<p><strong>EDITS</strong>: <em>Added Lemmas 2 and 3.</em></p>

<p>Here's a partial answer:  You can reach position $N$</p>

<ul>
<li>in $N$ moves using space $N^{O(\epsilon(N))}$, where $\epsilon(N) = 1/\sqrt{\log N}$.  (Lemma 1)</li>
<li>in $N^{1+\delta}$ moves using space $O(\log N)$ (for any constant $\delta&gt;0$) (Lemma 2).</li>
</ul>

<p>Also, we sketch a lower bound (Lemma 3): for a certain class of so-called <em>well-behaved</em> solutions, Lemma 1 is tight (up to constant factors in the exponent), and no such solution using poly-log space can reach position $N$ in time $O(N\,\text{polylog}\, N)$.</p>

<p><strong>Lemma 1.</strong> <em>For all $n$, it is possible to reach position $n$ in $n$ moves using space $$\exp(O(\sqrt{\log n}))~=~n^{O(1/\sqrt{\log n})}$$</em></p>

<p><em>Proof.</em> The scheme is recursive, as shown in the figure below. We use the following notation:</p>

<ul>
<li>$k$ is the number of levels in the recursion</li>
<li>$P(k)$ is the solution formed (with $k$ levels of recursion).</li>
<li>$N(k)$ is the maximum position reached by $P(k)$ (at time $N(k)$). </li>
<li>$S(k)$ is the space used by $P(k)$.</li>
<li>$L(k)$ is the number of <em>layers</em> used by $P(k)$, as illustrated below:</li>
</ul>

<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<a href="https://i.sstatic.net/FG4k7.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/FG4k7.png" alt="solution structure for lemma 1"></a></p>

<p>In the picture, time proceeds from top to bottom. The solution $P(k)$ does not stop at time $N(k)$, instead (for use in the recursion) it continues until time $2\,N(k)$, exactly reversing its moves, so as to return to a single pebble at time $2\,N(k)$.</p>

<p>The solid vertical lines partition the $L(k)$ layers of $P(k)$.  In the picture, $L(k)$ is five, so $P(k)$ consists of 5 layers.   Each of the $L(k)$ layers of $P(k)$ (except the rightmost) has two sub-problems, one at the top of the layer and one at the bottom, connected by a solid vertical line (representing a pebble that exists for that duration).  In the picture, there are five layers, so there are nine subproblems.  Generally, $P(k)$ is composed of $2\,L(k)-1$ subproblems. Each subproblem of $P(k)$ has solution $P(k-1)$.</p>

<p>The crucial observation for bounding the space is that, at any time, only two layers have "active" subproblems.  The rest contribute just one pebble each thus we have</p>

<ul>
<li>$S(k) \le L(k) + 2\,S(k-1)$, and</li>
<li>$N(k) = L(k) \cdot N(k-1)$</li>
</ul>

<p>Now, we choose $L(k)$ to fully determine $P(k)$.  I'm not sure if this choice is optimal, but it seems close: take $L(k) = 2^k$. Then the above recurrences give</p>

<ul>
<li>$S(k) \le k\cdot 2^k$, and</li>
<li>$N(k) = 2^{k(k+1)/2}$</li>
</ul>

<p>So, solving for $n=N(k)$,
we have $k \approx \sqrt {2 \log n}$
and $S(k) \approx \sqrt {2 \log n}\, 2^{\sqrt {2 \log n}} = \exp(O(\sqrt{\log n}))$.  </p>

<p>This takes care of all positions $n$ in the set $\{N(k) : k\in\{1,2,\ldots\}\}$.  For arbitrary $n$, trim the bottom of the solution $P(k)$ for the smallest $k$ with $N(k)\ge n$.  The desired bound holds because $S(k) / S(k-1) = O(1)$.  QED</p>

<hr>

<p><strong>Lemma 2.</strong> <em>For any $\delta&gt;0$, for all $n$, it is possible to reach position $n$ in $n^{1+\delta}$ moves using space $O(\delta 2^{1/\delta}\log n).$</em></p>

<p><em>Proof.</em> Modify the construction from the proof of Lemma 1 to delay starting each subproblem until the previous subproblem has finished, as shown below:</p>

