Less than 14 steps!

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You have two baskets,each one contains 15 small balls( 5 Yellow,5 Red,and 5 Blue),you should exchange each two balls from one basket with one ball from the other basket,according to these roles:

1- Yellow+Yellow....... should be exchanged with a red ball.

2- Red +Red..........should be exchanged with a Blue ball.

3- Blue +Blue..........should be exchanged with a Yellow ball.

4- Yellow +Red..........should be exchanged with a Blue ball.

5- Red + Blue........should be exchanged with a Yellow ball.

6- Blue + Yellow.....should be exchanged with a Red ball.

7- each of the above exchanges will be considered as one step.

Your aim is to get only ONE Yellow ball in one of these two baskets in less than 14 steps.

Have a nice trial :thumbsup:

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Posted · Report post

With each step you can reduce the number of balls in a basket by 1 (at the most) by excanging 2 balls for 1. Thus when you start with 15 balls in a basket you need at least 14 steps to reduce the number of balls down to 1. Not possible in less than 14 steps.

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Posted · Report post

If the aim is to get one yellow and any number of other colors in a basket, then 2 steps of applying first rule are enough. Assuming this is not the case.

If the aim is to get a basket with only one ball, and that ball be yellow, then at least 14 steps are required. Since every exchange decreases by 1 the number of balls in the source basket and increases by 1 the number of balls in the target basket, from 15 (initial) to 1 (aim), one would need at least 14 steps.

Starting from 5R5B5Y - 5R5B5Y the following 14 steps manage that
  1. 6R5B3Y - 4R5B7Y (YY R)
  2. 7R5B1Y - 3R5B9Y (YY R)
  3. 5R6B1Y - 5R4B9Y (RR B)
  4. 3R7B1Y - 7R3B9Y (RR B)
  5. 1R8B1Y - 9R2B9Y (RR B)
  6. 1R6B2Y - 9R4B8Y (BB Y)
  7. 1R4B3Y - 9R6B7Y (BB Y)
  8. 1R2B4Y - 9R8B6Y (BB Y)
  9. 1R0B5Y - 9R10B5Y (BB Y)
  10. 2R0B3Y - 7R10B7Y (YY R)
  11. 3R0B1Y - 6R10B9Y (YY R)
  12. 1R1B1Y - 9R9B9Y (RR B)
  13. 0R2B0Y - 10R8B10Y (YR B)
  14. 0R0B1Y - 10R10B9Y (BB Y)

And a slight variation on the puzzle, what if applying the same rule (no matter how many times) counts as a single step?

In the 14-step solution above, because the same rule is used consecutively, the solution would be counted as having 7 steps

  1. (YY R)x2,
  2. (RR B)x2,
  3. (BB Y)x4,
  4. (YY R)x2,
  5. (RR B)x1,
  6. (YR B)x1,
  7. (BB Y)x1

Not sure if there's a quicker solution with these rules.

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Posted · Report post

yes, it can be done in 2 moves with one yellow ball on a basket

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Posted · Report post

yes, it can be done in 2 moves with one yellow ball on a basket

Yes....thats right

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