D24 Contest possible?

Magic · Jun 20, 2011

Hi all,

As I was mentionning in my previous message in the It arrives section, I've been working on a special numbering of the D24.
/openfile/250778/photos/photo40611.jpg

It is a numbering that complies to a certain numbers of constraints that could be view as a mathematical puzzle. I was wondering if it was easy for someone from the Shapeways community to find out a solution to this problem. And perhaps a solution different from mine.

That's why I had the idea of creating a contest to challenge the community. The winner would be the first one to find a solution for numbering this D24 that complies to the given constraints, and would win a free print of this die with his own numbering.

In addition to the fact that it is very difficult for me to estimate the complexity of this challenge, I have two issues, and I would appreciate your opinion on them.

The first one is a "Intellectual Property" issue: what if this community member find the same solution as me (or a symmetric solution)? Would it be possible for me to sell my dice with my numbering without the winner arguing that this is *his* numbering? In fact I think I should send my own solution to a someone out of the contest (someone from Shapeways for example), or perhaps print my solution (which I have not done yet). But would it be enough?

The second one is a more practical issue: how do I offer a free copy of a print? Putting a zero markup is easy, but how to offer the print itself? I am ready to pay the material cost but not an extra shipping... If you are not in the same country this seems impossible for now...

Any feedback on those two points would be welcomed.
Thanks,

Magic

virtox · Jun 20, 2011

Cool, I'm game

I have no idea of the complexities involved, but building something to find a solution via genetic/evolutionary/fuzzy/straight algorithms should be fun

Regarding your concerns:
Uploading (as a new model) and printing your solution would at least provide prior art, and contrary to popular believe, that should be enough.

There are several services available for document/date registration.
(in the Netherlands, the tax offices do this)

Or you can send a registered letter to yourself which you don't open.
In case of dispute you let some one official open it.

You could also set up a simple terms and conditions, whereby entrants wave all right to their solution.

Depending on the price of the die (x), you could provide a voucher. Say you buy a voucher for $50, spent $50-x with it yourself and give the remaining code to the winner.

Cheers,

Stijn

Magic · Jun 20, 2011

Issue 1)
Yes, I guess printing one copy of my own numbering for myself would be the solution. The page where you upload it has also the date of uploaded that is clearly indicated. So it should be enough to demonstrate the anteriority...

Issue 2)
Prepaid Coupons... this is brillant!
I could adjust the markup so that the final price is $25: once the material paid, the extra cost would come back to me anyways (well, perhaps minus VAT?).
Actually I had never tried to figure out how Prepaid Coupons work.
Say, I am in France (well, I am in France) and I want to offer one of this coupon at someone in the US. How does that work?
Should I buy it for myself and then send it by regular mail to this person (or e-mail the code to him)? Or is there a way to indicate the mail address of the person you want to offer it to?

mctrivia · Jun 20, 2011

prepaid cupons are nothing more then a 5 digit code which the user types in a checkout. So you can just email them the code.

I do not see the difficulty of numbering the die. But from your photo I can tell you that the D24 I would like to make has a different numbering scheem then yours. Though the purpose is the same as you mentioned mind if I make it?

Magic · Jun 21, 2011

@Mctrivia Ahah! You want to take some advance on your competitors... No problem. I am curious to see if we end up with the same constraints...
On my side, I will model my own numbering tonight and probably order it.

@Virtox: about the complexity: I used an algorithm to number my D18. It was used to enumerate a subset of the possible numberings and checking the wanted conditions.
Say the algorithm took 1 second to solve the D18 problem (it actually tool more than that). To enumerate the solution for a D24, i would take 24x23x22x21x20x19 = 96,909,120 more time: that is more than 3 years. But more clever algorithms like the ones you mentionned may work much faster.

henryseg · Jun 21, 2011

Sounds fun (replying so I get updates on the thread!)

virtox · Jun 21, 2011

Yes that is an option too, but if the markup part is (a lot) bigger than the price of the die, you are essentially circumventing the $25 minimum shipping for Shapeways. (since most money flows back to you)
Hence my suggestion to send the left-overs of a $50 one..

mctrivia · Jun 21, 2011

Here is my version:

All sides of D24 add up to 24 since there is no such time as 2400or worse yet 2459

Magic · Jun 21, 2011

Ah yes, you are right... From 0 to 23 makes more sense for hours...
Well done also for the minutes: I like the D6 numbered from 00 to 50...
I will probably make a D60 numbered from 00 to 59, when I will have more time...

mctrivia · Jun 21, 2011

thought about D60 but it would have to be large and I didn't have the time to model it at the moment.

Magic · Jun 21, 2011

OK. So after designing, uploading and ordering my own solution, I would like to submit you this challenge: how to number my D24 from 1 to 24 so that any group of 4 adjacent faces sum to 50.
To understand better, let's see how the D24 is constructed:
/openfile/250778/photos/photo43234.jpg
In this image you can see a cube, a deltoidal icositetrahedron, and the final D24.
Basically, to construct my D24, I took a cube where each one of the 6 faces has been divided into 4. By moving some vertices I obtain the deltoidal icositetrahedron. And by intersecting this one with a sphere of the appropriate radius, I got the Truncated Sphere D24.

When you subdivide faces of the initial cube into 4, you introduce a first set of 6 group of 4 faces: they must sum to 50. In the drawing, one of these groups is represented in yellow.
But even if the the groups of 4 faces around the edges seem to play a different role in the cube and in the deltoidal icositetrahedron, you can see that in the truncated sphere they are totally equivalent. So I also want the 12 groups of 4 faces around each edge of the original cube to sum to 50. One of this group is represented in red in the drawing.

So here is what I am asking:
Number the 24 faces of my D24 from 1 to 24 (each number must be used only once) so that each one of the 18 groups of 4 adjacent faces (the 6 groups "on faces" and the 12 groups "around edges") sum to 50.
There is no condition on the group of 3 faces around a corner (actually they cannot sum to a constant: can you prove it?).

The first one to give a solution to this problem before July 3rd midnight (GMT) by answering this message, here in the forum, will win a $25 coupon to be exclusively used to buy the D24 with the winning numbering in Antique Bronze Glossy that I will make. The winner allows me to use the winning numbering if I decide to do so.

To help you understanding the puzzle, I attached a Excel file where you can enter a numbering in the first tab and you can see if the constraints are respected in the second tab.

Good luck!

DarrenAbbey · Jun 21, 2011

Is there a known valid solution to the puzzle you've presented, or are you using this contest as a method to find one?

mctrivia · Jun 21, 2011

he said he got one printed for himself. so there is a valid answer.

Does the answer have to comply with dice rules of opposites adding to 25? If so that is a way to cut down half the test points. you only can ignore half the dies numbering as there is no choice in what gets placed there. Reduces combinations to 12! instead of 24! Brute force would take half a day at the low estimate of 10,000 per second

henryseg · Jun 22, 2011

Here is the first solution that popped out of my code. There seem to be hundreds though. This doesn't have opposite sides add up to 25, although that doesn't seem to be a condition of the puzzle.

On the left is the numbers on the front side of the cube, on the right the numbers on the back.

TurtlesAreCool · Jun 22, 2011

I was about to comment that I think your solution isn't a solution, but I see now that may well be. It's as if you sliced a box into the pieces as shown, and the two can be put together without any rotation (I thought the back side had been rotated and was facing me). This makes the 12, 14, 22, 2 an adjacent edge and the 6, 18, 16, 10 an adjacent edge.

If I may ask, can you describe your algorithm for identifying the set of solutions?

henryseg · Jun 22, 2011

Assuming my code is right, there are 35552 solutions, ignoring rotation but counting reflection, or 17776 ignoring rotations and reflections. There don't seem to be any that have opposite faces adding to 25.

The algorithm is a brute force search, but with pruning. It checks to see if a partial list of the face numberings satisfy the conditions on the quadruples that have all four numbers specified, and only continues searching along that path if it does.

The python code is below. It uses a list 'v' of the specified numbers so far. The quadruples list gives the quadruples of indices to look into v corresponding to the faces of the die. I read the quadruples off of the picture attached at the end.

The input to the algorithm makes the assumption that the '1' in the first position, and I think making this assumption kills off all of the rotations and reflections, apart from the reflection that fixes that position.

