Proximity Inversion Functions on the Non-Negative Integers

1
Proximity Inversion Functions on the Non-Negative
Integers

• Presented By
• Brendan Lucier
• June 5, 2005
• CMS Summer 2005 Meeting
• Automatic Sequences and Related Topics

2
To be seen

• Introduction and Problem Statement
• A Solution
• Proving Correctness
• Future Work

3
To be seen

• Introduction and Problem Statement
• A Solution
• Proving Correctness
• Future Work

4
Our Problem

• Consider a puzzle where we must place points
  labeled (0, 1, 2, , n) on the real number line.
• The points must be placed so that, given distinct
  points labeled (k) and (ka), the distance
  between the points is at least 1/a. Denote the
  position of point k by f(k).
• We want to know what is the optimal (smallest)
  total amount of space required to place all of
  the points?

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Example First Six Values
Non-negative real number line
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Example First Six Values
0
f(0) 1

• Place point 0 somewhere lets say f(0) 1

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Example First Six Values
0
1
f(0) 1
f(1) 2

• Place point 0 somewhere lets say f(0) 1
• Distance between point 0 and point 1 must be at
  least 1/1 1 take f(1) 2

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Example First Six Values
2
0
1
f(0) 1
f(1) 2
f(2) 1/2

• Place point 0 somewhere lets say f(0) 1
• Distance between point 0 and point 1 must be at
  least 1/1 1 take f(1) 2
• Point 2 must be 1 away from point 1 and 1/2 away
  from point 0. Try f(2) 1/2

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Example First Six Values
2
3
0
1
f(0) 1
f(1) 2
f(2) 1/2
f(3) 3/2

• Place point 0 somewhere lets say f(0) 1
• Distance between point 0 and point 1 must be at
  least 1/1 1 take f(1) 2
• Point 2 must be 1 away from point 1 and 1/2 away
  from point 0. Try f(2) 1/2
• Continue in this fashion, placing six points

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Example First Six Values
2
3
0
1
4
f(0) 1
f(1) 2
f(2) 1/2
f(3) 3/2
f(4) 0

• Place point 0 somewhere lets say f(0) 1
• Distance between point 0 and point 1 must be at
  least 1/1 1 take f(1) 2
• Point 2 must be 1 away from point 1 and 1/2 away
  from point 0. Try f(2) 1/2
• Continue in this fashion, placing six points

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Example First Six Values
2
3
0
5
1
4
f(0) 1
f(1) 2
f(2) 1/2
f(3) 3/2
f(4) 0
f(5) 7/3

• Place point 0 somewhere lets say f(0) 1
• Distance between point 0 and point 1 must be at
  least 1/1 1 take f(1) 2
• Point 2 must be 1 away from point 1 and 1/2 away
  from point 0. Try f(2) 1/2
• Continue in this fashion, placing six points
• We have achieved a maximum of 7/3. Is this
  optimal for six values?

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Example First Six Values
2
3
0
5
1
4
f(0) 1
f(1) 2
f(2) 1/2
f(3) 3/2
f(4) 0
f(5) 7/3

• Place point 0 somewhere lets say f(0) 1
• Distance between point 0 and point 1 must be at
  least 1/1 1 take f(1) 2
• Point 2 must be 1 away from point 1 and 1/2 away
  from point 0. Try f(2) 1/2
• Continue in this fashion, placing six points
• We have achieved a maximum of 7/3. Is this
  optimal for six values?
• No, there is a better solution

1
2
3
4
0
5
f(1) 1/2
f(2) 3/2
f(3) 0
f(4) 1
f(0) 11/5
f(5) 2
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The Full Problem

• Suppose now we wish to place an infinite number
  of points
• We wish to find a way to do so in a bounded
  amount of space. What is the smallest amount of
  space required?

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Problem History

• Motivated by the Constraint Satisfaction Problem
• Given a set V of variables and a set C of
  constraints, find an assignment that satisfies
  all the constraints
• Binary Constraint Problem
• each constraint affects two variables
• Metric Space Binary Constraint Problem
• Generalization variables take values from a
  metric space
• Constraints are of the form Variables p and q
  must be assigned values that have distance at
  least cp,q from each other
• The Metric Space generalization is due to D. G.
  Fon-Der-Flaass, Real-valued Frequency
  Assignment, 1998.

