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diff --git a/doc/src/imap_uid.md b/doc/src/imap_uid.md deleted file mode 100644 index ecdd52b..0000000 --- a/doc/src/imap_uid.md +++ /dev/null @@ -1,203 +0,0 @@ -# IMAP UID proof - -**Notations** - -- $h$: the hash of a message, $\mathbb{H}$ is the set of hashes -- $i$: the UID of a message $(i \in \mathbb{N})$ -- $f$: a flag attributed to a message (it's a string), we write - $\mathbb{F}$ the set of possible flags -- if $M$ is a map (aka a dictionnary), if $x$ has no assigned value in - $M$ we write $M [x] = \bot$ or equivalently $x \not\in M$. If $x$ has a value - in the map we write $x \in M$ and $M [x] \neq \bot$ - -**State** - -- A map $I$ such that $I [h]$ is the UID of the message whose hash is - $h$ is the mailbox, or $\bot$ if there is no such message - -- A map $F$ such that $F [h]$ is the set of flags attributed to the - message whose hash is $h$ - -- $v$: the UIDVALIDITY value - -- $n$: the UIDNEXT value - -- $s$: an internal sequence number that is mostly equal to UIDNEXT but - also grows when mails are deleted - -**Operations** - - - MAIL\_ADD$(h, i)$: the value of $i$ that is put in this operation is - the value of $s$ in the state resulting of all already known operations, - i.e. $s (O_{gen})$ in the notation below where $O_{gen}$ is - the set of all operations known at the time when the MAIL\_ADD is generated. - Moreover, such an operation can only be generated if $I (O_{gen}) [h] - = \bot$, i.e. for a mail $h$ that is not already in the state at - $O_{gen}$. - - - MAIL\_DEL$(h)$ - - - FLAG\_ADD$(h, f)$ - - - FLAG\_DEL$(h, f)$ - -**Algorithms** - - -**apply** MAIL\_ADD$(h, i)$: - *if* $i < s$: - $v \leftarrow v + s - i$ - *if* $F [h] = \bot$: - $F [h] \leftarrow F_{initial}$ - $I [h] \leftarrow s$ - $s \leftarrow s + 1$ - $n \leftarrow s$ - -**apply** MAIL\_DEL$(h)$: - $I [h] \leftarrow \bot$ - $F [h] \leftarrow \bot$ - $s \leftarrow s + 1$ - -**apply** FLAG\_ADD$(h, f)$: - *if* $h \in F$: - $F [h] \leftarrow F [h] \cup \{ f \}$ - -**apply** FLAG\_DEL$(h, f)$: - *if* $h \in F$: - $F [h] \leftarrow F [h] \backslash \{ f \}$ - - -**More notations** - -- $o$ is an operation such as MAIL\_ADD, MAIL\_DEL, etc. $O$ is a set of - operations. Operations embed a timestamp, so a set of operations $O$ can be - written as $O = [o_1, o_2, \ldots, o_n]$ by ordering them by timestamp. - -- if $o \in O$, we write $O_{\leqslant o}$, $O_{< o}$, $O_{\geqslant - o}$, $O_{> o}$ the set of items of $O$ that are respectively earlier or - equal, strictly earlier, later or equal, or strictly later than $o$. In - other words, if we write $O = [o_1, \ldots, o_n]$, where $o$ is a certain - $o_i$ in this sequence, then: -$$ -\begin{aligned} -O_{\leqslant o} &= \{ o_1, \ldots, o_i \}\\ -O_{< o} &= \{ o_1, \ldots, o_{i - 1} \}\\ -O_{\geqslant o} &= \{ o_i, \ldots, o_n \}\\ -O_{> o} &= \{ o_{i + 1}, \ldots, o_n \} -\end{aligned} -$$ - -- If $O$ is a set of operations, we write $I (O)$, $F (O)$, $n (O), s - (O)$, and $v (O)$ the values of $I, F, n, s$ and $v$ in the state that - results of applying all of the operations in $O$ in their sorted order. (we - thus write $I (O) [h]$ the value of $I [h]$ in this state) - -**Hypothesis:** -An operation $o$ can only be in a set $O$ if it was -generated after applying operations of a set $O_{gen}$ such that -$O_{gen} \subset O$ (because causality is respected in how we deliver -operations). Sets of operations that do not respect this property are excluded -from all of the properties, lemmas and proofs below. - -**Simplification:** We will now exclude FLAG\_ADD and FLAG\_DEL -operations, as they do not manipulate $n$, $s$ and $v$, and adding them should -have no impact on the properties below. - -**Small lemma:** If there are no FLAG\_ADD and FLAG\_DEL operations, -then $s (O) = | O |$. This is easy to see because the possible operations are -only MAIL\_ADD and MAIL\_DEL, and both increment the value of $s$ by 1. - -**Defnition:** If $o$ is a MAIL\_ADD$(h, i)$ operation, and $O$ is a -set of operations such that $o \in O$, then we define the following value: -$$ -C (o, O) = s (O_{< o}) - i -$$ -We say that $C (o, O)$ is the *number of conflicts of $o$ in $O$*: it -corresponds to the number of operations that were added before $o$ in $O$ that -were not in $O_{gen}$. - -**Property:** - -We have that: - -$$ -v (O) = \sum_{o \in O} C (o, O) -$$ - -Or in English: $v (O)$ is the sum of the number of conflicts of all of the -MAIL\_ADD operations in $O$. This is easy to see because indeed $v$ is -incremented by $C (o, O)$ for each operation $o \in O$ that is applied. - - -**Property:** - If $O$ and $O'$ are two sets of operations, and $O \subseteq O'$, then: - -$$ -\begin{aligned} -\forall o \in O, \qquad C (o, O) \leqslant C (o, O') -\end{aligned} -$$ - -This is easy to see because $O_{< o} \subseteq O'_{< o}$ and $C (o, O') - C - (o, O) = s (O'_{< o}) - s (O_{< o}) = | O'_{< o} | - | O_{< o} | \geqslant - 0$ - -**Theorem:** - -If $O$ and $O'$ are two sets of operations: - -$$ -\begin{aligned} -O \subseteq O' & \Rightarrow & v (O) \leqslant v (O') -\end{aligned} -$$ - -**Proof:** - -$$ -\begin{aligned} -v (O') &= \sum_{o \in O'} C (o, O')\\ - & \geqslant \sum_{o \in O} C (o, O') \qquad \text{(because $O \subseteq - O'$)}\\ - & \geqslant \sum_{o \in O} C (o, O) \qquad \text{(because $\forall o \in - O, C (o, O) \leqslant C (o, O')$)}\\ - & \geqslant v (O) -\end{aligned} -$$ - -**Theorem:** - -If $O$ and $O'$ are two sets of operations, such that $O \subset O'$, - -and if there are two different mails $h$ and $h'$ $(h \neq h')$ such that $I - (O) [h] = I (O') [h']$ - - then: - $$v (O) < v (O')$$ - -**Proof:** - -We already know that $v (O) \leqslant v (O')$ because of the previous theorem. -We will now look at the sum: -$$ -v (O') = \sum_{o \in O'} C (o, O') -$$ -and show that there is at least one term in this sum that is strictly larger -than the corresponding term in the other sum: -$$ -v (O) = \sum_{o \in O} C (o, O) -$$ -Let $o$ be the last MAIL\_ADD$(h, \_)$ operation in $O$, i.e. the operation -that gives its definitive UID to mail $h$ in $O$, and similarly $o'$ be the -last MAIL\_ADD($h', \_$) operation in $O'$. - -Let us write $I = I (O) [h] = I (O') [h']$ - -$o$ is the operation at position $I$ in $O$, and $o'$ is the operation at -position $I$ in $O'$. But $o \neq o'$, so if $o$ is not the operation at -position $I$ in $O'$ then it has to be at a later position $I' > I$ in $O'$, -because no operations are removed between $O$ and $O'$, the only possibility -is that some other operations (including $o'$) are added before $o$. Therefore -we have that $C (o, O') > C (o, O)$, i.e. at least one term in the sum above -is strictly larger in the first sum than in the second one. Since all other -terms are greater or equal, we have $v (O') > v (O)$. |