<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<a href="https://i.sstatic.net/lH89u.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/lH89u.png" alt="solution structure for lemma 2"></a></p>

<p>Let $T(k)$ denote the time for the modified solution $P(k)$ to finish. Now at each time step, only one layer has a subproblem that contributes more than one pebble, so</p>

<ul>
<li>$S(k) \le L(k) + S(k-1)$,</li>
<li>$N(k) = L(k) \cdot N(k-1)$,</li>
<li>$T(k) = (2\,L(k)-1)\cdot T(k-1) \le 2\,L(k)\cdot T(k-1) \le 2^k N(k)$.</li>
</ul>

<p>Choosing $L(k) = 2^{1/\delta}$, we get</p>

<ul>
<li>$S(k) \le k 2^{1/\delta}$,</li>
<li>$N(k) = 2^{k/\delta}$,</li>
<li>$T(k) \le 2^k N(k)$.</li>
</ul>

<p>Solving for $S = S(k)$ and $T=T(k)$ in terms of $n=N(k)$, we have $k=\delta \log n$, and</p>

<ul>
<li>$S \le \delta 2^{1/\delta} \log n$, and</li>
<li>$T \le n^{1+\delta}$.  </li>
</ul>

<p>This takes care of all positions $n$ in the set $\{N(k) : k\in\{1,2,\ldots\}\}$.  For arbitrary $n$, trim the bottom of the solution $P(k)$ for the smallest $k$ with $N(k)\ge n$.  The desired bound holds because $S(k) / S(k-1) = O(1)$.  QED</p>

<hr>

<p>The solutions in the proofs of Lemmas 1 and 2 are <em>well-behaved</em>, in that, for sufficiently large $n$, for each solution $P(n)$ that reaches position $n$ there is a position $k\le n/2$ such that only one pebble is ever placed at position $k$, and the solution decomposes into a (well-behaved) solution $P(N-k)$ for positions $k+1,k+2,\ldots,n$ and two (well-behaved) solutions $P(k)$, each for positions $1,2,\ldots,k$, connected by the pebble at position $k$.  With an appropriate definition of <em>well-behaved</em>, let $V(n)$ denote the minimum <em>pebble volume</em> (the sum over time of the number of pebbles at each time) for any well-behaved solution. The definition implies that for sufficiently large $n$, for $\delta=1&gt;0$,
$$V(n) \ge \min_{k&lt;n} V(n-k) + \max(n/2, (1+\delta) V(k)).$$</p>

<p>I conjecture that for every sufficiently large $n$ there is a well-behaved solution that minimizes pebble volume.  Maybe somebody can prove it?  (Or just that some near-optimal solution satisfies the recurrence...)</p>

<p>Recall that $\epsilon(n) = 1/\sqrt{\log n}$.</p>

<p><strong>Lemma 3.</strong> <em>For any constant $\delta&gt;0$, the above recurrence implies $V(n) \ge  n^{1+\Omega(\epsilon(n))}$.</em></p>

<p>Before we sketch the proof of the lemma, note that it implies that any well-behaved solution that reaches position $n$ in $t$ steps has to take space at least $n^{1+\Omega(\epsilon(n))}/t$ at some step.  This yields corollaries such as:</p>

<ul>
<li>Lemma 1 is tight up to constant factors in the exponent (for well-behaved solutions).</li>
<li>No well-behaved solution can reach position $n$ in $n\,\text{polylog}\, n$ time steps using space $\text{polylog}\, n$.  (Using here that $n^{\Omega(\epsilon(n))} = \exp(\Omega(\sqrt{\log n})) \not\subseteq \,\text{polylog}\, n$.)</li>
</ul>

<p><em>Proof sketch.</em>  We show $2 V(n) \ge f(n)$ where $f(n) = n^{1+c\epsilon(n)}$ for some sufficiently small constant $c.$  We assume WLOG that $n$ is arbitrarily large, because by taking $c&gt;0$ small enough, we can ensure $2 V(n) \ge f(n)$ for any finite set of $n$ (using here that $V(n)\ge n$, say).</p>

