Optimal pebbling number of the square grid
Abstract
A pebbling move on a graph removes two pebbles from a vertex and adds one pebble to an adjacent vertex. A vertex is reachable from a pebble distribution if it is possible to move a pebble to that vertex using pebbling moves. The optimal pebbling number is the smallest number m needed to guarantee a pebble distribution of m pebbles from which any vertex is reachable. The optimal pebbling number of the square grid graph was investigated in several papers [1, 12, 9]. In this paper, we present a new method using some recent ideas to give a lower bound on . We apply this technique to prove that . Our method also gives a new proof for .
1 Introduction
Graph pebbling is a game on graphs. It was suggested by Saks and Lagarias to solve a number theoretic problem, which was done by Chung [2]. The main framework is the following: A distribution of pebbles is placed on the vertices of a simple graph. A pebbling move removes two pebbles from a vertex and places one pebble on an adjacent vertex. The goal is to reach any specified vertex by a sequence of pebbling moves. This may be viewed as a transportation problem on a graph where the cost of a move is one pebble. We begin with some notation needed to state our results.
Let be a simple graph. We denote the vertex and edge set of by and , respectively. A pebble distribution is a function from to the nonnegative integers. We say that has pebbles placed at the vertex under the distribution . We say that a vertex is occupied if and unoccupied otherwise. The size of a pebbling distribution , denoted , is the total number of pebbles placed on the vertices of .
Let be a vertex with at least two pebbles under , and let be a neighbor of . A pebbling move from to consists of removing two pebbles from and adding one pebble to . That is, a pebbling move yields a new pebbling distribution with and . We say that a vertex is reachable under the distribution if we can obtain, after a sequence of pebbling moves, a distribution with at least pebbles on . If we say simply that is reachable under . More generally, a set of vertices is reachable under the distribution if, after a sequence of pebbling moves, we can obtain a distribution with at least a total of pebbles on the vertices in .
A pebble distribution on is solvable if all vertices of are reachable under . A pebble distribution on is optimal if it is solvable and its size is minimal among all of the solvable distributions of . Note that optimal distributions are usually not unique.
The optimal pebbling number of , denoted by , is the size of an optimal pebble distribution. In general, the decision problem for this graph parameter is NPcomplete [6].
We denote with and the path and cycle on vertices, respectively. The Cartesian product of graphs and is defined in the following way: and if and only if and or and .
Let and be vertices of graph . denotes the distance between and , namely the number of edges contained in the shortest path between and . The distance neighborhood of contains the vertices whose distance from is exactly . We denote this set with .
The optimal pebbling number is known for several graphs including paths, cycles [1, 3, 4], caterpillars [8] and ary trees [11]. The optimal pebbling number of grids has also been investigated. Exact values were proved for [1] and [12]. The question for bigger grids is still open. The best known upper bound for the square grid can be found in [9]. Diagonal induced subgraphs of the square grid was studied in [10].
Instead of the square grid on the plane it is easier to work with the square grid on the torus. As the plane grid is a subgraph of this, any lower bound on the torus grid will also give a lower bound on the plane grid as well. It is well known that the torus grid is a vertex transitive graph, i.e. given any two vertices and of , there is some automorphism such that . Some of our statements will be stated for all vertex transitive graphs.
In this paper we present a new method giving a lower bound on the optimal pebbling number. We obtain as a lower bound for the optimal pebbling number of the square grid, which is better than the previously known bounds.
In Section 2 we show that the concept of excess — introduced in [12] — can be used to improve the fractional lower bound on the optimal pebbling number. The higher the total excess, the better the obtained bound on the optimal pebbling number is. The problem is that this method is not standalone, because excess can be zero and zero excess does not give us any improvement. Therefore the main objective of the rest of the paper is to give a lower bound on the excess using some other pebbling tools.
In Section 3 we study the concept of cooperation. Cooperation is the phenomenon which makes pebbling hard. We show there, that if cooperation can be bounded from above, then we can state a lower bound on the optimal pebbling number. We invent the tool called cooperation excess, which is a mixture of cooperation and excess. In this section we state and prove several small claims which will be required later to prove Lemma 4. This lemma is the essence of our work. It shows that if the total excess is small, then there is not much cooperation and if cooperation is huge, then the total excess is also large. Therefore in each case one of our two lower bounds works well.
