A Zero-Error Source Coding Solution to the Russian Cards Problem

Transcription

1 A Zero-Error Soure Coding Solution to the Russian Cards Problem ESTEBAN LANDERRECHE Institute of Logi, Language and Computation January 24, 2017 Abstrat In the Russian Cards problem, Alie wants to ommuniate her hand from a dek of ards to Bob through one publi announement suh that Cath does not learn her hand. In zero-error soure oding, Alie s goal is to ommuniate her information to Bob with the least amount of bits. Both problems share the goal of Bob learning Alie s information in one announement. However, they have additional objetives: in the Russian Cards ase it is seurity against Cath, for soure oding it is minimizing the ommuniation omplexity. At a first glane, these two additional properties are not mutually exlusive so we ask whether there is a solution that is both optimal and seure at the same time. This would permit us to model the Russian Cards problem as a zero-error soure oding problem to find the optimal seure announement and it will also enrih soure oding by adding the knowledge that it is also seure for an attaker that holds some orrelated information. Finally, we deal with the relationship between these two problems. To ahieve this, we use results of graph theory and design theory, in partiular blok graphs and Steiner systems. 1 Introdution There are many different approahes to the modelling of ommuniation in mathematis. A fruitful area of study in this area is measuring the amount of information exhanged to ommuniate the most by saying the least. Another approah seeks to reate ommuniation that an only be understood by the intended reipients. It is possible to ombine both goals in our modelling, to provide effiient and seure ommuniation. Zero-error soure oding is an example of the former approah whih has some similarities with the Russian Cards problem, a member of the latter. We will try to 1

2 ombine them to see if solutions for one also work for the other and to see what is neessary to ahieve this. Take a situation where three people (Alie, Bob and Cath) have a dek of ards. They randomly distribute the ards between themselves suh that they only know their ards. Suppose everything that is said an be heard and understood by all three people, is there a way for Alie to tell Bob her hand without Cath finding out? This is the question that the Russian Cards problem tries to answer. In essene, what we have is three people with probabilistially orrelated information in whih two of them attempt to ommuniate their knowledge without the information reahing the third person. This seems to suggest a onnetion with zero-error soure oding, whih tries to onstrut the minimal announement suh that Bob an know all of Alie s information. To further study this onnetion, we first define eah problem on its own. 1.1 The Russian Cards Problem The Russian Card problem as it appeared in the 2000 Mosow Mathematial Olympiad is as follows Suppose we have three partiipants- Alie, Bob and Cathwho have a dek with 7 ards represented by the numbers 0 to 6. The ards are dealt within the three in a random manner in a way that only eah player knows his or her ards. Both Alie and Bob reeive 3 ards while Cath is given the remaining one. Alie gets ards 0, 1 and 2, Bob gets 3, 4 and 5 and Cath gets 6. Alie and Bob want to know eah other s ards without Cath learning any of them. However, every ommuniation between Alie and Bob an be heard and understood perfetly by Cath and it is impossible to enrypt the message in a traditional way. Is there a way that they an ommuniate eah other s hand without Cath knowing the position of any ard she doesn t hold? What started as an Olympiad problem has been generalized for larger deks and different distributions of the ards. Other generalizations permit multiple announements, whih opens up a greater set of distributions that have a solutions. However, these examples no longer share the basi struture with zero-error soure oding, so we do no take them into aount. 1.2 Zero-error Soure Coding In zero-error soure oding we have the following: Alie holds some information x Ω where Ω is ommonly known by both Alie and Bob. She 2

