Journal of Graph Algorithms and Applications More on Generalized Jewels and the Point Placement Problem

The point placement problem is to determine the positions of a set of n distinct points, P = {p1, p2, p3,. .. , pn}, on a line uniquely, up to translation and reflection, from the fewest possible distance queries between pairs of points. Each distance query corresponds to an edge in a graph, called point placement graph (ppg), whose vertex set is P. The uniqueness requirement of the placement translates to line-rigidity of the ppg. In this paper we show how to construct in 2 rounds a line-rigid point placement graph of size 4n/3 + O(1) from certain small-sized graphs called 6:6 jewels. This improves an earlier result that used cycle-graphs on 5 vertices. More significantly, we improve the lower bound on 2-round algorithms from 17n/16 to 12n/11.


The problem
Let P = {p 1 , p 2 , ..., p n } be a set of n distinct points on a line L. In this paper, we address the problem of determining a unique placement (up to translation and reflection) of the p i 's on L, by querying distances between some pairs of points p i and p j , 1 ≤ i, j ≤ n.
The resulting queries can be represented by a point placement graph (ppg, for short), G = (P, E), where each edge e in E joins a pair of points p i and p j in P if the distance between these two points on L is known and the length of e, |e|, is the distance between the corresponding pair of points (Note the dual use of p i to denote a point on L as well as a vertex of G).
We will say that G is line rigid or just rigid when there is a unique placement for P .Thus, the original problem reduces to the construction of a line rigid ppg, G.
Let's take some simple examples to illustrate the ideas involved.Suppose we have just 3 points {p 1 , p 2 , p 3 } on a line whose positions we want to know.Three different ppgs, up to relabelling, are possible (omitting the trivial case when E = ∅) as shown in Fig. 1 below.Fig. 1(a) corresponds to the situation when the distance between a pair of points, say p 1 and p 2 , is known.For Fig. 1(b), the distances between 2 pairs of points, say {p 1 , p 2 } and {p 2 , p 3 }, are known.Fig. 1(c) is the ppg when all the pairwise distances are known.Clearly, for the ppg of Fig. 1(a) a unique placement is not possible since the point p 3 can be anywhere relative to p 1 and p 2 .The same is true of Fig. 1(b) -say we place p 1 and p 2 first, but then the position of p 3 relative to p 2 is ambiguous.However, a unique placement is possible for the ppg of Fig. 1(c) as long as the length of one edge is the sum or absolute difference of the lengths of the other two.Thus, if we first place p 1 and then place p 2 to p 1 's right, p 3 will be placed between p 1 and p 2 if the sum of its distances from p 1 and p 2 is |p 1 p 2 |, and to the left of p 1 or to the right of p 2 if the absolute difference of the distances is equal to |p 1 p 2 |.In other words, the ppg of Fig. 1(c) is rigid.
The last case suggests a simple algorithm for the unique placement of n points.Query the distance between two points, say p 1 and p 2 .The position of each of the remaining points p i , i ≥ 3 is determined by querying the distances from p i to p 1 and p 2 ; p i lies between p 1 and p 2 if the sum of the distances is equal to |p 1 p 2 |, and to the left of p 1 or to the right of p 2 if the difference of the distances is equal to |p 1 p 2 |.The corresponding ppg shown in Fig. 2 is then rigid.The number of queries made is 2n − 3, which is of the form αn + β.The principal goal is to make α as small as possible.With this in mind, let's look at the more complicated and illuminating case when we have 4 points.Many different ppg's are possible.We can dispense with those that have fewer than 4 edges since in these cases a unique placement is clearly not possible.Fig. 3 below shows the possible ppg's, up to relabelling, with 4 and 5 edges.The ppg of Fig. 3(a) is not rigid, for while the triangle formed by p 1 , p 2 and p 3 is rigid, the point p 4 can be placed to the left or right of p 3 , making the placement non-unique.The ppg of Fig. 3(b) is interesting in that if the two pairs of opposite edges are equal then there is no unique placement.This is easily seen by drawing the ppg as a rectangle as shown in Fig. 4(a) below and then giving a horizontal right shear to the top edge p 2 p 3 so that p 2 and p 3 lie on the same line as p 1 and p 4 , giving us the linear configuration shown in Fig. 4(b).A horizontal left shear produces the linear configuration shown in Fig. 4(c), which cannot be obtained from the linear configuration of Fig. 4(b) by translation and/or reflection.
The ppg of Fig. 3(c) is rigid since we have 2 triangles attached to the edge p 1 p 3 , each of which is rigid.Thus, it is the ppg of Fig. 3 This means that if we want to extend our previous algorithm for the unique placement of n points, by first placing two nodes, say, p 1 and p 2 on L and then building rigid quadrilaterals by querying distances from p 1 and p 2 with respect to two new nodes at a time, we must make sure that we meet the structural condition on the rigidity of each new quadrilateral.
To build a rigid ppg we need to do the queries in rounds.Here is a 2-round algorithm due to Damaschke [6].Let the number of points be n = 2b + 4, where b is a positive integer.In the first round, we make 2b + 3 distance queries represented by the edges in the graph in Fig. 5.There are b children p i (i = 3, ..., b + 2) rooted at p 1 and b + 2 children p j (j = b + 3, ..., 2b + 4) rooted at p 2 .In the second round, for each edge p 1 p i (i = 3, ..., b + 2) we find an edge p 2 p j rooted at p 2 satisfying the rigidity condition |p 1 p i | = |p 2 p j |.We can ensure this condition by having 2 extra edges at p 2 , in view of the following basic observation [7]: Observation 1 At most two equal length edges can be incident to any node in a ppg.
By Observation 1, there are at most 2 edges p 2 p j such that |p 1 p i | = |p 2 p j |.So, for each edge p 1 p i an edge p 2 p j will always be found such that |p 1 p i | = |p 2 p j |.Then for each i (i = 3, ..., b + 2), we query the distance p i p j to form a quadrilateral p 1 p i p j p 2 .It will be line rigid since |p 1 p i | = |p 2 p j |.It will fix the positions of p i and p j relative to p 1 and p 2 .For each of the 2 unused leaves p j , the distance p 1 p j is queried in the second round to form the triangle p 1 p j p 2 .It will fix the position of p j relative to p 1 and p 2 .
The number of queries made over the two rounds to construct this rigid ppg is 3b + 5, i.e., 3n/2 − 1.There are two noteworthy points: (a) we have reduced the value of α from 2 for the first algorithm to 3/2 for the second, and (b) there is a price for this -we have to query the edges in two rounds.
What if the number of points is greater than 6 but odd?Let n = 2b + 5, where b is a positive integer.We make an unique placement of the first 2b + 4 nodes using the above algorithm, and query the distances of the last odd node from any two nodes.Distance queries for this node can be made in either of the 2 rounds.

