizer for ? (or,
  optimizing permutation).
• The packing density of ? is the limiting value,
• (It always exists.)

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About packing densities

• Equivalently The packing density of ? is the
  largest number D
• such that there is a sequence of permutations
• of increasing size such that
• The ?n s dont have to be optimizers they
  just have to be close enough that they have the
  right limit.

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Describing the near-optimizers

• Heres how we describe good ?ns for 132

? Thats what the ?ns look like. The packing
density turns out to be
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Describing the near-optimizers

• Heres an easier picture

? The points line up along these lines.
This object is a probability measure on the unit
square. Lets formalize it.
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Measures on the Unit Square

• Let S be the unit square,
• S 0, 1 ? 0, 1 ? R2.
• A probability measure ? on S is a probability
  distribution on S.
• If a point is selected randomly according to ?,
  then ?(A) is the probability that the point is
  in A.

? is non-negative and countably additive. Also,
?(S) 1. All measures in this talk are
probability measures on S.
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The packing rate for a measure

• Let ? ?? ??m be a pattern, and let ? be a
  measure.
• If m points are selected randomly according to
  ?, then the packing rate of ? with respect to
  ? is the probability that the order type of
  their configuration is ?.
• We denote this packing rate by ? ( ?, ? ) .
• More precisely Let ?m ? ? ? ? ? ? be the
  product measure on S m S ? S ? ? S. Let
  A? ? Sm contain all m-tuples of points that form
  configurations with order type ?. Then
• ? ( ?, ? ) ?m ( A? ).

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order type of their configuration
If the points are ordered by increasing x
coordinates, x1 lt x2 lt lt xm, then the
order type of the configuration is the order type
of their y coordinates ( y1, y2, , ym ).
The order in which the points are selected is
irrelevant. If any two points have the same
x coordinate or the same y
coordinate, then the configuration has no order
type.
These four points have order type 2314.
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Examples of packing rates

• Example. Let ? be the uniform measure on S.
• Then all order types are equally likely. We
  therefore have
• ? ( 123, ? ) 1/6
• and, in general,
• ? ( ?, ? ) 1 / m!
• if ? has size m.

Points are drawn uniformly from the unit square.
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Examples of packing rates

• Example. Let ? be concentrated along the main
  diagonal.
• Then
• ? ( 123, ? ) 1
• and
• ? ( ?, ? ) 0
• for any other pattern of size 3.

It doesnt matter how probability is distributed
along the diagonal.
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Examples of packing rates

• Example. Let ? be concentrated along countably
  many diagonal segments as shown.
• Then
• ? ( 132, ? )
• This is the packing density of 132. No other
  measure gives a higher packing rate for 132.

This is the optimal measure for 132.
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Examples of packing rates

• Example. Let ? be uniform
• on a disk in S.
• WHAT IS ? ( 123, ? ) ?

Points are drawn uniformly from the disk.
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Optimal Measures

• Given a pattern ?, the optimal packing rate
  of ? (or just the packing rate of ? ) is
• ?(?) ( ? (?, ?) )
• where the supremum is taken over all measures.
• Any measure ? that realizes the supremum is
  called an optimal measure for ?.
• Does every pattern have an optimal measure ?

YES.
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Optimal Measures

• Is there always an optimal measure?
• Why not just find measures ?i whose packing
  rates approach the maximum, and take the limiting
  measure?
• (1) Its tricky to define limiting measures
• (2) Limits of measures dont always preserve
  packing rates

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Limits of measures

• If ?i is a measure for each i and ? is a
  measure, we say that
• ? lim ?i
• if
• for every continuous function f on S.
• With this definition, the set of probability
  measures on S forms a compact topological space.
• So, for any sequence ?i of measures, there is
  a subsequence ?i with a limit ? lim
  ?i.
• But the packing density of ? is not always equal
  to the limit of the packing rates of the ?is.

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Example limits dont preserve packing rates

• In this picture, ? lim ?i.
• But ? ( 132, ?i ) 1/6 for each ?i, while
  ? ( 132, ? ) 0.