Code:

quadruples = [[0,1,2,3],#start of front faces
              [0,3,4,5],
              [4,5,6,7],
              [4,7,8,9],
              [8,9,10,11],
              [0,1,8,11],#end of front faces
              [1,2,18,19],#start of connecting faces
              [2,3,21,22],
              [5,6,22,23],
              [6,7,13,14],
              [9,10,14,15],
              [10,11,17,18],#end of connecting faces
              [12,13,14,15],#start of back faces
              [12,15,16,17],
              [16,17,18,19],
              [16,19,20,21],
              [20,21,22,23],
              [12,13,20,23]]#end of back faces

def is_valid_quadruple(v, quadruple):
    sum = 0
    for i in quadruple:
        if i >= len(v): #value unknown as yet
            return True
        else:
            sum += v[i]
    if sum == 50:
        return True
    else:
        return False

def is_valid(v):
    for quadruple in quadruples:
        if not is_valid_quadruple(v,quadruple):
            return False
    return True

output = []

def search(input_v, remaining_numbers):
    for num in remaining_numbers:
        new_v = input_v[:]
        new_v.append(num)
        if is_valid(new_v):
            new_remaining_numbers = remaining_numbers[:]
            new_remaining_numbers.remove(num)
            if len(new_remaining_numbers) == 0:
                print new_v
                output.append(new_v)
            else:
                search(new_v, new_remaining_numbers)
    
search([1],[2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24])

henryseg · Jun 22, 2011

Magic: now your problem is to decide how to choose amongst the thousands of possibilities

The solution I posted above is a bit unsatisfactory in a way: there is a band of pairs of numbers going around the cube:

(1,24),(8,17),(12,13),(14,11),(21,4),(20,5),(15,10),(9,16)

Each pair adds up to 25. Then the remaining two faces of the cube have "orthogonally opposite" numbers adding to 25. This is a nice pattern in a way, but not that symmetric.

Does every solution look like this?

