Recent content by zheng004

If the vector only has 2 elements, i.e. A = [a1, a2]; B=[b1, b2]; C=[c1, c2], then we can store two values: v1 = (a1+a2-b1-b2) and v2 = (a1+a2+b1+b2). Then we can get dot (A, C) = 1/4 * ( (v2 - (b1+b2-c1-c2))^2 - (v1 + b1 + b2 - c1 - c2)^2). But if vector is lager than 2, I do not know how to...

Thanks for your reply. Unfortunately, b can not be overwritten as we need to use later. I know can we not get a unique solution from Ax = b as you mentioned, so I post the question here. This is problem abstracted from my research, any input is appreciated.

Thanks for your input. But in order to calculate ##\mathbf v^T \mathbf c##, we have to compute and store ##\mathbf v ##, but the problem is that ##\mathbf v ## is a vector and we do not have enough space to store it.

No, this is not a homework question. Very appreciated if you can help.

My general idea is to store some calculation/calculations between A and B, after replacing B with C, dot (A,C) can be generated from these values. But I can not figure out the details.

Ok, here is the context. The incoming large vectors in order are A, B, C. But spaces are limited, to receive vector C, we must delete A first. (B can not be deleted as will be used later.) The problem is how to get the dot (A, C). We are allowed to use extra but limited spaces, for examples...

We can store any value as long as using limited space.

Sorry, should be square of sum, i.e. store (a_1 + a_2 + ... + a_n - b_1 - b_2 - ... - b_n)^2 and (a_1 + a_2 + ... + a_n + b_1 + b_2 - ... + b_n)^2. These are the values so easy to store. Very appreciated if you can help. Thanks.

Thanks for your reply, it is not allowed to store A in B as B will be used later. Actually, deleting A is to save space.

Hi all, Suppose we have vectors coming in order as A, B and then C (but A must be deleted before C comes in). Then how to get the dot product between A and C? It is allowed to store some calculations of A before deleting elements of A, for example, we could store norm of A, dot(A, B) and etc...