Question: What number of comedies have more than 550 episodes?
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Answer: 0
SPARQL: SELECT (COUNT(DISTINCT ?e) AS ?count) WHERE { ?e <pred:instance_of> ?c . ?c <pred:name> "comedy" . ?e <number_of_episodes> ?pv . ?pv <pred:unit> "1" . ?pv <pred:value> ?v . FILTER ( ?v > "550"^^xsd:double ) .  }
Program:
function: FindAll
function: FilterNum
dependencies: 0
inputs: number of episodes
550
>
function: FilterConcept
dependencies: 1
inputs: comedy
function: Count
dependencies: 2