Visual walkthrough — Multiplexers (2 - 1, 4 - 1, n - 1)

We use a few symbols. Before any appear, here is the whole cast in plain words:

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Step 1 — The physical picture: a knob choosing one pipe

WHAT. Draw four water pipes feeding into one tap. A dial with a pointer selects which pipe's water reaches the single spout.

WHY start here. Every symbol below is a translation of this picture into electricity. If you can see the tap, you never lose the meaning of the algebra. "Multiplex" literally means many into one line — this drawing is that phrase.

PICTURE. The dial position (a number) chooses the pipe. The spout ignores every other pipe completely.

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Step 2 — Two pipes only: the 2:1 MUX, in pictures

WHAT. Shrink to the smallest interesting case: two data inputs $I_0$ and $I_1$, and one select bit $S$. When $S=0$ we want $Y=I_0$; when $S=1$ we want $Y=I_1$.

WHY the smallest case first. Everything bigger is copies of this. If you understand one select bit steering between two inputs, the 4:1 and n:1 are just "more of the same."

PICTURE. Two gates, like two doors. We need each door to open only in its own case.

The question the algebra must answer: "how do I write a switch that is ON only when $S=0$?" That is exactly what $\bar{S}$ is for.

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Step 3 — Turning "when $S=0$" into algebra: why we need $\bar{S}$

WHAT. We want a term that is $1$ precisely when $S=0$, so it can "wave through" $I_0$ then. And a term that is $1$ precisely when $S=1$, to wave through $I_1$.

WHY $\bar{S}$ and not something fancier. $S$ itself is $1$ when $S=1$ — that is our $I_1$ gate. But for $I_0$ we need the opposite on-condition. The one operation that flips a bit is NOT, written $\bar{S}$. It is the smallest tool that answers "am I in the $S=0$ world?" — so we use it, nothing heavier.

PICTURE. Watch the two gates open in opposite halves of the knob.

Now assemble the sentence. "Pass $I_0$ when $S=0$" is $\bar{S}\cdot I_0$. "Pass $I_1$ when $S=1$" is $S\cdot I_1$. Either one may be the winner, so we join them with OR ($+$):

$Y=\underset{\begin{array}{c}\text{ON when}\\ S=0\end{array}}{\underbrace{\bar{S}}}\underset{\begin{array}{c}\text{the pipe}\\ \text{it gates}\end{array}}{\underbrace{I_0}}+\underset{\begin{array}{c}\text{ON when}\\ S=1\end{array}}{\underbrace{S}}\underset{\begin{array}{c}\text{the pipe}\\ \text{it gates}\end{array}}{\underbrace{I_1}}$

Term-by-term: $\bar{S}{I}_0$ is "$I_0$ if we are in the $S=0$ world, else $0$." $S{I}_1$ is "$I_1$ if we are in the $S=1$ world, else $0$." The $+$ says "output whichever half is active." Because $\bar{S}$ and $S$ are never both $1$, only one half is ever nonzero — the promise from Step 1 is kept.

Let us check both cases exhaustively (only two exist):

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Step 4 — Two select bits build four "pointers": the minterm idea

WHAT. With two select bits $S_1$ (the high bit, MSB) and $S_0$ (the low bit), the knob has $2\times 2=4$ positions: $00,01,10,11$. Each spells a number $0,1,2,3$. We need one algebra-term that is $1$ for each position and $0$ everywhere else.

WHY. In Step 3 the "on-when-$S=0$" gate was $\bar{S}$. Now each of four positions needs its own on-gate. The gate for position "both bits as written" is the AND of each bit in its right flavour (barred if that bit is $0$, plain if $1$). That special product is a minterm.

PICTURE. A four-slot dial; each slot lights exactly one minterm lamp.

Read the lamps:

$S_1$	$S_0$	number $k$	minterm $m_k$ (its on-gate)	why barred/plain
0	0	0	${\bar{S}}_1{\bar{S}}_0$	both bits are $0$ → bar both
0	1	1	${\bar{S}}_1{S}_0$	$S_1=0$ bar it, $S_0=1$ plain
1	0	2	$S_1{\bar{S}}_0$	$S_1=1$ plain, $S_0=0$ bar it
1	1	3	$S_1{S}_0$	both $1$ → plain both

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Step 5 — Assemble the 4:1 MUX: gate each pipe with its pointer

WHAT. Give each input $I_k$ its own minterm $m_k$ as an on-gate, then OR everything.

