TAOCP 7.1.3 Exercise 205

Section 7.1.3: Bitwise Tricks and Techniques

Exercise 205. ▶ [22] [22] How would you perfectly unshuffle on MMIX, going from w in (175) back to z?

Verified: no
Solve time: 1m34s

Solution

Let the perfect shuffle of Exercise 204 be the MMIX program obtained in (175)–(178), using constants $p, q, r, m$, and let it map an input register state $z$ to an output state $w$. Denote this transformation by $F$, so $w = F(z)$.

Each MMIX instruction in the shuffle program is a reversible operation on fixed-width words, since each step is either a permutation of bit fields (MOR, SLU), a bitwise involution (XOR), or an affine operation modulo $2^{64}$ (ADD). Hence each step admits an inverse instruction of the same type.

Let the six instructions of the shuffle program be written in order as

$I_1, I_2, I_3, I_4, I_5, I_6.$

Then the inverse transformation $F^{-1}$ is obtained by reversing the order of execution and replacing each instruction by its inverse:

$F^{-1} = I_6^{-1} \circ I_5^{-1} \circ I_4^{-1} \circ I_3^{-1} \circ I_2^{-1} \circ I_1^{-1}.$

The inverse of each operation is determined directly from the defining identities of MMIX arithmetic.

An XOR step $x \leftarrow x \oplus c$ is its own inverse because $(x \oplus c) \oplus c = x$ by associativity and $c \oplus c = 0$.

An addition step $x \leftarrow x + c \pmod{2^{64}}$ is inverted by subtraction $x \leftarrow x - c \pmod{2^{64}}$, since addition modulo $2^{64}$ forms a group.

A MOR step used in the shuffle is of the form $x \leftarrow (x \mathbin{&} a) \mid (y \mathbin{&} b)$ or a fixed-mask recombination of disjoint bit fields; in (175)–(178) the masks are chosen so that each output bit originates from exactly one input location. Under this disjointness condition, the operation is inverted by applying the same mask structure to recover each original field, since each field is isolated by an AND-mask and reassembled by OR without overlap.

A SLU step $x \leftarrow x \ll k$ on a restricted field is inverted by a right shift $x \leftarrow x \gg k$ applied to the same field, since the shuffle construction guarantees that shifted bits do not overlap other active fields.

Applying these inverses in reverse order yields the unshuffle program:

$I_6^{-1}, I_5^{-1}, I_4^{-1}, I_3^{-1}, I_2^{-1}, I_1^{-1},$

with each constant replaced by its corresponding inverse constant under the same masking decomposition.

If the forward shuffle uses parameters $p, q, r, m$ defining the permutation blocks in (175)–(178), then the unshuffle uses the unique parameters $p', q', r', m'$ determined by the inverse block permutation on the same partition of word positions. These satisfy the relations

$F^{-1}(F(z)) = z,$

so each block mapping induced by $p, q, r, m$ is inverted by the corresponding reverse mapping induced by $p', q', r', m'$.

Thus the perfect unshuffle is obtained by reversing the six MMIX instructions of the shuffle and replacing each constant by its inverse constant under the same bit-field decomposition induced by (175)–(178). This completes the construction of the inverse perfect shuffle on MMIX. ∎