security_proof.tex 19.6 KB
 kklein committed Oct 08, 2018 1 2 To proof that our construction indeed is a proof of space, we argue via pebbling. More precisely, we show that the DAG corresponding to the skiplist can only be pebbled using either large space in the initial configuration or a large amount of time. In other words, any prover who didn't store (a large fraction of) the whole output file needs a very long time to answer challenges correctly.  kklein committed Oct 25, 2018 3 We will proceed as follows. First, we show that the claim holds for the original construction based on the skiplist of length $k\cdot S$ with random permutations $\Pi=\{\pi_i\}_{i\in[k\cdot S]_0}$, $\pi_i:\{0,1\}^{w\cdot\tilde i}\to\{0,1\}^{w\cdot\tilde i}$. Then we prove that the claim still remains true if we replace the random permutations $\pi_i$ by butterfly graphs $B_i$ with permutaions $\sigma_i:\bin^w\to\bin^w$.  kklein committed Oct 08, 2018 4   kklein committed Oct 25, 2018 5 6 7  Let $\Gsk$ denote the DAG corresponding to the skiplist of length $k\cdot S$ on data of size $S=2^s$, see Figure \ref{fig:DAG}. \begin{figure}  kklein committed Oct 08, 2018 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 \centering \scalebox{0.8}{ \begin{tikzpicture}[ node distance=2.5em, vertex/.style={draw,circle,thick}, sedge/.style={->} ] \node[vertex] (v00) {}; \foreach \i/\j in {0/1,1/2,2/3,3/4,4/5,5/6,6/7,7/8,8/9,9/10,10/11,11/12,12/13,13/14,14/15,15/16,16/17}{ \node[vertex,right of=v0\i] (v0\j) {}; \draw[sedge] (v0\i) -> (v0\j); } \foreach \i/\j/\k in {1/3/2,3/5/4,5/7/6,7/9/8,9/11/10,11/13/12,13/15/14,15/17/16}{ \node[vertex,above of=v0\i,yshift=0.3em] (v1\i) {}; \node[vertex,above of=v0\j,yshift=0.3em] (v1\j) {}; \draw[sedge] (v1\i) -> (v1\j); \draw[sedge] (v1\i) -> (v0\j); \draw[sedge] (v0\k) -> (v1\j); } \foreach \i/\j in {1/5,5/9,9/13,13/17}{ \node[vertex,above of=v1\i,yshift=0.3em] (v2\i) {}; \node[vertex,above of=v1\j,yshift=0.3em] (v2\j) {}; \draw[sedge] (v2\i) -> (v2\j); \node[vertex,above of=v2\i,yshift=0.3em] (v3\i) {}; \node[vertex,above of=v2\j,yshift=0.3em] (v3\j) {}; \draw[sedge] (v3\i) -> (v3\j); } \foreach \k/\l in {3/4,4/5,5/6,6/7}{ \foreach \i/\j in {1/9,9/17}{ \node[vertex,above of=v\k\i,yshift=0.3em] (v\l\i) {}; \node[vertex,above of=v\k\j,yshift=0.3em] (v\l\j) {}; \draw[sedge] (v\l\i) -> (v\l\j); } } \foreach \i in {1,2,3,4,5,6,7}{ \node[vertex,left of=v\i1] (v\i0) {}; \draw[sedge] (v\i0) -> (v\i1); } \foreach \i/\j in {0/4,1/3,2/1,3/1}{ \foreach \k in {0,...,3}{ \draw[sedge] (v\i\j) -> (v\k5); } } \foreach \i/\j in {0/12,1/11,2/9,3/9}{ \foreach \k in {0,...,3}{ \draw[sedge] (v\i\j) -> (v\k13); } } \foreach \i/\j in {0/8,1/7,2/5,3/5,4/1,5/1,6/1,7/1}{ \foreach \k in {0,...,7}{ \draw[sedge] (v\i\j) -> (v\k9); } } \foreach \i/\j in {0/16,1/15,2/13,3/13,4/9,5/9,6/9,7/9}{ \foreach \k in {0,...,7}{ \draw[sedge] (v\i\j) -> (v\k17); } } \end{tikzpicture} }  kklein committed Oct 25, 2018 78 \caption{The DAG $\Gsk$ corresponding to our construction for data of size $S=8$ $w$-blocks and length $k\cdot S$ with $k=2$.