Motivation
The motivation for studying this problem stems from the fact that it arises in diverse areas of research, to wit computational biology, learning theory, computational geometry, etc.
In learning theory [6] this problem is one of learning a set of points on a line non-adaptively, when learning has to proceed based on a fixed set of given distances, or adaptively when learning proceeds in rounds, with the edges queried in one round depending on those queried in the previous rounds.
The version of this problem studied in Computational Geometry is known as the turnpike problem.The description is as follows.On an expressway stretching from town A to town B there are several gas exits; the distances between all pairs of exits are known.The problem is to determine the geometric locations of these exits.This problem was first studied by Skiena et al. [11] who proposed a practical heuristic for the reconstruction.A polynomial time algorithm was given by Daurat et al. [8].
In computational biology, it appears in the guise of the restriction site mapping problem.Biologists discovered that certain restriction enzymes cleave a DNA sequence at specific sites known as restriction sites.For example, it was discovered by Smith and Wilcox [12] that the restriction enzyme Hind II cleaves DNA sequences at the restriction sites GTGCAC or GTTAAC.In lab experiments, by means of fluorescent in situ hybridization (FISH experiments) biologists are able to measure the lengths of such cleaved DNA strings.Given the distances (measured by the number of intervening nucleotides) between all pairs of restriction sites, the task is to determine the exact locations of the restriction sites.
The turnpike problem and the restriction mapping problem are identical, except for the unit of distance involved; in both of these we seek to fit a set of points to a given set of interpoint distances.As is well-known, the solution may not be unique and the running time is polynomial in the number of points.While the point placement problem, prima facie, bears a resemblance to these two problems it is different in its formulation -we are allowed to make pairwise distance queries among a distinct set of labeled points.It turns out that it is possible to determine a unique placement of the points up to translation and reflection in time that is linear in the number of points.

Prior work
Early research on this problem was reported in [10,9].In this paper, our first principal reference is [6], where it was shown that the jewel (Fig. 8) and K 2,3 are both line rigid, as also how to build large rigid graphs of density 8/5 (this is an asymptotic measure of the number of edges per node as the number of nodes go to infinity) out of the jewel.In a subsequent paper, Damaschke [7] proposed a randomized 2-round strategy that needs (1 + o(1))n distance queries with high probability and also showed that this is not possible with 2-round deterministic strategies.Our second principal reference is the work of [4] who improved many of the results of [6].Their principal contributions are the 3-round construction of rigid graphs of density 5/4 from 6-cycles and a lower bound on the number of queries necessary in any 2-round algorithm.

Our contribution
In this paper, we determine conditions under which generalizations of the jewel, called m : n jewels, remain line rigid.In [2] we showed how to construct in 2 rounds a line rigid ppg on n points, using an instance of a 5:5 jewel as the basic component.The number of edges queried during this construction is 10n/7 + O(1).In this paper we extend this result to 6:6 jewels, constructing in 2 rounds a line rigid ppg with 4n/3 + O(1) queries.This improves the result in [4] for constructing a ppg of the same size in 2 rounds using 5-cycles.We also improve substantially the lower bound on any 2-round algorithm from 17n/16 in [4] to 12n/11.

Generalized jewels
The examples described in the Introduction demonstrates well how small ppg's that are inherently rigid or rigid under some structural conditions can be glued together into a large rigid ppg.In this section we introduce a novel type of ppg, namely an m : n jewel, several copies of which we plan to glue together to form a large rigid ppg.
A generic m : n jewel consists of an m-vertex cycle C 1 and another n-vertex cycle C 2 that are joined by a strut going between two vertices Y (of C 1 ) and Z (of C 2 ), and hinged at a third common vertex, X (Fig. 6).An instance of an m : n jewel is obtained by the placement of the nodes that describe the cycles To attain our goal we need to determine the structural conditions that make a chosen instance of the m : n jewel line rigid.In this, the idea of a layer graph introduced by Chin et al. [4] comes in handy.We first choose two orthogonal directions x and y (actually, any 2 non-parallel directions will do).A graph G admits a layer graph drawing if the following 4 properties are satisfied: P1 Each edge e of G is parallel to one of the two orthogonal directions, x and y.
P2 The length of an edge e is the distance between the corresponding points on L.
P3 Not all edges are along the same direction (thus a layer graph has a twodimensional extent).
P4 When the layer graph is folded onto a line, by a rotation either to the left or to the right about an edge of the layer graph lying on this line, no two vertices coincide.
Chin et al. [4] proved the following result: Theorem 1 A ppg G is line rigid iff it cannot be drawn as a layer graph.
In the next section, we obtain structural conditions under which chosen instances of the m : n jewels remain rigid for small values of m and n by drawing them as layer graphs.Before we do that, we establish a few useful facts about the generic m : n jewel.The first is this.
Theorem 2 If cycles C 1 and C 2 , consisting of m and n nodes respectively, are line rigid then so is any m : n jewel made up of these two cycles.
Proof: Since C 1 and C 2 are rigid their respective vertices have unique linear layouts.Then in order for an m : n jewel to have a layer graph drawing these placements would have to be in the orthogonal directions x and y.Suppose the vertex Y is placed on the x-axis and the vertex Z on the y-axis, then the edge Y Z of the m : n-jewel is not parallel to either the x or the y direction.Hence, the m : n jewel cannot be drawn as a layer graph and must, therefore, be rigid.
As a direct consequence of the theorem we have the following corollary: Corollary 3 If an m : n jewel has a layer graph representation then in this representation at least one of C 1 or C 2 is a layer graph.
In order to obtain the structural conditions that make a cycle rigid, we draw all possible layer graph representations of it and find the structural conditions for the rigidity of each of these.The logical AND of all these conditions is our answer.The second corollary is this: Corollary 4 The union of the set of all the structural conditions that make C 1 rigid with those that make C 2 rigid, constitute a sufficient set of structural conditions that make an m : n jewel rigid.
We shall take this route in the next two sections to obtain the structural conditions for the rigidity of chosen instances of the m : n jewels for some small values of m and n.
It should be noted that a cycle with a fixed set of n x x-parallel edges and thus a fixed set of n y y-parallel edges can be drawn as a layer graph in different ways.They are all considered to be equivalent.For example, the three layer graph drawings of a 5-cycle in Fig. 7 are considered to be equivalent.From now on, for an equivalent class of layer graphs we shall draw just one of them -not all.We shall not use the term class either.By a particular layer graph, we shall mean the class of layer graphs that are equivalent to it.Thus, two layer graph drawings of an n-vertex cycle are distinct from each other if at least one edge has different orientations in the two graphs.As we shall resort to exhaustive enumerations of all the layer graph representations of a cycle, the following theorem [2] is useful for checking that we have the correct number.