Here are the first 100 solutions, listed in the order given in the previous post. There are obvious patterns of solutions that differ only very slightly.
[1, 8, 17, 24, 9, 16, 10, 15, 19, 7, 2, 22, 4, 5, 20, 21, 11, 14, 12, 13, 23, 3, 6, 18]
[1, 8, 17, 24, 9, 16, 10, 15, 19, 7, 2, 22, 4, 5, 20, 21, 13, 12, 14, 11, 23, 3, 6, 18]
[1, 8, 17, 24, 9, 16, 10, 15, 19, 7, 2, 22, 5, 4, 21, 20, 11, 14, 12, 13, 23, 3, 6, 18]
[1, 8, 17, 24, 9, 16, 10, 15, 19, 7, 2, 22, 5, 4, 21, 20, 13, 12, 14, 11, 23, 3, 6, 18]
[1, 8, 17, 24, 9, 16, 10, 15, 20, 6, 3, 21, 2, 7, 18, 23, 11, 14, 12, 13, 22, 4, 5, 19]
[1, 8, 17, 24, 9, 16, 10, 15, 20, 6, 3, 21, 2, 7, 18, 23, 13, 12, 14, 11, 22, 4, 5, 19]
[1, 8, 17, 24, 9, 16, 10, 15, 20, 6, 3, 21, 7, 2, 23, 18, 11, 14, 12, 13, 22, 4, 5, 19]
[1, 8, 17, 24, 9, 16, 10, 15, 20, 6, 3, 21, 7, 2, 23, 18, 13, 12, 14, 11, 22, 4, 5, 19]
[1, 8, 17, 24, 9, 16, 10, 15, 22, 4, 5, 19, 2, 7, 18, 23, 11, 14, 12, 13, 20, 6, 3, 21]
[1, 8, 17, 24, 9, 16, 10, 15, 22, 4, 5, 19, 2, 7, 18, 23, 13, 12, 14, 11, 20, 6, 3, 21]
[1, 8, 17, 24, 9, 16, 10, 15, 22, 4, 5, 19, 7, 2, 23, 18, 11, 14, 12, 13, 20, 6, 3, 21]
[1, 8, 17, 24, 9, 16, 10, 15, 22, 4, 5, 19, 7, 2, 23, 18, 13, 12, 14, 11, 20, 6, 3, 21]
[1, 8, 17, 24, 9, 16, 10, 15, 23, 3, 6, 18, 4, 5, 20, 21, 11, 14, 12, 13, 19, 7, 2, 22]
[1, 8, 17, 24, 9, 16, 10, 15, 23, 3, 6, 18, 4, 5, 20, 21, 13, 12, 14, 11, 19, 7, 2, 22]
[1, 8, 17, 24, 9, 16, 10, 15, 23, 3, 6, 18, 5, 4, 21, 20, 11, 14, 12, 13, 19, 7, 2, 22]
[1, 8, 17, 24, 9, 16, 10, 15, 23, 3, 6, 18, 5, 4, 21, 20, 13, 12, 14, 11, 19, 7, 2, 22]
[1, 8, 17, 24, 9, 16, 11, 14, 20, 7, 2, 21, 3, 6, 19, 22, 10, 15, 12, 13, 23, 4, 5, 18]
[1, 8, 17, 24, 9, 16, 11, 14, 20, 7, 2, 21, 3, 6, 19, 22, 13, 12, 15, 10, 23, 4, 5, 18]
[1, 8, 17, 24, 9, 16, 11, 14, 20, 7, 2, 21, 6, 3, 22, 19, 10, 15, 12, 13, 23, 4, 5, 18]
[1, 8, 17, 24, 9, 16, 11, 14, 20, 7, 2, 21, 6, 3, 22, 19, 13, 12, 15, 10, 23, 4, 5, 18]
[1, 8, 17, 24, 9, 16, 11, 14, 21, 6, 3, 20, 2, 7, 18, 23, 10, 15, 12, 13, 22, 5, 4, 19]
[1, 8, 17, 24, 9, 16, 11, 14, 21, 6, 3, 20, 2, 7, 18, 23, 13, 12, 15, 10, 22, 5, 4, 19]
[1, 8, 17, 24, 9, 16, 11, 14, 21, 6, 3, 20, 7, 2, 23, 18, 10, 15, 12, 13, 22, 5, 4, 19]
[1, 8, 17, 24, 9, 16, 11, 14, 21, 6, 3, 20, 7, 2, 23, 18, 13, 12, 15, 10, 22, 5, 4, 19]
