## Does Fourier transform help with exhaustive subtractions?

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• Hi Everybody, I have two sets (A and B) of n positive integers each, and I need to subtract each element of B from each element of A. Obviously, the
Message 1 of 5 , Nov 9, 2006
Hi Everybody,

I have two sets (A and B) of n positive integers each, and I need to
subtract each element of B from each element of A. Obviously, the
computation runs in O(n^2).

I was wondering if fast Fourier transfrom can be used to reduce the

I appreciate any feedbacks.

Kaveh
• I m not sure how this relates to primes, maybe the elements of A and B are all prime? *wishful thinking* Anyway, wouldn t this be an O(2n-1) = O(2n) = O(n)
Message 2 of 5 , Nov 9, 2006
I'm not sure how this relates to primes, maybe the elements of A and B
are all prime? *wishful thinking*

Anyway, wouldn't this be an O(2n-1) = O(2n) = O(n) operation? You could
add up all elements of B and then subtract that sum from each element of

BTW, its O(2n-1) because there are (n-1) additions performed in B, and
then there are n subtractions in A. Since Big-O doesn't care about
addends or constant multipliers, its really just O(n).

-David C.

Kaveh wrote:
>
>
> Hi Everybody,
>
> I have two sets (A and B) of n positive integers each, and I need to
> subtract each element of B from each element of A. Obviously, the
> computation runs in O(n^2).
>
> I was wondering if fast Fourier transfrom can be used to reduce the
> time complexity of this task.
>
> I appreciate any feedbacks.
>
> Kaveh
• Hello Kaveh, ... Consider the sets A = { N, 2N, 3N, ... N^2 }, B = { 0, 1, 2, ... N-1 }. The set obtained by subtracting each element of B from each element of
Message 3 of 5 , Nov 9, 2006
Hello Kaveh,

> I have two sets (A and B) of n positive integers each, and I need to
> subtract each element of B from each element of A. Obviously, the
> computation runs in O(n^2).
>
> I was wondering if fast Fourier transfrom can be used to reduce the
> time complexity of this task.

Consider the sets A = { N, 2N, 3N, ... N^2 }, B = { 0, 1, 2, ... N-1 }.
The set obtained by subtracting each element of B from each element of A
is { 1, 2, 3, ... N^2 }. In other words, the size of the resulting set can
be quadratic and thus the worst-case runtime of any algorithm producing
it is at least quadratic [*].

Thus, without having some additional restrictions on the structure of the
elements in those sets, the best you can hope for is a quadratic
algorithm.

Or did I misunderstood your problem statement?

Peter

[*] Assuming you're representing the set as an array of numbers.

--
[Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ] 134813278
• ... I think he s talking about making the nxm table of all the differences of (an element of A) - (an element of B). What I wonder about is what you want to do
Message 4 of 5 , Nov 9, 2006
On 11/9/06, David Cleaver <wraithx@...> wrote:
> Anyway, wouldn't this be an O(2n-1) = O(2n) = O(n) operation? You could
> add up all elements of B and then subtract that sum from each element of
> A. Or are there additional restrictions you're not telling us about?

I think he's talking about making the nxm table of all the differences
of (an element of A) - (an element of B).

What I wonder about is what you want to do with that table: sometimes
it's O(n+m) if you just, say, want the total; O(nm) if you want all
the elements; O(1) (but O(min(n,m)) for lookup) if you just want to
check whether a few particular numbers occur or not ...

--Joshua Zucker
• ... Hi Kaveh, I think you misunderstood Peter s point (quoted above). With fast multiplication, n digits times m digits, the result will only have n+m digits,
Message 5 of 5 , Nov 9, 2006
On 11/9/06, Peter Kosinar <goober@...> wrote:
> Consider the sets A = { N, 2N, 3N, ... N^2 }, B = { 0, 1, 2, ... N-1 }.
> The set obtained by subtracting each element of B from each element of A
> is { 1, 2, 3, ... N^2 }. In other words, the size of the resulting set can
> be quadratic and thus the worst-case runtime of any algorithm producing
> it is at least quadratic [*].

Hi Kaveh,
I think you misunderstood Peter's point (quoted above).

With fast multiplication, n digits times m digits, the result will
only have n+m digits, so Peter's argument says it takes at least
O(n+m) time just to store the resulting bignum! And indeed FFT mult
takes O(n ln n) which in general is way more than 2n.

With your subtraction problem, n numbers minus each of m numbers, the
result can have nm numbers, so he argues that since storing each
number takes O(1) time, the whole computation must take at least O(nm)
time. I think that argument sounds very convincing to me!

My solution, which takes O(1) time, depends on not storing all the
numbers, but rather on doing an O(n+m) lookup each time you want to
check if a number is in the set "A-B" that you want. Peter already
anticipated that, though, in his footnote which reminds us that the
O(nm) minimum is only true if you want to store all the nm
differences.

So, if you really need all the numbers in a list, Peter is right and
the minimum time is O(nm). If you can use some other way of
remembering the numbers, or if you know some other properties of the
numbers (like, say, the maximum is a lot less than nm, so you know
there can't be nm different differences among them) then there might
be an algorithm that exploits those other possibiities.

--Joshua Zucker
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