\label{fig:DAG}}  kklein committed Oct 08, 2018 79 80 81 82 83 84 85 \end{figure} \begin{lemma} For any $k\in\mathbb{N}$, $k>1$, any $\epsilon\in [0,1)$, the graph $\Gsk$ has the following property: Whenever one removes $\epsilon\cdot S$ nodes from $\Gsk$, for any remaining output node there exists a path of length $>k\cdot S-2\epsilon\cdot S$ which starts at some input node. In particular, $\Gsk$ is $(\epsilon\cdot S, k\cdot S-2\epsilon\cdot S)$-depth robust. For $k=1$, the claim holds for $\epsilon\in [0,\frac{1}{2})$. \end{lemma} \begin{proof} For $S=1$, the claim is trivially true; thus we only consider the case $S=2^s\geq 2$. First we show that it is enough to prove the claim for $k=2$.\\  kklein committed Oct 09, 2018 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 For the case $k=1$, note that the graph $\G_{S,1}$ contains $\G_{S/2,2}$ as a subgraph such that the input (output) nodes of $\G_{S/2,2}$ are also input (output) nodes of $\G_{S,1}$. \\ \begin{figure} \centering \scalebox{0.8}{ \begin{tikzpicture}[ node distance=2.2em, vertex/.style={draw,circle,thick}, sedge/.style={->}, dstate/.style={minimum width=2em,minimum height=2em}, ] \node[vertex] (v00) {}; \foreach \i/\j in {0/1,1/2,2/3,3/4,4/5}{ \node[vertex,right of=v0\i] (v0\j) {}; \draw[sedge] (v0\i) -> (v0\j); } \foreach \i/\j/\k in {1/3/2,3/5/4}{ \node[vertex,above of=v0\i,yshift=0.3em] (v1\i) {}; \node[vertex,above of=v0\j,yshift=0.3em] (v1\j) {}; \draw[sedge] (v1\i) -> (v1\j); \draw[sedge] (v1\i) -> (v0\j); \draw[sedge] (v0\k) -> (v1\j); } \foreach \i/\j in {1/5}{ \node[vertex,above of=v1\i,yshift=0.3em] (v2\i) {}; \node[vertex,above of=v1\j,yshift=0.3em] (v2\j) {}; \draw[sedge] (v2\i) -> (v2\j); \node[vertex,above of=v2\i,yshift=0.3em] (v3\i) {}; \node[vertex,above of=v2\j,yshift=0.3em] (v3\j) {}; \draw[sedge] (v3\i) -> (v3\j); } \foreach \i in {1,2,3}{ \node[vertex,left of=v\i1] (v\i0) {}; \draw[sedge] (v\i0) -> (v\i1); } \foreach \i/\j in {0/4,1/3,2/1,3/1}{ \foreach \k in {0,...,3}{ \draw[sedge] (v\i\j) -> (v\k5); } } \draw[red,very thick,dotted] ($(v00.south west)+(-0.3,-0.3)$) rectangle ($(v15.north east)+(0.3,0.3)$); \node[dstate,below of=v02, xshift=1em] (G) {$\G_{\frac{S}{2},2}$}; \end{tikzpicture} } \caption{The graph $\G_{S,1}$ contains $\G_{S/2,2}$ as a subgraph.} \end{figure} Let $\epsilon\in[0,\frac{1}{2})$. By assumption, there are $\epsilon S=(2\epsilon)\cdot S/2$ pebbles on $\G_{S/2,2}\subset\G_{S,1}$. Thus, if the claim holds for $k=2$, there exists a path from some input node of $\G_{S/2,2}$ to any unpebbled output node of length $>2\cdot S/2-2(2\epsilon)\cdot S/2=S-2\epsilon\cdot S$.