4:4 and 5:4 Jewels
The following observation is fundamental.A formal proof can be found in [6].
Two 4-cycles joined together as in Fig. 8 is called a jewel in [6].It is an instance of a generic 4 : 4 jewel that we will use in the rest of our discussion.To begin with, we prove the following theorem: Since XQZP is a 4-cycle evidently its layer graph can be a rectangle only.
Let the vertices of the line rigid 4-cycle XAY B lie on the x-parallel line  Thus, none of the two 4-cycles of the jewel has a layer graph representation.By Theorem 1 both the cycles are line rigid, and by Theorem 2 the 4:4 jewel is line rigid.
Unlike the 4:4 jewel of Fig. 8, the 5:4 jewel of Fig. 9 is not intrinsically line rigid.As a prelude to our discussion in the following sections, it is interesting to find the structural conditions (or simply conditions) that make it line rigid.
We first determine the conditions that make the cycle XABY C line rigid.By Theorem 5, there are five distinct layer graph representations of the 5cycle XABY C, shown in Fig. 10.As remarked earlier, each is a canonical representative of an entire class of layer graph representations; referring to Fig. 10( We note in passing that for each of the configurations in Figs.10(c)-10(e), we have an alternate condition that prevents its drawing as shown.Thus for example |XA| = ||CY |±|Y B|| also prevents the layer graph drawing of Fig. 10(c).With the help of the label mapping (X, C, Y, B, A) to (p 3 , p 4 , p 5 , p 1 , p 2 ) we can see that this condition encapsulates the 3 different conditions corresponding to the 3 equivalent layer graphs representations shown in Fig. 7.In such situations, whenever possible, we choose the simpler condition, unless the other one is more useful for the construction of a ppg. 3 Algorithm based on a 5:5 jewel We next consider the more complex case of the 5:5 jewel of Fig. 11.From now on, we will refer to it simply as the 5:5 jewel.By Theorem 5 there are exactly 5 distinct layer graph representations of a 5-cycle (see Fig. 10).Thus, the set of 5 distinct conditions in Lemma 1 are sufficient to ensure the line-rigidity of the 5-cycle XABY C.
Lemma 1 A 5-cycle XABY C is line rigid if its edges satisfy the following conditions: Proof: Omitted.A formal proof appears in [4].By Corollary 4, these 10 conditions collectively constitute a sufficient set of conditions for the line-rigidity of the 5:5 jewel.
Our goal is to glue several copies of the 5:5 jewel of Fig. 11 into a large ppg, as we did in the case of quadrilaterals in section 1.1.All of these will have a common strut Y Z.As each jewel will account for 7 new nodes in lieu of 10 new edge queries, we expect α to be 10/7.This indeed turns out to be the case.The challenge here is to design the ppg in such a way that the rigidity conditions are satisfied for every jewel.
The rigidity conditions for a cycle, in their current form, involve all its edges.This requires to query the lengths of all of its edges in the first round to check if the rigidity conditions are satisfied.This does not provide us with the flexibility of choice that we need to meet the rigidity conditions in a 2-round algorithm.The edge lengths may not satisfy the conditions.If any condition is not satisfied then the cycle and thus the whole jewel may not be line rigid because our set of conditions is sufficient (Theorem 2).Now, the 2-dimensional stretch of a layer graph gives a pointer -we can avoid involving one edge of a cycle from all the rigidity conditions for it.We shall avoid AB and P Q from the rigidity conditions for the two 5-cycles.Then the cycles will be rigid irrespective of the lengths of those edges.And the rigidity conditions for the cycles will involve all of their other edges.Again, in each rigidity condition we need to have at least one edge in it for which we can choose edge length, from among the options for edge lengths for that particular edge, that satisfies the condition.We shall provide options for choosing each of the edges Y B and ZQ.
There will be rigidity conditions of each cycle that will not involve these edges, i.e., Y B or ZQ.We cannot meet those rigidity conditions in a 2-round algorithm.We need to avoid other edge(s) from the rigidity conditions of a cycle and/or provide options for choosing edge(s) for a cycle.We shall avoid XC and XR from the rigidity conditions for the two cycles.Then we shall have options for choosing edges Y C and ZR to satisfy the rigidity conditions.
Thus, we shall avoid AB, P Q, XC and XR from the rigidity conditions.For each 5-cycle we shall replace each of its rigidity conditions that involve any of these edges.We shall replace that condition by a set of condition(s) that prevent the cycle from being drawn as the layer graph representation that corresponds to that condition.
Looking ahead slightly, Fig. 18 describes the structure of our proposed ppg.It has a pool of edges hanging from each end of the strut Y Z and a set of 2-pronged subgraphs.The lengths of the edges of this ppg are queried in the first round.In the second round, we join each 2-pronged subgraph to a pair of edges incident to Y and another pair of edges incident to Z to form a 5:5 jewel, making sure that all the rigidity conditions satisfied.
Over the rest of this section we show how to replace the rigidity conditions of the 5-cycle XABY C that involve XC and/or AB with rigidity conditions that exclude these edges.To replace a condition we shall find another set of conditions that prevents the drawing of the 5-cycle XABY C as a layer graph in the configuration corresponding to that condition.For example, to replace the condition |XC| = |Y B|, corresponding to the layer graph of Fig. 10(a), we shall find a set of conditions that prevent the drawing of the layer graph of the 5-cycle in the configuration of Fig. 10(a).
Our first attempt will be to use other edges in the layer graph drawing corresponding to a given rigidity condition involving XC and/or AB.If this does not suit our purpose, the basic strategy will be to embed the layer graph drawing corresponding to such a rigidity condition into all possible layer graph drawings of the 5:5 jewel and derive a rigidity condition from each such embedding.
The rigidity conditions that we will consider for replacement are:

