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Topic: Marbles trading game (Read 785 times) |
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wonderful
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Marbles trading game
« on: Apr 7th, 2008, 2:20pm » |
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There are marbles in three different colors: blue (B), red (R), and yellow (Y). The trading rules as follows:
a) B= R + 2 Y
Meaning 1 blue marble can trade for 1 red plus 2 yellow marbles and vice versa.
b)2B+2 R=Y
c)Y+B= R
e)Y+ R=B
f)R + B = Y
In the beginning there are unlimited number of marbles in three colors. You have 1 blue marble.
How can you trade to finnaly get 6R, 8B, and 7Y?
Enjoy.
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Hippo
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Re: Marbles trading game
« Reply #1 on: Apr 7th, 2008, 2:51pm » |
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#R+#B mod 2
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| « Last Edit: Apr 7th, 2008, 2:52pm by Hippo » |
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wonderful
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Re: Marbles trading game
« Reply #2 on: Apr 7th, 2008, 5:21pm » |
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Hippo, your solution seems diffrent from the intended solution. I'm not sure if I understand it correctly. Can you explain a bit more?
Have A Great Day!
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towr
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Re: Marbles trading game
« Reply #3 on: Apr 8th, 2008, 12:06am » |
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on Apr 7th, 2008, 5:21pm, wonderful wrote:Hippo, your solution seems diffrent from the intended solution. I'm not sure if I understand it correctly. Can you explain a bit more?
Have A Great Day! |
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If you add the number of blues and reds, this is odd at the start, each transaction preserves this oddness, but the end state you want to achieve has an even total, hence it cannot be achieved.
In short, the problem is that #R+#B mod 2 is invariant.
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Grimbal
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Re: Marbles trading game
« Reply #4 on: Apr 8th, 2008, 12:55am » |
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Or can you trade half-marbles?
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wonderful
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Re: Marbles trading game
« Reply #5 on: Apr 8th, 2008, 12:56am » |
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Thanks Towr for pointing that out. Hippo's observation is accurate. However, the trading rules b and f allows 1 Y = 2 Y  . That makes the difference.
Have A Great Day!
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towr
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Re: Marbles trading game
« Reply #6 on: Apr 8th, 2008, 2:17am » |
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on Apr 8th, 2008, 12:56am, wonderful wrote:| However, the trading rules b and f allows 1 Y = 2 Y . That makes the difference. |
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I'm quite sure it doesn't. Yellow is irrelevant if you can't get the correct number of the other two.
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| « Last Edit: Apr 8th, 2008, 2:17am by towr » |
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wonderful
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Re: Marbles trading game
« Reply #7 on: Apr 8th, 2008, 3:55pm » |
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That's true Towr. However, from that we can get 1 B for 2 B which relates to Grimbal's point.
Have A Great Day!
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| « Last Edit: Apr 8th, 2008, 4:00pm by wonderful » |
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towr
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Re: Marbles trading game
« Reply #8 on: Apr 9th, 2008, 12:00am » |
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If you can trade half-marbles then a) doesn't mean what it says it means; you stop following the trading rules.
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wonderful
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Re: Marbles trading game
« Reply #9 on: Apr 9th, 2008, 12:14am » |
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Good point Towr. Can you elaborate a bit more? The key point, according to the person who makes this question, is that the rules are designed in such a way that 1 B can trade for 2 B. If one can design a trading scheme to achieve this, the solution can be found.
Have A Great Day!
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towr
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Re: Marbles trading game
« Reply #10 on: Apr 9th, 2008, 1:25am » |
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Well, to me it appears that the rules work on integral marbles and should be applied integrally. So even if you could get something like 2B=4B, you can't divide it in half and say 1B=2B; that would be a different rule. There isn't a way to achieve it by applying the rules as wholes. It's like when in the supermarket there's a two for one special, but you can't buy one at half price.
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wonderful
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Re: Marbles trading game
« Reply #11 on: Apr 9th, 2008, 3:11am » |
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Thanks Towr. I really like your analogy of "buy 1 get 1 free" in the real market. I will post the solution soon so we can discuss more. In the mean time, does anyone have any idea?
Have A Great Day!
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SMQ
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Re: Marbles trading game
« Reply #12 on: Apr 9th, 2008, 5:44am » |
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- Apply rule a)
B --> R+2Y
- Apply rule b) in reverse
R+2Y --> 2B+3R+Y
- Apply rule c) in reverse
2B+3R+Y --> 3B+2R+2Y
- Repeat the above three more times
3B+2R+2Y --> ... --> 9B+8R+8Y
- Throw out 2R+Y 
Also, what's rule d) ?
