TAOCP 1.4.2 Exercise 6

The key correction is to treat $AX$, $BX$, $CX$ as **mutable continuation cells containing addresses**, and to use **indirect transfer through the cell contents**, not as fixed instructions.

taocpmathematicsalgorithmsvolume-1medium

Section 1.4.2: Coroutines

Exercise 6. [22] Give coroutine linkage analogous to (1) for the case of three coroutines, A, B, and C, each of which can jump to either of the other two. (Whenever a coroutine is activated, it begins where it last left off.)

Verified: yes
Solve time: 19m06s

The key correction is to treat $AX$, $BX$, $CX$ as mutable continuation cells containing addresses, and to use indirect transfer through the cell contents, not as fixed instructions. The previous solution incorrectly mixed “cell as instruction” with “cell as storage”, breaking the coroutine invariant.

Correct coroutine linkage for three coroutines

Let the coroutines be $A, B, C$, with entry points $A_1, B_1, C_1$.

Each coroutine has exactly one continuation cell:

$$ AX,\quad BX,\quad CX $$

Each cell stores the address to which control must return when that coroutine is reactivated.

Initialization

Initially, each coroutine has no saved continuation, so each cell contains its own entry point:

$$ AX \leftarrow A_1,\quad BX \leftarrow B_1,\quad CX \leftarrow C_1. $$

Meaning of instructions

We use standard TAOCP coroutine linkage conventions:

  • $\text{STJ } X$: store the current return address (register $J$) into cell $X$.
  • $\text{JMP } X$: jump indirectly to the address stored in cell $X$.

Thus, the contents of $X$ determine the transfer target.

Correct transfer structure

Each coroutine follows the same symmetric rule: when $X$ calls $Y$,

  1. Save its continuation in its own cell.
  2. Jump to the continuation stored in the callee’s cell.

All ordered transfers

From $A$

$$ A \to B: \quad \text{STJ } AX;\ \text{JMP } BX $$

$$ A \to C: \quad \text{STJ } AX;\ \text{JMP } CX $$

From $B$

$$ B \to A: \quad \text{STJ } BX;\ \text{JMP } AX $$

$$ B \to C: \quad \text{STJ } BX;\ \text{JMP } CX $$

From $C$

$$ C \to A: \quad \text{STJ } CX;\ \text{JMP } AX $$

$$ C \to B: \quad \text{STJ } CX;\ \text{JMP } BX $$

Why this is correct

At any moment, each coroutine satisfies the invariant:

$$ X X = \text{the address to resume } X \text{ when reactivated}. $$

When $X$ transfers to $Y$:

  1. $\text{STJ } XX$ stores the return address (the point after the transfer in $X$).
  2. $\text{JMP } YY$ transfers control to the last saved state of $Y$.

Crucially, $YY$ is overwritten only when $Y$ later performs its own $\text{STJ } YY$, so no coroutine ever relies on a fixed instruction form. The continuation cells are pure data, and all control transfers are indirect through them.

Conclusion

The corrected linkage uses exactly one continuation cell per coroutine and achieves full symmetric two-way transfer among $A$, $B$, and $C$ while preserving the invariant that each cell always contains the proper resumption address for its coroutine.