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**mbop****Member**- Registered: 2012-08-19
- Posts: 2

Find the solution for the following: There are 3 hats with 11, 6, and 4 marbles, respectively. On a single move we can remove one marble from one of the hats, one marble from some other hat, and place both removed marbles to the third hat. Can we reach a situation where there are 7 marbles in each hat after some number of moves? If so, show the way. If not, provide some argument why not?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,763

Hi mbop;

Welcome to the forum.

What happens when one hat has all the marbles? Does the game stop and we start again from the initial position? How about when a bag runs out of marbles?

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**mbop****Member**- Registered: 2012-08-19
- Posts: 2

If a hat has all the marbles then you must start again from the initial conditions of 11, 6 and 4.

If a hat becomes empty, you can refill it according to the same rules, but you can't have 'negative' marbles in a hat.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,526

I think itis impossible to get all three to be 7. I'm working on a proof.

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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**Phoolwati****Member**- Registered: 2012-08-20
- Posts: 1

What would be the next nos. in the no. series 1,2,12,12,30,.?.,.?.,

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,526

Hi Phoolwati

Why not make a new thread for your problem?

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: 3,848

Hi mbop,

I think you can't get 7 marbles in each hat from the 11,6,4 position.

I haven't got a nice proof for it...only an exhaustive one done by hand.

Working backwards from the trio 7,7,7, I checked all possible paths but couldn't find any move sequence that reached 11,6,4.

In the following worksheet, each of the 19 groups (which are the only ones possible) has a header trio under which are listed all possibilities (there can't be more than 3) from which the respective headers can be formed.

Starting with 7,7,7, all unique possibilities appear as headers in subsequently-created groups (thus maintaining the thread back to 7,7,7)...which exhausts after 19 groups.

Only the headers that appear in the worksheet connect to 7,7,7 (the moves are easily found via the worksheet)...and 11,6,4 isn't one of them.

So it can't be done.

```
7,7,7 8,8,5 9, 9,3 9,6,6 10,10,1 10,7,4 11,8,2 11,5,5 12, 9,0 12,6,3
----- ----- ------- ------ ------- ------ ------ ------ ------- ------
8,8,5 9,9,3 10,10,1 10,7,4 11, 8,2 11,8,2 12,9,0 12,6,3 13, 7,1 13,7,1
9,6,6 10, 7,4 7,7,7 11,5,5 12,6,3 9,6,6 10,10,1 13,4,4
8,8,5 9,9,3 10,7,4
======================================================================================
13,7,1 13,4,4 14,5,2 15,6,0 15,3,3 16,4,1 17,2,2 18,3,0 19,1,1
------ ------ ------ ------ ------ ------ ------ ------ ------
14,5,2 14,5,2 15,6,0 16,4,1 16,4,1 17,2,2 18,3,0 19,1,1 17,2,2
11,8,2 11,5,5 15,3,3 13,7,1 13,4,4 14,5,2 15,3,3 16,4,1
12,6,3
```

*Last edited by phrontister (2012-08-21 01:07:56)*

"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: 86,763

Hi phontister and anonimnystefy;

I believe it to be impossible. There is an incomplete proof of that on the net.

Hi Phoolwati;

Welcome to the forum. If you tack on to a post as you have done it is very likely

you will never get an answer. Please make a new thread in "Help Me' if you need halp and in "Puzzles and Games" if you already know the answer.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,395

hi Phoolwati

Couldn't find a new thread for your sequence.

I make it 1, 2, 12, 12, 30, 141, 467, 1177, ......

Did I bother to work this out? Must be losing my marbles.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,848

Hi,

I've got a formula for the '21 marbles in 3 hats' problem that with one quick calculation identifies whether or not it is possible for a particular combination to reach 7,7,7 after some number of moves.

For a test combination in hats A, B & C:

1. If (A-C) mod 3 = 0, it is possible.

2. If (A-C) mod 3 ≠ 0, it is impossible.

With 11,6,4 (the puzzle question) it's impossible, as (A-C) mod 3 = 1.

I think the formula works for all cases...but I might be wrong.

*Last edited by phrontister (2012-08-22 21:02:06)*

"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: 86,763

Hi phrontister;

That is an interesting idea. But 9,6,6 is 0 mod 3 and it is not possible.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,848

Hi Bobby,

A big hail storm woke me up!

9,6,6 to 7,7,7 is possible:

9,6,6 --> 8,5,8 and 8,5,8 --> 7,7,7.

9,6,6: A-1=8, B-1=5, C+2=8.

