Message #425

From: spel_werdz_rite <>
Subject: Discretions on the permutations of the 3^4
Date: Mon, 17 Sep 2007 03:58:29 -0000

Out of boredom, I chose to see for myself the number of positions on
the 3x3x3x3 and I kept getting half the value!

Here’s what I did:
24!x32!x16! - This is the total number of positions accounting all
n-sided stickers

(2^24)x(6^32)x(12^16) - The total number of orientations accounting
for all n-sided stickers

Now I divided numbers appropriately based on these findings:
2-I could not switch just two 2-colored stickers.
2-I could not switch just two 3-colored stickers.
2-I could not switch just two 4-colored stickers.
2-I could not flip just one 2-colored sticker.
2-I could flip just one 3-colored sticker, but only in three
orientations, not six.
3-I could flip just one 4-colored sticker, but only in four ways, not

So my end result was:
Which is half of the currently accepted value. I think the argument
lies where Eric Balandraud stated "Not all the permutations of the 24
2-colored and the 32 3-colored are possible. Only the permutations
that have the same parity on the 2-colored and the 3-colored … So
the number of positions reachable by just the 2-colored [and
3-colored] pieces is (24!x32!)/2"
He did mention that neither 2- or 3-colored stickers could switch just
two pieces, but he only divided by two once, not compensating for both
occurrences. The real value, I believe, should have been half of what
he gave us in that part of the equation.