## The problem

I have been playing with one of my toy projects and ended up having to shuffle text many times in a file that is around 10MB long. A naïve implementation chopping a string and gluing it together again ended up being painfully slow:

$O(n \cdot k)$

where n is the length of the string and k is the number of shuffling operations.

## A solution

It turns out that the problem can be efficiently solved using splay trees, where each node holds:

• the starting offset in the source string
• the length of the substring it represents
• the offset from the beginning of the substring taking into account all of the children on the left-hand-side
• total length of the substring taking into account all the children on the right-hand side

The find operation looks for a node containing the offset and splits it in two, if the node does not start with the offset.

 1 static Node *Find(uint32_t offset, Node *root)
2 {
3   if(!root)
4     return 0;
5   if(offset >= root->offset && offset < (root->offset + root->length))
6     return root;
7   if(offset < root->offset)
8     return Find(offset, root->left);
9   return Find(offset-root->offset-root->length, root->right);
10 }
11
12 static Node *STFind(uint32_t offset, Node *root)
13 {
14   Node *n = Find(offset, root);
15   Splay(n);
16
17   if(n->offset < offset) {
18     Node *newNode = new Node();
19     newNode->start  = n->start;
20     newNode->length = offset - n->offset;
21     n->start  = newNode->start + newNode->length;
22     n->length -= newNode->length;
23     newNode->left = n->left;
24     update(newNode);
25     n->left = newNode;
26     update(n);
27   }
28   return n;
29 }


The actual shuffling boils down to splitting and merging sub-trees and producing the result to traversing the whole tree in-order.

 1 void process(int i, int j, int k)
2 {
3   Node *n, *t, *left, *right = 0;
4   STSplit(left, n, i, pRoot);
5   if(j+1 < pS.length())
6     STSplit(n, right, j+1-i, n);
7   t = STMerge(left, right);
8   if(k < (t->offset + t->total_length))
9     STSplit(left, right, k, t);
10   else {
11     right = 0;
12     left = t;
13   }
14   pRoot = STMerge(left, n);
15   pRoot = STMerge(pRoot, right);
16 }


The amortized complexity is:

$O(k \cdot log(k) + n)$

See rope-splay.cpp.

## A test

It all runs pretty nicely for a test case with 10M characters and 100k shuffle operations:

]==> ./testgen 10000000 100000 > test
]==> time ./rope-naive < test > test-o-naive
./rope-naive < test > test-o-naive  386.17s user 66.19s system 99% cpu 7:32.86 total
]==> time ./rope-splay < test > test-o-splay
./rope-splay < test > test-o-splay  0.71s user 0.04s system 99% cpu 0.752 total
]==> diff test-o-splay test-o-naive