Replacing |XC| = |Y B|
This rigidity condition corresponds to the layer graph drawing of Fig. 10(a).
||XA| ± |AB|| = |Y C| is an alternate rigidity condition corresponding to the layer graph drawing in Fig. 10a) of the 5-cycle XABY C.However, it involves the edge AB that we wish to avoid.We shall find an alternate set of rigidity conditions.For this, we find all possible layer graph drawings of the 5:5 jewel in which the layer graph of Fig. 10(a) is embedded.Then we find conditions which prohibit those layer graph drawings.Consequently, those conditions will replace |XC| = |Y B|, because there will be no layer graph for the 5:5 jewel in which the layer graph of Fig. 10(a) is embedded.We shall follow this method whenever we cannot use any rigidity condition for a 5-cycle XABY C or XP QZR that involves some edges of the corresponding cycle only.We have the following lemma for the replacement of the current condition: Proof: We argue below that there are exactly 4 possible layer graph drawings of the 5:5 jewel in which the layer graph of Fig. 10(a) lies embedded.Two cases arise depending on the orientations of Y Z: • Y Z is horizontal (Fig. 12) Z is necessarily distinct from C, while Y Z and Y B are mutually perpendicular.Consider the edges on the path XRZ of the 5:5 jewel.If XR were vertical, then ZR would have to be horizontal, forcing R to coincide with C. Thus, XR must be horizontal and consequently, RZ must be vertical.
Next, we consider the edges on the path XP QZ.XP can be horizontal or vertical.If XP is horizontal then P Q must be vertical, else Q and R will coincide.This forces QZ to be horizontal giving us the layer graph of Fig. 12(a).
If XP is vertical, then P Q must be horizontal; otherwise, Q will coincide with C.This forces QZ to be vertical, giving us the layer graph of Fig. 12(b).
In these layer graphs, the edges Y C and Y Z are on a horizontal line CY Z, and are parallel to XR.The vertical edges XC and ZR connect the parallel edges.So, we must have |XC| = |ZR|.Thus, these layer graphs are not possible if |ZR| = |Y B|.
• Y Z is vertical (Fig. 13) It follows that there is no layer graph for the 5:5 jewel in which the layer graph in Fig. 10(a) of the 5-cycle XABY C is embedded if the edges of the jewel satisfy Eq. 3. Hence, the 5-cycle XABY C of the 5:5 jewel of Fig. 11 cannot be drawn as the layer graph of Fig. 10(a) if the edges of the jewel satisfy the conditions in Eq. 3. We argue below that there are exactly 12 possible layer graph drawings of the 5:5 jewel in which the layer graph of Fig. 10(e) lies embedded.There are 2 main cases to consider.
The path XRZ is made up of a vertical segment XR, followed by a horizontal segment ZR, else R will coincide with C. If we consider the path XP Q, by a similar argument when XP is horizontal P Q must be vertical.If QZ were vertical, then P would have to coincide with C. Thus, QZ is horizontal.This gives us the layer graph drawing of Fig. 14(a).
If XP is vertical, we can argue similarly as in the last paragraph that P Q must be horizontal and QZ vertical.This gives us the layer graph drawing of Fig. 14(b).• Y B and Y Z are collinear: 3 subcases arise depending upon the orientations of ZR and XR.
-ZR is perpendicular to Y B and Y Z, and XR is perpendicular to ZR (Fig. 15).In this case there are 4 distinct placements of the edges XP , P Q and QZ giving rise to 4 distinct layer graph drawings of the 5:5 jewel (Fig. 15    In the next section we show how to construct a composite ppg made up of 5:5 jewels such that all the 15 rigidity conditions listed above are satisfied for each of one these.

Algorithm
We use a pair of points {Y, Z} as reference points.We query the edge length |Y Z| and the pairwise distances of some other suitable nodes in the first round.All the nodes will be placed relative to Y and Z.Now we consider the second round.We select nodes in groups of 7 nodes each in such a way that the pairwise distances of the union of each group of nodes {X, A, B, C, P, Q, R} and {Y, Z} satisfy the conditions in Lemma 6.Then we query the remaining necessary pairwise distances of the union to form a 5:5 jewel.The jewel will be line rigid by Lemma 6 irrespective of the lengths of the edges AB, CX, P Q and RX, since no condition of the lemma involves any of these edges.The unused nodes are made line rigid by using triangle as the ppg.
Algorithm 1.First a bit of nomenclature.To indicate the affiliations of the vertices X, A, B, C, P, Q, R to different copies of a 5:5 jewel, we use the following indexing scheme: Let the number of points be n = 7b + 30, where b is a positive integer.In the first round, we make 6b + 29 distance queries represented by the edges in the graph in Fig. 18.There are 2b + 6 children B j (j = 1, ..., 2b + 6) rooted at Y and 2b + 22 children Q l (l = 1, ..., 2b + 22) rooted at Z.The remaining 3b nodes are organized into groups of 3 as (A i , X i , P i ) (i = 1, ..., b) and the distances |A i X i | and  In the second round, for each 2-link (A i X i , X i P i ) we find a pair of edges Y B j and Y B k , rooted at Y satisfying Conditions 1 and 2 of Lemma 6; next, we find a pair of edges ZQ m and ZQ l , rooted at Z satisfying Conditions 3 and 4 of Lemma 6.
Then for each i, (i = 1, ..., b), we query the distances Its edges will satisfy all the rigidity conditions of Lemma 6.
For each of the 6 unused leaves B j of the tree rooted at Y , we query the distance |B j Z| to form the triangle Y B j Z.Likewise, for each of the 22 unused leaves Q l of the tree rooted at Z we query the distance |Q l Y | to form the triangle Y Q l Z.
The following theorem establishes the correctness of our algorithm.

