Suppose we generate an array of \(500\) elements and look at the sums made by adjacent items. Then we build a frequency dictionary and look at how many ways we are able to

```
@time begin
X = [ rand(0:100) for _ in 1:500 ]
p1 = 1
p2(p, windowsize) = p + windowsize
fhist = Dict{Int64, Int64}()
for ws in 1:(length(X)-1)
for p in 1:(length(X)-ws)
lo = p
hi = p2(p, ws)
s = X[lo:hi] |> sum
if haskey(fhist, s)
fhist[s] += 1
else
fhist[s] = 1
end
end
end
bar([keys(fhist)...], [values(fhist)...])
end
```

© Daniel Marvin. Last modified: July 19, 2021.