3 wants to ommuniate x to Bob, who holds some orrelated information u, of whih Alie knows the form of (but not the ontent). The goal of a soure oding problem is for Alie to ommuniate x to Bob with the least amount of bits possible, taking advantage that x and u are orrelated. This is done by onstruting a onfusability graph G = (V, E) where the verties in V represent the possible values of x (that is eah element of Ω) and two verties v 1, v 2 are joined by an edge if and only if there is a ertain value of u suh that Bob would not be able to distinguish whether Alie holds v 1 or v 2. Given this graph, Alie must only find a minimal vertex oloring and ommuniate to Bob the olor of the vertex x. Bob ould now immediately distinguish whih is the orret vertex through the information enoded in u, as no verties of the same olor are onneted. The minimal olouring of this graph is an optimal solution for zero-error soure oding beause it is the smallest partition over the information that Alie holds whih is still informative to Bob. The oloring of the graph is agreed and indexed beforehand so the ommuniation onsists only of the index of the olor. If there are m olors, the announement will be of length log 2 m. 2 Informativity We have presented two distint problems from very different ontexts and with very different goals. However, they have a similar struture. In both we have an agent attempting to ommuniate his information to the other. While eah problem has additional desirable properties in the announement, they both agree on the fat that an announement must be informative for the reeiving agent. After ommuniation, the reeiver must always know exatly what information the sender holds. We bridge the two problems through this idea. It is very natural to see that the Russian Cards problem an be expressed as a zero-error soure oding problem with an extra ondition and a different goal. Ω is seen as all the possible hands for Alie, where x is the hand that she atually holds. Similarly, u beomes Bob s hand, whih is learly orrelated to Alie s hand. Alie an learly onstrut this graph, as she an think about all the possible hands that Bob an hold and build the edge set aordingly. In partiular this means that any solution that is informative to Bob in the Russian Cards problem is also informative in the soure oding ontext. 3

4 Looking for a onnetion between these two problems indiretly gives us a result for Russian Cards. The informativity of an announement an be tested in a onfusability graph, regardless of the way we onstrut the announement. If an announement ontains two hands that are onneted then there is at least one hand that Bob ould hold that will onsider both these hands as possible. Therefore, we must avoid this at all ost. The independene number of a graph represents the ardinality of the largest set of verties in the graph suh that all the verties are not onneted. Note that no announement an have two onneted verties, as that would mean that for a ertain ard deal, Bob would not be able to distinguish the two onneted hands. This gives us an upper bound on the size of an announement whih annot be higher than the independene number of the onfusability graph of that problem. If our goal is to onnet these two problems we will formally define what we mean by an announement and informativity. The presentation follows the one seen in [LFD15] for the Russian Cards problem, but generalized so that it an also be applied. It is important to note that while generally zeroerror soure oding literature does not expliitly treat an announement as a list of possible values for Alie s information, it an always be seen as suh. Definition 1. Let Ω N be the set of the possible information and A, B Ω be the information that Alie and Bob hold respetively, with A B =. An announement A ( Ω A ) is a list of possible values for A, with (Ω A ) representing all the subsets of Ω with ardinality equal to A. Definition 2. Let ( Ω A ) be the set of every possible information set that Alie may hold. An announement strategy S is a set of announements and a distribution funtion f over S suh that for every set A ( Ω A ) Alie will hoose an appropriate announement A S with the probability dependent on f Definition 3. Let A and B be the information that Alie and Bob hold respetively. An announement A is informative if after the announement Bob knows A, written in this way: A \ B = {A}. An announement strategy is informative if every announement in it is informative. In the Russian Cards setting, Ω represents the dek of ards and A, B (and C) are the hands of eah player. For the soure oding perspetive, eah piee of information is enoded in a natural number. This differs from the lassial presentation in the introdution, where Alie holds one element of Ω instead of a subset. It is easy to show that these two settings are equivalent, but we hoose to present it in this way to show the similiarity with the Russian Cards problem. We define the notion of an announement 4

5 strategy beause the hoie of an announement depends on the partiular information that Alie holds, however a strategy only depends on Ω and how the information is distributed. Therefore, we an define a strategy for any partiular deal of ards/soure oding problem. Note that we have a probability funtion to define whih announement Alie hooses for the ase that there is more than one announement for a partiular hand. In the soure oding paradigm, this funtion is unneessary as we are trying to minimize the total number of announements. It is natural to have only one possible announement per hand, otherwise we ould have redundany. As a matter of fat, the solution using oloring of graphs ensures that there an only be one announement per hand. 2.1 Example We will present an example to see that while the onept of informativity in both problems is equivalent, the problems are not. We will fous on the original problem with seven ards to show that the soure oding protool works. We will all this problem the (3, 3, 1) problem, as this is how the ards are distributed. While this is a Russian Cards problem, we interpret it as a soure oding problem and use the known solution for it. Then, we will see if it fulfills the requirements to be a good solution for the Russian ards problem. The way to solve a zero-error soure oding problem is through a onfusability graph. Alie onstruts a graph where eah vertex represents the information that she may hold (in this ase, every set of three ards) and joins every two verties whih ould be onfused by Bob if he held any valid hand (in this ase, valid refers both to Bob having three ards and not having any ard found in the verties). While we ould onstrut this graph vertex by vertex, there exists a lass of graphs alled Johnson Graphs whih represent exatly that. A Johnson graph J(n, k) is a graph where the verties are the k-element subsets of an n-element set and there is an edge between two verties if and only if they share a k 1 set. Lets take J(7, 3) and have every vertex represent one of Alie s possible hands. This is equivalent to the fat that two verties are onneted if they are indistinguishable for someone holding a n k 1-set. Therefore, the Johnson graph J(7, 3) is the onfusability graph for the (3, 3, 1) Russian ards problem, as Bob holds = 3 ards. Any oloring of this graph will give us a valid solution for the soure oding problem within. We use sage to find the following oloring: 5