Theorem 8
The ppg constructed by Algorithm 1 is line rigid.Theorem 9 10n/7 + 99/7 queries are sufficient to place n distinct points on a line in two rounds.
It is worth noting that our algorithm needs at least 37 points to work.When we have fewer points we can switch to the quadrilateral algorithm, described in the Introduction.The 2-round 5-cycle algorithm of Chin et al. [4] a total of 4/3n + 34/3 √ n queries for the placement of n points.Thus, our 5:5 jewel algorithm does better when n ≤ 4076.This provides the motivation for considering 6:6 jewels, which we do next.
4 Algorithm based on a 6 : 6 jewel The principal ideas underlying this algorithm are similar to the algorithm based on 5:5 jewel of the last section.So we will skip the repetitive details when there is no scope for confusion.Fig. 19 shows the ppg for an instance of the 6:6 jewel that we shall use in the construction of our composite ppg.For brevity we will refer to left cycle as C 1 and the right cycle as C 2 , and by 6:6 jewel we will mean the instance shown.By Corollary 4, the conjunction of these two sets of conditions constitutes a set of sufficient conditions for the line-rigidity of the 6:6 jewel above.

Finding substitutes
We would like to make the 6:6 jewel line rigid irrespective of the lengths of the edges AB, CD, P Q and RS as this allows us to query the remaining edges in such a way that the rigidity conditions are satisfied.Towards this goal, we reformulate 16 conditions (8 from each cycle) involving these edges with alternate sets of conditions, satisfying which we also satisfy the replaced ones.
We use the left cycle, C 1 = XABY CD, as a running example to demonstrate these replacements.

Replacing |AB| = |CD|
The layer graph for the 6-cycle C 1 corresponding to this condition is shown in Fig. 20(h).From the figure it is evident that we can replace this with the condition since this will also prevent the layer graph drawing of the cycle as in Fig. 20(h).

Replacing |XA| = |CD|
The layer graph of C 1 corresponding to this condition is shown in Fig. 20(f).To replace this condition we follow a similar strategy as for the 5:5 jewel, except for a small twist: we draw all possible layer graphs of the 6:6 jewel, excluding the chain XSRZ, in which the layer graph of Fig. 20(f) is embedded.The condition |XA| = |CD| is then amplified into the set of conditions that prevent the drawing of the layer graph representation of the 6-cycle corresponding to this condition (Fig. 20(f)).Two cases arise, depending on whether Y Z is horizontal or vertical.
• Y Z is horizontal: Here Z and X have different

Replacing |XD| = |AB|
The layer graph of the 6-cycle corresponding to this condition is as shown in Fig. 20(e).This layer graph is the same as that in Fig. 20(f

Rigidity Conditions
From conditions (4)-( 5) and Lemmas 7-13 we have the following lemma for the line-rigidity of the 6-cycle XABY CD of the 6:6 jewel of Fig. 19: Lemma 14 The 6-cycle XABY CD of the 6:6 jewel XABY CDP QZRS of Fig. 19 is line rigid if the edges of the jewel satisfy the following conditions: Similarly, we have the following lemma for the line-rigidity of the other 6cycle XP QZRS of the 6:6 jewel: Lemma 15 The 6-cycle XP QZRS of the 6:6 jewel XABY CDP QZRS of Fig. 19 is line rigid if the edges of the jewel satisfy the following conditions: By Corollary 4, the union of the two sets of conditions in Lemmas 14 and 15 constitutes a set of sufficient conditions for the line-rigidity of the 6:6 jewel of Fig. 19.Taking care of overlapping conditions between the two sets of conditions, we have 74 distinct conditions for the line-rigidity of the 6:6 jewel and hence the following lemma: Lemma 16 The 6:6 jewel XABY CDP QZRS of Fig. 19  In the next section we show how a composite ppg can be constructed by satisfying all the 74 conditions for each such jewel.