[1, 8, 17, 24, 9, 16, 11, 14, 22, 5, 4, 19, 2, 7, 18, 23, 10, 15, 12, 13, 21, 6, 3, 20]
[1, 8, 17, 24, 9, 16, 11, 14, 22, 5, 4, 19, 2, 7, 18, 23, 13, 12, 15, 10, 21, 6, 3, 20]
[1, 8, 17, 24, 9, 16, 11, 14, 22, 5, 4, 19, 7, 2, 23, 18, 10, 15, 12, 13, 21, 6, 3, 20]
[1, 8, 17, 24, 9, 16, 11, 14, 22, 5, 4, 19, 7, 2, 23, 18, 13, 12, 15, 10, 21, 6, 3, 20]
[1, 8, 17, 24, 9, 16, 11, 14, 23, 4, 5, 18, 3, 6, 19, 22, 10, 15, 12, 13, 20, 7, 2, 21]
[1, 8, 17, 24, 9, 16, 11, 14, 23, 4, 5, 18, 3, 6, 19, 22, 13, 12, 15, 10, 20, 7, 2, 21]
[1, 8, 17, 24, 9, 16, 11, 14, 23, 4, 5, 18, 6, 3, 22, 19, 10, 15, 12, 13, 20, 7, 2, 21]
[1, 8, 17, 24, 9, 16, 11, 14, 23, 4, 5, 18, 6, 3, 22, 19, 13, 12, 15, 10, 20, 7, 2, 21]
[1, 8, 17, 24, 9, 16, 13, 12, 22, 7, 2, 19, 4, 5, 20, 21, 10, 15, 14, 11, 23, 6, 3, 18]
[1, 8, 17, 24, 9, 16, 13, 12, 22, 7, 2, 19, 4, 5, 20, 21, 11, 14, 15, 10, 23, 6, 3, 18]
[1, 8, 17, 24, 9, 16, 13, 12, 22, 7, 2, 19, 5, 4, 21, 20, 10, 15, 14, 11, 23, 6, 3, 18]
[1, 8, 17, 24, 9, 16, 13, 12, 22, 7, 2, 19, 5, 4, 21, 20, 11, 14, 15, 10, 23, 6, 3, 18]
[1, 8, 17, 24, 9, 16, 13, 12, 23, 6, 3, 18, 4, 5, 20, 21, 10, 15, 14, 11, 22, 7, 2, 19]
[1, 8, 17, 24, 9, 16, 13, 12, 23, 6, 3, 18, 4, 5, 20, 21, 11, 14, 15, 10, 22, 7, 2, 19]
[1, 8, 17, 24, 9, 16, 13, 12, 23, 6, 3, 18, 5, 4, 21, 20, 10, 15, 14, 11, 22, 7, 2, 19]
[1, 8, 17, 24, 9, 16, 13, 12, 23, 6, 3, 18, 5, 4, 21, 20, 11, 14, 15, 10, 22, 7, 2, 19]
[1, 8, 17, 24, 10, 15, 9, 16, 18, 6, 3, 23, 4, 5, 20, 21, 12, 13, 11, 14, 22, 2, 7, 19]
[1, 8, 17, 24, 10, 15, 9, 16, 18, 6, 3, 23, 4, 5, 20, 21, 14, 11, 13, 12, 22, 2, 7, 19]
[1, 8, 17, 24, 10, 15, 9, 16, 18, 6, 3, 23, 5, 4, 21, 20, 12, 13, 11, 14, 22, 2, 7, 19]
[1, 8, 17, 24, 10, 15, 9, 16, 18, 6, 3, 23, 5, 4, 21, 20, 14, 11, 13, 12, 22, 2, 7, 19]
[1, 8, 17, 24, 10, 15, 9, 16, 19, 5, 4, 22, 2, 7, 18, 23, 12, 13, 11, 14, 21, 3, 6, 20]
[1, 8, 17, 24, 10, 15, 9, 16, 19, 5, 4, 22, 2, 7, 18, 23, 14, 11, 13, 12, 21, 3, 6, 20]
[1, 8, 17, 24, 10, 15, 9, 16, 19, 5, 4, 22, 7, 2, 23, 18, 12, 13, 11, 14, 21, 3, 6, 20]
[1, 8, 17, 24, 10, 15, 9, 16, 19, 5, 4, 22, 7, 2, 23, 18, 14, 11, 13, 12, 21, 3, 6, 20]
[1, 8, 17, 24, 10, 15, 9, 16, 21, 3, 6, 20, 2, 7, 18, 23, 12, 13, 11, 14, 19, 5, 4, 22]