6 000 {0, 2, 6}, {2, 3, 4}, {1, 2, 5}, {0, 4, 5}, {0, 1, 3} 001 {1, 5, 6}, {3, 4, 6}, {0, 3, 5}, {0, 1, 4} 010 {3, 4, 5}, {2, 5, 6}, {0, 3, 6}, {1, 4, 6}, {0, 2, 4}, {1, 2, 3} 011 {2, 4, 6}, {2, 3, 5}, {1, 4, 5}, {0, 5, 6}, {0, 3, 4}, {1, 3, 6}, {0, 1, 2} 100 {4, 5, 6}, {2, 3, 6}, {0, 2, 5}, {0, 1, 6}, {1, 3, 5}, {1, 2, 4} 101 {2, 4, 5}, {1, 2, 6}, {3, 5, 6}, {0, 4, 6}, {1, 3, 4}, {0, 2, 3}, {0, 1, 5} In the Olympiad problem, Alie held {0, 1, 2}. Therefore she would announe 011 as her hand is in the fourth announement. In this partiular ase, Cath will not be able to learn the position of any of the ards that either Alie or Bob hold. We know this beause it oinides with the projetive geometri solution seen in the following example, where Alie holds a line: This seems to show that the soure oding strategy is seure, but this is not the ase. Suppose instead of the original distribution of the ards, we have the following deal Alie: 1, 5, 6 Bob: 2, 3, 4 Cath: 0 In this ase, Alie would announe the following: {1, 5, 6}, {3, 4, 6}, {0, 3, 5}, {0, 1, 4} At first glane we an see that Cath learns the position of some of the ards, as no hand in the announement ontains 2. This lets Cath know that Alie annot hold 2, beause it does not appear in any of the possible hands for Alie. Beause Cath holds 0, she then knows Bob holds 2. It is even worse, as Cath an eliminate the last two hands whih shows her that Alie holds 6. Therefore, soure oding alone is not enough to solve the Russian Cards problem, as we need to add other restritions to ensure seurity of the strategy. 6

7 3 Seurity We have seen that solving the soure oding problem related to a Russian Cards problem is not enough to solve the original problem. A generi solution for the soure oding problem does not automatially ensure that Cath does not learn the position of a ard. However, there are many different minimal olorings for a graph and while we found one that does not work, there may be another whih does. For that, we must formalize the onept of seurity and see if it is refleted in some graph olorings. We want to ensure that, regardless of the ards that eah person holds, Cath does not learn the exat position of any ard she does not hold herself. This an be enoded in the following way. Definition 4. Let Ω be the set of ards, A an announement and the size of Cath s hand. If Y is a set of ards, let A \ Y be all the hands in A whih do not ontain an element of Y. We say A is seure if for all C ( Ω ) the following two properties hold The union of all the hands in A that avoid C is equal to Ω \ C, that is X i A\C X i = Ω \ C. The intersetion of all hands in A that avoid C is the empty set, that is X i A\C X i = The first property ensures that Cath does not learn the position of any ard in Alie s hand, the seond one ensures the same does not happen for any ard in Bob s hand. We an see that the inseure announement in the example fails both, as 2 {1, 5, 6} {3, 4, 6} and {1, 5, 6} {3, 4, 6} = {6}. As a matter of fat, no announement for this ase with four hands will be seure. This is beause in an announement every ard must appear in at least two announements. If it only appears in one, there might be the possibility that Cath holds a ard in the same hand and an then onlude that that ard is not held by Alie. The only way that Alie ould prevent this is by having that one ard only appear in the hand that she holds. However, this is not seure, as Cath will know that Alie s hand must be the one holding the ard that appears only one. If Alie held any other ard the announement would not be seure, as we have seen previously, whih means that Cath would know Alie s hand. That would mean that we must have at least 7 2 = 14 ards in the announement, but an announement with four hands has only 12. hene, by the pigeonhole priniple, this is 7

