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Asked: 2026-05-22T14:47:14+00:00

I could not solve this challenging problem: name(jack, math, 50). name(daniel, math, 60). name(jane,

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I could not solve this challenging problem:

name(jack, math, 50).
name(daniel, math, 60).
name(jane, phys, 70).
name(eto, comp, 73).

predicate : nameGrade(P, L, S). P is the list of people who are taking lesson L and whose grade is greater than S.

nameGrade([jack], math, 45). returns true

nameGrade([jack, daniel], math, 55). returns false. (because jack scored 50 which is less than 55)

nameGrade([], phys, 80). returns true

nameGrade(X, math, 70). returns X=[jack, daniel]

nameGrade([jack, daniel], math, X). returns X=50. (the less one).

Thank you.

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1 Answer

  1. Editorial Team

    You can use findall/3 to return a list of elements satisfying a predicate:

    nameGrade(P, L, S) :- is_list(P), name(_, _, S), findall(X, (name(X, L, S0), S0 >= S), P).
nameGrade(P, L, S) :- number(S), findall(X, (name(X, L, S0), S0 >= S), P).

    However, there are some contradictory goals in your requirement. For example, nameGrade([jack], math, 45) should fail because both jack and daniel take math and have higher scores than 45. nameGrade(X, math, 70) should return [] because no one takes math with a higher score than 70.

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