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Normalized Measures

• A measure ? on S is normalized if its
  projections on both
• the x and the y axis are uniform distributions.
  
• A limit normalized measures is normalized.
• If the measures ?i are all normalized and ?
  lim ?i then
• ? ( ?, ?i ) lim ? ( ?, ?i ) .
• Every measure can be normalized by monotone
  transformations of the x and y coordinates. This
  operation cant reduce packing rates.

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Every pattern has an optimizing measure

• Theorem For every pattern ?, there is a
  measure ? such that
• ? (?, ?) ( ? (?, ? ) )
  ?(?) .
• The measure ? can be chosen to be normalized.
  
• Proof Pick measures ?i whose packing
  densities approach the supremum. Replace them
  with normalized measures ?i. Find a
  subsequence that converges to a limit measure ?.
  Then ? is normalized, and is an optimizing
  measure for ?. //
• Question Is the normalized optimal measure for
  ? always unique?
• (Probably not, but it would be very nice if it
  were.)

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Packing rates Packing densities

• Theorem The optimal packing rate of any pattern
  is its packing density
• ? ( ? ) ? ( ? ).

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Packing rates Packing densities

• Why ? ( ? ) ? ? ( ? )
• Given any measure ?, just pick ?n randomly
  according to ?. Then on average the packing
  rate of ? in ?n is exactly ? (?, ? ).
• So, we can pick particular ?ns that achieve
  this average, and then we have
• lim ? ( ?, ?n ) ? ( ?, ? ).
• This forces ? ( ? ) ? ? ( ? ) .

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Packing rates Packing densities

• Why ? ( ? ) ? ? ( ? )
• Find a sequence of measures ?n with lim ? (
  ?, ?n ) ? ( ? ).
• For each ?n , construct a template measure
  (Presutti, PP2006)
• Now the limit of the template measures (or of
  some subsequence of them) is a measure ? with
  ? ( ?, ? ) ? ( ? ).
• So ? ( ? ) ? ? ( ? ).

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Summary

• For every pattern ?, there is a normalized
  optimizing measure ? that maximizes ? ( ?, ? ).
• For that measure, ? ( ?, ? ) ? ( ? ).
• So, to find the packing density, it suffices to
  find an optimizing measure.

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About 2413

• Why 2413 ?
• (1) Least-understood pattern of size 3 or 4
• (2) As far as possible from being layered
• What we know
• ? ( 2413 ) ? 6/64 0.09375 (because that bound
• works for any ? of size 4)
• ? ( 2413 ) ? 51/511 0.099804 (AAHHS, template
  of size 8)
• ? ( 2413 ) ? 0.104250980 (Presutti, weighted
  template
• of size 16)
• Upper bound
• ? ( 2413 ) ? 2/9 (AAHHS, not likely tight)
• Todays conjecture
• ? ( 2413 ) 0.10472422757673209041

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An optimal measure for 2413 ?

• So, what is an optimal measure for 2413?
• By symmetry and computational experience, we
  suspect a measure of this form

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Distribution need not be uniform
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Probability distribution along this line F(t)
F(t) fraction of this segments probability
that is in the leftmost fraction t of the segment
t0 .t1
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Symmetrical Four-Segment Measures

• (1) Probability is concentrated along the four
  segments shown
• (2) Measure has four-fold rotational symmetry
• (3) Measure is partially normalized combined
  mass of top and
• bottom segments is uniform on 1/4, 3/4, and
  similarly for
• side segments
• F(t) ( 1 F(1-t) ) 2t for t in 0, 1
• ? Values of F on 0, ½ determine all of F

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Examples of Symmetrical Four-Segment Measures

• Example F(t) t (for all t) --- probability
  is uniform along
• all four segments.
• ? packing rate is 6/64. (Not obvious!)
• Example F(t) 0 for t in 0, ½,
• 2t-1 for t in ( ½, 1
  --- probability is uniform
• along outer half of each segment
• ? packing rate is 6/64 ( four points form
  an instance
• of 2413 iff one point is on each segment )

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F
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Examples of Symmetrical Four-Segment Measures

• Example F(t) has slope 0 in 0, ¼ and
  ½, ¾
• slope 2 in ¼ ,
  ½ and ¾, 1 .
• ? packing rate is 51/512
• ( this is essentially the AAHHS estimate )

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Examples of Symmetrical Four-Segment Measures

• Example F(t) t2 for t in 0, 1
• ? packing density 349/3360 0.103869
• (Better than 51/511 0.099804,
• but not as good as the current record)
• This shows that there may be some advantage to
  an uneven distribution.