[1, 8, 17, 24, 10, 15, 9, 16, 21, 3, 6, 20, 2, 7, 18, 23, 14, 11, 13, 12, 19, 5, 4, 22]
[1, 8, 17, 24, 10, 15, 9, 16, 21, 3, 6, 20, 7, 2, 23, 18, 12, 13, 11, 14, 19, 5, 4, 22]
[1, 8, 17, 24, 10, 15, 9, 16, 21, 3, 6, 20, 7, 2, 23, 18, 14, 11, 13, 12, 19, 5, 4, 22]
[1, 8, 17, 24, 10, 15, 9, 16, 22, 2, 7, 19, 4, 5, 20, 21, 12, 13, 11, 14, 18, 6, 3, 23]
[1, 8, 17, 24, 10, 15, 9, 16, 22, 2, 7, 19, 4, 5, 20, 21, 14, 11, 13, 12, 18, 6, 3, 23]
[1, 8, 17, 24, 10, 15, 9, 16, 22, 2, 7, 19, 5, 4, 21, 20, 12, 13, 11, 14, 18, 6, 3, 23]
[1, 8, 17, 24, 10, 15, 9, 16, 22, 2, 7, 19, 5, 4, 21, 20, 14, 11, 13, 12, 18, 6, 3, 23]
[1, 8, 17, 24, 10, 15, 12, 13, 20, 7, 2, 21, 3, 6, 19, 22, 9, 16, 11, 14, 23, 4, 5, 18]
[1, 8, 17, 24, 10, 15, 12, 13, 20, 7, 2, 21, 3, 6, 19, 22, 14, 11, 16, 9, 23, 4, 5, 18]
[1, 8, 17, 24, 10, 15, 12, 13, 20, 7, 2, 21, 6, 3, 22, 19, 9, 16, 11, 14, 23, 4, 5, 18]
[1, 8, 17, 24, 10, 15, 12, 13, 20, 7, 2, 21, 6, 3, 22, 19, 14, 11, 16, 9, 23, 4, 5, 18]
[1, 8, 17, 24, 10, 15, 12, 13, 21, 6, 3, 20, 2, 7, 18, 23, 9, 16, 11, 14, 22, 5, 4, 19]
[1, 8, 17, 24, 10, 15, 12, 13, 21, 6, 3, 20, 2, 7, 18, 23, 14, 11, 16, 9, 22, 5, 4, 19]
[1, 8, 17, 24, 10, 15, 12, 13, 21, 6, 3, 20, 7, 2, 23, 18, 9, 16, 11, 14, 22, 5, 4, 19]
[1, 8, 17, 24, 10, 15, 12, 13, 21, 6, 3, 20, 7, 2, 23, 18, 14, 11, 16, 9, 22, 5, 4, 19]
[1, 8, 17, 24, 10, 15, 12, 13, 22, 5, 4, 19, 2, 7, 18, 23, 9, 16, 11, 14, 21, 6, 3, 20]
[1, 8, 17, 24, 10, 15, 12, 13, 22, 5, 4, 19, 2, 7, 18, 23, 14, 11, 16, 9, 21, 6, 3, 20]
[1, 8, 17, 24, 10, 15, 12, 13, 22, 5, 4, 19, 7, 2, 23, 18, 9, 16, 11, 14, 21, 6, 3, 20]
[1, 8, 17, 24, 10, 15, 12, 13, 22, 5, 4, 19, 7, 2, 23, 18, 14, 11, 16, 9, 21, 6, 3, 20]
[1, 8, 17, 24, 10, 15, 12, 13, 23, 4, 5, 18, 3, 6, 19, 22, 9, 16, 11, 14, 20, 7, 2, 21]
[1, 8, 17, 24, 10, 15, 12, 13, 23, 4, 5, 18, 3, 6, 19, 22, 14, 11, 16, 9, 20, 7, 2, 21]
[1, 8, 17, 24, 10, 15, 12, 13, 23, 4, 5, 18, 6, 3, 22, 19, 9, 16, 11, 14, 20, 7, 2, 21]
[1, 8, 17, 24, 10, 15, 12, 13, 23, 4, 5, 18, 6, 3, 22, 19, 14, 11, 16, 9, 20, 7, 2, 21]
[1, 8, 17, 24, 10, 15, 14, 11, 22, 7, 2, 19, 4, 5, 20, 21, 9, 16, 13, 12, 23, 6, 3, 18]
[1, 8, 17, 24, 10, 15, 14, 11, 22, 7, 2, 19, 4, 5, 20, 21, 12, 13, 16, 9, 23, 6, 3, 18]
[1, 8, 17, 24, 10, 15, 14, 11, 22, 7, 2, 19, 5, 4, 21, 20, 9, 16, 13, 12, 23, 6, 3, 18]
