maple_tree: introduce interfaces __mt_dup() and mtree_dup()

maple_tree: introduce interfaces __mt_dup() and mtree_dup()

Introduce interfaces __mt_dup() and mtree_dup(), which are used to
duplicate a maple tree.  They duplicate a maple tree in Depth-First Search
(DFS) pre-order traversal.  It uses memcopy() to copy nodes in the source
tree and allocate new child nodes in non-leaf nodes.  The new node is
exactly the same as the source node except for all the addresses stored in
it.  It will be faster than traversing all elements in the source tree and
inserting them one by one into the new tree.  The time complexity of these
two functions is O(n).

The difference between __mt_dup() and mtree_dup() is that mtree_dup()
handles locks internally.

Analysis of the average time complexity of this algorithm:

For simplicity, let's assume that the maximum branching factor of all
non-leaf nodes is 16 (in allocation mode, it is 10), and the tree is a
full tree.

Under the given conditions, if there is a maple tree with n elements, the
number of its leaves is n/16.  From bottom to top, the number of nodes in
each level is 1/16 of the number of nodes in the level below.  So the
total number of nodes in the entire tree is given by the sum of n/16 +
n/16^2 + n/16^3 + ...  + 1.  This is a geometric series, and it has log(n)
terms with base 16.  According to the formula for the sum of a geometric
series, the sum of this series can be calculated as (n-1)/15.  Each node
has only one parent node pointer, which can be considered as an edge.  In
total, there are (n-1)/15-1 edges.

This algorithm consists of two operations:

1. Traversing all nodes in DFS order.
2. For each node, making a copy and performing necessary modifications
   to create a new node.

For the first part, DFS traversal will visit each edge twice.  Let
T(ascend) represent the cost of taking one step downwards, and T(descend)
represent the cost of taking one step upwards.  And both of them are
constants (although mas_ascend() may not be, as it contains a loop, but
here we ignore it and treat it as a constant).  So the time spent on the
first part can be represented as ((n-1)/15-1) * (T(ascend) + T(descend)).

For the second part, each node will be copied, and the cost of copying a
node is denoted as T(copy_node).  For each non-leaf node, it is necessary
to reallocate all child nodes, and the cost of this operation is denoted
as T(dup_alloc).  The behavior behind memory allocation is complex and not
specific to the maple tree operation.  Here, we assume that the time
required for a single allocation is constant.  Since the size of a node is
fixed, both of these symbols are also constants.  We can calculate that
the time spent on the second part is ((n-1)/15) * T(copy_node) + ((n-1)/15
- n/16) * T(dup_alloc).

Adding both parts together, the total time spent by the algorithm can be
represented as:

((n-1)/15) * (T(ascend) + T(descend) + T(copy_node) + T(dup_alloc)) -
n/16 * T(dup_alloc) - (T(ascend) + T(descend))

Let C1 = T(ascend) + T(descend) + T(copy_node) + T(dup_alloc)
Let C2 = T(dup_alloc)
Let C3 = T(ascend) + T(descend)

Finally, the expression can be simplified as:
((16 * C1 - 15 * C2) / (15 * 16)) * n - (C1 / 15 + C3).

This is a linear function, so the average time complexity is O(n).

Link: https://lkml.kernel.org/r/20231027033845.90608-4-zhangpeng.00@bytedance.com
Signed-off-by: Peng Zhang <zhangpeng.00@bytedance.com>
Suggested-by: Liam R. Howlett <Liam.Howlett@oracle.com>
Cc: Christian Brauner <brauner@kernel.org>
Cc: Jonathan Corbet <corbet@lwn.net>
Cc: Mateusz Guzik <mjguzik@gmail.com>
Cc: Mathieu Desnoyers <mathieu.desnoyers@efficios.com>
Cc: Matthew Wilcox <willy@infradead.org>
Cc: Michael S. Tsirkin <mst@redhat.com>
Cc: Mike Christie <michael.christie@oracle.com>
Cc: Nicholas Piggin <npiggin@gmail.com>
Cc: Peter Zijlstra <peterz@infradead.org>
Cc: Suren Baghdasaryan <surenb@google.com>
Signed-off-by: Andrew Morton <akpm@linux-foundation.org>