Question

Is there an algorithm that with a given 2-3 tree T and a pointer to some node v in said tree, the algo can change the key of the node v so T would remain a legal 2-3 tree, in O(logn/loglogn) amortized efficiency?

Answer 1

No.

Assume it was possible, with the algorithm f, we will show we can sort an array with O(n*logn/loglogn) time complexity.

sort array A of length n:
(1) Create an  2-3 tree of size n, with no importance to keys. let it be T.
(2) store all pointers to nodes in T in a second array B.
(3) for each i from 0 to n:
   (3.1) f(B[i],A[i]) //modify the tree: pointer: B[i] new value: A[i]
(4) extract elements from T back to A inorder.

correctness:

After each activation of f the tree is legal. After finishing activating f on all elements of T and all elements of A, the tree is legal and contains all elements. Thus, extracting elements from A, we get back the sorted array.

complexity:

(1)Creating a tree [no importance which keys we put] is O(n) we can put 0 in all elements, it doesn’t matter

(2)iterating T and creating B is O(n)

(3)activating f is O(logn/loglogn), thus invoking it n times is O(n*logn/loglogn)

(4) extracting elements is just a traversal: O(n)

Thus: total complexity is O(n*logn/loglogn)

But sorting is an Omega(nlogn) problem with comparisons based algorithms. contradiction.

Conclusion: desired f doesn’t exist.