Non-Projective Example

Question

I'm looking for an example sentence with a non-projective dependency parse. It doesn't have to be in English, though such an example would be nice.

Answer 1

The paper, Non-projective Dependency Parsing using Spanning Tree Algorithms has a few examples of non-projective dependency trees.

Note that dependency graphs are (1) not a formalism and (2) not standardized, so each researcher may re-define what happens in dependencies. Some of those dependency graphs are not even trees. Refer Generating Typed Dependency Parses from Phrase Structure Parses.

Answer 2

Here's a different way to look at it: you draw the dependency tree above the sentence with each node aligned with the corresponding word in the sentence, draw a line between the words in the tree and their correspondent in the sentence, and check whether there are any lines intersecting.

The structure below is for the French sentence À qui veux-tu que Pierre parle ? (lit. To whom want you that Peter talks?, 'Who do you want Peter to talk to?'). This sentence is non-projective because à qui has sort of "emancipated" from its governor to position itself relatively to veux. The dependency between parle and à crosses three node projections.

Answer 3

Here is a short English example from the paper Non-Projective Dependency Parsing in Expected Linear Time:

Or, if you want it in Universal Dependencies,

Here is the LaTeX code to generate the latter:

\begin{dependency}
    \begin{deptext}[column sep=1.2em]
        A \& hearing \& is \& scheduled \& on \& the \& issue \& today \\
    \end{deptext}
    \deproot{4}{root}
    \depedge{2}{1}{det}
    \depedge{4}{2}{nsubj:pass}
    \depedge{4}{3}{aux:pass}
    \depedge{7}{5}{case}
    \depedge{7}{6}{det}
    \depedge[edge start x offset=-6pt]{2}{7}{nmod}
    \depedge[edge slant=6pt]{4}{8}{nmod:tmod}
\end{dependency}