Post me your hardest maths question you know
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Nice!
Is this actually the answer?
Of course not!
Associate
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Yes, you can certainly do it via a simplex approach, as anyone familiar with differential forms would have seen (I'm guessing from your replies you're such a person). However, this is a little fiddly, as the general approach is going to go into how you would do a general decomposition. Even when considering the 2d case, ie the area of a convex quadrilateral, its not a very elegant route, in my opinion (though its applications to differential geometry are elegant).
It might interest you to know (if you haven't checked) that the area of a convex quadrilateral can be expressed in terms of the norm of a cross product of the vectors defining the 'diagonals' of the quadrilateral. That's a general expression which doesn't involve breaking things into simplices. The method I came up with doesn't involve vector cross products but it does recover the component expression form. And it doesn't involve splitting the convex hull in question into 3-simplices. I don't know if the special case I described can be attacked in that manner in an elegant way but the general problem of the volume of a general convex hull is an open problem at present. My method, unsurprisingly, doesn't work for general convex hulls.
Soldato
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So basically 2n-3
Yeah. I like it my way though, doesn't use these new fangled mathematical equations such as "multiply" or, worse still, divide!
Nah one makes 10 profit, 1 makes 10 loss.
Left / Right
-20 / -20 Put into box
-30 / +30 Sold to left by right (money goes to right as he is the vendor, box to left).
+40 / 0 box contents
--------------
-10 / +10
Well beauty is in the eye of the beholder I guess!
If you identify a diagonal you have identified a choice of triangulation - so I fail to see the advantage. But elegance is subjective so it's all a much of a much-ness.
Sounds like interesting very interesting work. But this exchange probably isn't too interesting for most readers!
Amazingly, you are very close! The actual answer is 2N-4.N+(N-3)?
Edit: I used my fingers so could be crazily wrong
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1) N>3 problem
upper bound is N-1 + N-2 + N-3 ...
probably the answer is 2(N-1)
2) spaghetti problem
the answer is a random value from 1 to 1000.
upper bound is N-1 + N-2 + N-3 ...
probably the answer is 2(N-1)
2) spaghetti problem
the answer is a random value from 1 to 1000.
The extra one is to take into account women going off topic and not getting to the point
Victory (sort of) to the fingers!
I like the modification, but sadly the fingers would have to prove their guess is correct!The extra one is to take into account women going off topic and not getting to the point
Victory (sort of) to the fingers!
Incorrect I'm afraid. I really want to stress that the first problem is very, very difficult. A hint for the second problem:
When I tie two ends of spaghetti together, I either make a loop or I create a long piece of spaghetti. In either case, two ends are removed from the bowl.
Soldato
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47.2 X A MILLION!
It's not really that difficult. My basic guess was done on the simplest algorithm to share all the information, the method to get 2N-4 can quite easily be shown on fingers as well, after N=5 it gets more difficult I admit!
I'm afraid it really is. The problem has been well known since around 70s, and most professional mathematicians would find it very difficult. The problem for n=4 can easily be done, as could n=5,6,7 etc. using brute force. But the question doesn't specify what n is - your proof must hold for any n belonging to the infinite set {4,5,6,7,8,....}.
See, that's what's wrong with mathematicians, you over complicate things
I just came at it from a simple search pattern - then figure out the formula that matches the search! Bit like a Turing machine I guess. Although it's been so long since I did Turing machines I can't remember what use they were for, but I was quite good at picking them up!
As I'm sure you can show (with your fingers?!), given any finite sequence A,B,C,D,...,E of numbers, there are infinitely many different functions such that f(1)=A, f(2)=B, f(3)=C and so on. So "spotting a pattern" will never constitute a proof, unless you can then go on to prove that your guess is correct for all numbers.See, that's what's wrong with mathematicians, you over complicate things
For instance, if I worked out the problem in hand and got upper bounds of
4, 6, 8, 10, 12
for n = 4, 5, 6, 7, 8 gossiping women, then I might guess an upper bound was
f(n) = 2n - 4 + (n - 4)(n - 5)(n - 6)(n - 7)(n - 8)
which agrees with my list for n = 4, 5, 6, 7, 8. But of course we now know this guess (which is but one of infinitely many) is wrong.
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Given any finite sequence of numbers there is infinitely many formulae which will reproduce that sequence.See, that's what's wrong with mathematicians, you over complicate things
For instance, a particular problem with partitions of circles has, for the n=1,2,3,4,5 cases answers 1,2,4,8,16, which suggests f(n) = 2^(n-1). But that's wrong, the n=6 case is 31. See here.
Hence guessing the formula isn't proof, because you haven't proven that is actually what the n=500 or n = 5482385823954020 case follows. Most integers are very very big so saying "It works for the first 5 cases" or even "It works for the first trillion" isn't proof, its an hypothesis motivated by a tiny sample set.
/edit
Looks like w11tho and I were thinking along similar lines.