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**cubit****Member**- Registered: 2013-05-05
- Posts: 5

Not sure if this is an existing puzzle, my basic internet search has not found anything similar.

I came up with this puzzle while sitting in boring meeting at work!

Take a cube and divide each face into 3x3 squares.

On each cube face enter the numbers 1 - 9 in the squares without repeating such that the sum of all squares which share an edge or corner around the cube is equal.

Hint - there are 8 corners with 3 squares to be summed and 12 edges with 2 squares to be summed - all must be equal.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,811

Hi cubit,

I think I've got it (see image).

The cube is spread-eagled, with the bottom 3x3 square being the cube's base. Colour coding connects cells (note the alliteration) that share an edge or corner.

Nice puzzle!

And I couldn't think of anything better to do while sitting in a boring meeting at work either, than composing this puzzle or solving it.

*Last edited by phrontister (2013-05-07 15:21:45)*

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**cubit****Member**- Registered: 2013-05-05
- Posts: 5

phrontister wins!

Mathematically there should be solutions summing to 11, 12 and 13 - although I have done simulations with 11 and 12 my patience ran out trying to do the 13.

And I would guess there are many combinations of solutions.

Here is the challenging questions (which I dont know the answers) -

1. What are the number of solutions?

2. What are the minimal number of "fixed/seed" squares (squares which have already a number filled in and cannot be changed) that will guarantee only a single solution?

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,811

Hi cubit,

Sorry, but I have no idea how to answer those two challenges.

I managed to find a solution for sums=13, though (see image).

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,727

1. What are the number of solutions?

This has so far proved to be a very difficult question to answer. I am estimating that there is less than 1 chance in 2562890625 that a random cube has those properties.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,811

No wonder it took me a while! I'm so glad then that I got the solution on my last attempt before getting to the 2562890625th!

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,727

Hi;

There are less than 890939317237438025973104 solutions which is a very large number. I just do not know how many less.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,811

Hi cubit,

Mathematically there should be solutions summing to 11, 12 and 13

Yes, I see that...and they're the only ones.

Here's a solution for sums=11.

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**cubit****Member**- Registered: 2013-05-05
- Posts: 5

bobbym wrote:

Hi;

There are less than 890939317237438025973104 solutions which is a very large number. I just do not know how many less.

I am a little rusty, but I believe that each cube face can be arranged in 9! combinations, therefore there should be 9!^6 possible combinations across the entire cube.

This indeed is a very large number.

But since there are limitations to only summing to 11, 12 or 13 the number of combinations should be something much less.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,727

Yes there is (9!)^6 total possibilites but simulations suggest that number should be divided by at least 15^8, hence the smaller number of possible solutions. This is just a crude estimate based on a small pattern.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,811

cubit wrote:

But since there are limitations to only summing to 11, 12 or 13 the number of combinations should be something much less.

I used the following logic to seed some squares and reduce the options/combinations:

8 corners of 3 squares + 12 edges of 2 squares = 20 groups of similar sums.

Each face has the numbers 1 to 9, whose sum is 45.

The sum of all numbers on the cube is 45 x 6 = 270.

**For sum of edge and corner groups = 11**:-

Sum of edge and corner groups = 20 x 11 = 220

Sum of centres = Total sum - groups sum = 270 - 220 = 50

The six centre squares are {5,9,9,9,9,9}, {6,8,9,9,9,9}, {7,7,9,9,9,9}, {7,8,8,9,9,9} or {8,8,8,8,9,9}.

Ones not on edges (won't sum to 11).

**For sum of edge and corner groups = 12**:-

Sum of edge and corner groups = 20 x 12 = 240

Sum of centres = Total sum - groups sum = 270 - 240 = 30

The centre squares have many different combinations of six numbers from 1 to 9 (plus duplicates).

Ones and twos not on edges (won't sum to 12).

**For sum of edge and corner groups = 13**:-

Sum of edge and corner groups = 20 x 13 = 260

Sum of centres = Total sum - groups sum = 270 - 260 = 10

The six centre squares are {1,1,1,1,1,5}, {1,1,1,1,2,4}, {1,1,1,1,3,3}, {1,1,1,2,2,3} or {1,1,2,2,2,2,}.

Ones, twos and threes not on edges (won't sum to 13).

*Last edited by phrontister (2013-05-08 13:48:32)*

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,811

I successfully tested four scenarios where sums = 13, with results as per images.

The centre-square groups for this sum can only be {1,1,1,1,1,5}, {1,1,1,1,2,4}, {1,1,1,1,3,3}, {1,1,1,2,2,3} or {1,1,2,2,2,2,}, and my tests were:

- one from the {1,1,1,1,3,3} group (which is the image from post #4), and

- three from the {1,1,2,2,2,2,} group.

Testing covered:

1. Multiple solutions for sums = 13.

2. Different centre-square groups.

3. Same centre-square groups, with different layouts.

4. Same centre-square groups, with identical layouts.

I use an Excel spreadsheet to automatically:

- sum corners and edges;

- count each number's frequency;

- highlight numbers on a face that appear more than once.

That simplifies and speeds up the chasing of numbers around the grid in search of a solution.

*Last edited by phrontister (2013-05-08 13:47:48)*

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**cubit****Member**- Registered: 2013-05-05
- Posts: 5

Lots of great input, phrontister!

You must have a lot of boring meetings! ;-)

You are nearing the first challenge, I think.

What you have identified so far was very similar to what I did (determining the solutions are only 11,12 and 13; using excel to auto-sum the edges/corners; color-coding).

Although I hate to admit that I dont know how to attach images in these posts so I didnt provide the sample before.

Even if we dont get the real answer, suffice to say there are a lot of solutions (even though it does take some time to find them).

For the second challenge, I am interested to see if there is some minimal "seed" numbers that can guarantee the unique solution.

Maybe starting from the winning solution then "unfreezing" numbers until more than one solution can be created is one way to do it?

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,845

phrontister wrote:

For sum of edge and corner groups = 13:-

Sum of edge and corner groups = 20 x 13 = 260

Sum of centres = Total sum - groups sum = 270 - 260 = 10

The six centre squares are {1,1,1,1,3,3}, {1,1,1,2,2,3} or {1,1,2,2,2,2,}.

Ones, twos and threes not on edges (won't sum to 13).

Why not {1,1,1,1,1,5} and {1,1,1,1,2,4}?

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,811

anonimnystefy wrote:

Why not {1,1,1,1,1,5} and {1,1,1,1,2,4}?

Oops! Don't know how I missed those. Fixed now. Thanks!

Here is a link to a video of a search for a new solution from a solved position (Excel spreadsheet).

I initially uploaded this to youtube, whose transcoding killed the quality (as vimeo also did, but to a much lesser degree). So now I've used photobucket, which doesn't appear to have transcoded it at all and playback is as sharp as the original (I think). Just click on the fullscreen icon next to the loudspeaker icon to view without ads.

*Last edited by phrontister (2013-05-09 01:53:47)*

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,811

Hi cubit,

I'm still no closer to solving those two challenges and I think I'll let them be for now. Maybe if my enthusiasm batteries recharge I'll try again.

But it's been fun so far.

*Last edited by phrontister (2013-05-09 22:30:11)*

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**cubit****Member**- Registered: 2013-05-05
- Posts: 5

Good efforts!

If you regain your energy, please keep us posted!

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