The Magic Cafe Forum Index » » Magical equations » » Magical Cube 6x6x6 Filled with "0" and "1" - a problem! (0 Likes) Granite Golem New user St.-Petersburg, Russia 3 Posts  Posted: Jan 16, 2004 09:34 pm 0 Hello all! I have a question that exhausts me already 10 days, and I can't find a solution. Here I'll try to express it, and hope you give me a hint on how to create such a thing. There is an interesting feature about 6-digit binary numbers: if we write them all 64 in one sequence and count how many different numbers we have reading from left to right or from right to left, we'll see that there are exactly 36 variants! So we can "fold", so to say, 8x8 square to 6x6 square. We'll find here 8 symmetrical numbers: 000000 001100 010010 011110 100001 101101 110011 111111 And other 56 will be folded into 28, like 010011 and 110010 represent one "pattern". We can simply make a cube 6x6x6 and fill its columns with all these 36 numbers to read them from top to bottom and vice versa and receive our 64 numbers. Let's assume these 36 numbers "grow" from xOy plane along z axis, so reading in both positive and negative direction of the z axis we see 72 numbers, but while 8 of them are the same, we get 64; or, in other words, we get 2 sets of 36 6-digit numbers, and 8 numbers coincide in these sets, so we have all 64 when we sum these sets. But after seeing this, a question appeared in my mind: can we create such cube with 1 and 0 in every cell in such way, that while reading these 6-digit numbers along other axes - x and y - we see the same effect? I mean getting 2 sets which in sum give us all 64 numbers. I think it can be convenient to use 36 "invariants" along with these original 64 numbers. I found a way for calculating these invariants, so pairs of numbers (straight and mirrored) "fold" into 1 invariant, and there remains 8 symmetrical numbers with 1 invariant for each. Firstly, we must find two "halves" - p and q: p = (n mod 2) * 4 + int((n mod 4) / 2) * 2 + int((n mod 8) / 4), q = int(n / 8). Meaning of them: we read both "halves" from the center of each number, so EG for 010011 they will be: p = 6 (110) and q = 2 (010). For symmetrical number EG 100001 p = q = 4 (100). Then we calculate our invariant: I = max(p, q) * (max(p, q) + 1) / 2 + min(p, q). For mentioned 010011 we have I = 6 * 7 / 2 + 2 = 23. And for its reflection 110010 we have p = 2, q = 6 and I = 6 * 7 / 2 + 2 = 23. For 100001 it will be 4 * 5 / 2 + 4 = 14. Maximum is surely for 111111: 7 * 8 / 2 + 7 = 35, and minimum is 0 for 000000. So if we project numbers read from any side on that side, and then calculate their invariants, we must recieve all 36 of them on this side! And it must be right for all 6 sides of the cube. It is like magic squares really, and if one such thing can be found, it must be very beautiful and deep. A "Crystal", so to say. I tried to approach to this task from sifferent ways. I divided my cube into 8 lesser cubes (3x3x3 each), divided into 2 semi-cubes (3x6x6), or four quarters (3x3x6) and analyzed as I can, but so far found little. EG here are "integral" characteristics of this cube: a) there must be 108 "0" and 108 "1", while we have 216 cells, and thinking about balance. b) dividing our original cube by a plane into 2 parallelepipeds (3x6x6) (or semi-cubes) and examining triplets, there must be 72 of them, and while there are 8 variants of triplets, each triplet must be met 9 times. In one time it will be doubled with itself (so we get a symmetrical number), and 8 times it meets with other triplets. Each semi-cube has 36 triplets, so some must meet 4 times on it, and some - 5 times. c) If we examine 2 "inner" shears of our cube (I mean those which are near our dividing plane, 3rd and 4th if count from any vertex), we'll see that on each must be 18 "1" and 18 "0"; because 10 "folded" numbers have 11 in their middle, another 10 "folded" - 00, and the rest 16 have 10 of 01. So we must assure that we have 10 "11" on these inner shears and 10 "00". Sorry for such a long post, I hope the problem is clear for you. Of course if something is not evident, I'll try to answer your specifying questions. Thanks in advance. wayman Special user England - Sunderland 588 Posts  Posted: Jan 20, 2004 05:42 pm 0 What is its purpose? will it have a use? 0pus Inner circle New Jersey 1739 Posts  Posted: Jan 20, 2004 05:49 pm 0 Well, it frightens me. 0pus Granite Golem New user St.-Petersburg, Russia 3 Posts  Posted: Jan 20, 2004 06:30 pm 0 At least it seems to be very wonderful. Really, this cube is concerned with I Ching - you see, 6-digit binary numbers are hexagrams. There is well-known so-called "traditional sequence" of hexagrams, "folded" into square 6x6. I thought - why it reads only in one direction? Is there a configuration that allows to read all hexagrams along all 3 directions? So this thing I seek... Of what use is it? I don't know now its exact use, but I don't think that such a handsome thing will be of no use at all. I wrote a program to check variants and seek this cube, but I believe I need a super-computer to run my program and get a result quickly... I also think that with mathematics it is easier and quicker to find this cube, but I don't see now strict mathematical ways. The Magic Cafe Forum Index » » Magical equations » » Magical Cube 6x6x6 Filled with "0" and "1" - a problem! (0 Likes)
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