## Soln: three-way light control

#### Truth table

Rowxyz f
0 0000
1 0011
2 0101
3 0110
4 1001
5 1010
6 1100
7 1111
$f = \sum m (1, 2, 4, 7) \\= \bar{x} \bar{y} z + \bar{x} y \bar{z} + x \bar{y} \bar{z} + xyz$
$$\bar{x}$$ $$x$$
$$\bar{y}$$ $$y$$ $$\bar{y}$$
$$\bar{z}$$ 0 1 0 1
$$z$$ 1 0 1 0
$f = \sum m (1, 2, 4, 7) \\= \bar{x} \bar{y} z + \bar{x} y \bar{z} + x \bar{y} \bar{z} + xyz$

## Soln: multiplexer

Rowsx_1x_2 f
0 0000
1 0010
2 0101
3 0111
4 1000
5 1011
6 1100
7 1111
$$f = \sum m(2, 3, 5, 7)$$
$$\bar{s}$$ $$s$$
$$\bar{x_1}$$ $$x_1$$ $$\bar{x_1}$$
$$\bar{x_2}$$ 0 1 0 0
$$x_2$$ 0 1 1 1

$$f = sx_1 + \bar{s} x_2$$

$$f = sx_1 + \bar{s} x_2$$

## DeMorgan's theorem

• $$\overline{x \cdot y} = \bar{x} + \bar{y}$$
• $$\overline{x + y} = \bar{x} \cdot \bar{y}$$

## DeMorgan's theorem

• $$\overline{x \cdot y} = \bar{x} + \bar{y}$$
• $$\overline{x + y} = \bar{x} \cdot \bar{y}$$

## Prove that for two variable function

$$\overline{m}_0 = M_0$$

## Soln: Prove that for two variable function

$$\bar{m}_0 = M_0$$ \begin{align} m_0 &= \bar{x}_1 \bar{x}_2 & \text{ by definition of } m_0 \\ \implies \bar{m}_0 &= \overline{\bar{x}_1 \bar{x}_2} & \\ &= \bar{\bar{x}}_1 + \bar{\bar{x}}_2 & \text{ by DeMorgan's theorem } \\ &= x_1 + x_2 = M_0 & \text{ by definition of } M_0 \end{align}

## Prove that

If $$x + y = 1$$ and $$xy = 0$$ then $$\bx = y$$

## Soln: Prove that

If $$x + y = 1$$ and $$xy = 0$$ then $$\bx = y$$
\begin{align} &x + y = 1 \\ &\implies x \bx + y \bx = \bx & \text{ multiply both sides by } \bx \\ &\implies y \bx = \bx & \because x \bx = 0 \\ &\implies y \bx + y x = \bx + y x & \text{ add } y x \text{ to both sides } \\ &\implies y (\bx + x) = \bx & \text{by dist. prop. and } xy = y x = 0 \\ &\implies y = \bx & \end{align}

## Prove that for three variable function

$$\sum m(0, 2, 3, 7) = \prod M(1, 4, 5, 6)$$

## Soln: Prove that for three variable function

\begin{align} \sum m(0, 2, 3, 7) &= m_0 + m_2 + m_3 + m_7 \\ &= \overline{ m_1 + m_4 + m_5 + m_6} \\ &= \bar{m}_1 \bar{m}_4 \bar{m}_5 \bar{m}_6 \\ &= M_1 M_4 M_5 M_6 \\ &=\prod M(1, 4, 5, 6) \end{align}

## Find the minimum cost expression for

$$f(x, y, z) = \sum m(0, 2, 3, 4, 5, 6, 7)$$

## Solution

\begin{align} f &= \sum m(0, 2, 3, 4, 5, 6, 7) \\ &= \overline{\sum m(1)} \\ &= \prod M(1) \\ &= (x + y + \bz) \end{align}

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