@pykcel
A split is when you add a new line inside an existing written line (as if you split a paragraph in two).
n is the number of ops.
a(e) is the number of elem_ops located after elementary operation e at time of e.
b(e) is the number of elem_ops located before elementary operation e at time of e.
What do you think is fastest:
for each elementary operation e ordered by time, do a(e) operations.
At each split (relativeley rare ?), for each operation in the paragraph splitted (approx n/#para), do b(e) operations.
In other words: if s is the number of splits and L usual number of elem_ops in a paragraph
what is the fastest : O(na) or O(sLb) ?
I guess b is arround 3 times bigger than a because usually you write at the end of the file.
But I'm betting that s (number of splits) is sufficiently small such that s*L*3 < n <=> s*n/#para*3<n <=> s*3<#para.
Would you agree?
@pykcel A split is when you add a new line inside an existing written line (as if you split a paragraph in two). n is the number of ops. a(e) is the number of elem_ops located after elementary operation e at time of e. b(e) is the number of elem_ops located before elementary operation e at time of e. What do you think is fastest: for each elementary operation e ordered by time, do a(e) operations. At each split (relativeley rare ?), for each operation in the paragraph splitted (approx n/#para), do b(e) operations.
In other words: if s is the number of splits and L usual number of elem_ops in a paragraph what is the fastest : O(na) or O(sLb) ?
I guess b is arround 3 times bigger than a because usually you write at the end of the file.
But I'm betting that s (number of splits) is sufficiently small such that s*L*3 < n <=> s*n/#para*3<n <=> s*3<#para. Would you agree?