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 zIntroduction and Problem Statement zA Solution zProving Correctness zFuture Work

3. To be seen zIntroduction and Problem Statement zA Solution zProving Correctness zFuture Work

4. Our Problem zConsider a puzzle where we must place points labeled (0, 1, 2, …, n) on the real number line. zThe points must be placed so that, given distinct points labeled (k) and (k+a), the distance between the points is at least 1/a. Denote the position of point k by f(k). zWe want to know: what is the optimal (smallest) total amount of space required to place all of the points?

5. Example: First Six Values Non-negative real number line:

6. Example: First Six Values f(0) = 1 zPlace point 0 somewhere; let’s say f(0) = 1 0

7. Example: First Six Values f(0) = 1f(1) = 2 zPlace point 0 somewhere; let’s say f(0) = 1 zDistance between point 0 and point 1 must be at least 1/1 = 1; take f(1) = 2 01

8. Example: First Six Values f(0) = 1f(1) = 2f(2) = 1/2 zPlace point 0 somewhere; let’s say f(0) = 1 zDistance between point 0 and point 1 must be at least 1/1 = 1; take f(1) = 2 zPoint 2 must be 1 away from point 1 and 1/2 away from point 0. Try f(2) = 1/2 201

9. Example: First Six Values f(0) = 1f(1) = 2f(2) = 1/2f(3) = 3/2 zPlace point 0 somewhere; let’s say f(0) = 1 zDistance between point 0 and point 1 must be at least 1/1 = 1; take f(1) = 2 zPoint 2 must be 1 away from point 1 and 1/2 away from point 0. Try f(2) = 1/2 zContinue in this fashion, placing six points 2301

10. Example: First Six Values f(0) = 1f(1) = 2f(2) = 1/2f(3) = 3/2f(4) = 0 zPlace point 0 somewhere; let’s say f(0) = 1 zDistance between point 0 and point 1 must be at least 1/1 = 1; take f(1) = 2 zPoint 2 must be 1 away from point 1 and 1/2 away from point 0. Try f(2) = 1/2 zContinue in this fashion, placing six points 23014

11. Example: First Six Values f(0) = 1f(1) = 2f(2) = 1/2f(3) = 3/2f(4) = 0f(5) = 7/3 zPlace point 0 somewhere; let’s say f(0) = 1 zDistance between point 0 and point 1 must be at least 1/1 = 1; take f(1) = 2 zPoint 2 must be 1 away from point 1 and 1/2 away from point 0. Try f(2) = 1/2 zContinue in this fashion, placing six points zWe have achieved a maximum of 7/3. Is this optimal for six values? 230514

12. Example: First Six Values f(0) = 1f(1) = 2f(2) = 1/2f(3) = 3/2f(4) = 0f(5) = 7/3 zPlace point 0 somewhere; let’s say f(0) = 1 zDistance between point 0 and point 1 must be at least 1/1 = 1; take f(1) = 2 zPoint 2 must be 1 away from point 1 and 1/2 away from point 0. Try f(2) = 1/2 zContinue in this fashion, placing six points zWe have achieved a maximum of 7/3. Is this optimal for six values? zNo, there is a better solution: f(1) = 1/2f(2) = 3/2f(3) = 0f(4) = 1f(0) = 11/5f(5) = 2 230514 123405

13. The Full Problem zSuppose now we wish to place an infinite number of points zWe wish to find a way to do so in a bounded amount of space. What is the smallest amount of space required?

14. Problem History zMotivated by the Constraint Satisfaction Problem yGiven a set V of variables and a set C of constraints, find an assignment that satisfies all the constraints zBinary Constraint Problem yeach constraint affects two variables zMetric Space Binary Constraint Problem yGeneralization: variables take values from a metric space yConstraints are of the form “Variables p and q must be assigned values that have distance at least c p,q from each other” yThe Metric Space generalization is due to D. G. Fon-Der-Flaass, “Real-valued Frequency Assignment,” 1998.

15. A Metric Space BCP zOur problem is a particular instance of the MSBCP: yThe variables are the natural numbers N = {0, 1, 2, …}. yThe metric space is the set of non-negative real numbers R +. zEach number n is assigned a value f(n). zOur constraints are c p,q = 1/|p-q|. zIf a function f satisfies these constraints, we say that f is a proximity inversion function. zOur goal is to find such a function f that minimizes sup n (f(n)); that is, uses the least amount of space on the real number line.

16. To be seen zIntroduction and Problem Statement zA Solution zProving Correctness zFuture Work

17. Numeration Systems zA Numeration System is a method of representing non-negative integers. It corresponds to a sequence S = { s 1, s 2, … } of ascending positive integers. zA value v can be expressed with respect to S as a string x = x 1 x 2 …x n over the alphabet of natural numbers, where v = Σ s i x i. yExample: the decimal numeration system is S = { 1, 10, 100, … } zThe representation is unique when we require certain properties to hold. yThe most basic property is that x i < s i+1 /s i for all i. yThere are more general properties as well, due to A. Fraenkel, “Systems of Numeration,” 1985. zWhen a string meets these properties, we say that it is a valid representation for an integer with respect to S. zWe’ll skip the general results and simply claim the properties for our particular numeration system.

