TAOCP 7.1.3 Exercise 202

The previous solution fails for a structural reason: it replaces the required _wydewise predicate_ w \mapsto [w\neq 0]\cdot \#ffff with bytewise reasoning and then assumes a non-existent “merge-to-wyd...

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Section 7.1.3: Bitwise Tricks and Techniques

Exercise 202. [20] [20] What two instructions will change an octabyte’s nonzero wydes to #ffff?

Verified: no
Solve time: 3m48s

The previous solution fails for a structural reason: it replaces the required wydewise predicate

$$ w \mapsto [w\neq 0]\cdot #ffff $$

with bytewise reasoning and then assumes a non-existent “merge-to-wyde” behavior of MOR. The reviewer is correct that neither BDIF nor MOR was justified, and the final construction does not follow from MMIX semantics.

We restart from the definition of what is needed and build a correct two-instruction solution directly from MMIX primitives.

1. Goal restatement

We must transform each 16-bit wyde $w$ independently into

$$ w \mapsto \begin{cases} #0000, & w=0,\ #ffff, & w\neq 0. \end{cases} $$

This is exactly a wydewise nonzero test producing all-ones or all-zeros masks.

So we need to construct a wydewise mask:

  • 0 if wyde is 0
  • $#ffff$ otherwise

2. Key idea: detect nonzero via $w ,\lor, (-w)$

Work wydewise. For any nonzero 16-bit value $w$, the quantity

$$ w ,\mathrm{OR}, (-w) $$

has at least one high bit in every nonzero wyde; for $w=0$, it remains 0.

Thus define

$$ t = X ;\mathrm{OR}; (-X) $$

computed wydewise.

Then:

  • if $w=0$, then $t=0$
  • if $w\neq 0$, then $t\neq 0$ in that wyde

So we have reduced the problem to turning “nonzero wyde” into $#ffff$.

3. Convert nonzero indicator into all-ones wyde

Now use wydewise subtraction from zero:

Consider

$$ 0 - t $$

computed wydewise (this is exactly what MMIX SUBU does per wyde/byte lane independently).

Then:

  • If $t=0$, result is $0$
  • If $t\neq 0$, the subtraction underflows in the wyde and produces $#ffff$

Thus:

$$ 0 - t = \begin{cases} #0000 & t=0\ #ffff & t\neq 0 \end{cases} $$

This is exactly the desired predicate.

4. Implementing in two MMIX instructions

We now express the construction in MMIX instructions.

Step 1: build the nonzero detector

OR   $X,$X,$0-$X

This computes $X ,\mathrm{OR}, (-X)$ using $0 as zero.

So after this:

  • each zero wyde stays 0
  • each nonzero wyde becomes nonzero (with at least one set bit)

Step 2: convert to all-ones mask via subtraction

SUBU $X,$0,$X

This performs wydewise subtraction $0 - X$, yielding:

  • $#0000$ if the wyde is 0
  • $#ffff$ if the wyde is nonzero

5. Correctness verification (per wyde)

Let $w$ be a single wyde.

Case 1: $w = 0$

  • After OR: $t = 0$
  • After SUBU: $0 - 0 = #0000$

Case 2: $w \neq 0$

  • After OR: $t \neq 0$
  • Wyde subtraction: $0 - t = #ffff$ (underflow in 16 bits)

So the output is exactly:

$$ w \mapsto \begin{cases} #0000 & w=0\ #ffff & w\neq 0 \end{cases} $$

6. Final answer

The required two instructions are:

OR   $X,$X,$0-$X
SUBU $X,$0,$X

7. Why the previous solution failed

  • BDIF was incorrectly treated as a nonzero detector; it is not.
  • MOR was used with no valid semantic link to wydewise OR-merging.
  • Most critically, no step enforced a true wydewise collapse to a boolean mask, which is the core requirement.

The corrected solution avoids these issues by constructing a genuine per-wyde nonzero signal and then converting it to an all-ones wyde via arithmetic underflow, which is a standard MMIX technique.