<p>The lemma will follow inductively from the recurrence as long as, for all sufficiently large $n$, we have $f(n) \le \min_{k&lt;n} f(n-k) + \max(n, 2 f(k))$, that is, $f(n) - f(n-k) \le \max(n, (1+\delta) f(k))$ for $k&lt;n.$</p>

<p>Since $f$ is convex, we have $f(n) - f(n-k) \le k f'(n)$.  So it suffices if $k f'(n) \le \max(n, (1+\delta) f(k)).$</p>

<p>By a short calculation (using $f(n)/n = e^{c\sqrt{\log n}}$ and $f'(n)=(f(n)/n)(1+c/(2\sqrt{\log n})),$ and using a change of variables $x = \sqrt{\log k}$ and $y=\sqrt{\log n}$), this inequality is equivalent to the following:
for all sufficiently large $y$ and $x\le y$, $e^{cy}(1+c/(2y)) \le \max(e^{y^2 - x^2}, (1+\delta) e^{cx})$.  Since $1+z\le e^z$, and $e^z\le 1+2z$ for $z\le 1$, it suffices to show
$e^{cy + c/(2y)} \le \max(e^{y^2 - x^2}, e^{2\delta+cx}),$ that is,
$$cy + c/(2y) \le \max(y^2 - x^2, 2\delta+cx).$$</p>

<p>If $y \le x + 0.1\delta/c$, then $cy+c/(2y) \le cx + 0.2\delta$ (for large $y$) and we are done, so assume $y\ge x + 0.1\delta/c$.  Then $y^2-x^2 \ge 0.1y\delta/c$ (for large $y$), so it suffices to show
$$cy + c/(2y) \le 0.1 y\delta/c.$$
This holds for sufficiently small $c$ and large $y.$  QED</p>

CC BY-SA 3.0
2017-06-30T01:29:05.813Z
FWIW, I have a proof that there is always a near-optimal well-behaved solution, so the lower bound in Lemma 3 applies to all solutions.  It's a bit too involved to type in here -- if anybody is interested contact me by email (google "neal young computer science" for contact info).
Neal Young
CC BY-SA 3.0
2017-06-17T19:08:23.637Z
2017-06-20T06:44:31.777Z
2017-06-20T06:44:31.777Z
Neal Young
<p>In the <a href="https://en.wikipedia.org/wiki/Pebble_game" rel="noreferrer">pebble game</a> on a line there are N+1 nodes labelled 0 through N. The game starts with a pebble on node 0. If there is a pebble on node i, you can add or remove a pebble from node i+1. The goal is to place a pebble on node N, without placing many pebbles on the board at the same time and without taking too many steps.</p>

<p>The naive solution is to place a pebble on 1, then 2, then 3, and so forth. This is optimal in terms of the number of steps. It's not optimal in the maximum number of pebbles on the board at the same time: during the last step there are N pebbles on the board (not counting the one on 0).</p>

<p>A strategy that places fewer pebbles on the board at the same time is in <a href="https://arxiv.org/abs/quant-ph/9703009" rel="noreferrer">this paper</a>. They reach node N without exceeding $\Theta(\lg N)$ pebbles at a time, but at the cost of increasing the number of steps to $\Theta(n^{\lg_2 3})$. They toggle whether there's a pebble at position $N$ without leaving any other pebbles around by recursively toggling $N/2$, using that as a starting point to toggle $N$ with another recursive step, then toggling $N/2$ with a third half-sized recursive step to clear it.</p>

<p>I'm interested in the tradeoff between the maximum number of pebbles and the number of steps, under the assumption that pebble additions and removals can be done in parallel. By "parallel" I mean that each step can add or remove as many pebbles as desired, as long as each individual addition/removal is allowed and doesn't interact with the other moves being made. Specifically, if $A$ is the set of nodes we want to add or remove pebbles from, and $P$ is the set of nodes that had a pebble at the start of the step, then we can do all those additions and removals in a single step as long as $\{a-1 | a \in A \} \subseteq P - A$.</p>