Unfortunately, the proof of Lemma 4 is quite complicated. The third part of Section 3 and the whole Section 4 contain the parts of this proof. In Section 5 we show a general method which can be used to give a lower bound on the optimal pebbling number. This method relies on Lemma 4. Using this method we show that . We also present a new proof for .
2 Improving the fractional lower bound
The optimal pebbling number problem can be formulated as the following integer programming problem [13]:
Its fractional relaxation can be solved efficiently, and its solution is called the fractional optimal pebbling number, which gives a lower bound on the optimal pebbling number. Originally it was defined in a bit different way, but this is an equivalent definition. You can find the details of fractional pebbling in [13].
Notice that some vertices must be reachable in a solvable distribution if there is an unoccupied vertex. Optimal distributions usually contain many unoccupied and several reachable vertices. However, in some sense, , , or more reachability wastes the effect of pebbles. Also reachability induces larger waste than reachability. In order to measure this waste we use the notion called excess, which was introduced in [12].
Definition: Let be the greatest integer such that is reachable under . The excess of under is if is reachable and zero otherwise. It is denoted by .
We are interested in the total amount of waste, therefore we define the notation of total excess of , which is .
Definition: An effect of a pebble placed at is the following: .
Herscovici et al. proved that the fractional optimal pebbling number of a vertextransitive graph is , therefore it is a lower bound on the optimal pebbling number. The corollary of the next theorem improves this bound.
Theorem
If is a solvable distribution on , then
Proof
It is clear that if a vertex is reachable under , then it is mandatory that
. Summing these inequalities for all the vertices,
Exchange the summations on the left side and use the fact that is solvable on the right side,
Group the elements of the second sum according to the distance neighborhoods,
Corollary
If be a solvable distribution on a vertextransitive graph , then
Naturally, this bound is useless without a proper estimate of total excess. To say something useful about it we look at the optimal pebbling problem from a different angle.
3 Cooperation between distributions
In this section we talk about cooperation, which makes pebbling hard.
3.1 Pebbling cooperation
Definition: Let and be pebble distributions on graph . Now is the unique pebble distribution on which satisfies . and are disjoint when no vertex has pebbles under both distributions.
Definition: The coverage of a distribution is the set of vertices which are reachable under . We denote the size of this set with .
A natural idea to find small solvable distributions is finding a distribution with small size and huge coverage and make it solvable by placing some more pebbles.
In the rest of the section we assume that we add disjoint distributions and together. We would like to establish an upper bound using on . Similarly, we are interested in some relation between and .
Definition: A cooperation vertex is neither reachable under nor , but it is reachable under . We denote the number of such vertices with . A double covered vertex is reachable under both and , we denote the size of their set with .
The following claim is a trivial consequence of the definitions.
Claim
.
Definition: We say that a distribution is a unit, if only one vertex has pebbles under .
Units are the building blocks of pebble distributions in the following sense: Any distribution can be written as , where is a unit having pebbles at . Units have two main advantages over other distributions. Their coverage and total excess can be easily calculated:
Claim
3.2 Combining cooperation and excess
We would like to distinguish the sources of excess. Does it come from or or does it arise from the “cooperation of and ”?
Definition: The unit excess of , denoted by , is , where is a unit on containing exactly pebbles and all of them are placed at .
Definition: The cooperation excess of a vertex is . If it is positive, then we say that has cooperation excess.
Similarly, the cooperation excess between and is the total excess of minus the total excesses of and . Denote this with .
We have mentioned previously, that we can split any pebbling distribution into disjoint unit distributions. If we get unit distributions, then the application of Claim 3.1 and the definition of cooperation excess gives the following results.
Claim
Let be a pebbling distribution on and let be a disjoint decomposition of to unit distributions. Denote the elements of with . Now
(1) 
(2) 