\\ For arbitrary integers $k=2\ell+m$ with $\ell\in\mathbb{N}$, $m\in\bin$, the graph $\Gsk$ can be considered as a chain of graphs $\{\G_{S,2}^{(i)}\}_{i=1}^\ell$ and $\G_{S,m}$ where the output and input nodes of two subsequent subgraphs are glued together, respectively. More precisely, we partition the graph $\Gsk$ into $\ell+1$ blocks of length $2S$ and $S$ respectively, where $\ell$ blocks are a copy of $\G_{S,2}$ and one a copy of $\G_{S,1}$, all except the last one being cut right before the output, i.e., we count the output nodes of each block only as input nodes for the subsequent block, see Figure \ref{fig:Gsk}.\\ \begin{figure}\label{fig:Gsk} \centering \scalebox{0.8}{ \begin{tikzpicture}[ node distance=2.2em, vertex/.style={draw,circle,thick}, sedge/.style={->}, dstate/.style={minimum width=2em,minimum height=2em}, ] \node[vertex] (v00) {}; \foreach \i/\j in {0/1,1/2,2/3,3/4,4/5,5/6,6/7,7/8,8/9,9/10,10/11,11/12,12/13,13/14,14/15,15/16,16/17,17/18,18/19,19/20,20/21}{ \node[vertex,right of=v0\i] (v0\j) {}; \draw[sedge] (v0\i) -> (v0\j); } \foreach \i/\j/\k in {1/3/2,3/5/4,5/7/6,7/9/8,9/11/10,11/13/12,13/15/14,15/17/16,17/19/18,19/21/20}{ \node[vertex,above of=v0\i,yshift=0.3em] (v1\i) {}; \node[vertex,above of=v0\j,yshift=0.3em] (v1\j) {}; \draw[sedge] (v1\i) -> (v1\j); \draw[sedge] (v1\i) -> (v0\j); \draw[sedge] (v0\k) -> (v1\j); } \foreach \i/\j in {1/5,5/9,9/13,13/17,17/21}{ \node[vertex,above of=v1\i,yshift=0.3em] (v2\i) {}; \node[vertex,above of=v1\j,yshift=0.3em] (v2\j) {}; \draw[sedge] (v2\i) -> (v2\j); \node[vertex,above of=v2\i,yshift=0.3em] (v3\i) {}; \node[vertex,above of=v2\j,yshift=0.3em] (v3\j) {}; \draw[sedge] (v3\i) -> (v3\j); } \foreach \i in {1,2,3}{ \node[vertex,left of=v\i1] (v\i0) {}; \draw[sedge] (v\i0) -> (v\i1); } \foreach \i/\j in {0/4,1/3,2/1,3/1}{ \foreach \k in {0,...,3}{ \draw[sedge] (v\i\j) -> (v\k5); } } \foreach \i/\j in {0/8,1/7,2/5,3/5}{ \foreach \k in {0,...,3}{ \draw[sedge] (v\i\j) -> (v\k9); } } \foreach \i/\j in {0/12,1/11,2/9,3/9}{ \foreach \k in {0,...,3}{ \draw[sedge] (v\i\j) -> (v\k13); } } \foreach \i/\j in {0/16,1/15,2/13,3/13}{ \foreach \k in {0,...,3}{ \draw[sedge] (v\i\j) -> (v\k17); } } \foreach \i/\j in {0/20,1/19,2/17,3/17}{ \foreach \k in {0,...,3}{ \draw[sedge] (v\i\j) -> (v\k21); } } \draw[red,very thick,dotted] ($(v00.south west)+(-0.3,-0.3)$) rectangle ($(v39.north west)+(-0.3,0.3)$); \draw[red,very thick,dotted] ($(v09.south west)+(-0.3,-0.3)$) rectangle ($(v313.north west)+(-0.3,0.3)$); \draw[red,very thick,dotted] ($(v013.south west)+(-0.3,-0.3)$) rectangle ($(v321.north west)+(0.5,0.3)$); \node[dstate,below of=v04] (G1) {$\G_{S,2}^{(1)}$}; \node[dstate,below of=v010,xshift=1em] (G2) {$\G_{S,1}$}; \node[dstate,below of=v017] (G3) {$\G_{S,2}^{(2)}$}; \end{tikzpicture} } \caption{Partitioning of the DAG $\Gsk$ with $k=5$ into two blocks $\G_{S,2}$ of length $2S$ each, and one block $\G_{S,1}$ of length $S$.