Algorithm
It is interesting to note that the substitution mechanism has generated rigidity conditions on the strut Y Z (Condition 1 of Lemma 16).This implies that, unlike the case for a 5:5 jewel, we will need a pool of nodes, S, for which the pairwise distances of all pairs are known and from which we choose the end nodes of a strut in order to meet the rigidity conditions on Y Z.We have to choose the size of S carefully.Since there are 10 conditions on the length of an Y Z, from Observation 1 it follows that there must be at most 21 edges incident to the end-node Y , when we are looking for the other end-node Z of a strut.
However, if we use 22 designated nodes for the selection of Z for a particular Y it may happen that all the 6:6 jewels get attached to the same designated node Z.This hinders our goal of obtaining a better value for α than previously known.
Thus, we would like to attach the 6:6 jewels to the designated nodes in such a way that the number of jewels attached to two distinct nodes differ at most by 1.Let the valence of a node in this set be the number of times it is used as the end node of a strut to attach a jewel.The following lemma tells us how big S must be.
Lemma 17 A set S of 42 nodes is sufficient to ensure that the valence of two nodes in S differ by at most 1.
Proof: To build a 6:6 jewel we set, say Y , to a lowest valence node in S. Of the remaining 41 nodes, at most 20 nodes may not be chosen as Z because of the conditions on Y Z (Condition 1 of Lemma 16).From the remaining candidate nodes that satisfy the conditions on Y Z we set Z to the one with lowest valence.This way we can attach the first 11 6:6 jewels to 22 fixed nodes that do not have any 6:6 jewel attached to them.Thus, each of these 22 nodes will have valence 1; the rest are of valence 0.
Next we can keep selecting at least 1 node (as Y , say) from the 20 nodes of valence 0 until all are used up.The valence of this 20-node set will be raised to 1.If the second node is not found in that group we can choose it from the other group of valence 1, raising its valence to 2. Now, there is no node of valence 0, at least 22 nodes of valence 1 while the rest are of valence 2. Note that when both the nodes of a pair are chosen from the set of nodes of valence 0 the number of nodes of valence 1 is increased by 2. Consequently, the number of nodes of valence 1 remains even.When only one node is chosen from the set of nodes of valence 0, there being at least 22 nodes of valence 1 we can choose the other node from this set.This increases the number of nodes of valence 1 by 1 and decreases it by 1 at the same time.Thus, in both the cases there will always be an even number of nodes of valence 1.If there are more than 20 nodes of valence 1 we can choose pairs of nodes for Y and Z from nodes of valence 1 until exactly 20 nodes remain.Eventually, there will be 20 nodes of valence 1 and 22 of valence 2.
We shall show that we can attach the 6:6 jewels in such a way that at any point of time the fixed nodes will have at most 3 consecutive levels of valence.For this we use induction.Let us assume that there are 20 nodes of valence d and 22 nodes of valence d + 1.
We can choose at least 1 node from valence d nodes and at most 1 node from valence d + 1 nodes with a total of 2 to form a 6:6 jewel until there is no node of valence d.Then there will be at least 22 nodes of valence d + 1.The rest are of valence d + 2. As above we argue similarly to show that there will always be an even number of nodes of valence d + 1.
If there are more than 20 nodes of valence d + 1 we can use them in pairs as in the initial round above until the number of degree d + 1 nodes are exactly 20.The rest 22 will be of degree d + 2. Now the situation is the same as when we started except that the levels have increased by 1.
The set S of 42 nodes can be set as the vertices of 8 4:4 jewels hanging from a common strut.Since each 4:4 jewel is line rigid so is this configuration.We will call the nodes in S fixed since we can fix their placement on a line by querying the edges of this ppg.
From Condition 2 of Lemma 16 we see that we need 48 extra edges for the selection of an Y B that satisfies all the conditions on it as stated in the lemma.Similarly, by Conditions 3, 4 and 5 of Lemma 16 we need 98 extra edges for ZQ, 96 extra edges for Y C and 96 extra edges for ZR respectively.Thus, 98 extra edges at Y and Z will suffice to satisfy all the conditions on these edges.In addition to these extra 98 edges we need 2 more edges to accommodate the difference of 1 6:6 jewel that can be attached to them.Thus, we need a total of 100 extra edges at each of the 42 nodes of S.
The main idea underlying the algorithm below is to construct multiple copies of a 6:6 jewel over two rounds to ensure their rigidity.We use the set of nodes S as reference points.Any set of 42 points is chosen as S. The pair of nodes {Y, Z} that make up the strut Y Z (see Fig. 19) of a 6:6 jewel, is chosen from the set S. As part of the first round, a line rigid layout of S is fixed by attaching eight 4:4 jewels of Fig. 8 from a common strut.The common strut of the 4:4 jewels joins two nodes of S. Pairwise distances of some other suitable nodes are also queried in the first round.Now we consider the second round.Let S = P S be the complement of S. In the second round, the positions of all the nodes of S are fixed relative to the nodes in S by first selecting groups of 9 nodes each from S and placing them relative to a pair of nodes {Y, Z} of S. For this, we select a node Y ∈ S which has the lowest valence of 6:6 jewel of Fig. 19 and a 5-link (X, A, D, P, S).Then we select a node Z ∈ S such that it has the lowest valence of 6:6 jewel of Fig. 19 and that |Y Z| satisfies all the conditions of rigidity on it as stated in Condition 1 of Lemma 16.Thereafter, the nodes B, C, Q and R of S are selected such that the conditions of rigidity on |Y B|, |ZQ|, |Y C| and |ZR| as stated in respectively Conditions 2, 3, 4 and 5 of Lemma 16 are satisfied.Then we query the remaining necessary pairwise edge distances |AB|, |CD|, |P Q| and |RS| of the group to form a 6:6 jewel.The jewel will be line rigid by Lemma 16 irrespective of the lengths of the edges AB, CD, P Q and RS, since no condition of the lemma involves any of these edges.The unused nodes of S are made line rigid by using 4-cycle as the ppg.
Algorithm 2. As in Algorithm 1, we use the following indexing scheme: Let the total number of points be n.We attach b 6:6 jewels (Fig. 19) to each of 20 fixed nodes in S and b + 1 to the remaining 22.This gives us a total of 21b + 11 jewels.
In the first round, we make distance queries represented by the edges of the graph in Fig. 28.All the nodes Y u (u = 1, ..., 42) (or, Y v , v = 1, ..., 42) in the subgraph enclosed by the rectangle are fixed in the first round by using the 4:4 jewel of Fig. 8 as the ppg.There are 8 4:4 jewels (Fig. 8) attached to a common strut, 42 nodes and 65 edges in the subgraph.Each vertex Y u (u = 1, ..., 42) (or, Y v , v = 1, ..., 42) has 2b + 100 leaves to attach b or b + 1 6:6 jewels (Fig. 19).Since there will be 21b + 11 6:6 jewels we have 21b + 11 groups of 5 nodes (A i , D i , S i , P i , X i ) (i = 1, ..., 21b + 11).We query the distances |A i X i |, |D i X i |, |S i X i | and |P i X i |, (i = 1, ..., 21b + 11) in the first round.We will make a total of 168b + 4309 pairwise distance queries in the first round for the placement of n = 189b + 4297 points.A consequence of the last theorem is that our 6:6 jewel algorithm is better than the 5-cycle algorithm of Chin et al. [4] for n ≥ 11851.