<p>For example, consider the strategy that places a pebble at $i$ on step $i$, but marks the pebbles that are multiples of $\sqrt{N}$ as "checkpoints" and removes the highest-index pebble behind a pebbled checkpoint whenever possible. This strategy still reaches node N after $N$ steps, like the naive strategy, but reduces the maximum number of pebbles from $N$ to $2 \sqrt{N}$.</p>

<p>Are there parrallel-line-pebbling strategies that finish in $N$ steps with even lower asymptotic max-pebble complexity? What if we're willing to allow $O(N \lg N)$ steps? What are the "interesting" points, where the tradeoff between max-pebble and time is particularly good?</p>

CC BY-SA 3.0
2017-06-15T17:51:15.043Z
In each step, how many pebbles can you add or remove?  If only one, in your fourth paragraph, do you mean $O(N)$ total steps, rather than $N$?  When counting pebbles used, is it the maximum number on the board at any time during the sequence?
Neal Young
CC BY-SA 3.0
2017-06-15T18:08:49.590Z
@NealYoung In the parallel case, you can add and remove as many pebbles per step as you want. But if you affect position $k$ then there must have been a pebble at position $k-1$ present at the start of the step. I meant exactly N steps, but $O(N)$ is also interesting and of course included in $O(N \lg N)$. Yes, it's the maximum number of pebbles that matters.
Craig Gidney
CC BY-SA 3.0
2017-06-16T12:49:57.880Z
What about the original strategy but with "parallelization"? When we reach $N/2$ start clearing the first half in parallel; when reaching $3N/4$ start clearing the range $[N/2-3N/4]$ AND continue clearing the first half in parallel (at the time we place a pebble on $3N/4$ there is only $N/4$ pebbles remaining on the first half); and so on for $(2^k-1)N/2^k$ up to $N$. The pebble complexity should be the same:  $\Theta(\lg N)$, but with $N$ steps.
Marzio De Biasi
CC BY-SA 3.0
2017-06-16T14:17:09.687Z
@MarzioDeBiasi Why would the pebble complexity be the same? As far as I can tell, the recurrence relation would go from $F(n) = F(n/2) + 1 = O(lg(n))$ to $F(n) = 2F(n/2) + 1 = O(n)$.
Craig Gidney
CC BY-SA 3.0
2017-06-16T14:28:50.693Z
@CraigGidney: you're right ...
Marzio De Biasi
CC BY-SA 3.0
2017-06-17T06:52:02.247Z
Here's [a picture of an optimal play](http://imgur.com/a/Rl8rm) for reaching position N in N plays with N = 33 (space = 8).  (Found via integer linear programming.)  FWIW I'm guessing that the asymptotic optimal space will be $N^{o(1)}$, maybe something like $N^{\sqrt{(\log\log N) / \log N}}$.
Neal Young
CC BY-SA 3.0
2017-06-17T17:36:13.250Z
@NealYoung That looks an *awful lot* like Sierpinski's triangle... hmmmm.
Craig Gidney
CC BY-SA 3.0
2017-06-17T18:37:47.997Z
@NealYoung Ah, except Sierpinski's triangle has dimension 1.585, so even if was spread perfectly evenly the columns would have n^0.585 points, which is worse than the sqrt(n) solution.
Craig Gidney
<complexity><reversible-computing>
Parallel Pebble Game on a Line
Craig Gidney
# Parallel Pebble Game on a Line

(Asked by: username_0 on 2017-06-15)

In the [pebble game](https://en.wikipedia.org/wiki/Pebble_game) on a line there are N+1 nodes labelled 0 through N. The game starts with a pebble on node 0. If there is a pebble on node i, you can add or remove a pebble from node i+1. The goal is to place a pebble on node N, without placing many pebbles on the board at the same time and without taking too many steps.

The naive solution is to place a pebble on 1, then 2, then 3, and so forth. This is optimal in terms of the number of steps. It's not optimal in the maximum number of pebbles on the board at the same time: during the last step there are N pebbles on the board (not counting the one on 0).

A strategy that places fewer pebbles on the board at the same time is in [this paper](https://arxiv.org/abs/quant-ph/9703009). They reach node N without exceeding $\Theta(\lg N)$ pebbles at a time, but at the cost of increasing the number of steps to $\Theta(n^{\lg\_2 3})$. They toggle whether there's a pebble at position $N$ without leaving any other pebbles around by recursively toggling $N/2$, using that as a starting point to toggle $N$ with another recursive step, then toggling $N/2$ with a third half-sized recursive step to clear it.

