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Topic: Series of operations (Read 868 times) |
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wonderful
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Series of operations
« on: Mar 13th, 2008, 2:34am » |
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A series of operation is setup based on +, -, *, /, square and square root.
For instance: + 1, square, * 3 will transform the number 3 as follows:
3+1=4; 4^2=16, 16*3 = 48
Question: Find a series of operations that transform any interger number from 1 to 9 to zero. One trivial solution is * 0. Can you find another one?
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Aryabhatta
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Re: Series of operations
« Reply #1 on: Mar 13th, 2008, 7:33am » |
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Assuming square root returns the non-negative square root, the following seems to work:
-5, square, square root, -2, square, square root, -1 , square, square root, -0.5, square , square root, -0.5
I suggest we move this one to medium.
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towr
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Re: Series of operations
« Reply #2 on: Mar 13th, 2008, 8:24am » |
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Just using squares and subtraction is enough to reduce any set of numbers to zero
In this case, we could do
-5, ^2, -8, ^2, -25, ^2, -801, ^2, -284512.5, ^2, -54703362656.25
But there's quite a bit of choice
To reduce the set to one with at least one element less, pick any two distinct numbers, subtract their average from all numbers in the set, square all numbers in the set.
If for some perverse reason* the numbers need to remain integer throughout, multiply by two when there are no two numbers that give an integer average.
Or any other reason.
Aryabhatta's approach is much more efficient though, at least on sets of numbers in a arithmetic sequence.
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| « Last Edit: Mar 13th, 2008, 2:35pm by towr » |
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Grimbal
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Re: Series of operations
« Reply #3 on: Mar 13th, 2008, 9:41am » |
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Another same solution
-5, square, -8, square, -32.5, square, -632.25, square, -129600
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Eigenray
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Re: Series of operations
« Reply #4 on: Mar 13th, 2008, 2:33pm » |
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Here is a harder question: for which finite sets S is there a sequence of operations that takes S, and only S, to 0? Aryabhatta's solution shows that this is possible for any arithmetic progression. Of course there are non-arithmetic progressions for which this is possible too: for any sequence of operations, we can let S be the set of things which go to 0. But is there a way to tell whether such a function exists for a given S?
The only non-injective operation is squaring. So there must be a way to transform S to a set symmetric about 0 using addition, multiplication, and square rooting. If there is, we can square and recursively test the resulting set.
If |S| <= 4, we can always do it. The case |S|=1 or 2 are clear.
Case |S|=3: Translating, scaling and reflecting, we may assume S = {0,h,1} with 0<h 1/2. Now take square roots if necessary just until h > 1/4. If x 1/4, then  x 1/2, so we now have 1/4 < h 1/2. If h=1/2, we're done. Otherwise, let f(x) = x + {1+x} - 2{h+x}. Then f(0) = 1 - 2h < 0, while for x large, f(x) = (1/2 - h)x-1/2 + O(x-3/2) > 0. Therefore there is some x with f(x)=0, i.e., adding x and taking square roots, we get an arithmetic progression.
Case |S|=4: We can assume S = {0, h, k, 1}, 0<h<k<1. Take square roots until h > 1/2, then flip it around to assume 0 < h < k < 1/2. Let r=2-n, and f(x) = (h+x)r - xr - [(1+x)r - (k+x)r]. For n sufficiently large, f(0) = hr + kr - 1 > 0. And for large x, f(x) = r(h+k-1)xr-1 + O(xr-2) < 0. So we can find x such that f(x)=0, i.e., adding x, and taking square roots n times, we get S = {a, b, c, d}, where b-a = d-c. Subtracting (a+d)/2, we get {(a-d)/2, (2b-a-d)/2, (2c-a-d)/2 = (a+d-2b)/2, (d-a)/2}, so we can square and reduce to |S|=2.
What about |S|=5?
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wonderful
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Re: Series of operations
« Reply #5 on: Mar 13th, 2008, 4:11pm » |
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You guys are so smart!
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wonderful
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Re: Series of operations
« Reply #6 on: Mar 13th, 2008, 9:58pm » |
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A little harder version is as follows:
Using +, -,*, /, square, square root and the integer from 1 to 9 transform any integer number in [1,9] to zero.
For example, the solution such as:
- 5, square, square root, -2, square, square root, -1 , square, square root, -0.5, square , square root, -0.5
Doesn't meet the requirement since it contains 0.5
Can you do this?
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towr
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Re: Series of operations
« Reply #7 on: Mar 14th, 2008, 1:10am » |
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on Mar 13th, 2008, 9:58pm, wonderful wrote:
-5, square, square root, -2, square, square root, -1 , square, square root, *2, -1 (= -0.5*2), square , square root, -1 (= -0.5*2)
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wonderful
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Re: Series of operations
« Reply #8 on: Mar 14th, 2008, 6:31pm » |
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Great!
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