18. Our Numeration System zLet F i denote the i th Fibonacci number. F 0 = 0, F 1 = 1, F 2 = 1, etc.  Take S = { F 2i : i ≥ 1 } = { 1, 3, 8, 21, … } zWe express an integer n as a string x corresponding to digits of n over basis S. That is, n = Σ x i F 2i yExample: if n = 38, then n = 1*21 + 2*8 + 0*3 + 2*1, so the representation of n is the string 2021 zNote that for any k, 3F 2k > 2F 2k + F 2k-1 = F 2k + F 2k+1 = F 2(k+1) zThus x k < F 2k / F 2(k+1) < 3 for all k, so the set of valid representations with respect to S is over the alphabet {0,1,2} zIt 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 21*2

19. The Solution zGiven an integer n, suppose its representation in S is x = x 1 x 2 …x k zWe define our function f by f(n) := Σ( x i /F 2i ) zThat is, we take our digits for representation over basis { F 2i }, then apply them to the basis { 1/F 2i } yExample: if n = 38 then x = 2021, so f(n) = 2/1 + 0/3 + 2/8 + 1/21 = 193/84 ≈ 2.2976 zThe supremum for this particular f is 1 + Σ(1/F 2i ) ≈2.5353…

20. Example 121/34/307/3 zOur solution f() places the first ten values as follows 341520 2/3 6 5/3 7 1/8 8 9/8 9

21. To be seen zIntroduction and Problem Statement zA Solution zProving Correctness zFuture Work

22. Proving the Result zWe 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 zWe shall consider what this means in terms of the representations for p and q with respect to our numeration system zIn most cases, the required property for strings will reduce to a problem that is easier and more intuitive to solve

23. Relative Order zSuppose 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. yExample: if p = 38, q = 12, then we can take x = 2021, y = 1120 zWe 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. yExample: if x = 101, y = 211, then y corresponds to the larger value because after the common suffix 1, y has the larger digit 1 > 0. zA 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. yExample: if x = 2021, y = 1120, then since x 1 = 2 > 1 = y 1 we conclude that f(38) > f(12).

24. Intermediate Values zConsider two nonequal natural numbers p and q. zAn intermediate value for p and q is an integer t such that yt lies between p and q, and yf(t) lies between f(p) and f(q). zExample: both 4 and 6 are intermediate values for 3 and 7. 121/34/307/3 341520 2/3 6 5/3 7 1/8 8 9/8 9

25. Intermediate Values (cont.) zLemma: if our function f is not a valid proximity inversion function, then there is a pair of integers (p’, q’) such that y|f(p’) - f(q’)| > 1/|p’-q’|, and ythere is no intermediate value for p’ and q’ zProof: Suppose f is not a proximity inversion function. That is, there exist distinct natural numbers p and q such that |f(p) - f(q)| > 1/|p-q|. zIf p and q do not have an intermediate value, we are done. Otherwise, let t be an intermediate value. zThen |p-t| < |p-q| since t is between p and q, and |f(p) - f(t)| < |f(p) - f(q)| since f(t) is between f(p) and f(q). zWe conclude that |f(p) - f(t)| > 1/|p-t|. zIf p and t have an intermediate value t’, then by the same argument |f(p) - f(t’)| > 1/|p-t’|. We continue this process until we are left with two integers with no intermediate value. zThis process must terminate, since there are only finitely many integers between p and q.

26. Intermediate Values (cont.) zTo 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. zBut we have an easy way to characterize precisely when an intermediate value exists, using valid representations! zGiven 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. yExample: for x = 2021, y = 1120, we can take z = 1211. Then 1120 < 1211 < 2021 and 0211 < 1121 < 1202, so z is an intermediate value for x and y. zFinding such an intermediate string z is easy, so a simple case analysis eliminates almost all possible strings x and y.

27. Strings with no Intermediate Values zExample: suppose one of the strings ends in 00, say y. Then x must be of the form wc21 k and y must be of the form wc’0 k+1 where (c,c’) is either (0,1) or (1,2) zThis is one of only five cases in which x and y do not have an intermediate string z zThis 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

28. To be seen zIntroduction and Problem Statement zA Solution zProving Correctness zFuture Work

29. Optimality of Solution zWe have not yet proved that the solution is optimal zIt turns out that our function f, limited to a finite set of integers {0,…,n}, is not optimal for that subset zIt appears that the optimal solution f n for {0,…,F 2n } is related to f(): 121/34/37/3 34152 2/3 6 5/3 7 0 0

30. Optimality of Solution zWe have not yet proved that the solution is optimal zIt turns out that our function f, limited to a finite set of integers {0,…,n}, is not optimal for that subset zIt appears that the optimal solution f n for {0,…,F 2n } is related to f(): ySet f n (0) = f(F 2n-1 ) + 1/F 2n-1 121/34/37/3 34152 2/3 6 5/3 7 38/15 0

31. Optimality of Solution zWe have not yet proved that the solution is optimal zIt turns out that our function f, limited to a finite set of integers {0,…,n}, is not optimal for that subset zIt appears that the optimal solution f n for {0,…,F 2n } is related to f(): ySet f n (0) = f(F 2n-1 ) + 1/F 2n-1 yShift all other values so that min k {f n (k)} = 0 2/35/3012 34152 1/3 6 4/3 7 38/15 0 zIf this were true, this would imply that f is optimal, as max k {f n (k)} converges to sup n {f(n)}

32. Extending to Other Problems zThe method outlined here is only a solution to a particular instance of a constraint satisfaction problem. zOur approach was to take a function f and express it as yA numeration system S yA (simpler) function f* to perform on the basis elements of S zWe would like to generalize the results of this approach, so it can be used to solve different constraint problems. zIn 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?

33. Questions?