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Counting the auxiliary substitution and restriction pairs

For truncation order q, the recovery procedure considers U subset E subset {1,...,n} with |E|<=q. There are C_q=sum_{s=0}^{min(q,n)} 2^s*binom(n,s) pairs: choose E of size s, then any of its 2^s subsets U. Thus C_0=1 and C_q=3^n if q>=n. For 0<eta<=1, eta^q*C_q<=(1+2eta)^n, since eta^q<=eta^s for every retained s. If q=O(nR/Q) and 0<R/Q<1, choose eta=R/Q; taking logs gives log2 C_q=O(n*(R/Q)*log2(2Q/R)). This bound needs the stated order for q, not just small factor scopes.

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Updated 2026-11-07

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