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Finish the description of range checks in Appendix A.
Signed-off-by: Daira Hopwood <daira@jacaranda.org>
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@ -773,6 +773,7 @@ electronic commerce and payment, financial privacy, proof of work, zero knowledg
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\newcommand{\homomorphicPedersenCommitments}{\term{homomorphic Pedersen commitments}}
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\newcommand{\homomorphicPedersenCommitments}{\term{homomorphic Pedersen commitments}}
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\newcommand{\HomomorphicPedersenCommitment}{\titleterm{Homomorphic Pedersen Commitment}}
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\newcommand{\HomomorphicPedersenCommitment}{\titleterm{Homomorphic Pedersen Commitment}}
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\newcommand{\distinctXCriterion}{\term{distinct-$x$ criterion}}
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\newcommand{\distinctXCriterion}{\term{distinct-$x$ criterion}}
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\newcommand{\Nary}{\mbox{$N$-ary}}
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% Conventions
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% Conventions
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@ -876,6 +877,7 @@ electronic commerce and payment, financial privacy, proof of work, zero knowledg
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\newcommand{\grpneg}{\bigboxminus{1.8ex}}
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\newcommand{\grpneg}{\bigboxminus{1.8ex}}
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\newcommand{\suband}{\raisebox{-0.6ex}{\kern-0.06em\scalebox{0.65}{$\binampersand$}}}
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@ -9624,6 +9626,16 @@ Peter Newell's illustration of the Jubjub bird, from \cite{Carroll1902}.
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\intropart
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\intropart
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\section{Change History}
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\section{Change History}
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\subparagraph{2018.0-beta-29}
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\begin{itemize}
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\item No changes to \Sprout.
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\sapling{
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\item Finish \crossref{cctrange}.
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} %sapling
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\end{itemize}
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\introlist
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\subparagraph{2018.0-beta-28}
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\subparagraph{2018.0-beta-28}
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\begin{itemize}
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\begin{itemize}
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@ -10762,30 +10774,120 @@ Note that since $a$ and $c$ are provided in binary representation, their
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bit length $n$ is not limited by the field element size. We \emph{do not} assume
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bit length $n$ is not limited by the field element size. We \emph{do not} assume
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that the bits $a_\barerange{0}{n-1}$ are already boolean-constrained.
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that the bits $a_\barerange{0}{n-1}$ are already boolean-constrained.
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Suppose $c$ has $k$ bits set to $1$, and let $j_\barerange{0}{k-1}$ be the
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Define $\Pi_{m} = \sproduct{i=m}{n-1} (c_i = 0 \bor a_i = 1)$ for $m \in \range{0}{n-1}$.
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indices of those bits in ascending order. Let $t$ be the minimum of $k-1$ and
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Notice that for any $m < n-1$ such that $c_m = 0$, we have $\Pi_m = \Pi_{m+1}$,
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the number of trailing $1$ bits in $c$.
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and so it is only necessary to allocate separate variables for the $\Pi_m$
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such that $m < n-1$ and $c_m = 1$. Furthermore if $c_{\barerange{n-2}{0}}$ has
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$t > 0$ trailing $1$ bits, then we do not need to allocate variables for
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$\Pi_{\barerange{0}{t-1}}$ because those variables will not be used below.
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More explicitly:
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\introlist
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Let $\Pi_{n-1} = a_{n-1}$.
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Let $\Pi_{j_{k-1}} = a_{j_{k-1}}$. For $z \in \range{t}{k-2}$, constrain:
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\begin{formulae}
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For $i \from n-2 \downto t$,
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\item $\constraint{\Pi_{j_{z+1}}}{a_{j_z}}{\Pi_{j_z}}$
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\end{formulae}
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\introlist
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For $i \in \range{0}{n-1}$:
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\begin{itemize}
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\begin{itemize}
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\item if $c_i = 0$, constrain $\constraint{1 - \Pi_{j_z} - a_i}{a_i}{0}$ where $j_z$ is the least element of $j$ greater than $i$;
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\item if $c_i = 0$, then let $\Pi_i = \Pi_{i+1}$;
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\item if $c_i = 1$, then constrain $\constraint{\Pi_{i+1}}{a_i}{\Pi_i}$.
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\end{itemize}
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Then we constrain the $a_i$ as follows:
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\introlist
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For $i \from n-1 \downto 0$,
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\begin{itemize}
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\item if $c_i = 0$, constrain $\constraint{1 - \Pi_{i+1} - a_i}{a_i}{0}$;
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\item if $c_i = 1$, boolean-constrain $a_i$ as in \crossref{cctboolean}.
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\item if $c_i = 1$, boolean-constrain $a_i$ as in \crossref{cctboolean}.
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\end{itemize}
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\end{itemize}
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Note that the constraints corresponding to zero bits of $c$ are \emph{in place of}
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Note that the constraints corresponding to zero bits of $c$ are \emph{in place of}
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boolean constraints on bits of $a_i$.
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boolean constraints on bits of $a_i$.
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This costs $n + k - 1 - t$ constraints.
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This costs $n + k$ constraints, where $k$ is the number of non-trailing $1$ bits in
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$c_{\barerange{n-2}{0}}$.
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\todo{Explain why this works (see \url{https://github.com/zcash/zcash/issues/2234\#issuecomment-338930637}).}
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\introsection
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\begin{theorem} \label{thmrangeconstraints}
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Assume $c_{\barerange{0}{n-1}} \typecolon \bitseq{n}$ and $c_{n-1} = 1$.