Lower Bound for Two Rounds
In this section we revisit the adversarial argument given by [4] to establish a lower bound on 2-round algorithms.We show that a deeper analysis improves the lower bound substantially.
Let A denote any 2-round algorithm and B an adversary.The latter returns a value for the distance between any two points queried by A. B can also assign value to the distance between a pair of points not queried by A. While A's goal is to make as few distance queries as possible, B intends to keep the linear placement of the points as ambiguous as possible.
In the first round, A queries the distances between pairs of nodes corresponding to the edges E 1 of the ppg, G 1 = (V, E 1 ).In response, B returns queried edge-lengths consistent with the following 3-part strategy.
A vertex of degree 3 or more of a ppg (in particular G 1 ) will be called heavy.

S1.
The placement of all heavy nodes is fixed and the lengths of the edges incident to these nodes are set.

S2.
For each node of degree 2 that is connected to a node of degree 1, the length of one of the two edges incident to the degree 2 node is set to a fixed value c > 0.  This ensures that all the vertices of the prefix segment p 0 p 1 p 2 p 3 of the path is at a distance farther than l max away from p 0 .Clearly the remaining vertices on P k , however placed, will also be at a distance farther than l max .
The strategies adopted by B bound the lengths of maximal paths formed by degree 2 nodes in G 2 .The precise results are given in the next 3 lemmas.
Lemma 19 In G 2 , the length of a longest chain of consecutive edges from E 1 that terminates on a heavy node at each end of the chain is 4.
We first consider the case when both of p 0 and p k+1 are heavy nodes of G 1 .Given strategy S3 of B, if for an i < k, with i = 1 (mod 3), A attaches no edge to either p i or p i+1 in the second round then their positions will be ambiguous (Observation 2).Thus, the lemma is settled for this case.
Consider the case when p k+1 is of degree 1.In view of strategies S3 and S4 of B, A must attach an edge at p i or p i+1 in the second round, for i < k and i = 1 (mod 3), to make the placements of these nodes unambiguous (Observation 2).Thus, the lemma is settled for this case also.
Lemma 20 A maximal path P k of degree 2 nodes in G 2 that contains at least one edge of E 2 can have at most 2 consecutive edges of E 1 .
Proof: Let P k (k ≥ 2) be a maximal path of degree 2 nodes in G 1 ; p 0 and p k+1 are non degree 2 nodes adjacent to p 1 and p k respectively, one of which is of degree 1 in G 1 .
Suppose p 0 is of degree 1 in G 1 .In view of strategy S3 of B, if no edge is connected to either p i or p i+1 for some i = 1 (mod 3) then following strategy S4, B will set |p i p i+1 | = |p i−1 p i+2 | for one of those values of i in the second round.Thus, there must be an edge connected to either p i or p i+1 for all i = 1 (mod 3).In particular, A must add an edge to be incident to p 1 or p 2 (when i = 1).
If Thus, for both the cases, there will be at most 2 nodes of degree at most 2 at an end of a path of degree 2 nodes of G 1 , if the end node is of degree 1.The algorithm will place them in the second round by introducing edge/s to one or both of them.Thus, in a maximal path of degree 2 nodes in G 2 that contains at least one edge from E 2 there can be at most 2 consecutive edges from E 1 .
Lemma 21 The number of nodes in any maximal path of degree 2 nodes in G 2 is at most 3.
Proof: If a maximal path of degree 2 nodes of G 2 consists of edges from E 1 only then by Lemma 19 its length is at most 3. Now we consider maximal path of degree 2 nodes of G 2 that contains at least one edge from E 2 .In such a path there cannot be three consecutive edges from E 1 (Lemma 20).Suppose the number of degree 2 nodes in such a maximal path is 4. Let the nodes be p 1 , p 2 , p 3 and p 4 .Let p 0 and p 5 be heavy nodes adjacent to p 1 and p 4 respectively.Since any maximal path of degree 2 nodes in G 2 can have at most 2 consecutive edges from E 1 the edges p 0 p 1 , p 1 p 2 , p 2 p 3 , p 3 p 4 and p 4 p 5 can be from E 1 or E 2 in the following 5 combinations:   The density of a ppg, G = (V, E) is defined as the ratio |E|/n, where n = |V |.We establish the following lower bound on the density of a ppg constructed by any 2-round algorithm.

Theorem 12
The minimum density of any line rigid ppg for two round queries is at least 12/11.
Proof: Let each edge of G have weight 1, which we split evenly between the vertices in V that define it.If w i is the accumulated weight of the i-th vertex, clearly n i=1 w i = |E| so that n * min i {w i } ≤ |E|.Thus, min i {w i } is a lower bound on the density.
We can get a more precise estimate.Observe that a ppg has 2 types of nodes, heavy ones (already defined before) and nodes lying on maximal paths of degree 2 nodes that we call light nodes.If an edge joins two light nodes or two heavy nodes then the edge weight is divided equally between the nodes.Otherwise, the light node gets 1/2 + g of the weight and the heavy node 1/2 − g of the weight , where 0 ≤ g ≤ 1/2.
The density of a heavy node is at least 3(1/2−g).As for light nodes, we note that by Lemma 21 each maximal path of degree 2 nodes has length k, where k ≤ 3. The total edge weight of such a path is 2(1/2 + g) + (k − 1).Thus the average density of each node in such a path is 1 + 2g/k.It is minimum when k = 3.Thus, the density of a light node is at least 1 + 2g/3.
The minimum average density for all nodes in G 2 is thus

Figure 1 :
Figure 1: Some point placement graphs for 3 points

Figure 4 :
Figure 4: Point placement graph in the shape of a quadrilateral (a) with opposite edges being equal have 2 placements as shown in (b) and (c)

Figure 5 :
Figure 5: Query graph for first round in a 2-round algorithm using quadrilaterals

Figure 7 :
Figure 7: Equivalent layer graphs for a class of layer graphs of a 5-cycle

Theorem 6 Case 1
The 4:4 jewel of Fig. 8 is line rigid.Proof: We claim that cycles XAY B and XQZP are both line rigid.Let the edge Y Z is x-parallel.Three cases arise: The 4-cycle XAY B is line rigid, while the 4-cycle XQZP has a layer graph representation.

