Finding Three Transmissions is Hard

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Los Angeles, CA 90089, USA email: [email protected]. Abstract—We investigate a fundamental wireless broadcast problem called Index coding. We focus.
Globecom 2012 - Communication Theory Symposium

Finding Three Transmissions is Hard Arash Saber Tehrani

Alexandros G. Dimakis

Department of Electrical Engineering University of Southern California Los Angeles, CA 90089, USA email: [email protected]

Department of Electrical Engineering University of Southern California Los Angeles, CA 90089, USA email: [email protected]

Abstract—We investigate a fundamental wireless broadcast problem called Index coding. We focus on binary linear scalar index coding problems when it is a-priori known that the optimal solution requires three transmissions. For this case, we characterize the relation of the clique cover number of the side information graph G and the optimal index code. This allows us to show that even when there is a-priori knowledge that three transmissions can solve a given index coding problem, finding these transmissions is NP-hard. Another implication of our results is that for three solvable undirected problems, the benefit of interference alignment solutions is at most one compared to a simple clique cover.

I. Introduction We investigate a fundamental wireless broadcast problem called Index coding. The problem was introduced by Birk and Kol [1] as a stylized broadcasting problem with side information, initially motivated by a broadcast satellite application. In this setup, there is a single Base Station which has access to a set of packets, and a number of user receivers. Each user has one packet they desire and some other packets already known, as side information. This side information could be due to prior requests or perhaps wireless overhearing. Each receiver starts with a list of stored and desired packets and the broadcast model is that when the Base station transmits, all the users receive the transmitted packet without error. The index coding minimization objective is to satisfy all the users with the minimal number of broadcast transmissions. It can be easily seen that transmitting coded packets can be beneficial, as shown in the example of Fig. 1. The Base station is assumed to know the side information set of each user and has to decide on a set of broadcast transmissions. Despite its simplicity, this problem has been

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S U1 has: A has: C wants: B

U U21 has: B has: A wants: C

U3 has: B has: C wants: A

Fig. 1: An index coding example: Three users, each requesting one packet and having two others as side information. The Base station S has all three packets A, B, C and each transmission is within range for all three users. The Base station can exploit the broadcast benefit of the wireless channel to satisfy three requests in single transmission by sending the XOR of the three packets A + B + C .

proven tremendously challenging and theoretically deep. Bar-Yossef et al. [2] studied the problem and showed that the optimal linear solution is related to a rank minimization problem over a finite field [1], [3]. The computational problem of finding the minimum set of transmissions is NP-hard to find and to approximate [2], [4], [5]. Further, the work of [2] shows the connections of index coding to a well known open fundamental problem called the Shannon capacity of a graph [6], [7]. It was recently shown by El Rouayheb et al. [3] that any arbitrary network coding problem with multiple sources and receivers can be mapped to a properly constructed index coding instance. In addition, Index coding is a fundamental problem in understanding interference alignment [8]–[11] beyond the interference channel. Further, Wang [12] analyzed

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x1

u1

x1

u2

x2

u3

x3

1

x2

2

3

x3

Fig. 2: The undirected graph G and the bipartite models of the system shown in Fig. 1.

broadcast packet erasure problems that use closely related techniques. More concretely, we consider a base station that aims at transmitting a set of n bits x1 , · · · , xn ∈ {0, 1} to n different users u1 , · · · , un , while each user i wants the specific bit xi (with the same index i) for i ∈ {1, · · · , n}. Each user may have some prior side information, i.e., knows a subset of bits as side information. Furthermore, similar to [13], we consider the assumption of symmetric demands where if user i wants bit xi and knows bit xj , then user j who wants bit xj must know bit xi . The benefit of considering symmetric demands is that it allows us to represent the problem by an undirected graph G = (V, E) where each node i represents both user i and bit xi where user i wants bit xi and an edge between node i and j means that users i and j know bits xj and xi respectively. This symmetric demand assumption is, of course, limiting the set of index coding problems we consider. The most general representation of the problem is offered in [14], [15] where an instance of the problem is represented by a bipartite graph. In this more general problem, each bit can be demanded by more than one user and the number of bits and users need not be equal. Fig. 2 shows the undirected and bipartite graph models for the system shown in Fig. 1. Let x = [x1 , · · · , xn ] be the set of bits and for the set of indices S , let x(S) denote the projection of x on the coordinates in S . Further, let N (i) denote the set of neighbors of node i, i.e., the side information of user i. Then an index code C of length ` for graph G on n nodes is a set of codewords in {0, 1}` with 1 n an encoding function E : {0, 1} → {0, 1}` , and a set of n decoding functions D1 , · · · , Dn such that Di (E(x), x(N (i))) = xi for every user i. We call the code a scalar linear index code if the encoding function E(x) is linear function of x.