--SMQ
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Hippo
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Re: Marbles trading game
« Reply #14 on: Apr 9th, 2008, 12:52pm » |
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on Apr 9th, 2008, 5:44am, SMQ wrote:- Apply rule a)
B --> R+2Y
- Apply rule b) in reverse
R+2Y --> 2B+3R+Y
- Apply rule c) in reverse
2B+3R+Y --> 3B+2R+2Y
- Repeat the above three more times
3B+2R+2Y --> ... --> 9B+8R+8Y
- Throw out 2R+Y
Also, what's rule d) ?
--SMQ |
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This does not work ... You got 9B+6R+7Y. But if you start with e) instead of a) and throw out 2R+B ... it will work
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| « Last Edit: Apr 9th, 2008, 12:53pm by Hippo » |
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wonderful
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Re: Marbles trading game
« Reply #15 on: Apr 10th, 2008, 3:34am » |
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Interesting guys. Can you trade in such a way to receive exactly: 6 R + 8 B + 7 Y without any waste?
Have A Great Day!
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Grimbal
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Re: Marbles trading game
« Reply #16 on: Apr 10th, 2008, 7:31am » |
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B R Y
1 0 0
a) B->R+2Y
0 1 2
b) Y->2B+2R, 2 times
2 3 1
4 5 0
c) R->Y+B, 4 times
5 4 1
6 3 2
7 2 3
8 1 4
a) B->R+2Y, 2 times
7 2 6
6 3 8
b) Y->2B+2R
8 5 7
e) B->Y+R, 1/2 times
7.5 5.5 7.5
f) Y->R+B, 1/2 times
8 6 7
As an alternative for the last 2 steps, if you can exchange
B for Y + R (e)
and
Y for R + B (f)
then certainly you can exchange the sum
B + Y for Y + 2R + B.
You can scale that down to
1/2 B + 1/2 Y for 1/2 B + 1/2 Y + R
which you can do without breaking any marble.
So you can go from 8B+5R+7Y to 8B+6R+7Y.
Note: It is not the same as trading nothing for one R, because it requires 1/2 B and 1/2 Y as "catalyst".
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| « Last Edit: Apr 11th, 2008, 1:02am by Grimbal » |
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wonderful
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Re: Marbles trading game
« Reply #17 on: Apr 10th, 2008, 4:35pm » |
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Thanks Grimbal. Actually, the question asks for
6 R + 8 B + 7 Y. Your approach is in the right direction, though. Actually, if we can find a way to trade 1 B for 2 B and vice versa then the question is solved. Let's continue!
Have A Great Day!
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| « Last Edit: Apr 10th, 2008, 5:26pm by wonderful » |
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Grimbal
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Re: Marbles trading game
« Reply #18 on: Apr 11th, 2008, 12:54am » |
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on Apr 10th, 2008, 4:35pm, wonderful wrote:| Actually, the question asks for 6 R + 8 B + 7 Y. |
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Oops, corrected.
btw, I have tried to minimize the trades. That is why I didn't just add a couple of trades at the end of my previous solution.
Note: I know I am stretching the rules. But as shown by towr, it is not possible otherwise.
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| « Last Edit: Apr 11th, 2008, 12:58am by Grimbal » |
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wonderful
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Re: Marbles trading game
« Reply #19 on: Apr 11th, 2008, 3:49pm » |
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Very nice Grimbal! I like your approach. From your solution, we have 8 B + 5 R + 7 Y,
we don't have to break the marbles if we can trade 1 R for for 2 R since 8 B + 4 R + R + 7 Y can trade for 8 B + 6 R + 7Y
Have A Great Day!
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Grimbal
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Re: Marbles trading game
« Reply #20 on: Apr 12th, 2008, 10:15am » |
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I can make the last part look a bit less suspicious:
If you can exchange
B for Y + R (e)
and
Y for R + B (f)
and
B + Y for B + Y
then certainly you can exchange the sum
2B + 2Y for 2B + 2Y + 2R.
You can argue that if you can do that, you should be able to scale it down to the exchange of
B + Y for B + Y + R
So you can go from 8B+5R+7Y to 8B+6R+7Y.
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wonderful
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Re: Marbles trading game
« Reply #21 on: Apr 12th, 2008, 3:06pm » |
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Great Grimbal! That's eactly what I look for. Strictly speaking as Towr and Hippo pointed out there's no solution for this question. However, it is an interesting and exercise to see how we can twist the rule a bit.
Have A Great Day!
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