TAOCP 3.2.2 Exercise 5
Consider Algorithm M as defined in Section 3.
Exercise 5. ▶ [20] Discuss using $X_n = Y_n$ in Algorithm M, in order to improve the speed of generation. Is the result analogous to Algorithm B?
Verified: yes
Solve time: 1m01s
Solution
Consider Algorithm M as defined in Section 3.2.2. Let $(X_n)$ be a linear congruential sequence modulo $m = 2^e$, and let $(Y_n)$ be a sequence derived from $(X_n)$ by taking the $k$ most significant bits, that is,
$Y_n = \lfloor X_n / 2^{e-k} \rfloor, \quad 0 \le Y_n < 2^k.$
Algorithm M produces $Y_n$ using an auxiliary sequence $(X_n)$; each $Y_n$ requires computing $X_n$ and then extracting its high-order bits. The proposal in the exercise is to set $X_n = Y_n$ in the algorithm in order to avoid the computation of the full $X_n$ values.
If we do this, the recurrence for $X_n$ becomes
$X_{n+1} = \text{function of } X_n \bmod 2^k,$
because $Y_n$ takes values modulo $2^k$, and $X_n$ is now equal to $Y_n$. In other words, the generator is effectively reduced modulo $2^k$, and the computation of each $X_{n+1}$ requires only operations modulo $2^k$ rather than modulo $2^e$.
Let us analyze the consequences. Denote the original recurrence of $(X_n)$ as
$X_{n+1} = (a X_n + c) \bmod 2^e.$
If we set $X_n = Y_n$, then $X_n$ takes values modulo $2^k$; the recurrence becomes
$X_{n+1} = (a X_n + c) \bmod 2^k.$
This has the effect of reducing the modulus from $2^e$ to $2^k$, so that the period of the resulting sequence cannot exceed $2^k$, which is typically much shorter than the original period $2^e$. Consequently, the statistical properties of the sequence are degraded: the sequence is still linear, but it has lower potency and a shorter period, so it will fail the same tests that a low-order truncation of the original sequence fails.
From an implementation point of view, the speed of generation is improved, because each step only involves arithmetic modulo $2^k$, which can be implemented with simple bit masking or addition without carry beyond $k$ bits. In this sense, the result is analogous to Algorithm B, where one works directly with truncated values to speed up computation. Algorithm B also produces a faster sequence by operating on fewer bits, sacrificing the long period of the full generator in exchange for speed.
However, the analogy is not complete. In Algorithm B, the truncation occurs as a separate step after generating the full $X_n$, so the original sequence retains its full period internally; the output is only a truncated version. By contrast, setting $X_n = Y_n$ in Algorithm M reduces the modulus of the internal recurrence itself. Therefore, while speed is gained, the internal sequence itself is inherently shorter and less potent, not just the output.
In summary, using $X_n = Y_n$ in Algorithm M does improve computational speed because arithmetic is done modulo $2^k$ rather than $2^e$. The resulting generator behaves in a manner analogous to Algorithm B in that fewer bits are manipulated to produce each output. The critical distinction is that the period and potency of the sequence are reduced, so the sequence is statistically weaker than the original Algorithm M sequence.
This completes the proof.
∎