Then the optimal scalar linear broadcast rate is the minimum ` which can be achieved under all possible scalar linear codes. As shown in [2], [13], the rate of the best scalar linear index code β` (G) is equal to the graph functional minrank2 (G) defined as follows. Let the 0/1 matrix A fits the graph G. That is, i) aii = 1 for i = 1, · · · , n. ii) aij = 0 if {i, j} ∈ / E. Note that for {i, j} ∈ E , aij can be either 0 or 1. Then minrank2 (G) = min{rank(A) : A fits G}. It is shown in [2], [13], that for any index coding instance G, there exists a linear index code whose length equals minrank2 (G). This bound is optimal for all linear index codes for G, i.e., β` (G) = minrank2 (G). By representing an instance of index coding with symmetric demands by an undirected graph G, and showing the equality of β` (G) and minrank2 (G), bounds on minrank2 (G) can be directly applied to the scalar linear broadcast rate β` (G). It is shown in [16] that minrank2 (G) is bounded as follows: α(G) ≤ minrank2 (G) ≤ χ(G)

(1)

where G is the complement of G and α(G), χ(G), respectively, denote the independence number (size of the largest set of vertices in a graph, no two of which are adjacent) and the chromatic number (the smallest number of colors needed for coloring the graph with no two neighboring vertices having the same color). Recall that the chromatic number of G is the same as the clique cover number of G. Note that both upper and lower bounds are in general NP-hard to compute. Our goal in this paper is to analyze the graphs G whose optimal scalar linear rate is three, i.e., β` (G) = 3. We call them 3-linearly solvable graphs. We show that for such graphs, the bounds in (1) are within one bit. Specifically, for a 3-linearly solvable graph G, α(G) = 3 and χ(G) = 3 or 4. Further, using the above, we show that even if we know that minrank2 (G) = 3, finding the three linear transmissions, i.e., the linear encoding function E(x), is NP-hard. II. Our Results As explained, Let the undirected graph G = (V, E) represent an instance of index coding problem. Let β` (G) = K , i.e., the base station delivers all users their desired bits in K transmissions, each formed by a linear combination of some of the bits. Each

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transmission can be considered as a modulo two sum of bit subsets. We depict such a transmission by a sum of the elements of a vector v ∈ {0, 1}n with the support set V = {i : v(i) = 1}. The following intuitive lemma is the direct result of the previous definition: Lemma 1: Let β` (G) = K . User i for i ∈ {1, · · · , n} receives the bit it demands xi ∈ {0, 1} if and only if there exist values µk , ρh ∈ {0, 1} for k ∈ {1, · · · , K}, h ∈ N (i)

that satisfy the following linear equation: xi =

K X k=1

µk

n X

vk (j)xj +

j=1

X

ρh xh ,

(2)

h∈N (i)

where all the summations are over GF (2). From Lemma 1, it is apparent that for user (node) i to receive the bit it demands from v=