Figure 8 :Figure 9 :
Figure 8: An instance of a 4 : 4 jewel a) for example, other representations can be obtained by varying the position of A on the supporting line of XB.It is impossible to extend the layer graph representations of the 5-cycle XABY C shown in Figs.10(a) and (b) into a layer graph representation of the entire 5:4 jewel of Fig. 9 without one of the vertices P or Q coinciding with one of the vertices B or C.However, it is possible to extend each of the layer graph representations of Figs.10(c)-10(e) into a layer graph representation of our 5:4 jewel.The layer graph representations of Figs.10(c)-10(e) can be prevented by insisting on the condition |XC| = |AB|, |XA| = |Y B|, |Y C| = |AB| respectively.By Theorem 1, these collectively constitute a set of sufficient conditions for the line rigidity of the 5-cycle XABY C. For the 4-cycle XP ZQ the rigidity condition is |XP | = |ZQ| (Observation 2).Thus by Corollary 4, the set of sufficient conditions for the rigidity of the 5:4 jewel of Fig. 9 is {|XC| = |AB|, |XA| = |Y B|, |Y C| = |AB|, |XP | = |ZQ|}.

Figure 10 :Theorem 7
Figure 10: Different layer graph representations of a 5-cycle

Figure 11 :
Figure 11: An instance of the 5:5 jewel For the 5-cycle XP QZR these conditions are:

Lemma 2
The 5-cycle XABY C of the 5:5 jewel of Fig. 11 cannot be drawn as the layer graph of Fig. 10(a) if the edges of the jewel satisfy the following conditions:

Figure 12 :
Figure 12: Replacing the condition |XC| = |Y B| when Y B and Y Z are mutually perpendicular

Figure 13 :
Figure 13: Replacing the condition |XC| = |Y B| when Y B and Y Z are collinear

Figure 14 :
Figure 14: Replacing the condition |Y C| = |AB| when Y Z and Y B are perpendicular to each other.There is only one position for R.{|Y B| = ||XA| ± |XC||} is an alternate rigidity condition for the 5-cycle XABY C with the layer graph drawing as in (Fig.10(e)).This condition however involves the edge XC that we wish to avoid.For both the layer graph drawings of Fig.14, YB and Y Z being mutually perpendicular, the edges Y C and Y Z are on a line CY Z, and they are parallel to XR.So, we must have |XC| = |ZR|.Using this, we get the replacement rigidity condition {|Y B| = ||XA| ± |ZR||}.
(a)-(d)).In all the 4 layer graph drawings the edges Y Z and XR are horizontal and collinear, while the edge ZR is vertical and connects those two parallel edges.The edges Y B and XA are horizontal and collinear, while the edge AB is vertical and connects those two parallel edges.Y Z and Y B are collinear, and so are XR and XA.Therefore, we must have |AB| = |ZR| and the replacement rigidity condition for this subcase is |Y C| = |ZR|.

Figure 15 :Figure 16 :
Figure 15: Replacing the condition |Y C| = |AB| when Y B and Y Z are collinear, ZR is perpendicular to BY Z and XR is perpendicular to ZR -ZR is perpendicular to Y B and Y Z, and XR and ZR are collinear.In this case XP , P Q and QZ can be placed in 2 distinct configurations (Fig. 16).In these configurations of the jewel the 5-cycle XABY C cannot be drawn as a layer graph in the present configuration if ||XA| ± |XC|| = |Y B|.In both the configurations of the layer graph of the jewel Y C and XRZ are parallel, and both of XC and Y Z connect them.We must have |XC| = |Y Z|.We can rewrite the condition as ||XA| ± |Y Z|| = |Y B| for this subcase.

Figure 17 :Lemma 3
Figure 17: Replacing the condition |Y C| = |AB| when Y B and Y Z are collinear and ZR is collinear with them.XR can only be perpendicular to ZR.

Figure 18 :
Figure 18: Queries in the first round for a 5:5 jewel Figure 19: A 6 : 6 jewel By Theorem 5, the 6-cycle XABY CD has 16 different layer graph representations (Fig. 20), giving us the following 16 conditions for its line-rigidity,

Figure 20 :
Figure 20: layer graph representation of a 6-cycle

Figure 29 :
Figure 29: The residual parts of maximal paths of degree 2 nodes that will satisfy Step 2 p k+1 is of degree 1 then following strategy S3 the adversary sets |p k p k+1 | = |p k−2 p k−1 | in the first round.If A attaches no edge to either p k−1 or p k in the second round, then following S4, B sets |p k−1 p k | = |p k−2 p k+1 |.This makes the placements of the nodes p k−1 and p k ambiguous (Observation 2).Thus, A must attach an edge to p k−1 or p k to preempt B.
for the rectangular layer graph XQZP the diagonally opposite vertices X and Z lie on an x-parallel line collinear with Y Z. Assume otherwise.Now, X and Y are diagonally opposite vertices of the rectangle XAY B while X and Z are diagonally opposite vertices of the rectangle XQZP .Therefore a vertex of the 4-cycle XAY B must coincide with a vertex of the 4-cycle XQZP on an x-parallel line collinear with Y Z.As this violates property P4 that a layer graph should have, the cycles XQZP and XQZP cannot have simultaneous layer graph representations.
Theorem 10 The ppg constructed by Algorithm 2 is line rigid.Proof:The proof is similar to that of Theorem 8 for the line rigidity of the ppg constructed by Algorithm 1 and is omitted.The number of queries in the first and second rounds are 168b + 4309 and 84b + 2122 respectively.Thus, in 2 rounds a total of 252b + 6431 pairwise distances are to be queried for the placement of 189b + 4297 points.It is interesting to note that our algorithm would need at least 4486 points to work, which makes it reasonably practical.When we have fewer points we can use Algorithm 1 instead.Theorem 11 4n/3 + 2105/3 queries are sufficient to place n distinct points on a line in two rounds.