Parallelism and Time in Hierarchical Self-Assembly

Computer Science – Computational Complexity

Scientific paper

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previously was combined with this paper but they have now been split: http://arxiv.org/abs/1011.3493

Scientific paper

We study the role that parallelism plays in time complexity of Winfree's abstract Tile Assembly Model (aTAM), a model of molecular algorithmic self-assembly. In the "hierarchical" aTAM, two assemblies, both consisting of multiple tiles, are allowed to aggregate together, whereas in the "seeded" aTAM, tiles attach one at a time to a growing assembly. Adleman, Cheng, Goel, and Huang ("Running Time and Program Size for Self-Assembled Squares", STOC 2001) showed how to assemble an n x n square in O(n) time in the seeded aTAM using O(log n / log log n) unique tile types, where both of these parameters are optimal. They asked whether the hierarchical aTAM could allow a tile system to use the ability to form large assemblies in parallel before they attach to break the Omega(n) lower bound for assembly time. We show that there is a tile system with the optimal O(log n / log log n) tile types that assembles an n x n square using O(log^2 n) parallel "stages", which is close to the optimal Omega(log n) stages, forming the final n x n square from four n/2 x n/2 squares, which are themselves recursively formed from n/4 x n/4 squares, etc. However, despite this nearly maximal parallelism, the system requires superlinear time to assemble the square. We extend the definition of *partial order tile systems* studied by Adleman et al. in a natural way to hierarchical assembly and show that no hierarchical partial order tile system can build any shape with diameter N in less than time Omega(N), demonstrating that in this case the hierarchical model affords no speedup whatsoever over the seeded model. We strengthen the Omega(N) time lower bound for deterministic seeded systems of Adleman et al. to nondeterministic seeded systems. Finally, we show that for infinitely many n, a tile system can assemble an n x n' rectangle, with n > n', in time O(n^{4/5} log n), breaking the linear-time lower bound.

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