Inside–outside algorithm

The inside probability ${\displaystyle \beta _{j}(p,q)}$ is the total probability of generating words ${\displaystyle w_{p}\cdots w_{q}}$, given the root nonterminal ${\displaystyle N^{j}}$ and a grammar ${\displaystyle G}$:^[1]

${\displaystyle \beta _{j}(p,q)=P(w_{pq}|N_{pq}^{j},G)}$

The outside probability ${\displaystyle \alpha _{j}(p,q)}$ is the total probability of beginning with the start symbol ${\displaystyle N^{1}}$ and generating the nonterminal ${\displaystyle N_{pq}^{j}}$ and all the words outside ${\displaystyle w_{p}\cdots w_{q}}$, given a grammar ${\displaystyle G}$:^[1]

${\displaystyle \alpha _{j}(p,q)=P(w_{1(p-1)},N_{pq}^{j},w_{(q+1)m}|G)}$

Base Case:

${\displaystyle \beta _{j}(p,p)=P(w_{p}|N^{j},G)}$

General case:

Suppose there is a rule ${\displaystyle N_{j}\rightarrow N_{r}N_{s}}$ in the grammar, then the probability of generating ${\displaystyle w_{p}\cdots w_{q}}$ starting with a subtree rooted at ${\displaystyle N_{j}}$ is:

${\displaystyle \sum _{k=p}^{k=q-1}P(N_{j}\rightarrow N_{r}N_{s})\beta _{r}(p,k)\beta _{s}(k+1,q)}$

The inside probability ${\displaystyle \beta _{j}(p,q)}$ is just the sum over all such possible rules:

${\displaystyle \beta _{j}(p,q)=\sum _{N_{r},N_{s}}\sum _{k=p}^{k=q-1}P(N_{j}\rightarrow N_{r}N_{s})\beta _{r}(p,k)\beta _{s}(k+1,q)}$

Base Case:

${\displaystyle \alpha _{j}(1,n)={\begin{cases}1&{\mbox{if }}j=1\\0&{\mbox{otherwise}}\end{cases}}}$

Here the start symbol is ${\displaystyle N_{1}}$.

General case:

Suppose there is a rule ${\displaystyle N_{r}\rightarrow N_{j}N_{s}}$ in the grammar that generates ${\displaystyle N_{j}}$. Then the left contribution of that rule to the outside probability ${\displaystyle \alpha _{j}(p,q)}$ is:

${\displaystyle \sum _{k=q+1}^{k=n}P(N_{r}\rightarrow N_{j}N_{s})\alpha _{r}(p,k)\beta _{s}(q+1,k)}$

Now suppose there is a rule ${\displaystyle N_{r}\rightarrow N_{s}N_{j}}$ in the grammar. Then the right contribution of that rule to the outside probability ${\displaystyle \alpha _{j}(p,q)}$ is:

${\displaystyle \sum _{k=1}^{k=p-1}P(N_{r}\rightarrow N_{s}N_{j})\alpha _{r}(k,q)\beta _{s}(k,p-1)}$

The outside probability ${\displaystyle \alpha _{j}(p,q)}$ is the sum of the left and right contributions over all such rules:

${\displaystyle \alpha _{j}(p,q)=\sum _{N_{r},N_{s}}\sum _{k=q+1}^{k=n}P(N_{r}\rightarrow N_{j}N_{s})\alpha _{r}(p,k)\beta _{s}(q+1,k)+\sum _{N_{r},N_{s}}\sum _{k=1}^{k=p-1}P(N_{r}\rightarrow N_{s}N_{j})\alpha _{r}(k,q)\beta _{s}(k,p-1)}$