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A Metric Space BCP

• Our problem is a particular instance of the
  MSBCP
• The variables are the natural numbers N 0, 1,
  2, .
• The metric space is the set of non-negative real
  numbers R.
• Each number n is assigned a value f(n).
• Our constraints are cp,q 1/p-q.
• If a function f satisfies these constraints, we
  say that f is a proximity inversion function.
• Our goal is to find such a function f that
  minimizes supn(f(n)) that is, uses the least
  amount of space on the real number line.

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To be seen

• Introduction and Problem Statement
• A Solution
• Proving Correctness
• Future Work

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Numeration Systems

• A Numeration System is a method of representing
  non-negative integers. It corresponds to a
  sequence S s1, s2, of ascending positive
  integers.
• A value v can be expressed with respect to S as a
  string x x1x2xn over the alphabet of natural
  numbers, where v Ssixi .
• Example the decimal numeration system is S
  1, 10, 100,
• The representation is unique when we require
  certain properties to hold.
• The most basic property is that xi lt si1/si for
  all i.
• There are more general properties as well, due to
  A. Fraenkel, Systems of Numeration, 1985.
• When a string meets these properties, we say that
  it is a valid representation for an integer with
  respect to S.
• Well skip the general results and simply claim
  the properties for our particular numeration
  system.

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Our Numeration System

• Let Fi denote the ith Fibonacci number. F0 0,
  F1 1, F2 1, etc.
• Take S F2i i 1 1, 3, 8, 21,
• We express an integer n as a string x
  corresponding to digits of n over basis S. That
  is, n SxiF2i
• Example if n 38, then n 121 28 03
  21, so the representation of n is the string
  2021
• Note that for any k, 3F2k gt 2F2k F2k-1
• F2k F2k1
• F2(k1)
• Thus xk lt F2k / F2(k1) lt 3 for all k, so the set
  of valid representations with respect to S is
  over the alphabet 0,1,2
• It turns out that a string x in 0,1,2 is a
  valid representation if and only if it does not
  contain any subwords of the form 212

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The Solution

• Given an integer n, suppose its representation in
  S is x x1x2xk
• We define our function f by f(n) S(xi/F2i)
• That is, we take our digits for representation
  over basis F2i , then apply them to the basis
  1/F2i
• Example if n 38 then x 2021, so f(n) 2/1
  0/3 2/8 1/21 193/84 2.2976
• The supremum for this particular f is 1
  S(1/F2i) 2.5353

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Example

• Our solution f() places the first ten values as
  follows

3
4
1
5
2
0
6
7
8
9
1
2
1/3
4/3
0
7/3
2/3
5/3
1/8
9/8
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To be seen

• Introduction and Problem Statement
• A Solution
• Proving Correctness
• Future Work

22
Proving the Result

• We need to prove that f is a proximity inversion
  function that is, f(p) - f(q) 1/p-q holds
  for all natural numbers p and q
• We shall consider what this means in terms of the
  representations for p and q with respect to our
  numeration system
• In most cases, the required property for strings
  will reduce to a problem that is easier and more
  intuitive to solve

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Relative Order

• Suppose we have two natural numbers, p and q,
  represented w.r.t. S by strings x and y, padded
  with zeros to be the same length.
• Example if p 38, q 12, then we can take x
  2021, y 1120
• We can determine which of p or q is larger by
  scanning the digits of x and y from right to
  left. Whenever the strings first differ, the
  string with the larger digit corresponds to the
  larger number.
• Example if x 101, y 211, then y corresponds
  to the larger value because after the common
  suffix 1, y has the larger digit 1 gt 0.
• A similar property holds for f(p) and f(q). If
  we scan the two representations x and y from left
  to right we get the relative ordering of f(p)
  and f(q). This is not obvious, and the proof
  depends on our particular numeration system S.
• Example if x 2021, y 1120, then since x1 2
  gt 1 y1 we conclude that f(38) gt f(12).

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Intermediate Values

• Consider two nonequal natural numbers p and q.
• An intermediate value for p and q is an integer
  t such that
• t lies between p and q, and
• f(t) lies between f(p) and f(q).
• Example both 4 and 6 are intermediate values
  for 3 and 7.