8 impossible. We an generalize this last fat: If Cath has ards, eah ard must appear in at least + 1 hands in the announement. From here, we get that a seure announement must have at least (a + b + )( + 1) ards. This is neessary for our first ondition to hold, but not suffiient. We an see this by looking at the first announement in our previous example. {0, 2, 6}, {2, 3, 4}, {1, 2, 5}, {0, 4, 5}, {0, 1, 3} If Cath holds 0 she will know that Alie holds 2. Therefore we need stronger restritions to ensure seurity. To do this, we must first properly haraterize the graphs that we are onstruting. To onstrut the onfusability graph G = (V, E) indued by the Russian Cards problem (a, b, ) with V = ( Ω k ) and E V V Create a vertex for eah hand Alie an hold Join two verties if there is a hand that Bob ould hold suh that he ould not distinguish between the two hands. That is, Clearly G is regular and symmetri. (H, K) E H K a + Lemma 5. Let G = (V, E) be a graph indued by the Russian Cards problem (a, b, ). For any x V, x has i=1 ( a i )(b+ i ) neighbors. Proof. If we want to see how many verties y exist suh that x y = a + i for any i we hoose the i ards that we take away from x, whih gives us ( a i ) and the ards with whih we an replae them (b+ i ). By adding all the verties that differ from 1 to ards we get exatly i=1 ( a i )( ) b + i To ensure seurity, we have to prevent the situation where Cath an eliminate all hands in an announement that ontain a partiular ard. Beause we have to reate announements that work for every possible Cath hand, we have to make sure that if ards x and y appear together in a hand in the announement, there must be a hand in the same announement where x appears and y does not (and vieversa). However, as Cath s hands grow, so does this restrition. If Cath has two ards, then for every three ards 8

9 x, y and z in a hand in the announement there must be at least one hand that ontains x but not y and z. In general, if Cath has ards, for eah hand H in an announement, there must be at least + 1 other hands in the announement, eah one ontaining exatly one of the elements of H. This ondition is neessary to fulfill the seond property of seurity, as we defined it previously. To fulfill the first property, for any + 1-set ontained in a hand, there must be another hand in the announement suh that it ontains none of the ards in the set. Here, we run into a problem with our approah. Graph theory alone permits us to define loal relationships and not the relationships that we have just mentioned. Therefore, we annot represent these properties in the graph. Of ourse, we an find olorings and only aept one if these onditions are met, but we annot affirm whether there exists a minimal oloring that fulfills these properties without heking all of them. Therefore, purely graph theoreti tools are not enough in this ase and we need more robust mathematial strutures, in partiular Steiner systems, a type of blok designs. These systems are a subset of ( Ω a ) suh that every (a )-set appears exatly in one element of the system. This is enough to ensure that the previous properties hold. These designs have graphs assoiated to them and they happen to be equivalent to the graphs we have onstruted (when these designs exist). The Johnson Graph J(7, 3) is the blok graph of the 2-(7, 3, 1)-design. In [SS14], it is shown that an optimal announement strategy for a (a, b, ) deal is a large set of t-(a + b +, a, 1)-designs, with t = a. This is equivalent to the minimal oloring in suh graph, whenever the large set exists. Swanson & Stinson proved that the solution reated by t-(a + b +, a, 1)- designs is not only informative and optimal, but also seure. Not ontent with the definition of seurity that we presented, they also proved a stronger notion of seurity. They showed that for any set X of t ards, the probability that Alie holds X aording to Cath does not hange before and after the announement. This prevents Cath from gaining probabilisti information about Alie s hand, whih might be desirable depending on the importane of the privay of the information. They refer to this notion as t-perfet seurity. Definition 6. Let Ω be the set of ards, S an announement strategy and the size of Cath s hand. An announement A S for A ( Ω A ) is t-perfetly seure if for every t-set Y and every C ( Ω ), if Y C = then P(Y A C, A) P(Y A C) = 1 9