[1, 8, 17, 24, 10, 15, 14, 11, 22, 7, 2, 19, 5, 4, 21, 20, 12, 13, 16, 9, 23, 6, 3, 18]
[1, 8, 17, 24, 10, 15, 14, 11, 23, 6, 3, 18, 4, 5, 20, 21, 9, 16, 13, 12, 22, 7, 2, 19]
[1, 8, 17, 24, 10, 15, 14, 11, 23, 6, 3, 18, 4, 5, 20, 21, 12, 13, 16, 9, 22, 7, 2, 19]
[1, 8, 17, 24, 10, 15, 14, 11, 23, 6, 3, 18, 5, 4, 21, 20, 9, 16, 13, 12, 22, 7, 2, 19]
[1, 8, 17, 24, 10, 15, 14, 11, 23, 6, 3, 18, 5, 4, 21, 20, 12, 13, 16, 9, 22, 7, 2, 19]
[1, 8, 17, 24, 11, 14, 9, 16, 18, 5, 4, 23, 3, 6, 19, 22, 12, 13, 10, 15, 21, 2, 7, 20]
[1, 8, 17, 24, 11, 14, 9, 16, 18, 5, 4, 23, 3, 6, 19, 22, 15, 10, 13, 12, 21, 2, 7, 20]
[1, 8, 17, 24, 11, 14, 9, 16, 18, 5, 4, 23, 6, 3, 22, 19, 12, 13, 10, 15, 21, 2, 7, 20]
[1, 8, 17, 24, 11, 14, 9, 16, 18, 5, 4, 23, 6, 3, 22, 19, 15, 10, 13, 12, 21, 2, 7, 20]
[1, 8, 17, 24, 11, 14, 9, 16, 19, 4, 5, 22, 2, 7, 18, 23, 12, 13, 10, 15, 20, 3, 6, 21]
[1, 8, 17, 24, 11, 14, 9, 16, 19, 4, 5, 22, 2, 7, 18, 23, 15, 10, 13, 12, 20, 3, 6, 21]
[1, 8, 17, 24, 11, 14, 9, 16, 19, 4, 5, 22, 7, 2, 23, 18, 12, 13, 10, 15, 20, 3, 6, 21]
[1, 8, 17, 24, 11, 14, 9, 16, 19, 4, 5, 22, 7, 2, 23, 18, 15, 10, 13, 12, 20, 3, 6, 21]
[1, 8, 17, 24, 11, 14, 9, 16, 20, 3, 6, 21, 2, 7, 18, 23, 12, 13, 10, 15, 19, 4, 5, 22]
[1, 8, 17, 24, 11, 14, 9, 16, 20, 3, 6, 21, 2, 7, 18, 23, 15, 10, 13, 12, 19, 4, 5, 22]
[1, 8, 17, 24, 11, 14, 9, 16, 20, 3, 6, 21, 7, 2, 23, 18, 12, 13, 10, 15, 19, 4, 5, 22]
[1, 8, 17, 24, 11, 14, 9, 16, 20, 3, 6, 21, 7, 2, 23, 18, 15, 10, 13, 12, 19, 4, 5, 22]
[1, 8, 17, 24, 11, 14, 9, 16, 21, 2, 7, 20, 3, 6, 19, 22, 12, 13, 10, 15, 18, 5, 4, 23]
[1, 8, 17, 24, 11, 14, 9, 16, 21, 2, 7, 20, 3, 6, 19, 22, 15, 10, 13, 12, 18, 5, 4, 23]
[1, 8, 17, 24, 11, 14, 9, 16, 21, 2, 7, 20, 6, 3, 22, 19, 12, 13, 10, 15, 18, 5, 4, 23]
[1, 8, 17, 24, 11, 14, 9, 16, 21, 2, 7, 20, 6, 3, 22, 19, 15, 10, 13, 12, 18, 5, 4, 23]
[1, 8, 17, 24, 11, 14, 12, 13, 19, 7, 2, 22, 4, 5, 20, 21, 9, 16, 10, 15, 23, 3, 6, 18]
[1, 8, 17, 24, 11, 14, 12, 13, 19, 7, 2, 22, 4, 5, 20, 21, 15, 10, 16, 9, 23, 3, 6, 18]
[1, 8, 17, 24, 11, 14, 12, 13, 19, 7, 2, 22, 5, 4, 21, 20, 9, 16, 10, 15, 23, 3, 6, 18]
[1, 8, 17, 24, 11, 14, 12, 13, 19, 7, 2, 22, 5, 4, 21, 20, 15, 10, 16, 9, 23, 3, 6, 18]