Both and can be calculated easily. The “effect” of cooperation is calculated in the other, more complicated term. Lemma 4 is going to establish a connection between those quantities in a fruitful way.
3.3 Connection between cooperation and excess
Now let us consider an arbitrary graph , and let be the maximum degree of . In the rest of the section we assume that is a unit having pebbles only at vertex and its size is not zero. Now we state some basic claims about the recently defined objects.
Claim
Each cooperation vertex has a neighbor that has cooperation excess.
Proof
A cooperation vertex is not reachable under or . Therefore none of its neighbors is reachable under these distributions. On the other hand, is reachable under , hence there is a neighbor of which is reachable under this distribution. This means that has cooperation excess.
Definition: If a vertex is not a cooperation vertex and it does not have cooperation excess, then we call it cooperation free.
This name is a somewhat misleading, because these vertices can participate in cooperation in a sophisticated way. For an example see Figure 1.
Definition: A vertex is utilized by a pebbling sequence if there is a move in the sequence which removes or adds a pebble to the vertex. Let be the minimal number of cooperation vertices which are utilized by a pebbling sequence which satisfies that . If is not 2reachable under , then we say that .
Claim
If there is an available pebbling move which removes a pebble from a cooperation vertex , then either two neighbors of , say and , have cooperation excess at least 1 and and or a neighbor has cooperation excess at least 3, and .
Proof
The condition implies that can obtain two pebbles by some pebbling moves under . Consider a pebbling sequence which does this by utilizing cooperation vertices. Either moves the two pebbles to from two different neighbors and , or it can move both pebbles from the same neighbor . None of the neighbors are reachable under or , but , and has to be , and reachable under , respectively. This means that and have cooperation excess at least 1 and the cooperation excess of is at least 3. Furthermore, moves two pebbles to and or to , then it moves them to with some more moves. This shows that .
Claim
If the cooperation excess of a vertex is at least 3 and one of its neighbors, say , is a cooperation vertex, then there is a vertex that is adjacent to and .
Proof
does not have two pebbles under , otherwise can not be a cooperation vertex. obtains pebbles from its neighbors, so one of them, say , can get two pebbles by utilizing at most cooperation vertices. If is a cooperation vertex then a pebbling sequence resulting in two pebbles at utilizes more cooperation vertices than the sequence which does not make the final move from to .
Claim
If a vertex has cooperation excess, then it has a neighbor which has cooperation excess or reachable under or .
Proof
gets a pebble under , so a neighbor is reachable under . If is not 2reachable under of , then it has cooperation excess.
Remark 1
In fact, a stronger property holds. If a vertex gains an extra pebble by cooperation, then it can happen in two ways: A neighbor gained extra pebbles and it passes one of them. Or there are two or more neighbors of such that each of them can give some pebbles to , but these moves somehow blocks each other. The advantage of the cooperation is that some previously blocked moves can be done simultaneously. This is the way how cooperation free vertices can “help cooperation”.
3.4 Trajectories
Here we introduce a visualization of pebbling sequences, which is slightly different from the signature digraph used in several pebbling papers (i.e. in [6]).
Definition: The trajectory of a pebbling sequence , denoted by , is a digraph on the vertices of without parallel edges, where is a directed edge if and only if a pebbling move is contained in the sequence.
Definition: The size of a pebbling sequence is the total number of moves contained in it. We say that is a minimal pebbling sequence with property if its size is minimal among all pebbling sequences having property .
In the next proof, we need a lemma which is frequently used to solve pebbling problems. It is called NoCycle Lemma and proved in several papers [5, 6, 7]. We state this lemma in the language of this paper.
Lemma (NoCycle [7])
Let be a pebbling distribution on graph , and be an executable pebbling sequence. There is a subsequence whose trajectory does not contain directed cycles and for each vertex .
This implies the following corollary:
Corollary
If is a minimal pebbling sequence which moves pebbles to a vertex , then its trajectory is acyclic.
Claim
If has cooperation excess under , where , then is double covered.
Proof
The NoCycle lemma yields that we can move maximum number of pebbles to without removing a pebble from . We can move pebbles to , which means that we move here a pebble of while we keep the pebbles of , so is double covered.
The following definition will be crucial in the proof.
Definition: We say that a path is a coopexcess path, if each inner vertex of the path has cooperation excess.
Lemma
