Sustained Space Complexity
May 15, 2017 ยท Declared Dead ยท ๐ IACR Cryptology ePrint Archive
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Authors
Joel Alwen, Jeremiah Blocki, Krzysztof Pietrzak
arXiv ID
1705.05313
Category
cs.CR: Cryptography & Security
Citations
42
Venue
IACR Cryptology ePrint Archive
Last Checked
3 months ago
Abstract
Memory-hard functions (MHF) are functions whose evaluation cost is dominated by memory cost. MHFs are egalitarian, in the sense that evaluating them on dedicated hardware (like FPGAs or ASICs) is not much cheaper than on off-the-shelf hardware (like x86 CPUs). MHFs have interesting cryptographic applications, most notably to password hashing and securing blockchains. Alwen and Serbinenko [STOC'15] define the cumulative memory complexity (cmc) of a function as the sum (over all time-steps) of the amount of memory required to compute the function. They advocate that a good MHF must have high cmc. Unlike previous notions, cmc takes into account that dedicated hardware might exploit amortization and parallelism. Still, cmc has been critizised as insufficient, as it fails to capture possible time-memory trade-offs, as memory cost doesn't scale linearly, functions with the same cmc could still have very different actual hardware cost. In this work we address this problem, and introduce the notion of sustained-memory complexity, which requires that any algorithm evaluating the function must use a large amount of memory for many steps. We construct functions (in the parallel random oracle model) whose sustained-memory complexity is almost optimal: our function can be evaluated using $n$ steps and $O(n/\log(n))$ memory, in each step making one query to the (fixed-input length) random oracle, while any algorithm that can make arbitrary many parallel queries to the random oracle, still needs $ฮฉ(n/\log(n))$ memory for $ฮฉ(n)$ steps. Our main technical contribution is the construction is a family of DAGs on $n$ nodes with constant indegree with high "sustained-space complexity", meaning that any parallel black-pebbling strategy requires $ฮฉ(n/\log(n))$ pebbles for at least $ฮฉ(n)$ steps.
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