Magic · Jun 22, 2011

Wow! Henryseg, I am very impressed by the way you programmed the search. Congratulations! As I found my own solution manually (trials and fails with a good partial solution) I confess I did not even try to write a program because I though it would be too long to reach a valid solution. You demonstrated that I was wrong.

So in effect, there cannot be a solution where all opposite faces sum to 25. The reason is that, because of the constraints, you can find 6 constants so that the couples of faces that form "half-squares" sum to one of this six constant. In the example of Henryseg you can see for example that 8+1 = 7+2 = 6+3 = 5+4 = 9. 9 is one of the constants. Another will be necessarly 50-9, for the two other faces that complete the square. If you impose that opposite faces must sum to 25 then you impose that the constant K and 50-K must be equals. That is all the constants must be equals to 25. And this is only possible if all the numbers are equal, which is excluded. The constants in Henryseg solutions are: 9/41, 24/26 and 25/25.

By the way, you cannot impose that the sum of 3 faces around a corner of the original cube are constant, because this constant would be 300 (the sum of all the numbers from 1 to 24) divided by 8, which is not a whole number.

So Henryseg, congratulations again: if you want, I will design your solution or if you don't mind (since there are thousands of solutions) you can have mine.
By the way the constant in my solution are 13/37, 19/31 and 25/25.
I will send you the prepaid coupon code as soon as I receive it (but by PM

).

Now there are several questions that are still unsolved:
- Can we impose that the sum of the 3 faces around a corner of the original cube are 37 or 38? (or stay as close as possible from these two values)
- Can we have the constants equal to 23/27, 24/26 and 25/25?

(sorry Henry I did not read your latest post yet: we wrote at the same time

, I hope mine answer some questions).

Magic · Jun 22, 2011

OK, now I read your post

.
As I was saying in the previous post, you can try to pick a "more unique" solution by imposing:
- the 6 constants as close as possible to 25
- the sum of the corners as close as possible to 37 or 38

As far as the constants are concerned I think your first solution gives at the contrary the biggest possible range between the constants (9/41 is very different from 25/25).

henryseg · Jun 22, 2011

Let's figure out the "best" solution before printing anything

(Edit: fixed a bug in the code)

There are 1536 solutions with constants 23/27, 24/26,25/25 (I'll write this as 23,24,25 from now on), here's one of them:

[1, 22, 3, 24, 5, 20, 6, 19, 10, 16, 7, 17, 2, 21, 4, 23, 11, 14, 12, 13, 18, 8, 15, 9]

2048 is the largest size, which is (21, 24, 25). The smallest is 16, which both (11, 17, 24) and (13, 18, 21) have.

Here is the full list of possible constants:

(9, 21, 23),
(9, 21, 24),
(9, 21, 25),
(9, 23, 24),
(9, 23, 25),
(9, 24, 25),
(11, 17, 24),
(11, 17, 25),
(11, 18, 23),
(11, 19, 24),
(11, 19, 25),
(11, 23, 24),
(11, 24, 25),
(12, 19, 22),
(12, 19, 25),
(12, 22, 25),
(13, 17, 23),
(13, 17, 24),
(13, 17, 25),
(13, 18, 21),
(13, 18, 22),
(13, 18, 25),
(13, 19, 21),
(13, 19, 22),
(13, 19, 23),
(13, 19, 24),
(13, 19, 25),
(13, 20, 22),
(13, 20, 23),
(13, 20, 25),
(13, 21, 23),
(13, 21, 24),
(13, 21, 25),
(13, 22, 23),
(13, 22, 24),
(13, 22, 25),
(13, 23, 24),
(13, 23, 25),
(13, 24, 25),
(14, 19, 22),
(14, 19, 25),
(14, 22, 25),
(15, 19, 24),
(15, 19, 25),
(15, 20, 23),
(15, 21, 24),
(15, 21, 25),
(15, 23, 24),
(15, 24, 25),
(16, 21, 23),
(16, 21, 25),
(16, 22, 25),
(16, 23, 25),
(17, 21, 23),
(17, 21, 24),
(17, 21, 25),
(17, 23, 24),
(17, 23, 25),
(17, 24, 25),
(18, 21, 23),
(18, 21, 25),
(18, 22, 25),
(18, 23, 25),
(19, 21, 24),
(19, 21, 25),
(19, 22, 23),
(19, 22, 24),
(19, 22, 25),
(19, 23, 24),
(19, 23, 25),
(19, 24, 25),
(20, 22, 25),
(20, 23, 25),
(21, 23, 24),
(21, 23, 25),
(21, 24, 25),
(22, 23, 25),
(22, 24, 25),
(23, 24, 25)

I'll have a look around the corners next.

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