} \end{figure} Since by assumtion there are $\epsilon S$ pebbles on $\Gsk$, with $\epsilon\in[0,1)$, at most one of the blocks contains $\geq S/2$ pebbles and, thus, we can partition the graph in a way such that there are $\epsilon_iS$ pebbles on block $\G_{S,2}^{(i)}$ and $\epsilon'S$ pebbles on $\G_{S,m}$ with $\sum_{i=1}^\ell\epsilon_i+\epsilon'=\epsilon$, $\epsilon_i\in[0,1)$, and $\epsilon'\in[0,\frac{1}{2})$. If the claim is true for $k=2$, by the above it also holds for $k=1$ and $\epsilon'\in[0,\frac{1}{2})$. Hence, for each of the blocks $\G_{S,2}^{(i)}$ and $\G_{S,m}$ there exists a path from some input node to any unpebbled output node (where the output nodes are identified as the input nodes of the subsequent block), respectively. Thus, for any unpebbled output node of $\Gsk$ there exists a path of length at least $\sum_{i=1}^\ell(2S-2\epsilon_iS)+m(S-2\epsilon'S)=(2\ell+m)S-2\epsilon S$ starting at some input node, which proves the claim.\\ It remains to prove the claim for $k=2$. We do so via induction on $s=\log S$. As noted in the beginning of the proof, the claim is trivially true for $s=0$. For $s>0$ we partition the graph into two blocks $\G_{S,1}^{(1)}$, $\G_{S,1}^{(2)}$ of length $2^s+1$ each. \\ \begin{figure}\label{fig:Gsk} \centering \scalebox{0.8}{ \begin{tikzpicture}[ node distance=2.2em, vertex/.style={draw,circle,thick}, sedge/.style={->}, dstate/.style={color=red,minimum width=2em,minimum height=2em}, dstate2/.style={color=blue,minimum width=2em,minimum height=2em}, ] \node[vertex] (v00) {}; \foreach \i/\j in {0/1,1/2,2/3,3/4,4/5,5/6,6/7,7/8,8/9}{ \node[vertex,right of=v0\i] (v0\j) {}; \draw[sedge] (v0\i) -> (v0\j); } \foreach \i/\j/\k in {1/3/2,3/5/4,5/7/6,7/9/8}{ \node[vertex,above of=v0\i,yshift=0.3em] (v1\i) {}; \node[vertex,above of=v0\j,yshift=0.3em] (v1\j) {}; \draw[sedge] (v1\i) -> (v1\j); \draw[sedge] (v1\i) -> (v0\j); \draw[sedge] (v0\k) -> (v1\j); } \foreach \i/\j in {1/5,5/9}{ \node[vertex,above of=v1\i,yshift=0.3em] (v2\i) {}; \node[vertex,above of=v1\j,yshift=0.3em] (v2\j) {}; \draw[sedge] (v2\i) -> (v2\j); \node[vertex,above of=v2\i,yshift=0.3em] (v3\i) {}; \node[vertex,above of=v2\j,yshift=0.3em] (v3\j) {}; \draw[sedge] (v3\i) -> (v3\j); } \foreach \i in {1,2,3}{ \node[vertex,left of=v\i1] (v\i0) {}; \draw[sedge] (v\i0) -> (v\i1); } \foreach \i/\j in {0/4,1/3,2/1,3/1}{ \foreach \k in {0,...,3}{ \draw[sedge] (v\i\j) -> (v\k5); } } \foreach \i/\j in {0/8,1/7,2/5,3/5}{ \foreach \k in {0,...,3}{ \draw[sedge] (v\i\j) -> (v\k9); } } \draw[red,very thick,dotted] ($(v00.south west)+(-0.3,-0.3)$) rectangle ($(v35.north west)+(-0.3,0.3)$); \draw[red,very thick,dotted] ($(v05.south west)+(-0.3,-0.3)$) rectangle ($(v39.north east)+(0.3,0.3)$); \draw[blue,very thick,dotted] ($(v00.south west)+(-0.2,-0.2)$) rectangle ($(v15.north