Let be a vertex which is not double covered but it has cooperation excess. There is a coopexcess path between and a double covered vertex or there are at least two cooperation free vertices such that each of them is connected to by a coopexcess path. If is not 2reachable under both and , then these paths does not contain a vertex whose value is higher than .
Proof
Consider a pebbling sequence moving pebbles to utilizing cooperation vertices. Consider some path in the trajectory of connecting to . We can assume that the only sink in the trajectory path is . A cooperation vertex without cooperation excess can not be the tail of an arc which is contained in the trajectory, therefore each vertex in the trajectory is either cooperation free or it has cooperation excess.
If there is a path between and which is contained in the trajectory such that all vertices of this path have cooperation excess, then according to Claim 3.4 is double covered and this path is a coopexcess path. Otherwise, all of the paths which are contained in the trajectory contain cooperation free vertices.
In each such path let denote the cooperation free vertex which is the closest vertex to . If exist, then we have found 2 cooperation free vertices such that each of them is connected to by a coopexcess path.
In the remaining case there is only one such . Either it is a cut vertex in the trajectory or . Let be the set of vertices which are included in the trajectory. We divide to three sets , and in the following way:
We remove from the trajectory obtaining some components, then we place a vertex of to if is in the component containing , similarly we place to if it is in the component containing and place the remaining vertices to . Now we add to all of these sets. Let be the pebbling sequence containing all moves of which acts only on the vertices of . We define and similarly. The sources of the latter two sequences are only and vertices having pebbles under .
If is reachable under , then is empty ( is not double covered) and . We can replace with a pebbling sequence which does not use any pebbles of and . Therefore is an executable pebbling sequence and . must use a pebble of to do this, otherwise is executable under which is a contradiction. The trajectory of is connected, therefore there is a vertex which is double covered, furthermore each vertex in this trajectory is connected by a coopexcess path to , so we are done.
If is not reachable under , then . Thus, there is a minimal pebbling sequence which is executable under and . Clearly is not executable or . Both cases require that removes a pebble from a vertex contained in .
Let be the set of vertices from which removes a pebble. is executable under so these vertices are reachable under . Consider the trajectory of . If any vertex from is connected in the trajectory with a vertex contained in without passthrough , then each vertex in such a connecting path is 2reachable under , therefore it is cooperation free or has cooperation excess. So there is either an other cooperation free vertex connected by a coopexcess path to , or there is a coopexcess path between and which is connected to by a path in the trajectory of which does not contain , so that path has to contain a double covered or a cooperation free vertex, which is not .
The remaining case when separates all elements of from in the trajectory of .
Let be a maximal subset of which is executable without using the pebbles placed at , and let be the remaining subsequence. is not executable under or . Therefore moves more pebbles to than , but is executable under , thus , therefore has cooperation excess.
To prove the second claim, consider the paths we have found. If they were part of the trajectory of , therefore all of them are 2reachable under , so their value can not be higher than . Otherwise, the path consists of vertices from the trajectory of and some others whose value is zero, since they are 2reachable under .
The following claim is a trivial consequence of the definitions.
Claim
If contains at least two pebbles and it is double covered, then one of its neighbors is also double covered.
In the rest of the section we assume, that contains at least two pebbles, i. e. . Therefore we can use the previous claim.
Lemma
Assume that contains at least two pebbles. Then each double covered vertex is connected by a coopexcess path to an other double covered vertex or a cooperation free vertex. Furthermore each vertex of this coopexcess path is reachable under .
Proof
The previous claim handles the case when is , since the neighbor is connected to . So assume that .
Since is double covered, it is reachable from , so it is connected to by a path, whose vertices are reachable under . Therefore these vertices can not be cooperation vertices. If there is a vertex on this path which does not have cooperation excess, then the vertex closest to satisfies the conditions of the second type. Otherwise, has cooperation excess which means that it is double covered.
We are getting closer to establish a connection between the number of cooperation vertices and cooperation excess.
Definition: We call a subset of a block, if

