🧵 Gray Code — Reflected Form

Gray code is a binary sequence where each successive number differs by only one bit. This is useful in digital systems to minimize transition errors. But how do we reliably construct such a sequence — especially one that wraps around cyclically?

Let’s explore this from first principles, building intuition as we go.

🧩 Starting Construction

We begin with the base case: binary numbers of 1 digit.

0
1

This trivially satisfies the Gray code condition — only one bit changes between the two.

🧱 Expanding to the Next Digit

To expand to the next digit, we must now consider both the 0 MSB case and the 1 MSB case.

🔹 The 0 MSB Case

This is straightforward. We can simply append a 0 to the front of each existing number:

0 + 0 → 00
0 + 1 → 01

This preserves the 1-bit change property between these two numbers.

🔹 Entering the 1 MSB Case

Now we want to construct the 1 MSB case. But we must ensure that the transition from the last number in the 0 MSB case to the first number in the 1 MSB case still changes only one bit.

Since the MSB is the one that changes (from 0 to 1), we must ensure the remaining bits do not change. Thus, we keep the remaining bits from the last binary number within the 0 MSB case.

So if the last number in the 0 MSB case is 01, then the first number in the 1 MSB case must be 11.

This satisfies the 1-bit change requirement.

🔄 Preserving Transitions Within the 1 MSB Case

But we don’t just want the first transition to work — we want all transitions within the 1 MSB case to also change only one bit.

To do this, we realize that we can keep the remaining bits for the other numbers too — but read backwards from the last case to the first case in the 0 MSB case.

Why? Because 1-bit change not only happens forwards but also backwards.

So instead of:

1 + 0 → 10
1 + 1 → 11

We reverse the original sequence:

1 + 1 → 11
1 + 0 → 10

Now the full 2-bit Gray code is:

00
01
11
10

Each step changes only one bit. Even the transition from 10 back to 00 changes only the MSB — satisfying the wraparound condition.

🪞 Realization: This Is Just Reflection

We now realize that all of this is just an application of the reversal of the remaining bits within the previous cases.

This is the same as a reflection.

🔁 Wraparound Validity

Not only is this a reflection, but because the remaining bits of the last binary number match the remaining bits of the first binary number, they differ in only the MSB.

This means they differ in 1 bit — which satisfies Gray code again, achieving the wraparound feature as well.

✅ Final Construction Rule

Thus, a reliable way of generating Gray code with wraparound is:

Simply reflect all the current binary numbers, Assign 0 to the MSB of the first set, Assign 1 to the MSB of the second set, And that’s it.

This method guarantees:

• 1-bit transitions between all successive numbers
• Cyclic wraparound with only one bit change
• Recursive scalability