10 An announement strategy S is t-perfetly seure if every announement in it is t-perfetly seure. When the t is not diretly relevant, we will refer to this simply as perfet seurity. This gives us a powerful tool to ensure seurity in our announement. For that, we delve deeper into design theory, partiularly in the study of Steiner systems. We will first return to our example to see some examples of seurity. 3.1 Continued Example We go bak to the (3, 3, 1) ase. Note that we have ( 7 3 ) = 35 possible hands and that the independene number of the assoiated Johnson graph J(7, 3) is 7. Our first instint is that given that 7 divides 35, we might have a valid protool with five announements with seven ards eah. Unfortunately, this is impossible, as the hromati number of J(7, 3) is 6. Therefore, a protool must have at least six different possible announements. As a matter of fat, suh announement does exist and is the following 6-oloring of J(7, 3) (presented in [SS14]): {0, 1, 3}, {1, 2, 4}, {2, 3, 5}, {3, 4, 6}, {0, 4, 5}, {1, 5, 6}, {0, 2, 6} {0, 2, 3}, {1, 3, 4}, {2, 4, 5}, {3, 5, 6}, {0, 4, 6}, {0, 1, 5}, {1, 2, 6} {0, 2, 4}, {0, 3, 5}, {1, 2, 3}, {0, 1, 6}, {1, 4, 5}, {2, 5, 6} {0, 1, 2}, {2, 3, 4}, {4, 5, 6}, {1, 3, 5}, {0, 3, 6} {1, 2, 5}, {0, 5, 6}, {1, 4, 6}, {0, 3, 4}, {2, 3, 6} {3, 4, 5}, {0, 1, 4}, {0, 2, 5}, {2, 4, 6}, {1, 3, 6} Note that as antiipated, this is not perfetly seure, as we an see with the fourth announement. If Cath holds 6, the ards 1, 2 and 3 seem more likely than the rest as they appear twie in the redued list while the other ards appear only one. On the other hand, we have the geometri protool that was previously mentioned. This protool is perfetly seure for the (3, 3, 1) problem although it is pretty ineffiient as there are four possible announements per hand. However, we an optimize it by eliminating ertain redundant announements to have only two hoies per hand (also from [SS14]): 10

11 {2, 5, 6}, {2, 3, 4}, {1, 4, 5}, {1, 3, 6}, {0, 4, 6}, {0, 3, 5}, {0, 1, 2} {2, 5, 6}, {2, 3, 4}, {1, 4, 6}, {1, 3, 5}, {0, 4, 5}, {0, 3, 6}, {0, 1, 2} {3, 4, 5}, {2, 4, 6}, {1, 3, 6}, {1, 2, 5}, {0, 5, 6}, {0, 2, 3}, {0, 1, 4} {3, 4, 5}, {2, 4, 6}, {1, 5, 6}, {1, 2, 3}, {0, 3, 6}, {0, 2, 5}, {0, 1, 4} {3, 4, 6}, {2, 3, 5}, {1, 4, 5}, {1, 2, 6}, {0, 5, 6}, {0, 2, 4}, {0, 1, 3} {3, 4, 6}, {2, 3, 5}, {1, 5, 6}, {1, 2, 4}, {0, 4, 5}, {0, 2, 6}, {0, 1, 3} {3, 5, 6}, {2, 4, 5}, {1, 3, 4}, {1, 2, 6}, {0, 4, 6}, {0, 2, 3}, {0, 1, 5} {3, 5, 6}, {2, 4, 5}, {1, 4, 6}, {1, 2, 3}, {0, 3, 4}, {0, 2, 6}, {0, 1, 5} {4, 5, 6}, {2, 3, 6}, {1, 3, 4}, {1, 2, 5}, {0, 3, 5}, {0, 2, 4}, {0, 1, 6} {4, 5, 6}, {2, 3, 6}, {1, 3, 5}, {1, 2, 4}, {0, 3, 4}, {0, 2, 5}, {0, 1, 6} This announement is the best perfetly seure announement, with a ommuniation omplexity of 10 (4 bits) while our previous solution is 6 (3 bits). Here, there is a trade off between perfet seurity and optimality. It is important to note that none of these options would qualify as optimal in the soure oding paradigm, where the optimum is 5 (exept for the partiularity that the ommuniation omplexity of 5 and 6 are equivalent in bits). 4 Results in Steiner Systems In the proess of treating the Russian Cards problem as a zero-error soure oding problem, we found that in order to ensure seurity we would need tools stronger than the ones given to us by graph theory. The graphs we are interested in orresponded to blok graphs for Steiner systems, whenever these systems exist. We present a more formal treatment of these systems in the following work, noting that if we get a minimal oloring for our graph we will also get a large set of Steiner systems. Definition 7. A blok design (or simply, design) t-(v, k, r) is a set X of v elements and a olletion of k-subsets of X (alled bloks) suh that every t-subset of X appears in exatly r bloks. Definition 8. A Steiner system S(t, k, v) is a set X of v elements and a olletion of k-subsets of X (alled bloks) suh that every t-subset of X appears in exatly one blok. A Steiner system is equivalent to a t-(v, k, 1)-design. Definition 9. A large set of Steiner systems S(t, k, v) is a set of Steiner systems suh that every k-subset of X appears in exatly one Steiner system as a blok. 11

