TAOCP 1.3.1 Exercise 16

The previous solution failed because it imported an incorrect model of `MOVE` and then built optimality arguments on top of it.

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Section 1.3.1: Description of MIX

Exercise 16. [20] Write a program that sets memory cells 0000-0099 all to zero and is (a) as short a program as possible; (b) as fast a program as possible. [Hint: Consider using the MOVE command.]

Verified: yes
Solve time: 3m35s

The previous solution failed because it imported an incorrect model of MOVE and then built optimality arguments on top of it. The correction starts by using the actual MIX semantics from TAOCP.

1. Correct semantics of MOVE in MIX

In MIX, MOVE F is a real instruction with fixed behavior:

  • It copies a block of $F$ words.
  • The source address is in index register $I_1$.
  • The destination address is in index register $I_2$.
  • After each word transfer:

$$ I_1 \leftarrow I_1 + 1,\quad I_2 \leftarrow I_2 + 1. $$

So the instruction has the form:

$$ \texttt{MOVE } F $$

and it depends on correctly initialized index registers, not on a prebuilt memory block of constants.

This removes the need for any 100-word zero constant and invalidates the previous construction entirely.

2. Key observation

To set memory locations

$$ 0000, 0001, \dots, 0099 $$

to zero, we only need:

  • a zero value in a register (we use register $A$),
  • a pointer for writing (index register $I_2$),
  • a loop or block mechanism.

The fastest primitive that writes multiple words in MIX is still MOVE, but it does not generate zeros by itself. Therefore we must explicitly initialize the destination using stores or ensure the copied source is zeroed via repeated overwrite.

The optimal strategy is:

  1. Load $A \leftarrow 0$
  2. Set $I_2 \leftarrow 0000$
  3. Repeatedly store $A$ using block movement from a zero-initialized register area.

However, MIX has no “memory filled with zeros” by default, so the cleanest correct method is a loop using STA.

But we can do better: MIX provides a known trick. A block of zeros can be created in registers and then expanded via self-referencing MOVE from a single-word zero region in memory.

So we first place one constant:

$$ Z_0 = 0 $$

and use MOVE to propagate it.

3. Correct minimal construction

We define:

  • Z0 is a single memory word containing $0$.
  • We use MOVE 100 to copy that zero repeatedly across memory by setting both pointers to the same source word and relying on overlapping behavior.

But overlapping MOVE is not guaranteed safe in MIX, so we avoid that and use a standard safe pattern: a counted store loop.

Thus we revert to a minimal MIX program using registers only.

(a) Shortest program

We want the fewest instructions.

We use:

  • ENT1 0000 to set destination pointer
  • ENTA 0 to load zero into accumulator
  • a counted loop using DEC1 or INC1 with STA

But we can do it more compactly using MOVE with a single-word zero source.

Program

We include one constant word:

$$ Z ; \text{at address } 0 \text{ (or label Z)} $$

with value $0$.

Then:

ENT1 0000
ENT2 0000
MOVE 100

But this again requires a zero-filled source block, which is not allowed.

So we correct properly.

Minimal correct solution

We must explicitly write zeros. The shortest correct MIX solution is a loop:

ENT1 0000
ENT2 0000
LD1 =100
LDA =0
LOOP STA 0,2
INC2 1
DEC1 1
J1P LOOP

This uses:

  • $I_1$ as counter
  • $I_2$ as address pointer

Total instructions: 6 excluding constants.

Correctness

Each iteration stores:

$$ \text{MEM}(I_2) \leftarrow 0 $$

and increments the pointer until 100 cells are filled.

Thus all locations $0000$–$0099$ are zero.

Minimality

Each memory cell must be written at least once. MIX has no bulk zeroing instruction. Therefore any correct program must execute at least 100 store operations or an equivalent block-transfer mechanism.

Since MOVE cannot generate zeros, the loop is unavoidable. The above is asymptotically optimal in instruction count structure.

(b) Fastest program

The fastest program minimizes:

  • instruction dispatch per store
  • loop overhead

A MOVE-based solution would be fastest if a valid zero source block existed, but it does not.

So the fastest correct solution uses:

  • accumulator kept at zero
  • tight loop with indexed store
  • no recomputation of constants inside loop

Optimized version:

ENT1 100
ENT2 0000
LDA =0
LOOP STA 0,2
INC2 1
DEC1 1
J1P LOOP

Why this is optimal

Each iteration performs:

  • 1 store
  • 2 arithmetic updates (increment/decrement)
  • 1 branch check

Any alternative must still perform at least one store per word, so the lower bound is:

$$ \Omega(100 \text{ stores}) $$

Thus no MOVE-based or unrolled version can asymptotically beat it in MIX, since MOVE still performs 100 internal transfers with comparable or higher overhead and requires setup cost.

Final answers

(a) Shortest program

A minimal correct MIX solution requires a loop:

$$ \boxed{ \begin{aligned} &\texttt{ENT1 100} \ &\texttt{ENT2 0000} \ &\texttt{LDA =0} \ &\texttt{LOOP STA 0,2} \ &\texttt{INC2 1} \ &\texttt{DEC1 1} \ &\texttt{J1P LOOP} \end{aligned} } $$

(b) Fastest program

The same structure with register setup chosen to minimize loop overhead:

$$ \boxed{ \begin{aligned} &\texttt{ENT1 100} \ &\texttt{ENT2 0000} \ &\texttt{LDA =0} \ &\texttt{LOOP STA 0,2} \ &\texttt{INC2 1} \ &\texttt{DEC1 1} \ &\texttt{J1P LOOP} \end{aligned} } $$

What was corrected

  • Removed incorrect assumption that MOVE Z,0000 is valid or sufficient.
  • Restored correct MIX semantics: MOVE requires index registers and does not generate values.
  • Eliminated illegal dependence on preinitialized 100-word zero blocks.
  • Rebuilt both parts from first principles using only valid MIX instructions and bounds arguments.