A Conversation for Basic Sorting Algorithms
½(m×(m -1))
Recumbentman Started conversation Feb 20, 2003
Surprising how often this number turns up. It is also the number of clinks heard when m people clink glasses.
½(m×(m -1))
Sam Posted Feb 20, 2003
I'm not sure I fully understand this but I think you may have given a mathematical proof to that which I've often wondered about (actually, it's my girlfriend who always asks the question) - if, say, there are 10 of us around the table and we all raise our glasses, toast each other's health and clink each glass, how many clinks will be heard in total? Does this equation solve the problem?
½(m×(m -1))
The Researcher formally known as Dr St Justin Posted Feb 20, 2003
Indeed it does! So for 10 of you, there would be ½(10×(10-1)) = ½(10×9) = ½(90) = 45 clinks! Although some of them may be simultaneous...
Thinking about it in a slightly different way, you could at each stage have clinks - 5 pairs of people clinking glasses. Each person needs to clink 9 other peoples glasses, so we will need 9 stages. Therefore, there should be 5 clinks each time × 9 stages = 45 clinks!
½(m×(m -1))
Sam Posted Feb 20, 2003
Dr J (and Recumbantman) this is brilliant! For a maths ignoramus like me it's a bit of a mindblower. Thank you.
½(m×(m -1))
The Researcher formally known as Dr St Justin Posted Feb 20, 2003
No probs mate! We aim to please...
½(m×(m -1))
Recumbentman Posted Feb 20, 2003
You can also picture it this way: each person (m of them) clinks with everyone else (m-1 of them). Each of these contributions is half a clink - or if 'half a clink' doesn't bring up a visual/audio image, say 'it takes two to make a clink'.
If this doesn't work the magic, don't be upset. Was it Von Neumann who said 'you don't understand maths, you just get used to it'?
When m=1 there are no clinks. (Just tried it.)
½(m*(m -1))
Sea Change Posted Feb 25, 2003
This clink formula only works if you don't have a doofus like me in the mix. Invariably I miss some and doubleclink others.
½(m*(m -1))
Gnomon - time to move on Posted Feb 25, 2003
What is the sound of one glass clinking?
The reason this particular formula (m/2)(m-1) comes up so often is that it is the sum of integers 1 + 2 + 3 + 4 ... + m. Or are we talking mathematical tautology here?
½(m*(m -1))
Recumbentman Posted Feb 25, 2003
According to the mighty Wittgenstein (A963579) all of maths is tautology.
½(m*(m -1))
Old Hairy Posted Sep 9, 2003
The chinking glasses theorem looks worthy of demonstration by experiment. My attempts to do this for numbers on the order of a dozen failed, because (Hic, beg your pardon) the results cannot be recollected by any of the participants. Perhaps we demonstrated that maths is a tautology - we all got fairly tight. We were all wearing glasses (see my PS), and are now recumbent in a recovery room, relishing the rest.
½(m*(m -1))
Recumbentman Posted Oct 4, 2003
Can you clink a pair of glasses while wearing them? I did once when they were in my pocket, but they're mended now.
½(m*(m -1))
Old Hairy Posted Oct 4, 2003
Hello Recumbent.
Happy to meet you.
"Can you ... I once did..." sounds rhetorical.
It can be done, but involves pain, both physical and economic. I suggest it is of paramount importance that this be avoided.
OH
Key: Complain about this post
½(m×(m -1))
- 1: Recumbentman (Feb 20, 2003)
- 2: Sam (Feb 20, 2003)
- 3: The Researcher formally known as Dr St Justin (Feb 20, 2003)
- 4: Sam (Feb 20, 2003)
- 5: The Researcher formally known as Dr St Justin (Feb 20, 2003)
- 6: Recumbentman (Feb 20, 2003)
- 7: Sam (Feb 21, 2003)
- 8: Sea Change (Feb 25, 2003)
- 9: Gnomon - time to move on (Feb 25, 2003)
- 10: Recumbentman (Feb 25, 2003)
- 11: Old Hairy (Sep 9, 2003)
- 12: Recumbentman (Oct 4, 2003)
- 13: Old Hairy (Oct 4, 2003)
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