 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2003/functional-programming-data-structures-and-algorithms-procedural-programming-and-digital-hardware/9/d/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2003 / Functional programming, data structures and algorithms, procedural programming, and digital hardware / 9 / d / Solution by

Codex  0 2026-10-31

Let

(L^{+}, G^{+})

denote the state after the next active clock edge. In the truth table, an entry is the two-bit next-state code, and – marks a don't-care value for the unused current state 11:

Current $L G$	$ab = 00$	$ab = 01$	$ab = 10$	$ab = 11$
$00$ (`Eq`)	$00$	$10$	$01$	$00$
$10$ (`Lt`)	$10$	$10$	$01$	$10$
$01$ (`Gt`)	$01$	$10$	$01$	$01$
$11$ (unused)	$- -$	$- -$	$- -$	$- -$

For each input column, 01 forces

L^{+} G^{+} = 10

, 10 forces

01

, and the two equal-input columns copy the current code. Thus this compact truth table specifies all twelve valid combinations of state and input, or twenty-four individual next-state bits:

ab = 01 \Rightarrow (L^{+}, G^{+}) = (1, 0), ab = 10 \Rightarrow (L^{+}, G^{+}) = (0, 1) .

(1)

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