3
4
1
5
2
0
6
7
8
9
1
2
1/3
4/3
0
7/3
2/3
5/3
1/8
9/8
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Intermediate Values (cont.)

• Lemma if our function f is not a valid proximity
  inversion function, then there is a pair of
  integers (p, q) such that
• f(p) - f(q) gt 1/p-q, and
• there is no intermediate value for p and q
• Proof Suppose f is not a proximity inversion
  function. That is, there exist distinct natural
  numbers p and q such that f(p) - f(q) gt
  1/p-q.
• If p and q do not have an intermediate value, we
  are done. Otherwise, let t be an intermediate
  value.
• Then p-t lt p-q since t is between p and q,
  and f(p) - f(t) lt f(p) - f(q) since f(t) is
  between f(p) and f(q).
• We conclude that f(p) - f(t) gt 1/p-t.
• If p and t have an intermediate value t, then by
  the same argument f(p) - f(t)
  gt 1/p-t. We continue this process until we
  are left with two integers with no intermediate
  value.
• This process must terminate, since there are only
  finitely many integers between p and q.

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Intermediate Values (cont.)

• To prove that f is a proximity inversion
  function, it now suffices to show that our
  constraint holds for integers p and q that do not
  admit an intermediate value.
• But we have an easy way to characterize precisely
  when an intermediate value exists, using valid
  representations!
• Given two valid representations x and y, they
  will correspond to integers with an intermediate
  value precisely if exists a representation z that
  is lexicographically between x and y when read
  both forward and backward.
• Example for x 2021, y 1120, we can take z
  1211. Then 1120 lt 1211 lt 2021 and 0211 lt 1121 lt
  1202, so z is an intermediate value for x and y.
• Finding such an intermediate string z is easy, so
  a simple case analysis eliminates almost all
  possible strings x and y.

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Strings with no Intermediate Values

• Example suppose one of the strings ends in 00,
  say y. Then x must be of the form wc21k and y
  must be of the form wc0k1 where (c,c) is
  either (0,1) or (1,2)
• This is one of only five cases in which x and y
  do not have an intermediate string z
• This reduction makes the problem much more
  manageable. we can prove directly that strings of
  these forms correspond to integers that do not
  violate our constraints

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To be seen

• Introduction and Problem Statement
• A Solution
• Proving Correctness
• Future Work

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Optimality of Solution

• We have not yet proved that the solution is
  optimal
• It turns out that our function f, limited to a
  finite set of integers 0,,n, is not optimal
  for that subset
• It appears that the optimal solution fn for
  0,,F2n is related to f()

3
4
1
5
2
6
7
0
1
2
1/3
4/3
7/3
2/3
5/3
0
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Optimality of Solution

• We have not yet proved that the solution is
  optimal
• It turns out that our function f, limited to a
  finite set of integers 0,,n, is not optimal
  for that subset
• It appears that the optimal solution fn for
  0,,F2n is related to f()
• Set fn(0) f(F2n-1) 1/F2n-1

3
4
1
5
2
6
7
0
1
2
1/3
4/3
7/3
2/3
5/3
38/15
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Optimality of Solution

• We have not yet proved that the solution is
  optimal
• It turns out that our function f, limited to a
  finite set of integers 0,,n, is not optimal
  for that subset
• It appears that the optimal solution fn for
  0,,F2n is related to f()
• Set fn(0) f(F2n-1) 1/F2n-1
• Shift all other values so that minkfn(k) 0

3
4
1
5
2
6
7
0
2/3
5/3
0
1
2
1/3
4/3
38/15

• If this were true, this would imply that f is
  optimal, as maxkfn(k) converges to supnf(n)

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Extending to Other Problems

• The method outlined here is only a solution to a
  particular instance of a constraint satisfaction
  problem.
• Our approach was to take a function f and express
  it as
• A numeration system S
• A (simpler) function f to perform on the basis
  elements of S
• We would like to generalize the results of this
  approach, so it can be used to solve different
  constraint problems.
• In particular, can all metric space BSPs be
  solved this way? If not, what are the necessary
  and sufficient conditions on the distance
  constraints for such a solution to exist?

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Questions?