CF 104262C - Calibration Complications

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Solve time: 1m 2s
Verified: no

Solution

The five-letter word pairing scheme in Section 7.2.1.1 relies on masking a packed bitstring so that each mask isolates the lower portion of a word consisting of an integral number of fixed-size letter fields. In the standard construction each letter occupies $5$ bits, so a word is partitioned into blocks of size $5$, and the $j$th mask is designed to extract exactly the lowest $5(j+1)$ bits.

In that setting the correct masks have the form

$$m_j = z ,&, (2^{5j+5}-1),$$

since the binary number $2^{5j+5}-1$ consists of $5(j+1)$ consecutive $1$ bits, and therefore preserves exactly the lowest $5(j+1)$ bits of $z$ while clearing all higher bits.

The proposed modification replaces this by

$$m_j = z ,&, (25j+5 - 1).$$

Interpreting this in the intended bit-manipulation sense requires reading $25j+5$ as $2^{5j+5}$, since only a power of two produces a mask consisting of a contiguous block of low-order $1$ bits. Under this interpretation,

$$25j+5 - 1 = 2^{5j+5}-1,$$

so each $m_j$ extracts exactly the same set of low-order bits as in the original scheme.

The correctness of the pairing scheme depends only on the nesting property of the masks,

$$m_0 \subset m_1 \subset m_2 \subset \cdots,$$

meaning that each successive mask retains all bits retained by the previous one and adds a fixed additional block of $5$ bits. Both the original masks and the modified masks satisfy this property because

$$2^{5(j+1)}-1 = (2^{5j+5}-1) + 2^{5j+5},$$

which appends a disjoint higher block of $5$ bits.

Since all subsequent steps in the scheme depend only on consistent isolation of successive 5-bit fields and not on any arithmetic identity beyond these bit partitions, replacing the original masks by the modified ones leaves every intermediate extraction unchanged. Every grouped 5-bit letter remains aligned, and no cross-field interference occurs.

Therefore the algorithm produces exactly the same decomposition of the packed word into five-letter blocks and yields the same pairing results as the original construction.

This completes the proof. ∎