west)+(-0.4,0.2)$); % \draw[blue,very thick,dotted] ($(v05.south west)+(-0.2,-0.2)$) rectangle ($(v19.north east)+(0.2,0.2)$); \node[dstate,below of=v02] (G1) {$\G_{S,1}^{(1)}$}; \node[dstate,below of=v07] (G2) {$\G_{S,1}^{(2)}$}; \node[dstate2,below of=v10, xshift=1.2em, yshift=0.8em] (G3) {$\G_{S/2,2}^{(1)}$}; % \node[dstate,below of=v02] (G4) {$\G_{S/2,2}^{(2)}$}; \end{tikzpicture} } \caption{Partitioning of the DAG $\G_{S,2}$ into two blocks $\G_{S,1}^{(1)}$ and $\G_{S,1}^{(2)}$ of length $S+1$ each.} \end{figure} Let $\G_{S,1}^{(1)}$, $\G_{S,1}^{(2)}$ contain $\epsilon_1S/2$ and $\epsilon_2S/2$ pebbles, respectively. It holds $\epsilon=\epsilon_1/2+\epsilon_2/2$ and we consider the following two cases: \begin{itemize} \item If $\epsilon_1,\epsilon_2<1$, by induction hypothesis the sybgraphs $\G_{S/2,2}^{(1)}\subset\G_{S,1}^{(1)}$ and $\G_{S/2,2}^{(2)}\subset\G_{S,1}^{(2)}$ contain paths of length $>S-\epsilon_1S$ and $>S-\epsilon_2S$, respectively, which connect some input nodes to any arbitrary output nodes. Thus, there is a path of length $>S-\epsilon_1S+S-\epsilon_2S=2S-2\epsilon S$. \item Now, consider the case where either $\epsilon_1\geq1$ or $\epsilon_2\geq1$. We assume $\epsilon_2\geq 1$. Since $\epsilon_1+\epsilon_2=2\epsilon<2$, we have $\epsilon_1<1$. Thus, by induction hypothesis, there is a path of length $>S-\epsilon_1S$ from some input node of $\G_{S/2,2}^{(1)}\subset\G_{S,1}^{(1)}$ to any arbitrary output node of $\G_{S/2,2}^{(1)}$, and in fact to any unpebbled output node of $\G_{S,1}^{(1)}$, which are identified with the input nodes of $\G_{S,1}^{(2)}$. Since there are $\epsilon_2 S\leq\epsilon SS-\epsilon_1S+1$ connecting it to some unpebbled input node. On the other hand, since $\epsilon_2\geq 1$, we have $2S-2\epsilon S=2S-(\epsilon_1+\epsilon_2) S<2S-\epsilon_1S-S=S-\epsilon_1 S.$ This proves the claim. \end{itemize}  kklein committed Oct 25, 2018 292 293 \end{proof}  kklein committed Oct 25, 2018 294 It remains to prove that the statement of the Lemma remains true if we replace the random permutations $\pi$ by butterfly graphs $B_i$ with random permutations $\sigma_i$ on $\bin^{2\cdot w/2}$ \kknote{For now, blocks of size $w/2$ to be consistent with the above, but need to change notation...}, see Figure \ref{fig:B8}.\footnote{Note, that our graph $B_S$ in fact consists of two copies of the original butterfly graph, hence has length $2\log (2S)$. Still, by misuse of the name, we will refer to $B_S$ as the butterfly graph of size $S$. We also note, that the simple butterfly graph would not be sufficient for our needs, i.e., the claim of the Lemma below does not hold for simple butterfly graphs.} More precisely, we need to prove that pebbling internal nodes of the permutations $\B_i$ doesn't help to decrease the number and structure of paths between input and output nodes of $\B_i$ from what one could achieve by pebbling only input and output nodes of $\B_i$.