I'm interested in the tradeoff between the maximum number of pebbles and the number of steps, under the assumption that pebble additions and removals can be done in parallel. By "parallel" I mean that each step can add or remove as many pebbles as desired, as long as each individual addition/removal is allowed and doesn't interact with the other moves being made. Specifically, if $A$ is the set of nodes we want to add or remove pebbles from, and $P$ is the set of nodes that had a pebble at the start of the step, then we can do all those additions and removals in a single step as long as $\{a-1 | a \in A \} \subseteq P - A$.

For example, consider the strategy that places a pebble at $i$ on step $i$, but marks the pebbles that are multiples of $\sqrt{N}$ as "checkpoints" and removes the highest-index pebble behind a pebbled checkpoint whenever possible. This strategy still reaches node N after $N$ steps, like the naive strategy, but reduces the maximum number of pebbles from $N$ to $2 \sqrt{N}$.

Are there parrallel-line-pebbling strategies that finish in $N$ steps with even lower asymptotic max-pebble complexity? What if we're willing to allow $O(N \lg N)$ steps? What are the "interesting" points, where the tradeoff between max-pebble and time is particularly good?

**Tags:** complexity, reversible-computing

> **Comments:**
> **username_1**: In each step, how many pebbles can you add or remove?  If only one, in your fourth paragraph, do you mean $O(N)$ total steps, rather than $N$?  When counting pebbles used, is it the maximum number on the board at any time during the sequence?  
> **username_0**: @NealYoung In the parallel case, you can add and remove as many pebbles per step as you want. But if you affect position $k$ then there must have been a pebble at position $k-1$ present at the start of the step. I meant exactly N steps, but $O(N)$ is also interesting and of course included in $O(N \lg N)$. Yes, it's the maximum number of pebbles that matters.  
> **username_2**: What about the original strategy but with "parallelization"? When we reach $N/2$ start clearing the first half in parallel; when reaching $3N/4$ start clearing the range $[N/2-3N/4]$ AND continue clearing the first half in parallel (at the time we place a pebble on $3N/4$ there is only $N/4$ pebbles remaining on the first half); and so on for $(2^k-1)N/2^k$ up to $N$. The pebble complexity should be the same:  $\Theta(\lg N)$, but with $N$ steps.  
> **username_0**: @MarzioDeBiasi Why would the pebble complexity be the same? As far as I can tell, the recurrence relation would go from $F(n) = F(n/2) + 1 = O(lg(n))$ to $F(n) = 2F(n/2) + 1 = O(n)$.  
> **username_2**: @CraigGidney: you're right ...  
> **username_1**: Here's [a picture of an optimal play](http://imgur.com/a/Rl8rm) for reaching position N in N plays with N = 33 (space = 8).  (Found via integer linear programming.)  FWIW I'm guessing that the asymptotic optimal space will be $N^{o(1)}$, maybe something like $N^{\sqrt{(\log\log N) / \log N}}$.  
> **username_0**: @NealYoung That looks an *awful lot* like Sierpinski's triangle... hmmmm.  
> **username_0**: @NealYoung Ah, except Sierpinski's triangle has dimension 1.585, so even if was spread perfectly evenly the columns would have n^0.585 points, which is worse than the sqrt(n) solution.  

---

## 1 answers

### Answer by username_1 on 2017-06-17. Score: 5 | Accepted answer

**EDITS**: *Added Lemmas 2 and 3.*

Here's a partial answer: You can reach position $N$

* in $N$ moves using space $N^{O(\epsilon(N))}$, where $\epsilon(N) = 1/\sqrt{\log N}$. (Lemma 1)
* in $N^{1+\delta}$ moves using space $O(\log N)$ (for any constant $\delta>0$) (Lemma 2).

Also, we sketch a lower bound (Lemma 3): for a certain class of so-called *well-behaved* solutions, Lemma 1 is tight (up to constant factors in the exponent), and no such solution using poly-log space can reach position $N$ in time $O(N\,\text{polylog}\, N)$.

