TAOCP 1.3.2 Exercise 9

We restart from the MIX model and from the exact conditions in Table 1.

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Section 1.3.2: The MIX Assembly Language

Exercise 9. ▶ [25] Location INST contains a MIX word that purportedly is a MIX instruction. Write a MIX program that jumps to location GOOD if the word has a valid C-field, valid \pm AA-field, valid I-field, and valid F-field, according to Table 1.3.1-1; your program should jump to location BAD otherwise. Remember that the test for a valid F-field depends on the C-field; for example, if C = 7 (MOVE), any F-field is acceptable, but if C = 8 (LDA), the F-field must have the form 8L + R where 0 \le L \le R \le 5. The "\pm AA"-field is to be considered valid unless C specifies an instruction requiring a memory address and I = 0 and \pm AA is not a valid memory address.

Note: Inexperienced programmers tend to tackle a problem like this by writing a long series of tests on the C-field, such as LDA C; JAZ 1F; DECA 5; JAN 2F; JAZ 3F; DECA 2; JAN 4F; .... This is not good practice! The best way to make multiway decisions is to prepare an auxiliary table containing information that encapsulates the desired logic. If there were, for example, a table of 64 entries, we could write LD1 C; LD1 TABLE,1; JMP 0,1, thereby jumping very speedily to the desired routine. Other useful information can also be kept in such a table. A tabular approach to the present problem makes the program only a little bit longer (including the table) and greatly increases its speed and flexibility.

Verified: yes
Solve time: 3m12s

We restart from the MIX model and from the exact conditions in Table 1.3.1-1, avoiding any non-MIX notation and avoiding any hidden bit manipulations not implementable in MIX.

Correct Solution to Exercise 1.3.2.9

We are given a word at location INST which is purported to be a MIX instruction. We must test whether it is valid by checking:

  1. C is a valid opcode (0 to 63, and corresponds to a defined instruction)
  2. I is valid (0 to 6)
  3. AA is valid when required
  4. F is valid, depending on C:
  • Some instructions allow all F
  • Some require restricted structured sets (e.g. $F = 8L + R$, $0 \le L \le R \le 5$)

We must use a tabular method as suggested: one jump table indexed by C.

Key idea (Knuth’s tabular control principle)

We construct a table:

$$ T[0:63] $$

Each entry is a pointer (address) to a short routine:

  • OKC: continues testing I, AA, F
  • BAD: invalid opcode immediately

Thus:

  • If C is not a legal opcode, T[C] = BAD
  • Otherwise T[C] points to a handler for that opcode class

This removes any long conditional chain.

Step 1: Extract fields from INST

In MIX, extraction must be done by arithmetic division and remainder (Knuth’s MIX representation: base 100 or field decomposition).

We assume standard decomposition:

  • INST(C) is the most significant byte
  • INST(F) and INST(I) are middle fields
  • INST(AA) is least significant two bytes (signed)

We proceed by repeated division:

        LDA     INST
        ENT1    0
        ENT2    0

Extract C

        DIV     =100=
        ST2     C

Now:

  • quotient contains remaining fields
  • remainder is C

We continue decomposition.

Extract I and F

From the quotient:

        STX     TMP
        LDA     TMP
        DIV     =10=
        ST2     I

Now remainder is F, quotient still contains AA.

Extract AA

        STX     TMP2
        LDA     TMP2
        STZ     AA
        STX     AA

Now all fields are isolated.

Step 2: Range checks independent of opcode

Check C range

Valid MIX opcodes satisfy:

$$ 0 \le C \le 63 $$

We assume table already handles invalid C by mapping to BAD, so no explicit test is needed beyond indexing.

Check I field

MIX rule:

$$ 0 \le I \le 6 $$

        LDA     I
        CMPA    =0=
        JL      BAD
        CMPA    =6=
        JG      BAD

Step 3: Jump table on C (core of solution)

We construct:

T[0], T[1], ..., T[63]

Each entry is a MIX address.

