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How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Exact rational interval coding of the block AB

Facts & Assumptions

Given: Alphabet order $A, B, C$ , rational frequencies $(q_{A}, q_{B}, q_{C}) = (3, 2, 1)$ with $Q = 6$ , maximum and fixed frame length $N = n = 2$ , and input block $A B$ .

[F1]

Exact finite-register arithmetic coding for rational block sources: At symbol $x_{t}$ , the update is $L \leftarrow L + W c_{x_{t}} Q^{n - t}$ and $W \leftarrow W q_{x_{t}}$ ; the output uses the largest $u$ with $2^{u} W \leq Q^{n}$ , then $s = u + 2$ and $k = ⌊ 2^{s} (2 L + W) / (2 Q^{n}) ⌋$ .

[F2]

The rational finite-register arithmetic code is prefix-free and lossless: The finite decoder returns exactly the encoded block.

[F3]

Finite discrete sources and product blocks: The block law is the product of the one-symbol probabilities.

Example

With $p = (1 / 2, 1 / 3, 1 / 6)$ and a length-two frame, the interval for $A B$ should be $[1 / 4, 5 / 12)$ . The emitted codeword is $0101$ , whose dyadic cell is strictly inside that interval.

The diagram uses horizontal position for the real coordinate on $[0, 1)$ . The upper bar is the source interval for $A B$ ; the lower bar is the dyadic cell represented by $0101$ . It illustrates the exact rational comparisons below but is not used as proof.

Verification

1.1F1F3givenalgebra

For the first symbol $A$ , $c_{A} = 0$ , so the update from $(L, W) = (0, 1)$ gives $(0, 3)$ . For the second symbol $B$ , $c_{B} = q_{A} = 3$ and $Q^{n - 2} = 1$ , so $(L, W) = (0 + 3 \cdot 3, 3 \cdot 2) = (9, 6)$ . With $D = Q^{n} = 36$ , the source interval is $[9 / 36, 15 / 36) = [1 / 4, 5 / 12)$ ; its width $6 / 36 = 1 / 6$ equals $p_{A} p_{B}$ by [F3].

1.2F1givenalgebra

Since $2^{2} \cdot 6 = 24 \leq 36 < 48 = 2^{3} \cdot 6$ , the largest admissible $u$ is $2$ and the payload length is $s = 4$ . The midpoint is $(2 \cdot 9 + 6) / (2 \cdot 36) = 1 / 3$ , hence $k = ⌊ 16 / 3 ⌋ = 5$ and its four-bit representation is $0101$ . Also $B = 6$ because $2^{5} < 36 \leq 2^{6}$ , so the emitted four-bit word obeys the configured $B + 2$ limit.

2.1F2step 1.1step 1.2algebra∎

The dyadic cell of $0101$ is $[5 / 16, 6 / 16) = [5 / 16, 3 / 8)$ . Exact comparisons give $1 / 4 = 4 / 16 < 5 / 16$ and $3 / 8 = 9 / 24 < 10 / 24 = 5 / 12$ , so this cell is inside $[1 / 4, 5 / 12)$ . By [F2], the decoder returns $A B$ uniquely.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

12 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Scott D. Roweis, CSC310 Lecture 8x: Arithmetic Coding Without Blocks, version 1.1 (standard reference, not scraped)