**Lemma 1.** *For all $n$, it is possible to reach position $n$ in $n$ moves using space $$\exp(O(\sqrt{\log n}))~=~n^{O(1/\sqrt{\log n})}$$*

*Proof.* The scheme is recursive, as shown in the figure below. We use the following notation:

* $k$ is the number of levels in the recursion
* $P(k)$ is the solution formed (with $k$ levels of recursion).
* $N(k)$ is the maximum position reached by $P(k)$ (at time $N(k)$).
* $S(k)$ is the space used by $P(k)$.
* $L(k)$ is the number of *layers* used by $P(k)$, as illustrated below:

[![solution structure for lemma 1](https://i.sstatic.net/FG4k7.png)](https://i.sstatic.net/FG4k7.png)

In the picture, time proceeds from top to bottom. The solution $P(k)$ does not stop at time $N(k)$, instead (for use in the recursion) it continues until time $2\,N(k)$, exactly reversing its moves, so as to return to a single pebble at time $2\,N(k)$.

The solid vertical lines partition the $L(k)$ layers of $P(k)$. In the picture, $L(k)$ is five, so $P(k)$ consists of 5 layers. Each of the $L(k)$ layers of $P(k)$ (except the rightmost) has two sub-problems, one at the top of the layer and one at the bottom, connected by a solid vertical line (representing a pebble that exists for that duration). In the picture, there are five layers, so there are nine subproblems. Generally, $P(k)$ is composed of $2\,L(k)-1$ subproblems. Each subproblem of $P(k)$ has solution $P(k-1)$.

The crucial observation for bounding the space is that, at any time, only two layers have "active" subproblems. The rest contribute just one pebble each thus we have

* $S(k) \le L(k) + 2\,S(k-1)$, and
* $N(k) = L(k) \cdot N(k-1)$

Now, we choose $L(k)$ to fully determine $P(k)$. I'm not sure if this choice is optimal, but it seems close: take $L(k) = 2^k$. Then the above recurrences give

* $S(k) \le k\cdot 2^k$, and
* $N(k) = 2^{k(k+1)/2}$

So, solving for $n=N(k)$,
we have $k \approx \sqrt {2 \log n}$
and $S(k) \approx \sqrt {2 \log n}\, 2^{\sqrt {2 \log n}} = \exp(O(\sqrt{\log n}))$.

This takes care of all positions $n$ in the set $\{N(k) : k\in\{1,2,\ldots\}\}$. For arbitrary $n$, trim the bottom of the solution $P(k)$ for the smallest $k$ with $N(k)\ge n$. The desired bound holds because $S(k) / S(k-1) = O(1)$. QED

---

**Lemma 2.** *For any $\delta>0$, for all $n$, it is possible to reach position $n$ in $n^{1+\delta}$ moves using space $O(\delta 2^{1/\delta}\log n).$*

*Proof.* Modify the construction from the proof of Lemma 1 to delay starting each subproblem until the previous subproblem has finished, as shown below:

[![solution structure for lemma 2](https://i.sstatic.net/lH89u.png)](https://i.sstatic.net/lH89u.png)

Let $T(k)$ denote the time for the modified solution $P(k)$ to finish. Now at each time step, only one layer has a subproblem that contributes more than one pebble, so

* $S(k) \le L(k) + S(k-1)$,
* $N(k) = L(k) \cdot N(k-1)$,
* $T(k) = (2\,L(k)-1)\cdot T(k-1) \le 2\,L(k)\cdot T(k-1) \le 2^k N(k)$.

Choosing $L(k) = 2^{1/\delta}$, we get

* $S(k) \le k 2^{1/\delta}$,
* $N(k) = 2^{k/\delta}$,
* $T(k) \le 2^k N(k)$.

Solving for $S = S(k)$ and $T=T(k)$ in terms of $n=N(k)$, we have $k=\delta \log n$, and

* $S \le \delta 2^{1/\delta} \log n$, and
* $T \le n^{1+\delta}$.