K X

µk vk ,

k=1

with support set V , the following two conditions must be met i) i ∈ V . ii) User i should know every other bit in V \ i, i.e., V \ i ⊆ N (i) where as mentioned before, N (i) denotes the neighbors of node i which represents the prior side information of user i. Note that the second condition, given the undirected graph G, means that all other users j 6= i and j ∈ V \i know bit i since an undirected edge between nodes i and j means that i knows xj and j knows xi . Our main result is characterizing the relationship between the chromatic number of G, χ(G), and its optimal linear index code β` (G) for 3 solvable index coding problems (β` (G) = 3). This is presented in the following two theorems: Theorem 1: Given an instance of index coding problem G, if χ(G) = 3 then β` (G) = 3, hence the index coding problem is 3-linearly solvable. Theorem 2: Given a 3-linearly solvable instance of index coding G, β` (G) = 3, the clique cover number χ(G) is either 3 or 4. It is known that for the graph G, if minrank2 (G) is bounded, then χ(G) is bounded as well [17]. Theorem 2 shows that for the case minrank2 (G) = 3 we have χ(G) ≤ 4.

This result has the following implication: For any 3-solvable undirected index coding problem, the benefit of sophisticated interference alignment solutions [8] is at most one compared to a simple clique cover solution. Note that our results show that for an undirected instance of the index coding problem β` (G) = 2 if and only if χ(G) = 2. This is accomplished from the proof of Theorem 1. There we show that if β` (G) = 2 then χ(G) = 2. Further, if χ(G) = 2 then β` (G) = 2 since β` (G) = 1 means that α(G) = 1 which is only true for complete graphs with χ(G) = 1. Furthermore, since 2-coloring of a graph can be done in polynomial time, then finding the minrank2 (G) and the two transmissions can also be done in time polynomial in number of nodes and edges when β` (G) = 2. Another interesting implication of our characterization is a hardness result: we show that even if a given index coding problem is known to be 3-solvable, finding these three transmissions is computationally intractable. Theorem 3: Given an instance of index coding G, if β` (G) = 3 then finding the appropriate linear coding strategy is NP-hard. This theorem holds for the most general index coding problem, i.e., all instances that can be presented by the bipartite graph model. This is true, since the undirected graph model is the special case of the general bipartite model. We prove Theorem 3 by a reduction of 3- or 4-coloring of a 3-colorable graph [18] to scalar linear index coding. That is, we find a polynomial transformation that maps the solution of scalar linear index coding to 3- or 4- coloring. Note that Peeters in [17] has already shown that deciding whether a graph G has minrank2 (G) = 3 is NP-hard. Here we show that even if the answer to the above problem is yes, it is still hard to find the three transmissions which deliver the bits to their designated users. III. The Binary Linear Broadcast Rate β` (G) = 3 In this section we present the proof of Theorems 1 and 2. Proof of Theorem 1: Consider the undirected graph G with χ(G) = 3. As is well-known α(G) ≤ β` (G) ≤ χ(G). α(G), however, can not be 1 since the only graphs with α(G) = 1 are complete graphs which have

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χ(G) = 1. This contradicts our assumption χ(G) = 3. So, given χ(G) = 3 then β` (G) can either be 2 or 3. If β` (G) = 3 the proof is complete, so we only need to show that β` (G) cannot be 2. We do so by contradiction. Let β` (G) = 2, i.e., the base station delivers all users their designated bits in two transmissions each formed of a linear combination of subset of the bits as explained before. As a result, β` (G) = 2 means that the base station transmits the vectors1 v1 and v2 with supports V1 and V2 and then every user decode the bit it requires from v1 , v2 , or v1 + v2 . Note that V1 ∪ V2 = V since all nodes should receive the bit they require. We show that if this is the case, then χ(G) = 2 instead of three. We do so by showing that G can be partitioned into two cliques. Consider the set of vertices V1 \ V2 . Note that these nodes receive their bits (which also belongs to V1 \V2 ) through either v1 or v1 +v2 . In either case based on condition (ii), these nodes are adjacent to all other nodes in V1 \V2 , i.e., V1 \V2 is a clique. Similarly, V2 \ V1 is a clique. Consider now the nodes in V1 ∩ V2 . They receive their designated bits from either v1 or v2 . This is true since the support set of v1 +v2 is V \ (V1 ∩V2 ) and by condition (i), the node should belong to the support set of the transmission for it to be able to decode. The geometry of this system is shown in Fig. 3. Denote the nodes in V1 ∩ V2 who receive 1 and S 2 , respectively. their bit from v1 and v2 by S12 12 1 Note that based on condition (ii), every node in S12 is adjacent to all nodes in V1 , which include all nodes 2 are adjacent to all in V1 ∩ V2 . Further, nodes in S12 1 2 are the nodes in V2 . So (V1 \V2 )∪S12 and (V2 \V1 )∪S12 two cliques that partition the graph, which in return shows that χ(G) = 2. Fig. 3 shows the assignment of nodes to different cliques through assigning different color to each clique. Also, for more clarity, we have shown the edges incident to nodes in V1 \ V2 . This contradicts the assumption and the proof is complete.

