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**MathGuise****Banned**- Registered: 2009-06-25
- Posts: 6

Here is a Math problem I recently devised! It is called The 46656 Math Problem as a result from the Calculation 6 x 6 x 6 x 6 x 6 x 6 = 46656

The Problem involves two Players, Player A and Player B.

Player A selects Six Numbers,the Numbers can be any number from 1 to 6 for each of the Six Numbers,an example could be 123456 another example 333333 or another example 223336 etc. and then writes them down on paper,without Player B seeing the Six Hidden Numbers.

Player B then has to try and Guess the Six Hidden Numbers,using the least amount of Guesses possible,( One Guess would be Player B selecting any Six Numbers as a Group! Eg 112255 ) the only clue's Player A can give Player B as to what the Six Hidden Numbers might be,are if Player B Guesses a Number correct, then Player A will say One Number is correct,without saying which Number it is from the Six Hidden Numbers,and of course if Player B Guesses more correct then Player A will say how many are correct,but again not which Numbers are correct,Player B has to Guess all Six Hidden Numbers in the correct order!.

The interesting thing about this Problem is... Is there a Shortest way possible for Player B to Guess all Six Hidden Numbers in the correct order!

As an example if Player B was to make his first Guess 111111 and Player A had the Six Hidden Numbers as 221666 then Player A would say One Number correct,which in one way is good for Player B because he/She now Knows Player A only has One Number 1 as part of their Six Hidden Numbers,the hard part is working the correct order for all Six Numbers!

A better result of the first Guess by Player B would be if Player A said none correct! Because this now reduces the Problem down to 5 x 5 x 5 x 5 x 5 x 5 = 15625 possible ways! A massive reduction of 31031 But! Is this still the best way to start trying to solve the Problem?

I would just like to say a similar Game to this was once produced called MasterMind I am not sure if it is still made? The main difference with my version is that Player B is given less information as to what the correct order might be for the Six Hidden Numbers!

Any thoughts on this Problem will be much appreciated as I have been looking at it for a while now! And have not yet found what I think is the absolute Shortest Solution!...

Important!...Some more information concerning the Rules for this Problem / Game!

If Player A selects the Numbers 123456 as an example for their Six Hidden Numbers and Player B Guesses 654321 then the reply from Player A would be None correct!

This is just to make it clear that the idea is for Player B to Guess all of the Six Hidden Numbers in the correct order.

*Last edited by MathGuise (2009-08-07 00:16:23)*

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

There was a game like this, I believe it was called "Black box". The only difference is that it would tell you how many you got right (i.e. in the right place as well), and how many numbers you guessed but were in the wrong place.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**TheDude****Member**- Registered: 2007-10-23
- Posts: 361

First of all, I think you need to be more specific when you say "shortest solution". I would guess, although I'm not sure, that the algorithm with the best worst-case performance would be different from the algorithm with the best average performance.

Wrap it in bacon

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 105,671

Hi MathGuise;

You haven't mentioned what your methods are getting for an average number of guesses. Are you beating an average of 21 guesses?

Don't hold me to this because I have only done some tentative paper math (could be a hallucination) but it might be possible to do it in 12.

*Last edited by bobbym (2009-08-06 10:24:07)*

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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**MathGuise****Banned**- Registered: 2009-06-25
- Posts: 6

To Ricky

The Black box" Game you are referring to sounds exactly like the Game I mentioned called MasterMind I am not sure which one came first as the original!

I just found out about MasterMind that...The modern game with pegs was invented in 1970 by Mordecai Meirowitz, an Israeli postmaster and telecommunications expert, but the game resembles an earlier pencil and paper game called bulls and cows that may date back a century or more.

And this about Blackbox

Black Box is a game of "hide and seek" for one or two players, which simulates shooting rays into a black box to deduce the locations of atoms hidden inside. It was created by Eric Solomon. The board game was published by Waddingtons from the mid 1970s and by Parker Brothers in the late 1970s.

Black Box is played on a two-dimensional grid. The object of the game is to discover the location of objects ("atoms", represented by metal balls in the Waddingtons game and by yellow balls in the Parker Brothers version) hidden within the grid, by the use of the minimum number of probes ("rays"). The atoms are hidden by a person in a two-player game.

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To TheDude

The shortest solution" would be the most Efficient way for Player B to Guess the Six Hidden Numbers

as in my example in one Guess the Problem could be reduced very Efficiently from 6 x 6 x 6 x 6 x 6 x 6 = 46656 to 5 x 5 x 5 x 5 x 5 x 5 = 15625 possible ways!

