0
$\begingroup$

Given a complex vector $V$ of length $n^2$. Each complex entry in the vector is of size (number of digits or bits required to express the complex number) $c$ for some constant $c$.

Is it always possible to create two vectors $V_a$ and $V_b$ (each of length $n$) with each entry of size $kc$ for some constant $k$ such that, the 'all pair product vector' using the ordered entries in $V_a$ and $V_b$ results in vector $V$ (if we take the first $c$ digits after calculating each product pair).

Improve this question
$\endgroup$
6
  • $\begingroup$ Presumably the "size" of a vector is meant to be its dimension? But what is the "size" of an entry? Its magnitude? The terminology here is unclear to me. $\endgroup$
    – Michael Engelhardt
    Commented Nov 4, 2022 at 12:59
  • $\begingroup$ @MichaelEngelhardt Size of a vector - vector length (updating). Size of an entry - number of bits or digits (I couldn't figure out an accurate word for this) $\endgroup$
    – xyz
    Commented Nov 4, 2022 at 13:01
  • $\begingroup$ I don't understand the expression "number of bits required to express". How are you turning bits into complex numbers? $\endgroup$
    – Ben McKay
    Commented Nov 4, 2022 at 17:19
  • $\begingroup$ its a rough idea. One way is to have 2 bits for sign (for each real and imaginary part), and 2 binary strings of some fixed size (one for complex another for real). Plus 2 more position counters of fixed size, that tell the position of decimal in each real and imaginary part respectively. $\endgroup$
    – xyz
    Commented Nov 4, 2022 at 17:52
  • $\begingroup$ the main point i was trying to suggest was that the size of each complex entry has an upper bound. $\endgroup$
    – xyz
    Commented Nov 4, 2022 at 17:54

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .