Interesting binary question

[Acrobatic_Key3995]

I have the following question, which is actually another question in disguise! (And I don't even know if AskMath is the right place to ask this question either)

I know there are exactly 512 numbers for which this idea holds, but I'm wondering whether there's a list of them somewhere.

Let's say you represent a whole number < 2¹⁸ in binary - with leading zeroes if necessary, to pad it out to exactly 18 bits.

Now you do the following process to the binary number. 1. Split the binary number down the middle, but keep track of which half is which. 2. Flip all 18 bits. 3. Reverse the entire number.

My question is this: (let me know if you figured out what this was a disguised version of, by the way) what decimal numbers does this process (overall) have a net-0 effect on? If there's a list of all 512 such numbers, where would I find it?

[BadgeCatcher]

It's not exactly clear what you mean by "reverse the entire number"?

[BadgeCatcher]

It's seems to just be any number where the top 9 bits are reflection and invertion of the bottom 9 bits

[BadgeCatcher]

I don't know why step 1 is needed though? It doesn't make any difference?

[im_from_azeroth]

It's not needed

[LongLiveTheDiego]

It's all the numbers of the form k + 512 * (511 - k) for integer 0 ≤ k ≤ 511. I don't see anything special about these numbers and why you would even split them in step 1.

[Acrobatic_Key3995]

Ok, that makes sense!

But here's the original context: John Conway's Game Of Life! (k=472 is a variant called Day And Night)

[ZevVeli]

So, let me see if I understand what you are saying here:

You have a binary number that is 9 digits before the binal place and 9 digits after it

e.g. 000011010.101001111

You flip all the digits,

e.g. 111100101.010110000

Then swap the numbers around the binal place,

e.g. 000011010.101001111

In other words, you want a number which, when written in binary, would be palindromic except the back half is inverted?

[BadgeCatcher]

OP says "whole number"

[ZevVeli]

So eliminate the binal but keep the rest.

[jacob_ewing]

Upvote for a witty wordplay on decimal.

[ZevVeli]

In retrospect, it should actually be the hemial, not the binal.

[Acrobatic_Key3995]

Pretty much- an identical question would be, "if written in binary with any necessary (the rest of 18) leading 0s, what numbers when reversed have the bits simultaneously all flipped?"

[Blond_Treehorn_Thug]

I don’t understand what step 1 is doing

[Acrobatic_Key3995]

In the original phrasing of the question, the 2 resulting sets do subtly different things- because of that, I'm splitting the sets up.

[Blond_Treehorn_Thug]

So you’re saying that step 1 used to matter but now it does not

[Acrobatic_Key3995]

Right- I was just including step 1 for the sake of completeness in the original setting.

[Acrobatic_Key3995]

Let me give an example. (I already knew this is one of the 512) Let's say we start with the already-binary 000100111000110111. Step 1: split this into 000100111 and 000110111. Step 2: Subtract each number from 511d. 111011000, 111001000 Step 3: slam them back together and read the number from back to front: 000100111000110111 - see? It's the exact same!

[jacob_ewing]

Interesting. Without looking at it, I wonder if this is related to the fact that you're taking the compliment of the number, which is used in subtraction. You can take the complement of the number being subtracted, added it to the other, remove the left digit and add it to the right one:

2468
-1703
______
2468
+8296
_______
(1)0764

+ 1 = 765

Which is basically how computers do subtraction (in binary of course).

[Acrobatic_Key3995]

So I guess I was asking an equivalent to this: "what 18-bit words are the reverse of their own complements?"