From 21d2ce3ff6069f3e89544d83dc96b1394c6b1e03 Mon Sep 17 00:00:00 2001
From: Brandon Rozek <rozekbrandon@gmail.com>
Date: Fri, 18 Jun 2021 01:22:10 -0400
Subject: [PATCH] New Post

---
 content/blog/comparatorlogicgate.md | 53 +++++++++++++++++++++++++++++
 1 file changed, 53 insertions(+)
 create mode 100644 content/blog/comparatorlogicgate.md

diff --git a/content/blog/comparatorlogicgate.md b/content/blog/comparatorlogicgate.md
new file mode 100644
index 0000000..814ad3f
--- /dev/null
+++ b/content/blog/comparatorlogicgate.md
@@ -0,0 +1,53 @@
+---
+title: "Comparator Logic Gate"
+date: 2021-06-18T01:09:45-04:00
+draft: false
+tags: []
+---
+
+This post is heavily derived from the Wikipedia post on [Digital Comparators](https://en.wikipedia.org/wiki/Digital_comparator) and therefore can be distributed under the Creative Commons Attribution-ShareAlike 3.0 license.
+
+To compare two binary numbers, we need to find a way to check if two single bits are equal or if one is greater than the other.
+
+## Checking Single Bit Equality
+
+Let's construct the truth table:
+
+| A    | B    | $A = B$ |
+| ---- | ---- | ------- |
+| 0    | 0    | 1       |
+| 1    | 0    | 0       |
+| 0    | 1    | 0       |
+| 1    | 1    | 1       |
+
+The only two ways this evaluates to true is if
+
+- A is 1 and B is 1
+- A is 0 and B is 0
+
+Therefore,
+$$
+(A = B) \equiv AB + \bar{A}\bar{B} \equiv \overline{(A \oplus B)}
+$$
+
+## Single Bit Comparator
+
+To check if one bit is greater than the other, let us first look at the truth table:
+
+| A    | B    | $\bar{B}$ | $A > B$ |
+| ---- | ---- | --------- | ------- |
+| 0    | 0    | 1         | 0       |
+| 1    | 0    | 1         | 1       |
+| 0    | 1    | 0         | 0       |
+| 1    | 1    | 0         | 0       |
+
+The only way for this to evaluate to true is if A is 1 and B is 0. That is $(A > B) \equiv A\bar{B}$
+
+
+
+## L-Bit Comparator
+
+To compare an entire bitstring, we start from the most significant bit and check to see if one bit is greater than the other. If not, it will then check the next bit while confirming that all the previous bits were the same. For a 3-bit comparator, the logic will look like the following:
+$$
+(A_3A_2A_1 > B_3 B_2 B_1) \equiv A_3\bar{B_3} + \overline{(A_3 \oplus B_3)}A_2\bar{B_2} + \overline{(A_3 \oplus B_3)}\overline{(A_2 \oplus B_2)}A_1\bar{B_1} 
+$$