each pair of vertices in is connected by a coopexcess path,

it contains a vertex having cooperation excess
and it is maximal with these properties.
Notice that the intersection of two blocks can not contain a vertex having cooperation excess.
Lemma
Each block either

contains at least two double covered vertices, or

contains one double covered vertex and one cooperation free vertex, or

contains at least two cooperation free vertices.
Proof
Later we generalize the notion of blocks, so that we keep the properties of 3.4. The following statement will be useful for this.
Lemma
If a vertex having cooperation excess is adjacent to a cooperation vertex such that , then there are vertices and , such that each of them is either double covered or cooperation free and they are connected to by coopexcess paths containing only vertices whose values are smaller than .
Proof
has a cooperation vertex neighbor, therefore is not 2reachable under or . According to Lemma 3.4 there is a double covered vertex or there are two cooperation free vertices who are connected to by a coopexcess path containing only vertices whose values at most . In the latter case we are done. Since the double covered vertex is connected to an other double covered or cooperation free by a coopexcess path containing vertices whose M value is zero, according to Lemma 3.4. The concatenation of these two coopexcess paths fulfills criteria.
4 Connection between total cooperation excess, number of cooperation vertices and maximum degree
In this section we prove a crucial lemma. Unfortunately, the proof requires quite a lot of effort, including many small claims.
Lemma
Let be an arbitrary pebbling distribution on and be a unit having at least two pebbles, such that does not contain a pebble at . Now we have
This lemma gives a connection between the total cooperation, the total number of double covered vertices and total cooperation excess. The proof would be relatively easy if the effect of a pebble would appear close to the location of the pebble. The example on Figure 2. shows, that unfortunately this is not always true.
Another difficulty arises from the fact that a cooperation vertex can have cooperation excess. For such an example see Figure 3. To prove Lemma 4 we get rid of such vertices one by one.
So to prove the lemma we will change the graph in several steps. In the new graph it will be easier to isolate these effects.
We introduce a sequence of auxiliary graphs , whose vertices are vectors of four coordinates. The first and fourth coordinate is always an integer, while the other coordinates are binary. We denote the vertices of these graphs with underlined letters and the th coordinate of vertex with . We encode the parameters of the investigated pebbling problem in the auxiliary graph and in the coordinates in the following way:
is isomorphic to . The first coordinate of each vertex is the amount of the cooperation excess of the corresponding vertex. The second coordinate is 1 iff the corresponding vertex is a cooperation vertex. The third coordinate is 1 when the vertex is double covered. Finally, the last coordinate is , i.e. the minimum number of cooperation vertices have to be utilized by a pebbling sequence to obtain 2 pebbles at . So is representation of the original configuration, the labels give the values of the various quantities that we are interested in.
The other graphs in the sequence will be obtained from by applying certain operations recursively, until we finally obtain with some useful properties. It is important to note that although the labels of are obtained from the pebbling distribution on , this will not be true any more for the other auxiliary graphs. We are not trying to change the graph and the pebbling distribution and then obtain the new labels from these. We just apply the transformation on the abstract, labeled graphs.
Now we translate the properties of the pebbling distribution to properties of .
Definition: We call a path in an path, if each inner vertex of satisfies . We say that is an block iff

there is a vertex such that ,

if , then there is an path which connects them, and
and is maximal to these properties.
Note that the concept of path and block are generalizations of coopexcess path and block, respectively. In this language, the statement of Lemma 4 can be formulated as:
We state four properties of which will be inherited to later auxiliary graphs. The significant properties are the first and the last. The other two are technical ones which will help the proof of inheritance stated in Claim 4.
Claim
The following statements hold for :

If then one of the following two cases hold:

there exist a which is adjacent to , and , or

there are vertices and such that they are neighbors of , and are both positive, and .


If and has a neighbor such that , then there is a , which is adjacent to and .

Let be a vertex whose first and second coordinates are both positive. If is a neighbor of such that and , then there are vertices and such that each of them is connected to by paths containing only vertices having their fourth coordinate smaller than , and their third coordinate is either or the first and second coordinates of them are .