12 We know that these designs exist whenever we have the divisibility ondition by [Kee14], and that the onstrution is not trivial. However, we do not know if the large sets exist. For example, designs of the form S(2, 7, 3) do exist (the Fano plane). However, if we had a large set of S(2, 7, 3) systems, we would have a strategy for the (3, 3, 1) Russian Cards problem. However, as we have seen previously, this strategy does not exist. THis is onsistent with results in design theory presented in [SS14]. Lemma 10. A Steiner system S(t, k, v) has (v t ) ( k t ) bloks. Proof. Every t-subset of our set of v elements must appear in the system, so a design must have ( v t ) elements distributed within its bloks of k elements. Every blok fits ( k t ) elements, so there must be ( v t ) /( k t ) for eah t-subset to appear exatly one. Lemma 11. A large set of Steiner systems S(t, k, v) has ( v t k t ) designs. Proof. Every k-subset of our set of v elements must appear in the large set, so a large set must have ( v k ) elements distributed within its designs of ( v t ) /( k t ) elements. Every design fits ( v t ) elements, so there must be (v k ) ( v t ) for eah /( k t ) k-subset to appear exatly one. We an easily redue this to ( v t k t ). It is important to know what the [SS14] refers to by an optimal strategy. We showed that the optimal announement strategy for (3, 3, 1) has 6 possible announements. However this learly is not a large set of Steiner systems. Therefore, an optimal strategy in this ase refers not to the solution with lowest m, but speifially with the solution (whih may or may not exist) equal to the following bound, presented in [SS14]. Lemma 12. Suppose a > and there exists a strategy for Alie that is informative for Bob. Then the number of announements m is bound by ( ) b + 2 m. Definition 13. Let (a, b, ) be the sizes of Alie, Bob and Cath s hands. We say an announement strategy is design-optimal if it ontains exatly ( b+2 ) announements. With this, we present the following theorem from [SS14] 12

13 Theorem 14. Suppose that a >. A design-optimal (a, b, )-strategy for Alie that is informative for Bob is equivalent to a large set of t-(a + b +, a, 1)-designs, where t = a. To find a solution for the Russian Cards problem, we first generate the onfusability graph G and then we find the hromati number χ(g). An optimal informative announement strategy has χ(g) possible announements, any strategy with less announements would either have uninformative announements or not have relevant announements for ertain hands. If we want this minimal announement to be t-perfetly seure then the graph must orrespond to the blok graph of a Steiner system. By [Kee14], a Steiner system S(t, k, v) exists if and only if ( k i r i ) divides (v i r i ) for every 0 i r 1. This means that there are ertain Russian Cards problems that annot be haraterized by designs. While the graph exists, it does not orrespond to a blok graph and therefore we annot apply these results to it. This does not mean a seure solution does not exist, only that it is not a design and it is not t-perfetly seure. Even if a Steiner system exists, the large set might not exist. There are no general proofs for the existene of large sets, but results exists for partiular values, for example k = 3. The following theorem provides a neessary and suffiient ondition to see if large sets of the form S(a, a, a + b + ) exists. Theorem 15. Take 0 < < a b N suh that ( a i r i ) divides (a+b+ i r i ) for every 0 i r 1. There is a large set of S(a, a, a + b + ) Steiner systems if and only if the hromati number of the graph G indued by the Russian Cards problem (a, b, ) is equal to ( b+2 ) Proof. Suppose suh design exists. By Theorem 14 it is an optimal solution to the Russian Cards problem (a, b, ). Beause a large set ontains ( b+2 ) designs, then we an olor G by ( b+2 ) olors and we annot olor it with less beause the design is optimal. By Theorem 14 we know that every informative announement strategy with ( b+2 )-oloring is a large set of S(a, a, a + b + ) Steiner systems. With this result we know that, for given (a, b, ), if χ(g) is larger than ( b+2 ), suh large sets annot exist. While this result is not new, we have built it from the onnetion of the Russian Cards problem and zero-error soure oding. Design theory permits us to onnet graph olorings with our notion of seurity, although unfortunately they do not work for all examples. The lak of existene of a large set of designs means that we have to take deisions regarding what we think is more important: optimality or seurity. 13

