How about this, based on enumerating binary trees:
e←{⍵,⊂f ⍵}
f←{↑,/(⊂⍵)g¨⍳≢⍵}
g←{↑,/,(⍵⊃⍺)∘.h ⍵⊃⌽⍺}
h←{(⊂'(',⍺),¨'+-×÷',¨⊂(⍵,')')}
t1←,⊂,'4' ⍝ vec of charvecs representing trees of size 1
t←,⊂t1 ⋄ t←e⍣4⊢t
(t1 t2 t3 t4 t5)←t
Now t5 is a list of parenthesised expressions of "size" 5, i.e.
containing 5 4s. At this point in Dyalog APL I would do:
⊃t5/⍨7={11::0 ⋄ ⍎⍵}¨t5 ⍝ 11::0 means return 0 if ⍎ signalled DOMAIN ERROR
(4+((4+(4+4))÷4))
(4-((4-(4×4))÷4))
(4+(((4+4)+4)÷4))
(4+(((4×4)-4)÷4))
(((4+(4+4))÷4)+4)
(((4×(4+4))-4)÷4)
((((4+4)+4)÷4)+4)
((((4+4)×4)-4)÷4)
((((4×4)-4)÷4)+4)
I am sure there is a simple equivalent in GNU APL.
For bonus marks I guess you might want to remove from this list any
expressions that are equivalent due to + and × being commutative and
associative.
Jay.
On 24 January 2017 at 15:47, Elias Mårtenson <loke...@gmail.com> wrote:
> My daughter had this maths problem, and quote it from memory:
>
> Given five 4's, separated by the mathematical functions +, -, × and ÷ form
> the number 7. The problem also allowed her to use parentheses.
>
> I decided to write an APL expression to solve this, and this is what I came
> up with:
>
> d ← {⍵÷⍺}
> (↑¨ 7 = ⍎¨ v) / v ← {"4 " , ↑,/ {{⍵," 4 "}¨⍵} ⍵}¨ ,↑∘.,/4 ⍴ ⊂'+-×d'
>
> Now, this program finds two solutions, which is good enough for me. However,
> my program doesn't find solutions that require parentheses.
>
> Problem 1: Can anyone simplify my program?
> Problem 2: Can anyone come up with a solution that also checks for
> parentheses?
>
> Regards,
> Elias
