Given an infinitely-long sequence $$L = V \cdot V \cdot V \cdots$$ ...that is the repeated concatenation of $$V = (v_1, \dots, v_{2^n - 1}), v_i \in \Bbb{F}_2^n$$ ...a sequence of $2^n - 1$ vectors where all subsequences of length $n$ form a basis for $\Bbb{F}_2^n$ (and this rule "wraps around" or "crosses" the boundary of a concatenation; see this question and answer for how this sequence can be constructed and how there is such a sequence for all $n$).

Then what is the expected length $k$ of a subsequence $$S = (v_{r(i, 1)}, \dots, v_{r(i, k)})$$ ...that starts at a random location $i$ in $L$ and continues by skipping each subsequent vector with probability $p$ (i.e. including each subsequent vector with probability $1 - p$) until its vectors span $\Bbb{F}_2^n$?

I think this defines $r(i, j)$ to get the skipping action: $$r(i, j) = \begin{cases} i, & j = 1 \\ r(i, j) + 1, & X < p \\ r(i, j - 1) + 1, & \mbox{otherwise} \end{cases}$$ ...where $X$ is a uniform distribution from 0 to 1. This should make $r(i, j)$ a monotonically increasing function that either increases by one from the previous value or randomly recurses to find another value to increase by (in effect randomly skipping an index with probability $p$). If $p = 0$ then it just gives integers at least as large as $i$; higher values of $p$ mean that a vector in $L$ is more likely to be skipped.

For example, here's the first part of an $L$ for $\Bbb{F}_2^3$: $$v_1 = (0, 0, 1)$$ $$v_2 = (0, 1, 0)$$ $$v_3 = (1, 0, 0)$$ $$v_4 = (0, 1, 1)$$ $$v_5 = (1, 1, 0)$$ $$v_6 = (1, 1, 1)$$ $$v_7 = (1, 0, 1)$$ $$v_8 = v_1 = (0, 0, 1)$$ $$v_9 = v_2 = (0, 1, 0)$$ $$\vdots$$

Notice how if we let $S' = (v_1, v_2)$ and we're looking for another vector to span the field but due to the random process we skip $v_3$, then chances are still good that we'll soon find another vector to finish the sequence. For example, although $v_4$ won't do, $v_5, v_6, v_7, v_{3 + 2^n - 1} = v_{10} = v_3$, and any vectors offset in $L$ from those by a multiple of $2^n - 1 = 7$ are perfectly suitable to finish the sequence. But $v_4$ is more likely to be included than the later vectors so $k$ is more likely to be larger since $v_4$ is effectively a waste.

Also, how does the expected $k$ compare if we follow the same procedure but replace $L$ with $L'$, an infinite sequence of vectors drawn (uniformly) randomly from $\Bbb{F}_2^n$?

I'm not really sure where to start with the maths. In the case of $L'$ (random vectors) the likelihood of selecting a vector that (along with $n - 1$ previously-selected linearly-independent vectors) spans the field feels like it ought to be around 50% (since exactly one dimension will be missing; this is regardless of $p$ since $L'$ is random anyway). Indeed with some software I can tell empirically that the average value of $k$ is ~$1.65 + n$ and that the values of $k$ follow what looks like a Poisson distribution (I tested for $10 \leq n \leq 200$).

  • $\begingroup$ knowing that the $k$ previous vectors span a subspace of dimension $m$, does the probability that the $k+1$th vector be contained in that subspace depends on all the previous vectors, or only on $k$ and $m$ ? $\endgroup$ – reuns Apr 23 '16 at 0:16
  • $\begingroup$ @user1952009 I'm not quite sure, but do you think that's similar to math.stackexchange.com/a/171995/284627? $\endgroup$ – Matt Thomas Apr 23 '16 at 0:43
  • $\begingroup$ I didn't read everything but it seems under an uniform distribution the dimension of the spanned subspace by $k+1$ vectors doesn't depend on each of the $k$ previous vectors, but only on the dimension of the subspace they span. here you are not using a uniform distribution, and I'm not sure to understand your filtered Poisson process in $\mathbb{F}^n_2$. $\endgroup$ – reuns Apr 23 '16 at 0:49
  • $\begingroup$ @user1952009 I made an edit to try and clarify what's happening with the random process. Does that help? $\endgroup$ – Matt Thomas Apr 23 '16 at 1:49

For your example $L$, we can do the calculation explicitly. Triples of vectors that sum to zero are always in relative positions $(0,1,3)$. This is the only relevant property, so there's complete symmetry and it doesn't matter where we start. Also, note that once a vector is in the space we've already spanned (whether we've selected this specific vector or not), we can ignore it when it's selected.

First we take an expected number $\frac1{1-p}$ of steps to select the first vector. Label the remaining vectors in order after the selected one $v_0$ to $v_5$. The next vector selected will be $v_k$ with probability

$$ q_k=p^k(1-p)\left(1+p^6+p^{2\cdot6}+\cdots\right)=\frac{p^k(1-p)}{1-p^6} $$

after an expected number

$$ 1+k+7\left(1-p^6\right)\left(0\cdot1+1\cdot p^6+2\cdot p^{2\cdot6}+\cdots\right)=1+k+\frac{7p^6}{1-p^6} $$

of steps, so this selection takes an expected number

$$ \sum_{k=0}^5\frac{p^k(1-p)}{1-p^6}\left(1+k+\frac{7p^6}{1-p^6}\right)=\frac{1-p^7}{(1-p)\left(1-p^6\right)} $$

of steps.