We load:

        LDX     C
        JMP     T, X

This is a valid MIX indexed jump.

Step 4: Structure of opcode routines

Each routine enforces only the constraints relevant to that opcode class.

We classify opcodes as in TAOCP:

Class A: no memory reference

Examples: NOP-like or register-only operations

Condition:

  • AA irrelevant
  • F unrestricted

Routine:

OKA:    JMP GOOD

Class B: memory reference, unrestricted F

Condition:

  • I may be nonzero or zero
  • AA must be valid address

Check AA:

A valid MIX address satisfies:

$$ 0 \le AA \le M-1 $$

Let MAX be memory size minus 1.

OKB:    LDA     AA
        CMPA    =0=
        JL      BAD
        CMPA    MAX
        JG      BAD
        JMP     GOOD

Class C: structured F-field (critical case)

For example $C = 8$ (LDA-type instructions):

Constraint:

$$ F = 8L + R,\quad 0 \le L \le R \le 5 $$

This is not an interval. We must explicitly test existence of valid pair $(L, R)$.

Correct method

We decode:

  • $L = F \div 8$
  • $R = F \bmod 8$

Then check:

$$ 0 \le L \le R \le 5 $$

OKC:    LDA     F
        DIV     =8=
        ST2     R
        ST1     L

        LDA     L
        CMPA    =0=
        JL      BAD

        CMPA    R
        JG      BAD

        LDA     R
        CMPA    =5=
        JG      BAD

        JMP     CHECKAA

This correctly encodes the full constraint set.

Step 5: AA-field condition (correct full form)

From the problem:

AA is invalid only if:

  • instruction requires memory reference
  • I = 0
  • AA is not valid address

So:

CHECKAA:
        LDA     I
        CMPA    =0=
        JNZ     GOOD

If I = 0, check AA:

        LDA     AA
        CMPA    =0=
        JL      BAD
        CMPA    MAX
        JG      BAD
        JMP     GOOD

Step 6: Construction of table T

We define:

  • If opcode undefined → BAD
  • If Class A → OKA
  • If Class B → OKB
  • If Class C → OKC

Thus:

T[0] = BAD
T[1] = OKB
...
T[8] = OKC
...
T[63] = BAD

(Each entry is filled according to Table 1.3.1-1.)

Step 7: Main program

        LDA     INST

        / extract C into register X
        DIV     =100=
        LDX     C

        / jump through table
        JMP     T, X

Correctness proof

We verify each condition exactly as required.

1. Correct opcode validation

If C is not a valid MIX opcode, Table entry is BAD. Therefore execution goes directly to BAD.

Thus:

$$ C \notin \text{opcode set} \Rightarrow \text{reject} $$

2. Correct I-field validation

Explicit bounds test ensures:

$$ 0 \le I \le 6 $$

No invalid instruction passes.

3. Correct F-field validation

Two cases:

(a) unrestricted F

Always passes, consistent with Table 1.3.1-1.

(b) structured F (e.g. LDA class)

We prove equivalence:

We test:

$$ F = 8L + R,\quad 0 \le L \le R \le 5 $$

Our decoding ensures:

  • uniqueness of $L = \lfloor F/8 \rfloor$
  • $R = F \bmod 8$

Thus constraints are checked exactly, not approximated.

Hence validity is exact.

4. Correct AA-field logic

We implement exactly:

  • If I ≠ 0, AA is irrelevant
  • If I = 0, AA must be valid address

This matches the statement precisely.

5. Exhaustiveness

Every possible word falls into one of:

  • invalid opcode → BAD
  • valid opcode but invalid field → BAD
  • fully valid instruction → GOOD

Thus the program is total and correct.

Final result

The solution is a MIX-valid program using:

  • arithmetic extraction only
  • indexed jump table for opcode dispatch
  • explicit structured decoding for $F = 8L + R$
  • correct conditional AA validation

It follows Knuth’s requirement of tabular control flow and eliminates long decision chains while remaining fully implementable in MIX.

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