Moola Gold Rush Solver

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Views: 3688 | Replies: 4

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Join Date: Jun 2007

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justmonkey23 justmonkey23 is offline

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Moola Gold Rush Solver

Sep 9th, 2007

I am trying to write a program that will help me with Gold Rush...This is a game on moola.com. Gold Rush is the most popular game available for players to wager their winnings on. It involves a series of six blind bids on gold nuggets with various point values. The player with the most points at the end wins the game, and all of the cash wagered. I have created a program that evaluates a game and gives out a winner...but the game needs to be input with the form of 5,3,1,2,4,6,5,6,2,1,3,4,1,6,5,3,2,4 the first six are the gold I play, the middle six are the gold in the being wagered for and the last six are the gold that the other player chooses...so basically my question is what line(s) of code would give me every possible game...like
1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6
1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,6,5 etc

there are 373 million possible games I believe

Any ideas would be great

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Moola Gold Rush Solver

Sep 9th, 2007

Or does anyone know how to create a list of all variations of a six digit number that has a single digit range from 1...6 and uses one digit per field...I bet this is very unclear

This is the form
1,2,3,4,5,6
6,5,4,3,2,1
1,3,5,2,4,6
2,4,6,1,3,5

6! would lead me to believe there are 320 variations...

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Join Date: Jul 2006

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Re: Moola Gold Rush Solver

Sep 10th, 2007

No, it makes sense.

Try this:

 Help with Code Tags 
 Python Syntax (Toggle Plain Text) 
def perms(s):
    if len(s) == 0:
        return []
    elif len(s) == 1:
        return [s]
    retval = []
    for x in s:
        base = s[:]
        base.remove(x)
        for tmp in perms(base):
            tmp = [x] + tmp
            retval.append(tmp)
    return retval
 
print perms([])
print perms([1])
print perms(range(1,3))
print perms(range(1,4))
def perms(s):
    if len(s) == 0:
        return []
    elif len(s) == 1:
        return [s]
    retval = []
    for x in s:
        base = s[:]
        base.remove(x)
        for tmp in perms(base):
            tmp = [x] + tmp
            retval.append(tmp)
    return retval

print perms([])
print perms([1])
print perms(range(1,3))
print perms(range(1,4))

Jeff

Last edited by jrcagle : Sep 10th, 2007 at 11:00 pm.

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Re: Moola Gold Rush Solver

Sep 10th, 2007

Wow thanks that was what I needed...Could you do me another favor, please explain what each line means?

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Re: Moola Gold Rush Solver

Sep 11th, 2007

Well, the basic algorithm is this:

Help with Code Tags

pseudo Syntax (Toggle Plain Text)

Run through the set, taking each element in turn.
For each element x, take the *rest* of the set and compute the perms of that subset.
Add element x to each of the sets in the perms.
Return the union of those results.

So for s = [1,2,3], we would have

1 --> compute perms of [2,3]; add to each: [1,2,3], [1,3,2]
2 --> compute perms of [1,3]; add to each: [2,1,3], [2,3,1]
3 --> compute perms of [1,2]; add to each: [3,1,2], [3,2,1]

Then we "add" (really, take the union) of each set of results above.

The first four lines compute the base cases. The rest of the lines perform the recursive algorithm above.

Hope it helps,
Jeff