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Julia: update benchmark code #178

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Julia: update benchmark code #178

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559a498
Remove type-level memoization from Julia fibonacci implementation
non-Jedi Dec 4, 2024
d2c1737
Optimize julia loop implementation
non-Jedi Dec 4, 2024
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18 changes: 4 additions & 14 deletions fibonacci/julia/code.jl
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Original file line number Diff line number Diff line change
@@ -1,15 +1,5 @@
# base case 1
fibonacci(::Val{0}) = 0
# base case 2
fibonacci(::Val{1}) = 1
# general case
fibonacci(::Val{n}) where n = fibonacci(Val(n-1)) + fibonacci(Val(n-2))
fibonacci(n) = n < 2 ? n : fibonacci(n-1) + fibonacci(n-2)
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@bddicken bddicken Dec 6, 2024

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This needs to go back to having an if / case for 0 and 1.

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@non-Jedi non-Jedi Dec 6, 2024

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Is (n == 0 || no == 1) ? n : fibonacci(n-1) + fibonacci(n-2) okay, or does it need to be the full

n == 0 ? 0 :
    n == 1 ? 1 :
    fibonacci(n-1) + fibonacci(n-2)


let
u = parse(Int,ARGS[1])
r = 0
for i ∈ 1:u
r += fibonacci(Val(i))
end
println(r)
end
main(u) = println(sum(fibonacci, 1:u))
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@bddicken bddicken Dec 6, 2024

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Can you explain if / how this is equivalent to the loop?

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@non-Jedi non-Jedi Dec 6, 2024

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The sum function just calls mapreduce with addition as the reducing function. So this is equivalent to mapping fibonacci to all elements in the range from 1 to u and then adding them up.

If you're prioritizing making the implementations more uniform between languages rather than using more idiomatic language constructs, going back to a loop is just fine. Let me know.


isinteractive() || main(parse(Int, ARGS[1]))
15 changes: 8 additions & 7 deletions loops/julia/code.jl
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Original file line number Diff line number Diff line change
@@ -1,12 +1,13 @@
let
function main(u)
a = zeros(Int,10^4)
u = parse(Int,ARGS[1])
r = rand(1:10^4)
@inbounds for i ∈ 1:10^4
@inbounds for j ∈ 1:10^5
a[i] = a[i] + j%u
for i ∈ 1:10^4
for j ∈ 1:10^5
a[i] += j%u
end
a[i] = a[i] + r
a[i] += r
end
println(a[r])
end
end

isinteractive() || main(parse(Int, first(ARGS)))