Abstract
In this paper, we consider the problem of partitioning active users in a manner that facilitates multi-access without collision. The setting is of a noisy, synchronous, Boolean, multi-access channel where active users (out of a total of users) seek to access. A solution to the partition problem places each of the users in one of groups (or blocks) such that no two active nodes are in the same block. We consider a simple, but non-trivial and illustrative case of active users and study the number of steps used to solve the partition problem. By random coding and a suboptimal decoding scheme, we show that for any , where and are positive constants (independent of ), and can be arbitrary small, the partition problem can be solved with error probability , for large . Under the same scheme, we also bound from the other direction, establishing that, for any , the error probability for large ; again and are constants and can be arbitrarily small. These bounds on the number of steps are lower than the tight achievable lower-bound in terms of for group testing (in which all active users are identified, rather than just partitioned). Thus, partitioning may prove to be a more efficient approach for multi-access than group testing.
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