Proof of Theorem 2: We need to show that if β` (G) = 3 then χ(G) is either 3 or 4. Note that χ(G) can not be two since then the base station could achieve β` (G) = 2 by sending the two sums of the bits in the two cliques. Let v1 , v2 , and v3 with supports V1 , V2 , and V3 be the three transmissions of the base station where V1 ∪ V2 ∪ V3 = V since all users must

v1

v1

v2

V1

V2 v1 + v 2

v2

v1 + v 2

Fig. 3: Assignment of nodes in V1 \V2 , V2 \V1 , and V1 ∩ V2 to two cliques which are depicted with different colors, namely, yellow and blue.

receive their bits by these three transmissions. That is, the users must decode the bits they require from one of the linear combinations v1 , v2 , v3 , v1 + v2 , v1 + v3 , v2 + v3 , or v1 + v2 + v3 . Consider the partition of nodes into seven disjoint sets of nodes V1 \ (V2 ∪ V3 ), V2 \ (V1 ∪ V3 ), V3 \ (V1 ∪ V2 ), V1 ∩ V2 \ V3 , V1 ∩ V3 \ V2 , V2 ∩ V3 \ V1 ,

and V1 ∩ V2 ∩ V3 . Note that the nodes in each of these sets can only recover the bits they want from linear combination of 1 messages whose support includes that set. Do recall that each node is adjacent to all other nodes in its partition. For example, nodes in V1 \ (V2 ∪ V3 ) may recover their bits from v1 , v2 , v1 +v2 , or v1 +v2 +v3 . This is shown in Fig. 4. Taking the intersection of the support of these four vectors and using condition (ii), we see that for any node i ∈ V1 \ (V2 ∪ V3 ), V1 \ ((V2 ∪ V3 ) ∪ i) ⊆ N (i) which means that all nodes in V1 \ (V2 ∪ V3 ) are adjacent, i.e., V1 \ (V2 ∪ V3 ) is a clique. Similarly, we can show that other 6 partitions are also cliques. To complete the proof, we show that the graph can be partitioned into four cliques. We do so by considering the four cliques, namely,

1

As mentioned before, by transmitting the vector we mean transmitting the modulo two sum of the elements in the vector.

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C1 = (V1 ∩ V2 ) \ V3 , C2 = (V1 ∩ V3 ) \ V2 , C3 = (V2 ∩ V3 ) \ V1 , and C4 = V1 ∩ V2 ∩ V3 ;

v1

v1 + v 2

v2

v1 + v 2

C1 v1 + v 3 !3

i=1

V1

v2 + v 3

V2 C4

vi

!3

i=1

C2

C3 V3

v3

!3

i=1

v1 + v 3

vi

vi

v2 + v 3

Fig. 4: Assignment of nodes in V1 \ (V2 ∪ V3 ), V2 \ (V1 ∪ V3 ), and V3 \ (V1 ∪ V2 ) to the four cliques C1 , C2 , C3 , and C4 based on the linear combination of transmissions the nodes decode from.

