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A full-period LCG can alternate its low bit

Example

Take $w = 3$ , $a = 5$ , $c = 1$ , and seed $x_{0} = 0$ . The recurrence visits every residue modulo $8$ before returning to $0$ , yet its least significant bit alternates at every step.

Facts & Assumptions

Given: The LCG recurrence with modulus $8$ , multiplier $5$ , increment $1$ , and a seed chosen uniformly from the eight states.

[F1]

Under $a \equiv 1 (m o d 4)$ and odd increment, every seed has full period $2^{w}$ , but a uniform seed does not make adjacent states independent. The power-of-two mixed LCG has full period but dependent adjacent states

[F2]

The state update is $x_{n + 1} = [a x_{n} + c]_{2^{w}}$ . Linear congruential generator modulo a power of two

[F3]

The uniform law on eight states assigns mass $1 / 8$ to each state. The uniform probability space on a nonempty finite set

Verification

1.1givenF1F2algebra

Direct iteration gives $0, 1, 6, 7, 4, 5, 2, 3, 0$ . These are the eight residues modulo $8$ , so the cycle is full. The parameters satisfy $5 \equiv 1 (m o d 4)$ and $1$ is odd, also verifying [F1]'s hypotheses.

2.1F1F2F3step 1.1algebra∎

Reducing [F2] modulo $2$ gives $x_{n + 1} \equiv x_{n} + 1 (m o d 2)$ , because $5 \equiv 1 (m o d 2)$ and $c = 1$ . Thus the low bit alternates deterministically. By [F3], a uniform seed gives each state mass $1 / 8$ ; since the update is a permutation, each time marginal is uniform. The adjacent pair lies on the eight-point graph of the recurrence rather than the 64-point product space, so the pair is dependent.

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Sources

J. Jung, Monte-Carlo Methods lecture notes, §3.1 and Theorem 3.6 (standard reference, not scraped)