Retrieval check answer. Four chunks — one per octet (192, 168, 1, 100) — the same move as
lesson 2’s FBI · CIA · NSA · IRS: recode twelve-ish raw symbols into a handful of familiar units, and
working memory’s fixed slot count stops being the bottleneck.
The computation
Chunk A (one use, 1 day ago): B = ln(1^−0.5) = ln(1) = 0.
Chunk B (three uses, 2/4/8 days ago) — sum the power-law term for each use, then take the log:
- 2 days: 2^−0.5 = 1/√2 ≈ 0.7071
- 4 days: 4^−0.5 = 1/√4 = 0.5000
- 8 days: 8^−0.5 = 1/√8 ≈ 0.3536
Sum ≈ 1.5607. B = ln(1.5607) ≈ 0.445.
Chunk B is more active than chunk A — despite every single one of its uses being older than A’s only use. That’s the answer that isn’t naive, and it’s worth sitting with why: the misconception this lesson targets is treating d as something that ticks up with each retrieval (“used once, so it’s decayed once”). It doesn’t. d is a fixed exponent (0.5) applied independently to the elapsed time of each individual use; what changes with more uses is that you get more terms to sum, not a different value of d. A single fresh use produces one term close to its maximum (t^−d shrinks as t grows, so small t → a term near 1); three older uses each produce a smaller term, but three smaller terms summed can still exceed one larger term — which is exactly what happened here: chunk A’s single term is 1^−0.5 = 1, while chunk B’s three terms sum to 0.7071 + 0.5 + 0.3536 = 1.5607. B’s sum is larger before the log is even applied, so ln(1.5607) ≈ 0.445 comfortably beats ln(1) = 0.
The general lesson: frequency compounds; recency alone doesn’t dominate it. A chunk retrieved several times, even if none of those retrievals is recent, can out-rank a chunk retrieved once yesterday, because activation is driven by the sum of use-events, each independently decaying, not by whichever single use happens to be most recent. This is also why ACT-R’s model of learning through repetition works at all: every additional retrieval — including a deliberate one, which is exactly what next lesson’s retrieval practice amounts to — adds one more term to this sum, and the sum is what determines how retrievable something ultimately is.
For your harness: a memory or cache-ranking policy that scores entries by “was this used recently” (a single boolean or single timestamp) is implicitly assuming the single-use formula. A policy that sums frequency-weighted, independently-decaying terms per access event — closer to this full formula — correctly keeps a frequently-hit-but-not-recently-hit key warm, the way chunk B stayed more retrievable than a single very recent lookup.
Where this goes: every additional use in this sum is, mechanically, a retrieval. Next lesson asks what kind of retrieval counts — and finds that not all “uses” are equal, which is exactly the design principle behind why this whole puzzle path makes you produce an answer before showing you one.