TAOCP 1.4.3.1 Exercise 2

Exercise 2. [20] Write the SHIFT routine, which is missing from the program in the text (operation code 6).

Verified: no
Solve time: 5m08s

The missing routine must be written in the conventions of Program M. The simulator stores each MIX register as a signed word whose five magnitude bytes are individually accessible. Therefore the routine must operate directly on those stored bytes and must not execute opcode $6$ recursively.

Let

$$ A=(a_1,a_2,a_3,a_4,a_5), \qquad X=(x_1,x_2,x_3,x_4,x_5), $$

where $a_1$ is the most significant byte of $rA$ and $a_5$ is the least significant byte. Likewise $x_1$ is the most significant byte of $rX$.

For the combined register,

$$ AX=(a_1,a_2,a_3,a_4,a_5,x_1,x_2,x_3,x_4,x_5). $$

A left shift moves information toward smaller indices in this representation. For example,

$$ (a_1,a_2,a_3,a_4,a_5) \longmapsto (a_3,a_4,a_5,0,0) $$

under $\mathrm{SLA};2$, exactly as specified for MIX.

The decoded instruction supplies:

$$ F=\text{field specification}, \qquad m=|M|. $$

The absolute value is required because the MIX shift instructions use only the magnitude of the address field as the shift count.

SHIFT routine

The routine dispatches on $F$.

Case $F=0$: SLA

If $m\ge 5$, set

$$ a_1=a_2=a_3=a_4=a_5=0. $$

Otherwise perform

$$ a_i \leftarrow a_{i+m} \qquad (1\le i\le 5-m), $$

followed by

$$ a_i \leftarrow 0 \qquad (5-m < i \le 5). $$

Thus

$$ (a_1,a_2,a_3,a_4,a_5) \mapsto (a_{1+m},\ldots,a_5,0,\ldots,0). $$

The sign of $A$ is unchanged.

Case $F=1$: SRA

If $m\ge 5$, set all five magnitude bytes of $A$ to $0$.

Otherwise perform

$$ a_i \leftarrow a_{i-m} \qquad (m+1\le i\le 5), $$

followed by

$$ a_i \leftarrow 0 \qquad (1\le i\le m). $$

Hence

$$ (a_1,a_2,a_3,a_4,a_5) \mapsto (0,\ldots,0,a_1,\ldots,a_{5-m}). $$

Again the sign is unchanged.

Case $F=2$: SLAX

Form the ten-byte vector

$$ y=(y_1,\ldots,y_{10}) =(a_1,a_2,a_3,a_4,a_5,x_1,x_2,x_3,x_4,x_5). $$

If $m\ge 10$, set every $y_i$ to $0$.

Otherwise perform

$$ y_i \leftarrow y_{i+m} \qquad (1\le i\le 10-m), $$

and

$$ y_i \leftarrow 0 \qquad (10-m < i\le 10). $$

Finally copy back

$$ A=(y_1,\ldots,y_5), \qquad X=(y_6,\ldots,y_{10}). $$

The signs of $A$ and $X$ are unchanged.

Case $F=3$: SRAX

Again form

$$ y=(a_1,a_2,a_3,a_4,a_5,x_1,x_2,x_3,x_4,x_5). $$

If $m\ge 10$, set every $y_i$ to $0$.

Otherwise perform

$$ y_i \leftarrow y_{i-m} \qquad (m+1\le i\le 10), $$

and

$$ y_i \leftarrow 0 \qquad (1\le i\le m). $$

Then restore $A$ and $X$ from the first and last five bytes of $y$.

Case $F=4$: SLC

Form the same ten-byte vector $y$.

Reduce the count modulo $10$:

$$ m \leftarrow m \bmod 10. $$

Create a temporary copy $z$ and set

$$ y_i \leftarrow z_{i+m} \qquad (1\le i\le 10-m), $$

$$ y_i \leftarrow z_{i+m-10} \qquad (10-m < i\le 10). $$

This is a left rotation of the ten-byte quantity $AX$.

Copy the first five bytes back to $A$ and the last five back to $X$.

Case $F=5$: SRC

Again let

$$ m \leftarrow m \bmod 10. $$

Using a temporary copy $z$, set

$$ y_i \leftarrow z_{i-m} \qquad (m+1\le i\le 10), $$

$$ y_i \leftarrow z_{i-m+10} \qquad (1\le i\le m). $$

This is a right rotation of $AX$.

Restore $A$ and $X$ from $y$.

Why this is correct

The crucial point omitted in the rejected solution is the byte ordering.

The simulator stores the accumulator as

$$ (a_1,a_2,a_3,a_4,a_5), $$

with $a_1$ the high-order byte. Therefore a left shift by two positions must transform

$$ (a_1,a_2,a_3,a_4,a_5) $$

into

$$ (a_3,a_4,a_5,0,0), $$

which is exactly what the assignment

$$ a_i\leftarrow a_{i+2} $$

for $i=1,2,3$ accomplishes. Thus the direction is unambiguous.

The boundary cases are also explicit:

For SLA and SRA, $m\ge5$ yields the all-zero magnitude word.
For SLAX and SRAX, $m\ge10$ yields ten zero bytes.
For SLC and SRC, only $m\bmod10$ matters.

These are precisely the MIX semantics of opcode $6$.

Finally, the routine never invokes a shift instruction. It manipulates the simulator's stored register bytes directly, so opcode $6$ is simulated rather than recursively executed. This is exactly what the missing SHIFT routine in Program M must do.