BCD-to-seven-segment decoder, segments a and e

A BCD-to-seven-segment decoder takes a 4-bit BCD digit $ABCD$ ($A$ = MSB) and drives the seven segments a–g of a display so that the decimal digit 0–9 is shown. Inputs 1010–1111 never occur, so they are don't cares. (Common-cathode display: a 1 lights the segment.)

   --a--
  |     |
  f     b
  |--g--|
  e     c
  |     |
   --d--

Truth table for segments a and e

So $a = \Sigma m(0,2,3,5,6,7,8,9) + d(10\text{–}15)$ and $e = \Sigma m(0,2,6,8) + d(10\text{–}15)$.

K-map for a

        CD=00 01  11  10
AB=00     1   0   1   1
AB=01     0   1   1   1
AB=11     X   X   X   X
AB=10     1   1   X   X

Groups: $A$ (rows 11, 10 with don't cares), $C$ (columns 11, 10), $BD$ (cells 5, 7, 13, 15), $B'D'$ (corners 0, 2, 8, 10).

$$a = A + C + BD + B'D'$$

K-map for e

        CD=00 01  11  10
AB=00     1   0   0   1
AB=01     0   0   0   1
AB=11     X   X   X   X
AB=10     1   0   X   X

Groups: $B'D'$ (corners 0, 2, 8, 10) and $CD'$ (column 10: 2, 6, 14, 10).

$$e = B'D' + CD' = D'(B' + C)$$

Circuits

a:  A --------------------+
    C --------------------+
    B --+[AND]------------+--[4-in OR]--> a
    D --+[   ]            |
    B'--+[AND]------------+
    D'--+[   ]

e:  B' --+--[OR]--+
    C  --+--[  ]  +--[AND]--> e
    D' -----------+--[   ]

$e$ needs only one OR gate, one AND gate and an inverter for $D$ ($B'$ is shared with segment a). Check: digit 6 (0110): $a = 0+1+0+0 = 1$, $e = D'(B'+C) = 1\cdot(0+1) = 1$, correct.