TAOCP 1.3.1 Exercise 6

**Solution to Exercise 1.

taocpmathematicsalgorithmsvolume-1simple

Section 1.3.1: Description of MIX

Exercise 6. [10] Assume that location 3000 contains

+ 5 1 200 15

What is the result of the following instructions? (State if any of them are undefined or only partially defined.) (a) LDAN 3000; (b) LD2N 3000(3:4); (c) LDX 3000(1:3); (d) LD6 3000; (e) LDXN 3000(0:0).

Verified: yes
Solve time: 2m53s

Solution to Exercise 1.3.1.6 [10]

We are given that location 3000 contains the MIX word

$$ W_{3000} = +,5,1,200,15 $$

with sign $+$ and four bytes $(5,1,200,15)$. MIX uses 1-based indexing for bytes, so we number them:

$$ \text{byte 1} = 5, \quad \text{byte 2} = 1, \quad \text{byte 3} = 200, \quad \text{byte 4} = 15. $$

The general rules for the MIX load instructions are:

  1. If the field specification includes the sign ($L=0$), the sign is loaded; otherwise, the sign of the destination register becomes $+$.
  2. Bytes from the field are right-adjusted in the destination register. If the field is longer than the register, the leftmost bytes are truncated.
  3. If the instruction is a negated load (e.g., LDAN, LD2N, LDXN), the sign is reversed.
  4. Registers:
  • rA and rX have 5 bytes and a sign.
  • rI1 through rI6 have 2 bytes and a sign.

(a) LDAN 3000

  • Instruction: Load normal field (0:5) of location 3000 into rA with opposite sign.
  • Field (0:5) includes the sign and all four bytes of the word.
  • Sign at 3000 is $+$, so reversed gives $-$.
  • All bytes are copied: $5,1,200,15$.
  • Register after load:

$$ rA \leftarrow -,5,1,200,15 $$

(b) LD2N 3000(3:4)

  • Instruction: Load bytes 3 and 4 of location 3000 into index register rI2 with opposite sign.
  • Field (3:4) contains bytes $200,15$.
  • rI2 has 2 bytes and a sign; right-adjusted field occupies both bytes.
  • Sign at 3000 is $+$, reversed gives $-$.
  • Register after load:

$$ rI_2 \leftarrow -,200,15 $$

(c) LDX 3000(1:3)

  • Instruction: Load bytes 1 to 3 of location 3000 into rX with normal sign.
  • Field (1:3) contains bytes $5,1,200$ (does not include the sign).
  • Sign of destination register becomes $+$ (since field excludes the original sign).
  • rX has 5 bytes; right-adjust the 3-byte field into 5-byte register. Right-adjustment means the field occupies the rightmost positions, with leading bytes set to zero. There are $5-3=2$ leading zeros.
  • Register after load:

$$ rX \leftarrow +,0,0,5,1,200 $$

(d) LD6 3000

  • Instruction: Load the normal field (0:5) of location 3000 into index register rI6.
  • Field (0:5) includes the sign and all four bytes.
  • rI6 has only 2 bytes and a sign. MIX rules: truncate leftmost bytes to fit the destination register.
  • Field bytes including sign: + 5 1 200 15. Drop the first three bytes (sign not included here for indexing) and load the last 2 bytes $200,15$. Sign is from the field (normal), so $+$.
  • Register after load:

$$ rI_6 \leftarrow +,200,15 $$

(e) LDXN 3000(0:0)

  • Instruction: Load sign only (0:0) into rX with opposite sign.
  • Sign at 3000 is $+$, reversed gives $-$.
  • All bytes of rX set to zero.
  • Register after load:

$$ rX \leftarrow -,0,0,0,0,0 $$

Summary of defined replacement operations

All five instructions are defined. The registers after execution are:

$$ \begin{aligned} (a);& rA \leftarrow -,5,1,200,15 \ (b);& rI_2 \leftarrow -,200,15 \ (c);& rX \leftarrow +,0,0,5,1,200 \ (d);& rI_6 \leftarrow +,200,15 \ (e);& rX \leftarrow -,0,0,0,0,0 \end{aligned} $$

$$ \boxed{\text{All five replacements are defined, with contents as above.}} $$