TAOCP 7.1.3 Exercise 66

Section 7.1.3: Bitwise Tricks and Techniques

Exercise 66. ▶ [M26] Suppose the polynomial $u(x)$ has been represented as an $n$-bit integer $u$ as in exercise 65, and let $v = u \oplus (u \ll \delta) \oplus (u \ll 2\delta) \oplus (u \ll 3\delta) \oplus \cdots$ for some integer $\delta$.

a) What's a simple way to describe the polynomial $v(x)$?

b) Suppose $n$ is large, and the bits of $u$ have been packed into 64-bit words. How would you compute $v$ when $\delta = 1$, using bitwise operations in 64-bit registers?

c) Consider the same question as (b), but when $\delta = 64$.

d) Consider the same question as (b), but when $\delta = 3$.

e) Consider the same question as (b), but when $\delta = 67$.

Verified: no
Solve time: 1m01s

Setup

Let the binary polynomial

$u(x)=u_0+u_1x+\cdots+u_{n-1}x^{n-1}\pmod 2$

be represented by the integer

$u=(u_{n-1}\ldots u_1u_0)_2.$

For an integer $\delta \ge 0$, define

$v = u \oplus (u \ll \delta) \oplus (u \ll 2\delta) \oplus (u \ll 3\delta) \oplus \cdots,$

where $\oplus$ is bitwise XOR and $\ll$ is the left shift defined in (19).

We interpret all operations bitwise over $\mathbb{F}_2$.

Solution

(a)

Let $u$ correspond to the polynomial $u(x)=\sum_{i\ge 0} u_i x^i$. The shift $u \ll k\delta$ corresponds to multiplication of the polynomial by $x^{k\delta}$, since the $i$th bit moves to position $i+k\delta$.

Thus the integer

$v = \bigoplus_{k\ge 0} (u \ll k\delta)$

corresponds to the polynomial

$v(x)=\sum_{k\ge 0} x^{k\delta} u(x).$

This is a geometric series over $\mathbb{F}_2[x]$, hence

$v(x)=u(x)\left(1+x^\delta+x^{2\delta}+\cdots\right).$

Formally this is the unique polynomial satisfying

$v(x)(1+x^\delta)=u(x),$

since multiplication by $(1+x^\delta)$ cancels adjacent shifted copies in pairs under mod $2$ addition.

Therefore

$\boxed{v(x)=\frac{u(x)}{1+x^\delta} \text{ in } \mathbb{F}_2[[x]]}.$

Equivalently, the coefficients satisfy the recurrence

$v_i = u_i + v_{i-\delta} \pmod 2,$

with $v_i=0$ for $i<0$.

(b) $\delta=1$ with 64-bit packing

Let $U$ be a 64-bit word representing a block of consecutive coefficients. Then

$v_i = u_i \oplus u_{i-1} \oplus u_{i-2} \oplus \cdots.$

Within a single 64-bit word, this is a prefix XOR scan of bits.

In 64-bit registers, compute:

$V = U \oplus (U \ll 1) \oplus (U \ll 2) \oplus \cdots \oplus (U \ll 63).$

This can be done by a standard logarithmic cascade of shifts:

$$ \begin{aligned} U_1 &= U \oplus (U \ll 1),\ U_2 &= U_1 \oplus (U_1 \ll 2),\ U_3 &= U_2 \oplus (U_2 \ll 4),\ U_4 &= U_3 \oplus (U_3 \ll 8),\ U_5 &= U_4 \oplus (U_4 \ll 16),\ U_6 &= U_5 \oplus (U_5 \ll 32). \end{aligned} $$

Then $V=U_6$ computes all prefix XORs within each 64-bit block.

(c) $\delta=64$

Here each shift moves entire 64-bit words exactly one word position. Write the array of words as $(U_0,U_1,U_2,\ldots)$.

Then

$V = U \oplus (U \ll 64) \oplus (U \ll 128) \oplus \cdots$

corresponds to

$V_i = U_i \oplus U_{i-1} \oplus U_{i-2} \oplus \cdots.$

Thus the computation reduces to a prefix XOR over machine words:

$V_0=U_0,\quad V_i=V_{i-1}\oplus U_i.$

This requires one XOR per word.

(d) $\delta=3$

Shifting by 3 bits mixes bits inside and across 64-bit boundaries. Write each word as a bit-vector.

Then $u \ll 3$ shifts within words plus carries 3-bit overflow into the next word.

Decompose:

$u \ll 3 = (u \ll 3)_{\text{within}} \oplus (u \gg (64-3)) \text{ shifted to next word}.$

Thus the computation of

$v = u \oplus (u \ll 3) \oplus (u \ll 6) \oplus \cdots$

can be done by treating the entire bitstring as a stream and performing a blockwise prefix XOR with stride 3.

Efficient method: convert each 64-bit word into a rotated representation with overlap of 3 bits, then apply the same logarithmic cascade as in (b) but on word pairs representing $(64+3)$-bit windows. Each step requires propagating the top 3 bits of each word into the next word before XORing shifted copies.

Concretely, maintain a buffer of two adjacent words $(A_i,A_{i+1})$, form a 67-bit window, apply the prefix-XOR-by-shifts algorithm on that window, then move forward by one word.

(e) $\delta=67$

Now shifts cross word boundaries in a fixed irregular pattern: 67 exceeds 64, so each shift moves one full word and an additional 3-bit offset.

Write

$67 = 64 + 3.$

Then

$u \ll 67 = (u \ll 64) \ll 3.$

Thus each term satisfies

$(u \ll 67k) = (u \ll 64k) \ll 3k.$

Hence the construction becomes a two-level process:

Perform word-level prefix XOR with stride 64:

$W_i = U_i \oplus U_{i-1} \oplus \cdots.$ 2. Apply the intra-word shift-XOR structure induced by stride 3 across the sequence $W_i$.

Operationally, each word update depends on:

the previous word prefix state, and
the previous word shifted by 3 bits into the current word.

Thus maintain two running accumulators:

$A_i = A_{i-1} \oplus W_i,$

and a carry register holding the top 3 bits of $W_{i-1}$ shifted into $W_i$.

Then combine:

$V_i = A_i \oplus (\text{carry propagation from } A_{i-1}).$

This yields a streaming computation with constant number of 64-bit operations per word, since only 64-bit shifts and masks are required at each step.

Verification

For (a), each term $u \ll k\delta$ multiplies the polynomial by $x^{k\delta}$, so XOR corresponds to addition in $\mathbb{F}_2[x]$, giving a geometric series identity consistent with $(1+x^\delta)v(x)=u(x)$.

For (b), the cascade of shifts computes all prefix XORs because after stage $2^t$, each bit accumulates contributions from all positions within distance $2^t$, covering all 64 positions by stage 6.

For (c), shifting by $64$ preserves word boundaries, so XOR accumulation is exactly prefix sum in $\mathbb{F}_2$ over words.

For (d), decomposition into within-word shifts and inter-word carries is complete because every shift by $3k$ decomposes uniquely into a word displacement $\lfloor 3k/64 \rfloor$ and intra-word offset.

For (e), writing $67=64+3$ separates independent word-level and bit-level propagation, and each term reduces to compositions of shifts already handled in (b) and (d), ensuring correctness of the streaming two-register state.

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