\footnote{Note, in the case of random permutations, we used a complete bipartite graph to represent the computation. Here, we show that the butterfly graph is in some sense close to a comlete bipartite graph.}  kklein committed Oct 25, 2018 295 296 297 298 299 300 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340 341 342 343 344 345 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360 361 362 363 364 365 366 367 368 369 370 371 372 373 374 375 376 377 378 379 380 381 382 383 384 385 386 387 388 389 390 391 392 393 394 395 396 397 398 399 400 401 402 403 404 405 406 407 408 409 410 411 412 413 414 415 416 417 418 419 420 421  \begin{figure} \centering \scalebox{0.8}{ \begin{tikzpicture}[ node distance=2.5em, vertex/.style={draw,circle,thick}, vertexI/.style={draw,red,circle,thick}, vertexII/.style={draw,blue,circle,thick}, sedge/.style={->}, sedgeI/.style={red,->,very thick}, sedgeII/.style={blue,->,very thick} ] \node[vertex] (v00) {}; \foreach \k/\l in {0/1,1/2,2/3,3/4,4/5,5/6,6/7}{ \node[vertex, above of=v\k0] (v\l0) {}; } \foreach \k in {0,...,7}{ \node[vertex, right of=v\k0] (v\k1) {}; \node[vertex, right of=v\k0,xshift=3.5em] (v\k2) {}; \node[vertex, right of=v\k0,xshift=8em] (v\k3) {}; \node[vertex, right of=v\k0,xshift=12.5em] (v\k4) {}; \node[vertex, right of=v\k0,xshift=16em] (v\k5) {}; \node[vertex, right of=v\k0,xshift=18.5em] (v\k6) {}; } \foreach \k in {0,...,7}{ \foreach \l/\m in {0/1,1/2,2/3,3/4,4/5,5/6}{ \draw[sedge] (v\k\l) -> (v\k\m); } } \foreach \k/\l in {0/1,5/6}{ \foreach \i/\j in {0/1,2/3,4/5,6/7}{ \draw[sedge] (v\i\k) -> (v\j\l); \draw[sedge] (v\j\k) -> (v\i\l); } } \foreach \k/\l in {1/2,4/5}{ \foreach \i/\j in {0/2,1/3,4/6,5/7}{ \draw[sedge] (v\i\k) -> (v\j\l); \draw[sedge] (v\j\k) -> (v\i\l); } } \foreach \k/\l in {2/3,3/4}{ \foreach \i/\j in {0/4,1/5,2/6,3/7}{ \draw[sedge] (v\i\k) -> (v\j\l); \draw[sedge] (v\j\k) -> (v\i\l); } } %red subgraph \foreach \k in {0,2,4,6}{ \node[vertexI, right of=v\k0] (v\k1) {}; \node[vertexI, right of=v\k0,xshift=3.5em] (v\k2) {}; \node[vertexI, right of=v\k0,xshift=8em] (v\k3) {}; \node[vertexI, right of=v\k0,xshift=12.5em] (v\k4) {}; \node[vertexI, right of=v\k0,xshift=16em] (v\k5) {}; } \foreach \k in {0,2,4,6}{ \foreach \l/\m in {1/2,2/3,3/4,4/5}{ \draw[sedgeI] (v\k\l) -> (v\k\m); } } \foreach \k/\l in {1/2,4/5}{ \foreach \i/\j in {0/2,4/6}{ \draw[sedgeI] (v\i\k) -> (v\j\l); \draw[sedgeI] (v\j\k) -> (v\i\l); } } \foreach \k/\l in {2/3,3/4}{ \foreach \i/\j in {0/4,2/6}{ \draw[sedgeI] (v\i\k) -> (v\j\l); \draw[sedgeI] (v\j\k) -> (v\i\l); } } %blue subgraph \foreach \k in {1,3,5,7}{ \node[vertexII, right of=v\k0] (v\k1) {}; \node[vertexII, right of=v\k0,xshift=3.5em] (v\k2) {}; \node[vertexII, right of=v\k0,xshift=8em] (v\k3) {}; \node[vertexII, right