and then adding nodes from other partitions to these cliques so that the outcomes remain cliques. Fig. 4 depicts these four cliques with the assignment of other nodes to them. Beginning with V1 \ (V2 ∪ V3 ), as mentioned before, these nodes can recover their bits from v1 , v1 + v2 , v1 +v3 , or v1 +v2 +v3 . Add the first group, i.e., those nodes which decode by v1 , to C1 , the second group to C3 , the third group to C1 , and the last group to 1 C4 . Going to V2 \(V1 ∪V3 ), these nodes can decode their bits by v2 , v1 + v2 , v2 + v3 , or v1 + v2 + v3 . Similar to above, add the first group to C3 , the second group to C3 , the third group to C2 , and the last group to C4 . Finally, for V3 \(V1 ∪V2 ), we can easily check to see that these nodes recover their bits from v3 , v1 + v3 , v2 + v3 , or v1 + v2 + v3 . And again, add the first group to C2 , the second group to C1 , the third group to C2 , and the last group to C4 . In short, add all nodes which decode by v1 or v1 + v3 to C1 , all nodes which decode by v3 or v2 + v3 to C2 , all nodes that decode by v2 or v1 + v2 to C3 , and finally all nodes which decode by v1 + v2 + v3 to C4 . For more clarity, Fig. 4 depicts

the assignment of nodes to different cliques based on the linear combination of transmissions they decode from. It also shows the internal connections of nodes in V1 \ (V2 ∪ V3 ) and external neighbors of nodes in V1 \(V2 ∪V3 ) that decode by v1 . Further note that the edges drawn in grey depict part of side information of these nodes which are not used in the decoding of the demanded packets from v1 . Note that one might think that a node in V1 ∩ V2 ∩ V3 that decodes with v1 might not be connected to a node in V2 \ (V1 ∪ V3 ) that decodes with v1 + v2 + v3 . But this is not the case since the node in V2 \(V1 ∪V3 ) must know all the nodes in V1 ∩ V2 ∩ V3 to be able to decode with v1 + v2 + v3 including those that decode with v1 . So the two nodes are adjacent. It is easy to check that the nodes from different partitions that are assigned to Ci for i ∈ {1, 2, 3, 4} are adjacent to each other so 3 ≤ χ(G) ≤ 4. Further, C4 would be an empty set if no node decode its packet from v1 + v2 + v3 and V1 ∩ V2 ∩ V3 = ∅ which gives χ(G) = 3. So, the proof is complete. Before ending this section, let us point out that the clique cover presented in the proof is not unique and one can find other alternative clique covers on G. IV. Hardness of Scalar Linear Coding of Size Three The by-product of our results as mentioned before is the proof of hardness of finding the three linear transmissions while we know β` (G) = 3. We prove the hardness of three linear transmissions by reduction of 3- or 4-coloring of 3-colorable graphs into scalar linear coding. As is shown in [18], 3- or 4-coloring of 3-colorable graphs are NP-hard. That is, we find a constructive transformation from every instance of 3-colorable graphs to an instance of index coding problem with β` (G) = 3 and show that the three linear scalar transmissions can be mapped to a 3- or 4-coloring of the 3-colorable graph. Proof of Theorem 3: Consider the three colorable graph G, i.e., χ(G) = 3. Let G denote the compliment of G. Since χ(G) = 3 by Theorem 1 β` (G) = 3. Assuming that we know the three linear transmissions which can deliver all the packets to the nodes of G, i.e., we know vectors v1 , v2 , and v3 . Then by the same method we described in proof of Theorem 2, we can find four cliques in G which can be transformed into a 4-coloring of G. Note that as

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shown in [18], 4-coloring of 3-colorable graphs is NPhard, so finding three scalar linear transmissions on instances of index coding problem with β` (G) = 3 is also NP-hard and the proof is complete. Note that the result of Theorem 3 is also applicable to the general index coding problem represented by a bipartite graph as introduced in [14]. The transformation from a undirected graph G to the bipartite model is explained in [14]. So finding the three scalar linear transmissions is NP-hard for the general instance of index coding problem. An alternative way to interpret this result is as follows. Consider the instance of the problem G and assume we know minrank2 (G) = 3. Then by Theorem 3 finding the basis of A which minimizes the minrank2 is NP-hard. Note that it is known in general that solving minrank2 (G) is NP-hard, however, we show here that it is NP-hard to find a matrix completion, even when a very low rank one exists. References

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