Question: What number of comedies are produced by Alex Gibney?
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Answer: 0
SPARQL: SELECT (COUNT(DISTINCT ?e) AS ?count) WHERE { ?e <pred:instance_of> ?c . ?c <pred:name> "comedy" . ?e <producer> ?e_1 . ?e_1 <pred:name> "Alex Gibney" .   }
Program:
function: Find
inputs: Alex Gibney
function: Relate
dependencies: 0
inputs: producer
backward
function: FilterConcept
dependencies: 1
inputs: comedy
function: Count
dependencies: 2