Each block contains either

two vertices with third coordinate , or

two vertices with first and second coordinates , or

one vertex with third coordinate and one vertex with first and second coordinates .

This claim is equivalent to the following, previously proven, statements with the new notation: Claim 3.3 (8), Claim 3.3 (9), Lemma 3.4 (10) and Lemma 3.4 (11).
We will obtain from by applying one of two transformations. Then we repeat this until it is possible to apply at least one of the transformations. Both transformations will preserve , , the fourth coordinate of each vertex and , the maximum degree in the graph. The objective of the transformations is to replace vertices satisfying (i.e. it has cooperation excess and it is a cooperation vertex) with (one ore more) vertices satisfying . From this point, we call these vertices saturated vertices. Both transformations will increase the number of vertices in the auxiliary graph.
Let be a vertex where such that its fourth coordinate is maximal among these vertices. By Claim 4 (8) there are two cases.
Case 1: If has a neighbor such that and , then we apply the following transformation to :
Transformation 1

Choose a neighbor of such that its fourth coordinate is minimal among all neighbors of .

Let be the set of ’s neighbors without where the product of the first and the second coordinate is positive.

Delete and add three vertices , and , such that and . , and . Connect with , and .

Delete each element of and add two vertices and and set the coordinates as: , , , , and . We connect to and to and to each original neighbor of .

We connect the neighbors of which are not included in to .

Set .

If , then add an extra vertex and connect it only with . Set its vector to .
In other words this transformation replaces each saturated neighbor of (excluding a chosen ) with two vertices such that one of them is a leaf with zero first coordinate and the other one is act as the original vertex, but its second coordinate is zero. To handle the increased degree of , we triple it. Also, if is saturated then we add the additional vertex. Note that this can be done when . If we have to handle this case in a slightly different way.
Case 2: If has two neighbors such that their first coordinates are positive and their fourth coordinates are strictly less than , then we apply the second transformation:
Transformation 2

We choose neighbors and whose fourth coordinate is minimal among all neighbors and .

We delete and add vertices and . We set the coordinates of these vectors as: , , , , and .