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To bobbym

Solving the problem in 21 is Good! 12 is of course much better!?

I am keeping my best method silent at the moment..as I am seeing if any one else will find the same fast Efficient way ( maybe not the Best? )

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**TheDude****Member**- Registered: 2007-10-23
- Posts: 361

MathGuise wrote:

To TheDude

The shortest solution" would be the most Efficient way for Player B to Guess the Six Hidden Numbers

as in my example in one Guess the Problem could be reduced very Efficiently from 6 x 6 x 6 x 6 x 6 x 6 = 46656 to 5 x 5 x 5 x 5 x 5 x 5 = 15625 possible ways!

That still doesn't answer my question. Let's say you have 2 different solutions to this problem. Out of the 46556 possible combinations, on average Solution A will take 15 steps. However, in the worst case scenario it will take 30 steps to finally get the answer. Solution B's worst case scenario takes only 25 steps, but on average it takes 20 steps to get the answer. Which solution would you consider to be the shortest or most efficient?

Wrap it in bacon

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

I think I can get a worst case of 15 guesses.

I'll show a game where the answer is 654321, since I believe that's the combination that this method is worst at.

First guess 5 repdigits to see how many of each digit is in the combination.

111111 (1)

222222 (1)

333333 (1)

444444 (1)

555555 (1)

Here we see that there are one of each of 1, 2, 3, 4, 5, and so there must also be one 6.

To get the first digit, we try:

122222 (1)

Note that if the first digit was a 1, then the 2 would be found somewhere else and so this guess would score 2. Also, if the first digit is 2, then this guess would score 0. In all other cases it scores 1.

We make the next two guesses using similar logic:

344444 (1)

566666 (0)

The first of these guesses means that the first digit isn't 3 or 4, and the second shows that it is a 6.

Continuing to find the second digit:

612222 (2)

634444 (2)

Guess #1 eliminates 1 and 2, Guess #2 eliminates 3 and 4, so 5 is all the second digit can be.

Now we do the same thing to find the rest of the digits.

651222 (3)

653444 (2)

654122 (4)

654312 (4)

By this point we know what the combination is, so depending on the rules, we might be allowed to finish here.

If not though, here's the fifteenth and final guess:

654321 (6)

If there are more than one of some digit, that doesn't cause this method any problems. It actually gets easier then, because that means that there are some digits that don't appear and so we have less things to guess for each position.

Why did the vector cross the road?

It wanted to be normal.

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**MathGuise****Banned**- Registered: 2009-06-25
- Posts: 6

How Can We Score The Best Most Efficient Method For Solving The 46656 Math Problem

One Part of the Scoring could be to use Standard Deviation! If we find that the Most Efficient Average is (15)

after many Test's, that is to Guess all Six of the Hidden Numbers in the correct order,from sample Test's.

Below is an example of three Players X,Y and Z who have competed against each other ten times,all three Score the same Total and Mean Average,but Z Wins because he/she has a Zero Standard Deviation Score!

It would be obvious that X had some luck with a Score of 10 in one Game,and even though Y Scored with a 13,14, and 14 they also had some high Scores of 17 and 17.

Is this a fair Scoring Method or are there better ones?

X Y Z

1 16 15 15

2 16 17 15

3 15 17 15

4 10 13 15

5 16 14 15

6 16 14 15

7 15 15 15

8 15 15 15

9 16 15 15

10 15 15 15

sum 150 150 150

Mean Average 15 15 15

Standard

Deviation 1.7320 1.1832 0

*Last edited by MathGuise (2009-08-13 09:06:07)*

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**quittyqat****Member**- Registered: 2009-04-08
- Posts: 1,213

There is a game called Vault Breaker on my brother's iPod Touch, but:

1. You have to set the difficulty.

2. The numbers are allowed to be digits 1-9 instead of 1-6.

3. It tells you how many you got right but in the wrong places, but not which ones.

4. It is easier on the easiest level and harder on the hardest.

I'll be here at least once every month. XP

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**MathGuise****Banned**- Registered: 2009-06-25
- Posts: 6

The problem with only using Standard Deviation to Score which player has the most Efficient method,for solving this problem over a series of 10 Test's etc. ,is that if another Player let's say Player Z2 was to Score 14,14,14,14,14,14,14,14,14,14 then they would have the same Standard Deviation Score as the Best Player Z equal to zero! But Z2 would have a Total of 140 and a mean Average of 14 making Player Z2 the Best Player with the most Efficient method.

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