Voting systems

In the context of digital circuits, voting systems refer to electronic or logical implementations designed to process multiple inputs and determine a result based on a majority rule or predefined logic.

Example: Committee voting system

We will design an electronic voting system for the Executive Committee of a football team. This committee has four members: the President, the Secretary, the Treasurer, and the Vocal. Each may vote Yes (1) or No (0).

The approval rules are:

1. A proposal is approved when three or more members vote in favour.
2. In case of a tie (2 in favour and 2 against), the President's vote decides. If the President votes "Yes", the proposal is approved; if he votes "No", it is rejected.

We will design a logic circuit with four inputs ($P$, $S$, $T$, $V$) and one output ($A$) that indicates whether the proposal is approved.

Define the input variables ($1$ means "Yes", $0$ means "No"):

• $P$: vote of the President
• $S$: vote of the Secretary
• $T$: vote of the Treasurer
• $V$: vote of the Vocal

The output variable is $A$ (Approved). $1$ means approved, $0$ means rejected.

We will analyse under which assumptions the proposal will be approved; we will call these the approval conditions and they are derived from the approval rules.

Condition 1: Three or more votes in favour.

We identify all combinations where the sum of P, S, T, V is 3 or 4.

• Approved by three votes in favour
  • $P=0$, $S=1$, $T=1$ and $V=1$ --> ($\overline{P}STV$)
  • $P=1$, $S=1$, $T=1$ and $V=0$ --> ($PST\overline{V}$)
  • $P=1$, $S=1$, $T=0$ and $V=1$ --> ($PS\overline{T}V$)
  • $P=1$, $S=0$, $T=1$ and $V=1$ --> ($P\overline{S}TV$)
• Approved by four votes in favour
  • $P=1$, $S=1$, $T=1$ and $V=1$ --> ($PSTV$)

The proposal will be approved if any of these circumstances is met (logical OR). Condition 1 is expressed in Boolean algebra as:

$$C_1=\overline{P}STV+PST\overline{V}+PS\overline{T}V+P\overline{S}TV+PSTV$$

Condition 2: Tie resolved in favour of the President.

We must identify all combinations with two votes for and two against. Of these, those with $P=1$ are of interest.

• Approved
  • $P=1$, $S=1$, $T=0$ and $V=0$ --> ($PS\overline{T}\overline{V}$)
  • $P=1$, $S=0$, $T=1$ and $V=0$ --> ($P\overline{S}T\overline{V}$)
  • $P=1$, $S=0$, $T=0$ and $V=1$ --> ($P\overline{S}\overline{T}V$)
• Not approved
  • $P=0$, $S=1$, $T=1$ and $V=0$
  • $P=0$, $T=0$, $S=1$ and $V=1$
  • $P=0$, $S=0$, $T=1$ and $V=1$

The proposal will be approved if any of these assumptions hold. Condition 2 is expressed as:

$$C_2=PS\overline{T}\overline{V}+P\overline{S}T\overline{V}+P\overline{S}\overline{T}V$$

The output $A$ will be $1$ if condition $C_1$ or condition $C_2$ is satisfied. The Boolean function is a logical OR of these two conditions:

$$A=C_1+C_2$$

And therefore the expression for $A$, as a function of $P$, $S$, $T$ and $V$, is:

$$A=\overline{P}STV+PST\overline{V}+PS\overline{T}V+P\overline{S}TV+PSTV+PS\overline{T}\overline{V}+P\overline{S}T\overline{V}+P\overline{S}\overline{T}V$$

We build the truth table with all possible cases:

$P$	$S$	$T$	$V$	Votes in favour	$C_1$	$C_2$	$A$
0	0	0	0	0	0	0	0
0	0	0	1	1	0	0	0
0	0	1	0	1	0	0	0
0	0	1	1	2	0	0	0
0	1	0	0	1	0	0	0
0	1	0	1	2	0	0	0
0	1	1	0	2	0	0	0
0	1	1	1	3	1	0	1
1	0	0	0	1	0	0	0
1	0	0	1	2	0	1	1
1	0	1	0	2	0	1	1
1	0	1	1	3	1	0	1
1	1	0	0	2	0	1	1
1	1	0	1	3	1	0	1
1	1	1	0	3	1	0	1
1	1	1	1	4	1	0	1

The expression for $A$ as a function of $P$, $S$, $T$ and $V$ is quite long. A logical simplification with Boolean algebra would be complex. Alternatively we can use a Karnaugh map for the output function $A$, with inputs $P$, $S$, $T$ and $V$ grouped.

TV PS	00	01	11	10
00	0	0	0	0
01	0	0	1	0
11	1	1	1	1
10	0	1	1	1

The next step is to group the '1's into four groups, which we mark with different colours.

TV PS	00	01	11	10
00	0	0	0	0
01	0	0	1	0
11	1	1	1	1
10	0	1	1	1

The fixed variables in the blue group are $P=1$ and $T=1$, which translates to the term $PT$ in the solution.

$$A=PT+···$$

TV PS	00	01	11	10
00	0	0	0	0
01	0	0	1	0
11	1	1	1	1
10	0	1	1	1

The fixed variables in the yellow group are $P=1$ and $V=1$, add the term $PV$ to the solution.

$$A=PT+PV+···$$

TV PS	00	01	11	10
00	0	0	0	0
01	0	0	1	0
11	1	1	1	1
10	0	1	1	1

The green colour grouping has in common $P=1$ and $S=1$, add the term $PS$ to the solution

$$A=PT+PV+PS+···$$

TV PS	00	01	11	10
00	0	0	0	0
01	0	0	1	0
11	1	1	1	1
10	0	1	1	1

The fixed variables in the green group have in common $P=1$ and $S=1$, so we add the term PS to the solution

$$A=PT+PV+PS+···$$

TV PS	00	01	11	10
00	0	0	0	0
01	0	0	1	0
11	1	1	1	1
10	0	1	1	1

Finally, in the red group the fixed variables are $S=1$, $T=1$ and $V=1$, add the term $STV$ to the solution.

$$A=PT+PV+PS+STV$$

This is the final simplified Boolean expression, because all the '1's have already been accounted for in some group.

The digital circuit derived from this expression is the following:

Exercises on Jutge.org: Introduction to Digital Circuit Design

• Executive Committee
• Two or more ones
• 4-input majority voter
• Majority of 5 inputs

Remember that to access the exercises and for the Judge to evaluate your solutions you must be enrolled in the course. You will find all instructions here.

Xavier Casas, Francesc Madrid
Lliçons.jutge.org
© Universitat Politècnica de Catalunya, 2026