Question: What city's population is more than 7800 in the year 2012 and country is Zimbabwe?
Choices: Bulawayo
Kingston
Harare
Georgetown
Cairo
Dresden
Berlin
Espoo
Cork
Trieste
SPARQL: SELECT DISTINCT ?e WHERE { ?e <pred:instance_of> ?c . ?c <pred:name> "city" . ?e <population> ?pv . ?pv <pred:unit> "1" . ?pv <pred:value> ?v . FILTER ( ?v > "7800"^^xsd:double ) . [ <pred:fact_h> ?e ; <pred:fact_r> <population> ; <pred:fact_t> ?pv ] <point_in_time> ?qpv . ?qpv <pred:year> 2012 . ?e <country> ?e_1 . ?e_1 <pred:name> "Zimbabwe" .  }
Program:
function: FindAll
function: FilterNum
dependencies: 0
inputs: population
7800
>
function: QFilterYear
dependencies: 1
inputs: point in time
2012
=
function: FilterConcept
dependencies: 2
inputs: city
function: Find
inputs: Zimbabwe
function: Relate
dependencies: 4
inputs: country
backward
function: FilterConcept
dependencies: 5
inputs: city
function: And
dependencies: 3
6
function: What
dependencies: 7
Answer: Bulawayo