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Define $A_m := \ssum{i=m}{n-1} a_i \mult 2^i$ and $C_m := \ssum{i=m}{n-1} c_i \mult 2^i$.
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For any\, $m \in \range{0}{n-1}$, $A_m \leq C_m$ iff the restriction of the above
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constraint system to $i \in \range{m}{n-1}$ is satisfied. Furthermore the system
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at least boolean-constrains $a_{\barerange{0}{n-1}}$.
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\end{theorem}
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\begin{proof}
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For $i \in \range{0}{n-1}$ such that $c_i = 1$, the corresponding $a_i$ are
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unconditionally boolean-constrained. This implies that the system
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constrains $\Pi_i \in \bit$ for all $i \in \range{0}{n-1}$. For $i \in \range{0}{n-1}$
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such that $c_i = 0$, the constraint $\constraint{1 - \Pi_{i+1} - a_i}{a_i}{0}$
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constrains $a_i$ to be $0$ if $\Pi_{i+1} = 1$, otherwise it constrains $a_i \in \bit$.
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So all of $a_{\barerange{0}{n-1}}$ are at least boolean-constrained.
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To prove the rest of the theorem we proceed by induction on decreasing $m$,
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i.e.\ taking successively longer prefixes of the big-endian binary representations
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of $a$ and $c$.
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Base case $m = n-1$: since $c_{n-1} = 1$, the constraint system has
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just one boolean constraint on $a_{n-1}$, which fulfils the theorem since
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$A_{n-1} \leq C_{n-1}$ is always satisfied.
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Inductive case $m < n-1$:
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\begin{itemize}
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\item If $A_{m+1} > C_{m+1}$, then by the inductive hypothesis the constraint system
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must fail, which fulfils the theorem regardless of the value of $a_m$.
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\item If $A_{m+1} \leq C_{m+1}$, then by the inductive hypothesis the constraint system
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restricted to $i \in \range{m+1}{n-1}$ succeeds. We have
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$\Pi_{m+1} =
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\sproduct{i=m+1}{n-1} (c_i = 0 \bor a_i = 1) =
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\sproduct{i=m+1}{n-1} (a_i \geq c_i)$.
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\begin{itemize}
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\item If $A_{m+1} = C_{m+1}$, then $a_i = c_i$ for all $i \in \range{m+1}{n-1}$ and
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so $\Pi_{m+1} = 1$.
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Also $A_m \leq C_m$ iff $a_m \leq c_m$. \\
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When $c_m = 1$, only a boolean constraint is added for $a_m$ which fulfils the theorem. \\
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When $c_m = 0$, $a_m$ is constrained to be $0$ which fulfils the theorem.
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\item If $A_{m+1} < C_{m+1}$, then it cannot be the case that $a_i \geq c_i$
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for all $i \in \range{m+1}{n-1}$, so $\Pi_{m+1} = 0$. \\
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This implies that the constraint on $a_m$ is always equivalent to
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a boolean constraint, which fulfils the theorem because $A_m \leq C_m$ must
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be true regardless of the value of $a_m$.
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\end{itemize}
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\end{itemize}
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\vspace{-2ex}
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This covers all cases.
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\end{proof}
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Correctness of the full constraint system follows by taking $m = 0$ in the above theorem.
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The algorithm in \crossref{ccteddecompressvalidate} uses range checks with
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$c = \ParamS{r}-1$ to validate compressed Edwards points. In that case $n = 255$ and
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$k = 132$, so the cost of each such range check is $387$ constraints.
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\nnote{It is possible to optimize the computation of $\Pi_{\barerange{t}{n-2}}$ further.
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Notice that $\Pi_m$ is only used when $m$ is the index of the last bit of a
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run of $1$ bits in $c$. So for each run of $N$ $1$ bits, it is sufficient to compute
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an \Nary{} AND: $R = \sproduct{i=0}{N-1}{X_i}$. This can be computed in $3$ constraints
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for any $N < \ParamS{r}$; boolean-constrain the output $R$, and then add constraints
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\vspace{1ex}
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\begin{tabular}{@{\tab}l@{\;\;}l}
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$\constraint{N - \ssum{i=0}{N-1}{X_i}}{\mathsf{inv}}{1-R}$ &to enforce that
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$\ssum{i=0}{N-1}{X_i} \neq N$ when $R = 0$; \\[2ex]
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$\constraint{N - \ssum{i=0}{N-1}{X_i}}{R}{0}$ &to enforce that
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$\ssum{i=0}{N-1}{X_i} = N$ when $R = 1$. \\
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\end{tabular}
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\vspace{-1ex}
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where $\mathsf{inv}$ is witnessed as $\Big(N - \ssum{i=0}{N-1}{X_i}\Big)^{\!-1}$ if $R = 0$
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or is unconstrained otherwise.
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In fact the last constraint is not needed in this context because it is sufficient to
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compute an upper bound on each $\Pi_m$ (i.e.\ it does not benefit a malicious prover to
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witness $R = 1$ when the result of the AND should be $0$).
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So the cost of computing $\Pi$ variables for an arbitrarily long run of $1$ bits can be
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reduced to $2$ constraints. For example, for $c = \ParamS{r}-1$ the overall cost would
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be reduced to $255 + 68 = 323$ constraints.
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These optimizations are not used in \Sapling.}
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\introsection
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\introsection
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