We connect only with . In contrast, we connect with all neighbors of except .
Both transformation can be seen on Figure 4.
Claim
Both transformations preserve , and if .
Claim
Both transformations decrease the number of saturated vertices.
Claim
If the statements of Claim 4 hold for an auxiliary graph, then they hold for the new one obtained by applying one of the above defined transformations.
Proof
We say that a vertex is created by the th transformation if is a vertex of and is a vertex of . In this situation we say that is a descendant of . A vertex is involved in a transformation if either it is created by that or its vector is changed by it.
Notice that the transformations keep the fourth coordinates of the vertices and if two vertices are descendants of the same vertex, then their fourth coordinates are the same.
(8): If is a saturated vertex in , then it is not created by the th transformation and it was saturated in also. If none of its neighbors were involved in the last transformation, then the property is clearly holds. Therefore assume the opposite. Assume that had a neighbor in , such that and in .
If is contained in also, then the th transformation did not change . In that case and are adjacent in and we are done.
Otherwise the th transformation created some descendants of .
If it was transformation 1 then a descendant of is connected to and either its first coordinate equals or it is . In the first case we are done and the latter can happen if and only if acted as in that transformation. However, this is not possible, because this would mean that acted as but by (9) and the choice of in Transformation 1, therefore which is a contradiction.
The remaining case is that two descendants of are created by Transformation 2. Since can not be or in Transformation 2, therefore it is adjacent to in and the first coordinate of this vertex equals .
Now assume that there are neighbors and in such that their fourth coordinates are smaller than and . We may assume that , otherwise we obtain the previous case. Therefore neither nor can act as in Transformation .
If is contained in , then it is still adjacent to . If is replaced with some descendants by the th transformation, then one of its descendants keep its first coordinate and that one is connected to . Like in the previous case it cannot happen that and in Transformation 1 or and in Transformation 2. We can state the same for .
(9): If is contained in then in also. Therefore according to (9) there is a which is adjacent to and . Either or one of its descendants is adjacent to in , therefore we are done.
Otherwise is a descendant of a vertex . has a neighbor whose fourth coordinate is at most . There are several cases:
First case: in Transformation , where . If we remove from the neighborhood of and add we obtains the neighborhood of . Therefore has a neighbor whose fourth coordinate is not bigger.
Second case: in Transformation . had the smallest fourth coordinate among the neighbors of , thus and and are adjacent in .
Third case: in Transformation . Since and according to (9).
(10): Transformation 1 keeps the paths, because it keeps connectivity and the first coordinate becomes zero only at leaves. Transformation 2 destroys some paths but all of them contain the saturated vertex which was handled by the transformation and whose fourth coordinate was at least .
(11): Transformation 1 does not split an block, furthermore it keeps the number of vertices whose third coordinate is one and whose first and second coordinate are both zero in each block.
Transformation 2 either does not split an block and keeps the investigated quantities, or it splits an block to two blocks. But (10) guarantees that both blocks contain enough vertices whose third coordinate is one or both first and second coordinates are zero.
Claim
If Claim 4 holds for and there is a saturated vertex, then at least one of the two transformations can be applied to .
The first proposition of Claim 4 guarantees it.
Corollary
There is an , such that there is no saturated vertex in , furthermore , .
Lemma
Definition: Let be an block in an auxiliary graph. We say that a vertex of is inner vertex if its first coordinate is positive, otherwise it is called a boundary vertex.
Claim
Consider an block of an auxiliary graph . Let the number of the boundary and the number of inner vertices denoted by and , respectively. If holds for each then is satisfied.
Proof
Proof by induction: The base case is an block with one inner vertex. This block is the closed neighborhood of the only inner vertex, therefore the number of boundary vertices is at most . Now we assume that for any the inequality is true. Let . We take a spanning tree of the inner vertices and consider a leaf vertex . If we set to zero, then becomes a boundary vertex and at most boundary vertices, which are neighbors of , are dropped from the block. The number of inner vertices is decreased by one, and the number of boundary vertices is decreased by at most . Using the induction hypothesis the proof is completed.
Proof of Lemma 4: Consider an block and a boundary vertex of . Either or . Thus we have: where is the number of vertices in whose first two coordinates are zero. It is also clear that . Combining these observations and the previous claim:
Claim 4 implies that Claim 4 (11) holds for . This guarantees that . Therefore
Proof of Lemma 4: We distinguish three cases depending on .
Case 1:
Let be the auxiliary graph which we obtained from by applying transformations until it does not contain any more saturated vertices. The last lemma holds for each block, therefore:
Only boundary vertices can be included in multiple blocks, and the first and second coordinate of a boundary vertex is zero, thus:
Using Claim 4 we obtain
Case 2:
If the graph consists of multiple connected components we may restrict our attention to the component containing the unit. Let be the number of double covered vertices. We first verify the lemma in the case . In this case the graph consists of a matching and isolated vertices. Thus, we must have , and we must show that . If the unit, , is isolated the result is trivial suppose the unit is in an edge . If , then . If , the . Suppose that and set . We have and
This completes the proof in the case. If then we may assume that the graph is a path or a cycle. In this case we have so we must show . However, it is easy to see that in a path or a cycle every cooperation vertex is adjacent to a double covered vertex and, moreover, that double covered vertex is on the path between the cooperation vertex and (possibly itself). It follows that there are at least as many double covered vertices as cooperation vertices, as desired.
Case 3:
We remind the reader that the problem with the case is that it is not possible to add in transformation 1.
is needed when is saturated in . In that case Transformation 1 handles and substitutes it with unsaturated descendants. If the degree of any of the descendants of is smaller than , then we can make adjacent to this vertex and the problem is eliminated. Otherwise has three neighbors: , and . The one whose fourth coordinate is minimal among them is , also and both and are saturated vertices.
Now we make one of ’s descendants saturated. We have to make sure that (8) holds for this saturated descendant therefore we have to make a few new transformations.
Case 1: (8) a) holds for