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 / e / Solution

Arrange each four-variable Karnaugh map in Gray code order $LG=00,01,11,10$ down the rows and $ab=00,01,11,10$ across the columns. For $L^{+}$ the row strings are 0100, 0100, ----, 1110; for $G^{+}$ they are 0001, 1011, ----, 0001. The maps are drawn alongside the state-transition diagram above.

In the $L^{+}$ map, group the whole ab=01 column to obtain $\neg a\wedge b$. Group rows 11,10 with columns 00,01 to obtain $L\wedge \neg a$. Group those same rows with columns 01,11 to obtain $L\wedge b$. The unused-state row permits these larger groups. Symmetrically, in the $G^{+}$ map group column ab=10, rows 01,11 with columns 11,10, and those same rows with columns 10,00 (wrapping around the map). This givesThese expressions drive the D inputs of the two D flip-flops. With $a=b$, both copy their stored bit. With $ab=01$, they force Lt; with $ab=10$, they force Gt. The equations also specify a harmless extension at unused code 11: equal inputs retain it and unequal inputs recover to a valid comparison state. Correct reset is still required. This extension was chosen using don't-cares; the circuit is not claimed to recover from every invalid state without reset.