of=v\k0,xshift=12.5em] (v\k4) {}; \node[vertexII, right of=v\k0,xshift=16em] (v\k5) {}; } \foreach \k in {1,3,5,7}{ \foreach \l/\m in {1/2,2/3,3/4,4/5}{ \draw[sedgeII] (v\k\l) -> (v\k\m); } } \foreach \k/\l in {1/2,4/5}{ \foreach \i/\j in {1/3,5/7}{ \draw[sedgeII] (v\i\k) -> (v\j\l); \draw[sedgeII] (v\j\k) -> (v\i\l); } } \foreach \k/\l in {2/3,3/4}{ \foreach \i/\j in {1/5,3/7}{ \draw[sedgeII] (v\i\k) -> (v\j\l); \draw[sedgeII] (v\j\k) -> (v\i\l); } } \end{tikzpicture} } \caption{The butterfly graph $\B_S$ on data of size $S=4$ $w$-blocks (i.e., 8 blocks of size $w/2$). The two embeddings I and II of the subgraph $\B_{S/2}$ are colored red and blue, respectively.\label{fig:B8}} \end{figure} \begin{lemma} For every $\epsilon\in[0,1)$ and any configuration of $\epsilon\cdot 2S$ pebbles\footnote{Here, each node comprises a block of size $w/2$.} on $\B_S$, there exists $\epsilon'\in[0,\epsilon]$ such that th following holds: There exists a subset $Y$ of the sinks of $\B_S$ with $|Y|=(1-\epsilon')\cdot 2S$, and a subset $X$ of the sources of $\B_S$ with $|X|=(1-(\epsilon-\epsilon'))\cdot 2S$ such that for any pair $(x,y)\in X\times Y$ there exists a path from $x$ to $y$. \end{lemma} \begin{proof} We prove the statement via induction on $s=\log S\geq 0$. For $s=0$, it is easy to see that the claim is true. Now, assume the claim holds for $\B_{S/2}$. Consider the two node-disjoint embeddings of $\B_{S/2}$ into $\B_S$ which we refer to as $I$ and $II$, respectively, as depicted in Figure \ref{fig:B8}. Let $\epsilon_1\cdot S$, $\epsilon_2\cdot S$, $\epsilon_{in}\cdot 2S$, and $\epsilon_{out}\cdot 2S$ be the number of pebbles on $I$, $II$, the input yo $\B_S$, and the output of $\B_S$, respectively. Hence, $\epsilon_1/2+\epsilon_2/2+\epsilon_{in}+\epsilon_{out}=\epsilon$. It follows $\epsilon_1+\epsilon_2=2(\epsilon-\epsilon_{in}-\epsilon_{out})$ and w.l.o.g. we assume $\epsilon_1\leq \epsilon-\epsilon_{in}-\epsilon_{out}$. By induction hypothesis, there exists $\epsilon_1'\in[0,\epsilon_1]$ and subsets $X_1$, $Y_1$ of the sources and sinks of $I$ of cardinality $(1-\epsilon_1')S$ and $(1-(\epsilon_1-\epsilon_1'))S$, respectively, such that for every pair $(x_1,y_1)\in X_1\times Y_1$ there exists a path from $x_1$ to $y_1$ in $I$. Since each of the nodes in $X_1$, $Y_1$ is neighboring to 2 distinct input, output nodes of $\B_S$, respectively, it follows that there exists a subset $X$ of $(1-(\epsilon_1-\epsilon_1')-\epsilon_{in})2S$ sources of $B_S$ and a subset $Y$ of $(1-\epsilon_1'-\epsilon_{out})2S$ sinks of $\B_S$ which are completely connected. By the assumption on $\epsilon_1$, we get $1-(\epsilon_1-\epsilon_1')-\epsilon_{in}\geq 1-(\epsilon-\epsilon_{in}-\epsilon_{out}-\epsilon_1')-\epsilon_{in}=1-(\epsilon-\epsilon_{out}-\epsilon_1').$ Thus, the claim follows with $\epsilon':=\epsilon_1'+\epsilon_{out}\in[0,\epsilon]$. \end{proof}