8,5,8: A-1=7, B+2=7, C-1=7.

If you have a look at my post #7 you can trace the path of those moves.

Back to bed.

"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: 86,763

Hi phrontister;

I think I totally misunderstood your point. I am sorry for that, have a good rest.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,848

Hi Bobby,

All rested up now, thanks.

I've altered my post #10 to clarify my point.

The formula isn't a proof, of course, but I composed it from a proof that I worked out. I ran short of time last night and only just quickly posted the formula as I haven't thought of a good way of expressing the proof yet. I'll post it once I clear my thinking about it.

Off to work first, though. Catchya later.

*Last edited by phrontister (2012-08-22 11:52:36)*

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,848

*EDIT: Ignore this post (replaced by post #19 & #20).*

I haven't got around to writing down the proof yet, because another thought that I couldn't leave alone popped into my head and I had to deal with that first (the thought, that is, not my head).

My formula in post #10 is only good for identifying/rejecting groups relating to the 7,7,7 stream, and I got to wondering if I could adapt it to test any group against another to see if they were from the same stream.

This one does it, I think:

For any combinations totalling 21 that are distributed in hats A, B & C, two groups are from the same stream (ie, one can reach the other after some number of moves) if

The two different subscripts denote the two different groups.

Examples:

11,7,3 can never reach 6,7,8 because

13,6,2 can reach 8,7,6 because

Here's a sequence of moves (there are several other options too) to show how that one can be done:

13,6,2 --> 12,5,4 --> 11,4,6 --> 10,6,5 --> 9,5,7 --> 8,7,6.

As for the OP's question, 11,6,4 can never reach 7,7,7 because

The order of the two groups in the formula is irrelevant for this exercise.

*Last edited by phrontister (2012-08-27 01:07:29)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,763

Hi phrontister;

See if I am right with this:

(11,6,4)->(10,5,6)->(9,7,5)->(8,6,7)->(7,8,6)->(6,7,8)

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,848

*EDIT: Ignore this post (replaced by post #19 & #20).*

Oops! I got lost in my messy scribblings and typed in 11,6,4 instead of 11,7,3. Sorry about that.

I'd mistakenly said that

whereas the result of the mod is 0...which, according to my formula, is why your moves worked out.

I've corrected my post #15.

And I think you'll find that 11,7,3 can't reach 6,7,8...according to my formula result, anyway:

*Last edited by phrontister (2012-08-27 01:07:48)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,763

Hi phrontister;

Looks pretty good now!

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,848

Hi Bobby,

Posts #15 to #18 are now redundant.

When working on my proof (see my next post) I discovered that the simple '(A-C) Mod 3' applies to all cases...which I mentioned near the end of my proof.

Should we delete those posts (including this one) to unclutter the thread?

*Last edited by phrontister (2012-08-27 00:52:08)*

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,848

Ok. Here's my proof. I hope I can get the idea across.

There are three hats A, B & C (always in that order).

At each move the hats' contents increase/decrease by the amounts from one of the following three (the maximum number possible) options:

Hats A | B | C

------------------

Option1 -1 | -1 | +2

Option2 -1 | +2 | -1

Option3 +2 | -1 | -1

The sum of the difference (ie, not absolute, but including the sign) between the three possible hat pairs AB, AC & BC can be used to determine connectivity between different groups of hat contents (eg, the puzzle's groups 11,6,4 and 7,7,7).

Those sums are (A-B)+(A-C)+(B-C), resulting in:

Option1: (-1-(-1))+(-1-2)+(-1-(-1)) = -6

Option2: (-1-2)+(-1-(-1))+(2-(-1)) = 0

Option3: (2-(-1))+(2-(-1))+(-1-(-1)) = 6

From this we can get the following formula that is common to all three Options (it has a common result: "0", in this case):

When applying that formula to different groups of hat contents, the results, which are either "0", "1" or "2", must be identical for those groups to be in the same stream. Any groups that satisfy that condition can reach each other after some number of moves.

Applying this formula to the two puzzle groups yields the following different results:

"0" for 7,7,7 and "1" for 11,6,4...which means that 11,6,4 can't reach 7,7,7 as the formula results aren't identical.

Hope that makes sense!

*Last edited by phrontister (2012-08-27 11:36:09)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,763

Hi phrontister;

If it is incorrect, I can not see where. I have been unable to find a counter example so congratulations

on the first proof.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,848

Thanks, Bobby!

Tricky little puzzle, that was. Took me ages to find something to hang my hat on!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,763

Hi phrontister;

It is very clever. Thanks for providing it.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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