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A Conjecture for the Packing Density of 2413

This talk presents a conjecture on the packing density of 2413 permutations, exploring optimal measures and recursion in maximizing packing rates. The concept of near-optimizers for permutations is discussed, along with examples of packing rates and optimal measures for various patterns.

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A Conjecture for the Packing Density of 2413

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  1. 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

  2. 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…

  3. 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,

  4. About packing densities… • We’re 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.)

  5. 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 don’t have to be optimizers; they just have to be close enough that they have the right limit.

  6. Describing the near-optimizers • Here’s how we describe good n‘s for 132:  That’s what the n‘s look like. The packing density turns out to be

  7. Describing the near-optimizers • Here’s an easier picture:  The points line up along these lines. This object is a probability measure on the unit square. Let’s formalize it.

  8. 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.

  9. 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 ).

  10. “…order type of their configuration…” If the points are ordered by increasing x coordinates, x1 < x2 < … < 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.

  11. 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.

  12. 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 doesn’t matter how probability is distributed along the diagonal.

  13. 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.

  14. Examples of packing rates • Example. Let  be uniform • on a disk in S. • WHAT IS ’ ( 123,  ) ? Points are drawn uniformly from the disk.

  15. 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.

  16. 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) It’s tricky to define limiting measures • (2) Limits of measures don’t always preserve packing rates

  17. 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 i’s.

  18. n is concentrated on [ 1/(n+1), 1/n ]2  is a point mass at the origin Example – limits don’t preserve packing rates • In this picture,  = lim i. • But ’ ( 132, i ) = 1/6 for each i, while ’ ( 132,  ) = 0.

  19. 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 can’t reduce packing rates.

  20. 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.)

  21. Packing rates = Packing densities • Theorem: The optimal packing rate of any pattern is its packing density: • ’ (  ) =  (  ).

  22. 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 n‘s that achieve this average, and then we have • lim  ( , n ) = ’ ( ,  ). • This forces  (  )  ’ (  ) .

  23. 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 ’ (  )   (  ).

  24. 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.

  25. 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) • Today’s conjecture: •  ( 2413 ) ≈ 0.10472422757673209041…

  26. 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:

  27. Distribution need not be uniform

  28. Probability distribution along this line: F(t) F(t) = fraction of this segment’s probability that is in the leftmost fraction t of the segment t=0 ……………………….t=1

  29. 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

  30. 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 )

  31. F

  32. 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 )

  33. F

  34. 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.

  35. 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

  36. 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

  37. 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:

  38. 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.

  39. 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.)

  40. So what is F ? • The above formula gives a negative value when t < ¼ - 2J. So the real formula is • F(t) = 0 if t < 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…

  41. So what is F ? • …so finally, F(t) is given by • F(t) = 0 when t < 0.14779091617675321550… • otherwise.

  42. Best packing rate ? • This F gives • ’ ( ,  ) ≈ 0.10472339512772223636… • which is a new lower bound for the packing density of 2413. • Compare last year’s value, from weighted template: 0.104250980…

  43. Recursion bubbles • But wait! • This plan leaves about 14% of each segment empty.

  44. 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 !

  45. Shouldn’t the recursion bubble be bigger? • Shouldn’t it? The measure was optimal before we introduced recursion. Now, there’s more advantage to drawing points from the box than there was before. At the margin, isn’t 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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