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Packing rate calculated from F

• Theorem. Let ? be a symmetrical four-segment
  measure with the distribution along each segment
  given by F.
• Then the packing rate is given by

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How would one prove such a formula?

• There are three possibilities for 2413-patterns
  arising from ?

One point per segment Two, one, one
Two pairs
works only if right dot is above left dot, and
bottom dot right of top dot
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Chance of getting type 1

• Probability of an occurrence of the first
• type
• (1) Probability that points are in four
• different segments 6/64
• (2) Probability that right point is above
• left point

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Chance of getting type 1

• (3) Probability that bottom point is to
• right of top point Same as (2)
• (4) Therefore Probability of type 1
• occurrence of 2413
• Calculate type-2 and type-3 probabilities
  similarly add to get theorem.

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Calculus of Variations

• Write
• How do we choose F to maximize this value ?
• Theorem Let . If F
  maximizes the expression above, then
• whenever F is not constrained at t.
• (That is, whenever the slope of F is not 0 or
  2.)

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So what is F ?

• The above formula gives a negative value when t
  lt ¼ - 2J. So the real formula is
• F(t) 0 if t lt t
  ¼ - 2J,
• formula above if t ? t
• The above formula makes F dependent on J,
  which is an integral of F itself. The resulting
  condition determines J and t
• t 0.14779091617675321550
• J 0.05110454191162339225

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So what is F ?

• so finally, F(t) is given by
• F(t) 0 when t lt 0.14779091617675321550
• otherwise.

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Best packing rate ?

• This F gives
• ? ( ?, ? ) 0.10472339512772223636
• which is a new lower bound for the packing
  density of 2413.
• Compare last years value, from weighted
  template 0.104250980

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Recursion bubbles

• But wait!
• This plan leaves about 14 of each segment empty.

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With the recursion bubble

• Recall packing rate so far
  0.10472339512772223636
• With one recursion bubble on each segment, the
  packing rate becomes
• 
  0.10472417578055968289
• A MASSIVE IMPROVEMENT !

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Shouldnt the recursion bubble be bigger?

• Shouldnt it? The measure was optimal before we
  introduced recursion. Now, theres more
  advantage to drawing points from the box than
  there was before. At the margin, isnt that a
  reason for increasing the size of the bubble?
• So, make the bubble bigger.
• But then the low end of F is inconsistent. We
  need to add another recursion bubble to get F
  back on track.
• And then another, and another, and another

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Best result with two bubbles

• One bubble
• bubble ends at t .1477
• packing density 0.10472417578055968289
• Two bubbles
• first bubble expands to t .15141
• second bubble ends at t .15352
• packing density 0.10472422757673209041

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How much proof, how much conjecture?

• We conjectured that the optimal measure is a
  symmetrical four-segment measure, with recursion
  bubbles inserted.
• We proved that the optimal four-segment measure
  is the one given, with ?(?,?)
  0.10472339512772223636.
• We proved that adding a single recursion box to
  each segment increases the packing density to
  0.10472417578055968289.
• (So, that becomes a proven lower bound for the
  packing density.)
• We calculated (using Mathematica Maximize) that
  the optimum with two recursion boxes is
  0.10472422757673209041.
• We conjecture that the optimal measure contains
  an infinite string of recursion bubbles at each
  end of each segment, plus an interleaved section
  in the middle.

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What next?

• Prove some of this stuff.
• Now that we know how rich optimal measures can
  be,
• go hunting for more good examples.