TAOCP 7.1.3 Exercise 204

The failure in the previous construction comes from a false invariant: masking with x \,\&\, 0xFFFFFFFF00000000 does not produce a normalized 32-bit quantity.

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Section 7.1.3: Bitwise Tricks and Techniques

Exercise 204. ▶ [22] [22] What are the amazing constants p, q, r, m that achieve a perfect shuffle with just six MMIX commands? (See (175)–(178).)

Verified: no
Solve time: 4m26s

The failure in the previous construction comes from a false invariant: masking with

$$ x ,&, 0xFFFFFFFF00000000 $$

does not produce a normalized 32-bit quantity. It leaves the upper tetrabyte shifted by 32 bits, and this shift propagates into the multiplication, destroying the wyde alignment.

The correct fix is simple: never multiply a shifted 64-bit block as if it were a normalized 32-bit word. Instead, explicitly extract the two 32-bit tetrabytes in aligned form before applying SWAR dilation.

1. Decomposition of the octabyte

Write the octabyte $x$ as two genuine 32-bit words:

$$ x = x^{(0)} + 2^{32} x^{(1)}, $$

where

$$ x^{(0)} = x_3x_2x_1x_0,\quad x^{(1)} = x_7x_6x_5x_4 $$

in byte notation.

The goal is to form

$$ y = x_3x_7x_2x_6x_1x_5x_0x_4, $$

i.e., each wyde contains one byte from each tetrabyte:

$$ y^{(j)} = x_j + 2^8 x_{j+4}, \quad 0 \le j \le 3. $$

2. Correct SWAR idea

We use the standard byte-dilation constant

$$ r = 0x0001000100010001, $$

which maps

$$ (a_0,a_1,a_2,a_3) \mapsto (a_0,0,a_1,0,a_2,0,a_3,0) $$

at the byte level inside 16-bit lanes.

To place the second tetrabyte into the high byte of each wyde, we use

$$ m = 2^8 r = 0x0100010001000100. $$

3. Correct extraction of tetrabytes

We must extract true 32-bit values:

$$ p = 0x00000000FFFFFFFF,\quad q = 0xFFFFFFFF00000000. $$

Define:

$$ a = x ,&, p = x^{(0)}, $$

$$ b = (x ,&, q) \gg 32 = x^{(1)}. $$

This step is essential: the right shift removes the fatal 32-bit displacement identified in the review.

Now both $a$ and $b$ are properly normalized 32-bit integers.

4. Byte dilation

Apply SWAR multiplication separately:

Lower tetrabyte

$$ t = a \cdot r. $$

This expands each byte $x_j$ into the low byte of wyde $j$.

Upper tetrabyte

$$ u = b \cdot m. $$

Since $m = 2^8 r$, this places each byte $x_{j+4}$ into the high byte of wyde $j$.

No cross-lane interference occurs because:

  • each byte is $< 2^8$,
  • nonzero coefficients in $r$ and $m$ are spaced by 16 bits,
  • carries cannot propagate across wyde boundaries.

5. Combination

Form:

$$ y = t ;|; u. $$

Then each wyde satisfies:

$$ y^{(j)} = x_j + 2^8 x_{j+4}, \quad j=0,1,2,3. $$

Thus:

  • wyde 0: $x_0 + 2^8 x_4$
  • wyde 1: $x_1 + 2^8 x_5$
  • wyde 2: $x_2 + 2^8 x_6$
  • wyde 3: $x_3 + 2^8 x_7$

which is exactly the required perfect shuffle.

6. Constants

A valid choice is:

$$ \boxed{ p = 0x00000000FFFFFFFF,\quad q = 0xFFFFFFFF00000000,\quad r = 0x0001000100010001,\quad m = 0x0100010001000100 } $$

Note that $q$ is used only for conceptual symmetry; operationally, it is combined with the shift.

7. Six-MMIX-instruction realization

A minimal MMIX sequence is:

  1. $a \leftarrow x ,&, p$
  2. $b \leftarrow x ,&, q$
  3. $b \leftarrow b \gg 32$
  4. $a \leftarrow a \cdot r$
  5. $b \leftarrow b \cdot m$
  6. $y \leftarrow a ,|, b$

This is exactly six instructions, matching the exercise requirement.

8. Correctness summary

  • Both tetrabytes are explicitly normalized before arithmetic.
  • SWAR multiplication by $r$ and $m$ performs byte dilation without interference.
  • The 32-bit shift removes the structural error in the previous solution.
  • Each wyde cleanly combines one byte from each half-word in the required order.

This yields a correct perfect shuffle in six MMIX instructions.