14 5 Conlusion and Further Researh We have found a onnetion between the Russian Cards problem and zeroerror soure oding. Both problems are seen extensions of the same problem in whih Alie wants to ommuniate information to Bob, who holds some orrelated information. However, eah problem adds an additional ondition: seurity in Russian ards and minimal ommuniation omplexity for soure oding. We have seen that these two onditions are equivalent if the Steiner system S(a, a, a + b + ) exists (in the soure oding paradigm, Cath s hand represents Bob s unertainty). We also saw that in other ases, this is not true and we must make a hoie between seurity and optimality. This means that while the two problems are similar and oasionally equivalent, it is not always the ase that an optimal solution will be seure. While we saw that we an present a Russian Cards problem in the form a soure oding problem, the opposite is not always true. In partiular, Russian Cards problems have very well behaved onfusability graphs (regular, symmetri). However, it ould be interesting to see if the onept of seurity ould be extended to all of soure oding, seeing where seurity does not affet optimality. By adding an additional restrition to soure oding we reate a new problem in whih optimality of ommuniation is not the only onern, whih might help model situations in whih we do not want the method of data ompression itself prevents the information from being ompromised. If there is no perfetly seure minimal oloring, then we are faed with one hoie of what we value the most: optimality or seurity. For the first ase, we know that every oloring will be informative, so we must look for the smallest (referring to the number of olors) seure oloring. However, this oloring would not be perfetly seure. If we want perfet seurity we will need a oloring that leads to an equitable protool and for that we need all the sets to be of the same size. This means that the oloring must have a number of olors that divides the number of hands. We would then rejet many possible olorings, some of whih may have less olors, but not a number that divides the number of possible hands. A natural line to ontinue the researh is finding whether optimal announements preserve stronger notions of seurity (but weaker than perfet seurity), like ε-strong seurity presented in [LFD15], where the quotient of probabilities of a hand should be between 1 ε and 1 + ε. If we are more interested in minimizing the number of announements, it might be interesting to onsider announement strategies in whih the distri- 14

15 bution funtion hanges depending on Alie s hand. If a S(a, a + b +, a) system does not exist, we know that there is no oloring for the onfusability graph suh that it entails a perfetly seure announement. Therefore, there is no perfetly seure strategy in whih every hand appears exatly one, at least some hands appear in more than one announement. This entails two possibilities, either some hands appear in more announements than others or they all appear the same number of times. The seond ase would hardly be optimal, so it ould be interesting to fous on what Swanson and Stinson all non equitable strategies to find strategies that an be seure and have a low number of possible announements. Referenes BJL99. Thomas Beth, Dieter Jungnikel, and Hanfried Lenz. Design theory, volume 69. Cambridge University Press, Kee14. Peter Keevash. The existene of designs. arxiv preprint arxiv: , LFD15. Esteban Landerrehe and David Fernández-Duque. A ase study in almost-perfet seurity for unonditionally seure ommuniation. Designs, Codes and Cryptography, pages 1 24, OK98. RW. SS14. Alon Orlitsky and J. Körner. Zero-error information theory. IEEE Transations on Information Theory, 44(6): , Todd Rowland and Eri W. Weisstein. Steiner system. mathworld.wolfram.om/steinersystem.html. From MathWorld A Wolfram Web Resoure. Colleen M Swanson and Douglas R Stinson. Combinatorial solutions providing improved seurity for the generalized russian ards problem. Designs, Codes and Cryptography, 72(2): ,