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## #1 2010-07-06 10:04:42

timinator
Member
Registered: 2010-07-06
Posts: 2

First off, this is not for school or for marks but is very important to a personal project.

I'm trying to find a general formula to solve the following problem with any numbers

Problem:

I have a 3 slot machine that,when spun, can land on 10 different objects(a,b,c,d,e,f,g,h,i,j)

This slot machine has 5 different levers. And by inspection i have figured out how much each lever spins each slot. [For example lever 1 spins slot 1 from object a to object f (changes it 5 times), slot 2 from object g to object i(2 spots), and slot three from object c to a(8spots)]

Each lever does something different but i know what each one does.

I need to find what levers i should pull, and in what order to get 3 of the object of my choice. [for example do i pull lever 1 twice and lever 3 once to get (c c c)]

I don't have specific numbers because i need to be able to use this for any number.

Thanks

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## #2 2010-07-06 11:06:03

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Hi timinator;

Your info seems to be based on moduli. Provided you know what each lever does for each slot you should be able to predict where it will land next.

for example do i pull lever 1 twice and lever 3 once to get (c c c)

Unfortunately you have not provided me with how lever 3 works in relation to the 3 slots. Therefore I can't say. You also have not made it clear what position they start in.

Describe to me in full how everything works. Then prediction will be possible.

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #3 2010-07-06 11:32:54

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

Depending on what exactly the levers do, it might not always be possible to get the slots to do what you want. (Trivial example: If all 5 levers move every slot 0 places, you can't get anything to happen)

But in general, the strategy would be to try to work out combinations of levers that only affect one slot. The ultimate goal should be to get a combination that moves the first slot one place, and 2 more combinations that do the same for the second and third slot.

I'll use my calculator to make some random lever functions and see what I can do with them.

A: (2,5,0)
B: (7,6,7)
C: (9,8,7)
D: (4,9,5)
E: (3,6,4)

A looks like it could be useful, since it doesn't affect the third slot.
We should be looking for sets that add up to 10 for one of the slots, since that would mean that combination doesn't change that slot.

For example, pulling B and E would leave the first slot alone, and pulling D twice wouldn't change the third.

Eventually we'll find something useful.

Here, pulling A once and D twice has a final result of (0,3,0).
It's good that the second slot is moved 3 places, because that means that repeating this combination can get us any position we want on the second slot.

We can now effectively ignore the second slot in our calculations, since we have full power over it anyway.

Here's the important stuff:

A: (2,0)
B: (7,7)
C: (9,7)
D: (4,5)
E: (3,4)

A only affects one slot, which is good, but we can only get half of the positions from it, so more work is needed.

Bingo! B and E give a result of (0,1), which is exactly what we're looking for.
Now that we can ignore the third slot as well, the last part is easy. For example, B three times moves the first slot once.

We've now shown theoretically that any position can be made, and to find how to get to a specific one, you backtrack through the working we've just done.

Let's say the slots start at (a,a,a), and you want them to get to (f,c,j).
In number form, this is equivalent to (5,2,9).

Since we solved it last, we'll get the first slot to the right position first.
According to the calculations, this is done by pulling B 3*5 = 15 times.
However, pulling any lever 10 times will result in nothing, so we can actually pull B just 5 times if we want.

The slots now sit at (f,a,f). Next we sort out the third.
We want to move it four times, so we do (BE)x4.

This results in (f,i,j).

Finally we need to move the second slot forward four times.

So an overall command for (a,a,a) -> (f,c,j) would be A6B9D2E4.

Note that there may be (and probably are) ways to do this involving less lever pulls.
To try to improve it, I'd try to find combinations of levers that affect nothing overall. Then you can add or take away those combinations to a solution to get less overall pulls.

Why did the vector cross the road?
It wanted to be normal.

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## #4 2010-07-06 11:41:58

timinator
Member
Registered: 2010-07-06
Posts: 2