WHY. Same sentence as the 2:1, just four clauses: "$I_0$ if we're in position 0, or $I_1$ if position 1, or …". Each clause is $m_k\cdot I_k$; only one clause is ever nonzero (Step 4), so exactly one pipe reaches $Y$.

PICTURE. Four AND gates fanning into one OR gate — the shape of every MUX.

$Y=\underset{m_0}{\underbrace{{\bar{S}}_1{\bar{S}}_0}}{I}_0+\underset{m_1}{\underbrace{{\bar{S}}_1{S}_0}}{I}_1+\underset{m_2}{\underbrace{S_1{\bar{S}}_0}}{I}_2+\underset{m_3}{\underbrace{S_1{S}_0}}{I}_3$

Term-by-term: each brace is a pointer (a minterm) that is $1$ only for its own knob setting; the $I_k$ next to it is the pipe it lets through; the $+$ signs OR the four clauses, but since only one pointer fires, only one pipe survives.

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Step 6 — The pattern that scales: the general n:1 MUX

WHAT. For $n$ select bits there are $2^n$ knob positions and $2^n$ inputs. Every position has its own minterm; gate each input with it and OR them.

WHY. Nothing new — Steps 3–5 already are the pattern. We just write it compactly with a summation sign $\sum$, which means "add all these clauses."

PICTURE. The universal AND-array → OR shape, drawn for general $k$.

$Y={\sum }_{k=0}^{2^n-1}\underset{\begin{array}{c}\text{minterm of the }n\\ \text{select bits}=k\end{array}}{\underbrace{m_k}}\cdot \underset{\begin{array}{c}\text{the input at}\\ \text{address }k\end{array}}{\underbrace{I_k}}$

Term-by-term: ${\sum }_{k=0}^{2^n-1}$ says "OR together one clause for every address from $0$ to $2^n-1$." $m_k$ is the fingerprint that fires only when the knob spells $k$. $I_k$ is the pipe at that address. This block of minterms is exactly what a decoder produces — a MUX is a decoder feeding an OR gate.

And the count of select lines, since $n$ bits address $2^n$ inputs, inverts to: $n=\lceil {\log }_{2}N\rceil (\text{for }N\text{ inputs}).$

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Step 7 — The degenerate and edge cases (never leave a gap)

WHAT. Push the formula to its extremes so no reader hits an unshown scenario.

WHY. A derivation you can trust must survive its own boundaries.

PICTURE. Three tiny corner cases side by side.

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The one-picture summary

Every step collapses into a single claim: a MUX is minterms (a decoder) choosing pipes, ORed at one spout. The picture below packs the whole derivation — knob → minterm fingerprints → AND-gate pipes → single OR → $Y$.

Recall Feynman retelling — the whole walkthrough in plain words

Imagine a water tap over four pipes with a dial. We wanted to write that tap in the language of $0$s and $1$s. First we did the tiny two-pipe tap: we needed a switch that's ON only when the dial says $0$, so we invented $\bar{S}$ ("not $S$"), and the switch for the dial-says-$1$ side is just $S$. Each switch waves its own pipe through, and we say "either one" with a $+$ (OR): $Y=\bar{S}{I}_0+S{I}_1$. Then for four pipes the dial has two bits; for each of the four positions we made a fingerprint — a minterm — that lights up for exactly one position by barring the bits that are $0$ and leaving the $1$s plain. We paired every fingerprint with its pipe and ORed them all. Because only one fingerprint ever lights, only one pipe ever comes out — the tap's promise is kept. Writing "add a clause for every address" with a $\sum$ gives the general formula, and it survives all the corners: no selects means a plain wire, all-equal inputs means a constant, and an enable bit $E$ can shush the whole thing by multiplying everything by zero.

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Connections

• Multiplexers (2 - 1, 4 - 1, n - 1) — the parent topic these pictures derive.
• Decoders — the minterm/fingerprint generator living inside every MUX.
• Demultiplexers — the mirror image: one pipe, many spouts.
• Boolean Algebra Minterms — the fingerprints of Step 4.
• Shannon Expansion — factoring a variable out to shrink onto a smaller MUX.
• Tri-state Buffers — how enable becomes high-Z on a shared bus.
• Combinational Logic Design — the MUX as a universal function block.