Now we've spanned some triple, and if we renumber such that the triple is at $(0,1,3)$, the new index $j$ of the vector just selected depends on $k$ as follows:

\begin{array}{c|c} k&j\\\hline 0&1\\ 1&3\\ 2&3\\ 3&0\\ 4&1\\ 5&0\\ \end{array}

Denote the expected number of steps for selecting the third vector after selecting (renumbered) vector $j$ as the second vector by $x_j$. Then we have

\begin{align} x_0&=1+x_1\;,\\ x_1&=1+p(1+x_3)\;,\\ x_3&=1+p(1+p(1+p(1+x_0)))\;. \end{align}

Substituting the second equation into the first yields $x_0=2+p(1+x_3)$, and substituting that into the third yields $x_3=1+p(1+p(1+p(3+p(1+x_3))))$, and thus

\begin{align} x_3&=\frac{1+p+p^2+3p^3+p^4}{1-p^4}\;,\\ x_1&=\frac{1+2p+p^2+p^3+2p^4}{1-p^4}\;,\\ x_0&=\frac{2+2p+p^2+p^3+p^4}{1-p^4}\;. \end{align}

Thus the selection of the third vector is expected to take

$$ (q_3+q_5)x_0+(q_0+q_4)x_1+(q_1+q_2)x_3\\ = \frac{1-p}{\left(1-p^6\right)\left(1-p^4\right)}\left(\left(p^3+p^5\right)\left(2+2p+p^2+p^3+p^4\right)+\\\left(p^0+p^4\right)\left(1+2p+p^2+p^3+2p^4\right)+\left(p^1+p^2\right)\left(1+p+p^2+3p^3+p^4\right)\right)\\ =\frac{1+2p+p^2+4p^3+5p^4+4p^5+p^6+2p^7+p^8}{\left(1-p^6\right)\left(1+p^2\right)} $$

steps. So the expected total number of steps is

$$ \frac1{1-p}+\frac{1-p^7}{(1-p)\left(1-p^6\right)}+\frac{1+2p+p^2+4p^3+5p^4+4p^5+p^6+2p^7+p^8}{\left(1-p^6\right)\left(1+p^2\right)}\\ =\frac{3+4p+5p^2+8p^3+9p^4+8p^5+4p^6+4p^7+2p^8}{\left(1-p^6\right)\left(1+p^2\right)}\;. $$

Here's code I used to test this result.

The messiness of the result already for this simple case suggests that you may not get a simple closed form for general $n$. In any case you'd first need to show that the probability doesn't depend on the choice of $L$; as Jyrki pointed out in the answer you linked to, different generators of $\Bbb{F}_{2^n}^*$ can be used to generate different sequences.

The case of vectors uniformly randomly drawn from $\mathbb F_2^n$ that you want to compare with is relatively straightforward to treat. If you've already spanned a $k$-dimensional space, $2^k$ vectors are no longer useful, so the probability of gaining a dimension is $(1-p)\left(2^n-2^k\right)/\,2^n$. Thus the expected number of steps is

$$ \frac1{1-p}\sum_{k=0}^{n-1}\frac{2^n}{2^n-2^k}\;. $$

For $n=3$, this is

$$ \frac{94}{21}\cdot\frac1{1-p}\;. $$

But that's not really a fair comparison, since you not only ordered the vectors but also removed the zero vector. So perhaps we should compare with uniformly randomly drawing from $\mathbb F_2^n\setminus\{0\}$. The zero vector has to be taken out of both $2^n$ and $2^k$, so now we get

$$ \frac1{1-p}\sum_{k=0}^{n-1}\frac{2^n-1}{2^n-2^k}\;, $$

and for $n=3$ this is

$$ \frac{47}{12}\cdot\frac1{1-p}\;. $$

Here's a plot that compares the probabilities in the three cases.

Note that the two methods without the zero vector have the same residue $-\frac{47}{12}$ at $p=1$, which makes sense since if you skip almost all vectors it no longer matters whether you ordered them. By contrast, at $p=0$ your method is of course certain to complete in exactly $3$ steps whereas the random method takes almost $4$.

  • $\begingroup$ I wish I could +2 your answer! P.S. Do you mean the residues are -47/12 at p=1? $\endgroup$ – Matt Thomas Apr 24 '16 at 3:10
  • $\begingroup$ @Matt: I do, thanks; fixed. $\endgroup$ – joriki Apr 24 '16 at 7:26
  • $\begingroup$ One more thing: how does the probability of gaining a dimension depend on $p$ in the case of randomly selecting vectors from $\Bbb{F}^n_2$? $\endgroup$ – Matt Thomas Apr 25 '16 at 1:04
  • $\begingroup$ @MattThomas: I'm not sure I understand the question. It's in the answer: $(1-p)\left(2^n-2^k\right)/\,2^n$; that's how it depends on $p$. Perhaps you meant "Why does it depend on $p$?" Because you gain a dimension if a) you don't skip a vector and b) then draw a useful vector. You said in the question that you wanted to follow the same procedure, just replace $L$ by a randomly uniformly drawn list $L'$. If you didn't mean to include the skipping part of the procedure, you can just leave out the factor $1-p$. $\endgroup$ – joriki Apr 25 '16 at 5:25
  • $\begingroup$ Sorry, let me rephrase: in the case of vectors uniformly randomly drawn from $\Bbb{F}^n_2$, why does it matter whether $p=0.999\dots$ or whether $p=0$, since one random vector is as likely to give us another dimension as another? I guess what I mean is the skipping part of the procedure probably isn't relevant in the random case, so I'll do as you suggest and leave out the factor $1 - p$ $\endgroup$ – Matt Thomas Apr 25 '16 at 6:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.