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About Peter Rowlett
I am interested in mathematics education & maths promotion.
More information on my website peterrowlett.net.
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Wednesday, 25 January 2012
Puzzle from Maths Jam Nottingham: Jon's coloured balls
At Maths Jam Nottingham January 2012, Jon brought this puzzle.
You have three pairs of coloured balls - 2 each of red, white and blue. Within each pair one ball is heavy and one is light but you do not know which. All three heavy balls are equally heavy and all three light balls are equal weight too.
Q: What is the minimum number of weighings needed to identify each ball?
A: 1 is fairly clearly impossible. 3 is trivial (weigh each pair separately). So, in order for us to have an interesting puzzle the answer must be 2.
Q: How can it be done in 2?
For a while Christian Perfect has suggested that Maths Jam local organisers write blog posts based on what went on at their meetings. Rather than write a wholesale account of what happened at each meeting once a month I have decided to drop occasional puzzles from Nottingham Maths Jam meetings into this blog. I will tag these posts so you can search for "mathsjam" and find them.
It is important in problem solving that you have an honest attempt before reading a solution. Once someone has shown you the solution you are forever robbed of the chance to have that experience (in future you will half-remember the solution rather than reason it out) so it is important that you attempt this puzzle before reading the solution. For this reason I will post the solution separately. I will post this as a 'Page' since I don't think they appear in the blog stream so you have a reduced chance of inadvertently stumbling upon it. Jon's coloured balls solution.
N.B. I assume the puzzles written about are old puzzles. They are brought to Maths Jam meetings, or half remembered at the time, by attendees. If I have done something wrong by posting a puzzle here please tell me and I will be happy to correct the mistake.
You have three pairs of coloured balls - 2 each of red, white and blue. Within each pair one ball is heavy and one is light but you do not know which. All three heavy balls are equally heavy and all three light balls are equal weight too.
Q: What is the minimum number of weighings needed to identify each ball?
A: 1 is fairly clearly impossible. 3 is trivial (weigh each pair separately). So, in order for us to have an interesting puzzle the answer must be 2.
Q: How can it be done in 2?
For a while Christian Perfect has suggested that Maths Jam local organisers write blog posts based on what went on at their meetings. Rather than write a wholesale account of what happened at each meeting once a month I have decided to drop occasional puzzles from Nottingham Maths Jam meetings into this blog. I will tag these posts so you can search for "mathsjam" and find them.
It is important in problem solving that you have an honest attempt before reading a solution. Once someone has shown you the solution you are forever robbed of the chance to have that experience (in future you will half-remember the solution rather than reason it out) so it is important that you attempt this puzzle before reading the solution. For this reason I will post the solution separately. I will post this as a 'Page' since I don't think they appear in the blog stream so you have a reduced chance of inadvertently stumbling upon it. Jon's coloured balls solution.
N.B. I assume the puzzles written about are old puzzles. They are brought to Maths Jam meetings, or half remembered at the time, by attendees. If I have done something wrong by posting a puzzle here please tell me and I will be happy to correct the mistake.
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5 comments:
You seem to be applying something akin to the anthropic principle to puzzles lately, i.e., using the fact that you're being told the puzzle to infer something about its solution.
Now I must concoct a puzzle which takes advantage of this fact to confound you!
PS: hooray for more MathsJam blogging!
and in re solutions: I forgot last month that I'd installed a spoiler-hiding plugin on my blog. This month solutions will be hidden until you ask to see them.
I think this is true for a lot of puzzles. There's no possible interest in saying "Weigh each pair individually. Done." The anthropic aspect comes from the fact someone has chosen to show you this puzzle (cf., say, problems that arise in a real-world context). You're right that if you're going to get all meta on me you could design a puzzle to exploit this!
Wouldn't the minimum number be 0?
You can drop the balls a set distance and measure the speed at which they fall.
Or is this cheating?
Yes that's cheating! You also, presumably, can't tell which is heavier as you're lifting them to put them on and off the scales.
See my piece on constrained reality in puzzles.
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