How to solve the following kid puzzle by using only digits from 1 to 9 exactly one time :
With the following code, I found 20 solutions if we take into account operators priority, 119 without.
type NUM = BigDecimal
val NUM = BigDecimal //> NUM : math.BigDecimal.type = scala.math.BigDecimal$@52ed6f9
def fctA(ds: List[NUM]) = {
val List(a, b, c, d, e, f, g, h, i) = ds
a + NUM(13) * b / c + d + NUM(12) * e - f - NUM(11) + g * h / i - NUM(10)
} //> fctA: (ds: List[Vietation.NUM])scala.math.BigDecimal
def fctB(ds: List[NUM]) = {
val List(a, b, c, d, e, f, g, h, i) = ds
(((((((((((a + NUM(13)) * b) / c) + d) + NUM(12)) * e) - f) - NUM(11)) + g) * h) / i) - NUM(10)
} //> fctB: (ds: List[Vietation.NUM])scala.math.BigDecimal
val digits = List(1, 2, 3, 4, 5, 6, 7, 8, 9).map(x => NUM(x))
//> digits : List[scala.math.BigDecimal] = List(1, 2, 3, 4, 5, 6, 7, 8, 9)
def isSolution(fct:List[NUM]=>NUM)(candidate:List[NUM]):Boolean = {
val r = fct(candidate)
r == 66 && r.scale == 0
} //> isSolution: (fct: List[Vietation.NUM] => Vietation.NUM)(candidate: List[Viet
//| ation.NUM])Boolean
digits
.permutations
.filter(isSolution(fctA))
.size //> res0: Int = 20
digits
.permutations
.filter(isSolution(fctB))
.size //> res1: Int = 119
The solutions are :
List(3, 2, 1, 5, 4, 7, 8, 9, 6)
List(3, 2, 1, 5, 4, 7, 9, 8, 6)
List(5, 2, 1, 3, 4, 7, 8, 9, 6)
List(5, 2, 1, 3, 4, 7, 9, 8, 6)
List(5, 3, 1, 7, 2, 6, 8, 9, 4)
List(5, 3, 1, 7, 2, 6, 9, 8, 4)
List(5, 4, 1, 9, 2, 7, 3, 8, 6)
List(5, 4, 1, 9, 2, 7, 8, 3, 6)
List(5, 9, 3, 6, 2, 1, 7, 8, 4)
List(5, 9, 3, 6, 2, 1, 8, 7, 4)
List(6, 3, 1, 9, 2, 5, 7, 8, 4)
List(6, 3, 1, 9, 2, 5, 8, 7, 4)
List(6, 9, 3, 5, 2, 1, 7, 8, 4)
List(6, 9, 3, 5, 2, 1, 8, 7, 4)
List(7, 3, 1, 5, 2, 6, 8, 9, 4)
List(7, 3, 1, 5, 2, 6, 9, 8, 4)
List(9, 3, 1, 6, 2, 5, 7, 8, 4)
List(9, 3, 1, 6, 2, 5, 8, 7, 4)
List(9, 4, 1, 5, 2, 7, 3, 8, 6)
List(9, 4, 1, 5, 2, 7, 8, 3, 6)