Additional Constructions to Solve the Generalized Russian Cards Problem using Combinatorial Designs

Colleen M. Swanson, Douglas R. Stinson

In the generalized Russian cards problem, we have a card deck $X$ of $n$ cards and three participants, Alice, Bob, and Cathy, dealt $a$, $b$, and $c$ cards, respectively. Once the cards are dealt, Alice and Bob wish to privately communicate their hands to each other via public announcements, without the advantage of a shared secret or public key infrastructure. Cathy should remain ignorant of all but her own cards after Alice and Bob have made their announcements. Notions for Cathy's ignorance in the literature range from Cathy not learning the fate of any individual card with certainty (weak $1$-security) to not gaining any probabilistic advantage in guessing the fate of some set of $\delta$ cards (perfect $\delta$-security). As we demonstrate, the generalized Russian cards problem has close ties to the field of combinatorial designs, on which we rely heavily, particularly for perfect security notions. Our main result establishes an equivalence between perfectly $\delta$-secure strategies and $(c+\delta)$-designs on $n$ points with block size $a$, when announcements are chosen uniformly at random from the set of possible announcements. We also provide construction methods and example solutions, including a construction that yields perfect $1$-security against Cathy when $c=2$. We leverage a known combinatorial design to construct a strategy with $a=8$, $b=13$, and $c=3$ that is perfectly $2$-secure. Finally, we consider a variant of the problem that yields solutions that are easy to construct and optimal with respect to both the number of announcements and level of security achieved. Moreover, this is the first method obtaining weak $\delta$-security that allows Alice to hold an arbitrary number of cards and Cathy to hold a set of $c = \lfloor \frac{a-\delta}{2} \rfloor$ cards. Alternatively, the construction yields solutions for arbitrary $\delta$, $c$ and any $a \geq \delta + 2c$.

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