Question 5 of 5: 4-Bit Hex → Braille Converter

Question 5: 4-Bit Hex → Braille Converter (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A 4-bit active-HIGH hex value $X_3X_2X_1X_0$ (0–F) and the fully-specified 16-pattern Braille dot table above (K,L,M,N, all active-HIGH, no don’t-cares — every one of the 16 hex digits has a defined pattern).

Find. (a) minimized SOP equations for K, L, M, N. (b) a 2-input-NAND-only realization of K and M.

Approach. Read the four output columns straight off the given dot table (one minterm list per output), K-map/Quine–McCluskey-minimize each of the four fully-specified 4-variable functions, brute-force-verify every minimized expression against all 16 rows, then convert K’s and M’s SOP into an all-NAND network by AND→double-NAND and OR→NAND-of-inverted-inputs.

1. Part (a), Step 1 — tabulate the four outputs.

   [Figure not reproduced: Fig. Q5 — the 16 given Braille dot patterns (filled = raised dot), redrawn from the source legend as a verification aid. See the official exam paper.]

   Hex digit → Braille dot pattern (K L M N)

   Hex (X3X2X1X0)	K	L	M	N
   0 (0000)	1	1	1	1
   1 (0001)	0	1	0	1
   2 (0010)	1	0	1	1
   3 (0011)	1	1	0	1
   4 (0100)	1	1	1	0
   5 (0101)	1	0	0	1
   6 (0110)	0	1	1	1
   7 (0111)	1	0	1	0
   8 (1000)	0	1	1	0
   9 (1001)	1	0	0	1
   A (1010)	0	1	0	1
   B (1011)	1	0	1	1
   C (1100)	1	1	0	1
   D (1101)	1	1	1	0
   E (1110)	0	1	1	0
   F (1111)	0	0	0	0

2. Part (a), Step 2 — K-map minimization. Each output is a fully-specified (no don’t-care) 4-variable function; grouping the 1-cells of each of the four K-maps (Quine–McCluskey) gives $$K = X_2\overline{X_1} + \overline{X_3}\,\overline{X_2}\,\overline{X_0} + \overline{X_3}X_1X_0 + X_3\overline{X_2}X_0$$ $$L = \overline{X_3}\,\overline{X_2}X_0 + X_2\overline{X_0} + X_3\overline{X_0} + X_3X_2\overline{X_1} + \overline{X_3}\,\overline{X_2}\,\overline{X_1}$$ $$M = \overline{X_3}\,\overline{X_0} + \overline{X_3}X_2X_1 + \overline{X_2}\,\overline{X_1}\,\overline{X_0} + X_3\overline{X_2}X_1X_0 + X_3X_2\overline{X_1}X_0 + X_2X_1\overline{X_0}$$ $$\boxed{N = \overline{X_2}X_0 + \overline{X_2}X_1 + \overline{X_3}\,\overline{X_2} + \overline{X_3}\,\overline{X_1}X_0 + \overline{X_3}X_1\overline{X_0} + X_3X_2\overline{X_1}\,\overline{X_0}}$$ Each is checked term-by-term against its own 16-row truth column (zero mismatches); no simplification beyond this is available since all 16 minterms are defined (no don’t-cares to exploit) for a Hex-to-Braille map with this much digit reuse (recall 9=5, A=1, B=2, C=3, D=4, E=8 in the given table, which is exactly what keeps each K-map from reducing to something visually simpler).

Part (b) — K and M with 2-input NAND gates only. K has 4 product terms (2–3 literals each) and M has 6 (3–4 literals each). Build every product term as a chain of 2-input NAND+inverter pairs (each pair is exactly a 2-input AND: $\overline{\overline{xy}}=xy$), then combine the resulting terms in a binary tree of NAND-of-inverted-inputs gates (each such gate is exactly a 2-input OR: $\overline{\bar a\bar b}=a+b$ by De Morgan). Both substitutions are exact, so the finished networks compute $K$ and $M$ using nothing but 2-input NAND gates (plus the input-side inverters, themselves NAND gates with tied inputs).