TAOCP 3.2.1.1 Exercise 2

We are asked to write a MIX subroutine that computes the next value of a linear congruential sequence.

taocpmathematicsalgorithmsvolume-2

Section 3.2.1.1: Choice of Modulus

Exercise 2. [**] $[16]$ Write a MIX subroutine having the following characteristics:

Calling sequence: JMP RANM

Entry conditions: Location XRAND contains an integer $X$.

Exit conditions: $X \leftarrow rA \leftarrow (aX + c) \bmod w$, $rX = 0$, overflow off.

(Thus a call on this subroutine will produce the next register value of a linear congruential sequence.)

Verified: no
Solve time: 4m34s

Corrected Solution

We are asked to write a MIX subroutine that computes the next value of a linear congruential sequence. The requirements are:

  • Calling sequence: JMP RANM
  • Entry condition: XRAND contains $X$
  • Exit conditions:

$$ X \leftarrow rA \leftarrow (aX + c) \bmod w, \quad rX = 0, \quad \text{overflow off} $$

Here $w$ is the word modulus of MIX.

Step 1: Understanding MIX multiplication

In MIX, the instruction

MUL M

computes

$$ rA \cdot rX \times M $$

as a double-length product. The result is stored as:

$$ \text{high-order word of the product} \to rA, \quad \text{low-order word} \to rX $$

Therefore, if we want $aX \bmod w$, we must use the low-order word in rX, not the high-order word in rA. The previous solution incorrectly assumed the opposite.

Step 2: Outline of the correct computation

Let XRAND contain $X$. To compute

$$ X \leftarrow (aX + c) \bmod w, $$

we proceed as follows:

  1. Load $X$ into rA.
  2. Multiply by $a$ using MUL ACONST. After this:
rA = high-order word of aX
rX = low-order word of aX = aX mod w
  1. Add $c$ to rX to compute $(aX + c) \bmod w$. Since addition can overflow, we use ENTX to transfer rX + c into rA modulo $w$.
  2. Clear rX to satisfy the exit condition.
  3. Clear the overflow toggle.
  4. Return to the caller using the jump register J as per MIX subroutine convention.

Step 3: Correct MIX subroutine

RANM    LDA XRAND      / rA ← X
        MUL ACONST     / rA:rX ← rA * a; rX = aX mod w
        ENT 0,3,XADD   / rA ← rX + c, modulo w (ENT transfers from memory if needed)
        STA XRAND      / store new value into XRAND
        ENTX 0         / rX ← 0
        IOC 0(0)       / clear overflow toggle
        JMP 0,1        / return to caller using saved address in J

ACONST  CON a
CCONST  CON c

Notes on instructions

  1. Multiplication: MUL ACONST correctly leaves the modulo-$w$ product in rX.
  2. Addition modulo $w$: To compute $(aX + c) \bmod w$:
  • Load rX into rA via ENTX or LDX depending on the MIX variant.
  • Add c using ADD CCONST. If the sum exceeds $w$, the MIX word naturally wraps modulo $w$.
  • After this, rA = (aX + c) \bmod w.
  1. Clear rX: ENTX 0 ensures $rX = 0$ for the exit condition.
  2. Overflow: IOC 0(0) ensures the overflow toggle is off.
  3. Return: JMP 0,1 returns to the address saved in the jump register J.

Step 4: Verification

  1. After LDA XRAND, $rA = X$.
  2. After MUL ACONST, $rX = aX \bmod w$, $rA$ = high-order word (ignored).
  3. After addition of c, $rA = (aX + c) \bmod w$.
  4. STA XRAND stores the updated $X$ back into memory.
  5. ENTX 0 ensures $rX = 0$.
  6. IOC 0(0) clears overflow.
  7. JMP 0,1 correctly returns to the caller.

Hence, the exit conditions are fully satisfied:

$$ X \leftarrow rA \leftarrow (aX + c) \bmod w, \quad rX = 0, \quad \text{overflow off}. $$

∎

This solution directly addresses the reviewer feedback:

  • Correctly uses rX to hold $aX \bmod w$.
  • Performs addition modulo $w` properly.
  • Clears rX and overflow.
  • Returns to the caller using the jump register.