This takes care of all positions $n$ in the set $\{N(k) : k\in\{1,2,\ldots\}\}$. For arbitrary $n$, trim the bottom of the solution $P(k)$ for the smallest $k$ with $N(k)\ge n$. The desired bound holds because $S(k) / S(k-1) = O(1)$. QED

---

The solutions in the proofs of Lemmas 1 and 2 are *well-behaved*, in that, for sufficiently large $n$, for each solution $P(n)$ that reaches position $n$ there is a position $k\le n/2$ such that only one pebble is ever placed at position $k$, and the solution decomposes into a (well-behaved) solution $P(N-k)$ for positions $k+1,k+2,\ldots,n$ and two (well-behaved) solutions $P(k)$, each for positions $1,2,\ldots,k$, connected by the pebble at position $k$. With an appropriate definition of *well-behaved*, let $V(n)$ denote the minimum *pebble volume* (the sum over time of the number of pebbles at each time) for any well-behaved solution. The definition implies that for sufficiently large $n$, for $\delta=1>0$,
$$V(n) \ge \min\_{k

I conjecture that for every sufficiently large $n$ there is a well-behaved solution that minimizes pebble volume. Maybe somebody can prove it? (Or just that some near-optimal solution satisfies the recurrence...)

Recall that $\epsilon(n) = 1/\sqrt{\log n}$.

**Lemma 3.** *For any constant $\delta>0$, the above recurrence implies $V(n) \ge n^{1+\Omega(\epsilon(n))}$.*

Before we sketch the proof of the lemma, note that it implies that any well-behaved solution that reaches position $n$ in $t$ steps has to take space at least $n^{1+\Omega(\epsilon(n))}/t$ at some step. This yields corollaries such as:

* Lemma 1 is tight up to constant factors in the exponent (for well-behaved solutions).
* No well-behaved solution can reach position $n$ in $n\,\text{polylog}\, n$ time steps using space $\text{polylog}\, n$. (Using here that $n^{\Omega(\epsilon(n))} = \exp(\Omega(\sqrt{\log n})) \not\subseteq \,\text{polylog}\, n$.)

*Proof sketch.* We show $2 V(n) \ge f(n)$ where $f(n) = n^{1+c\epsilon(n)}$ for some sufficiently small constant $c.$ We assume WLOG that $n$ is arbitrarily large, because by taking $c>0$ small enough, we can ensure $2 V(n) \ge f(n)$ for any finite set of $n$ (using here that $V(n)\ge n$, say).

The lemma will follow inductively from the recurrence as long as, for all sufficiently large $n$, we have $f(n) \le \min\_{k

Since $f$ is convex, we have $f(n) - f(n-k) \le k f'(n)$. So it suffices if $k f'(n) \le \max(n, (1+\delta) f(k)).$

By a short calculation (using $f(n)/n = e^{c\sqrt{\log n}}$ and $f'(n)=(f(n)/n)(1+c/(2\sqrt{\log n})),$ and using a change of variables $x = \sqrt{\log k}$ and $y=\sqrt{\log n}$), this inequality is equivalent to the following:
for all sufficiently large $y$ and $x\le y$, $e^{cy}(1+c/(2y)) \le \max(e^{y^2 - x^2}, (1+\delta) e^{cx})$. Since $1+z\le e^z$, and $e^z\le 1+2z$ for $z\le 1$, it suffices to show
$e^{cy + c/(2y)} \le \max(e^{y^2 - x^2}, e^{2\delta+cx}),$ that is,
$$cy + c/(2y) \le \max(y^2 - x^2, 2\delta+cx).$$

If $y \le x + 0.1\delta/c$, then $cy+c/(2y) \le cx + 0.2\delta$ (for large $y$) and we are done, so assume $y\ge x + 0.1\delta/c$. Then $y^2-x^2 \ge 0.1y\delta/c$ (for large $y$), so it suffices to show
$$cy + c/(2y) \le 0.1 y\delta/c.$$
This holds for sufficiently small $c$ and large $y.$ QED

> **Comments:**
> **username_1**: FWIW, I have a proof that there is always a near-optimal well-behaved solution, so the lower bound in Lemma 3 applies to all solutions.  It's a bit too involved to type in here -- if anybody is interested contact me by email (google "neal young computer science" for contact info).  

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