Denote the plain Kolmogorov complexity of a string $u$ by $C(u)$. Now let $u$ be a string of length $n$ with $C(u) \ge n - O(1)$ and suppose $u = u_1 \cdots u_{\log n}$, a subdivision of the string. Further suppose for each $1 \le i \le \log n$ we have $|u_i| = n/\log n$. Show that $$ C(u_i) \ge n / \log n - O(1) $$ for all $1 \le i \le \log n$.
Any ideas how to solve this? If I have a algorithm to compute $u_i$, then I can describe the rest $u_1 \cdots u_{i-1} u_{i+1} \cdots u_{\log n}$ literally, but to compose $u$ out of them I need the information where to place $u_i$ (i.e. as the $i$-th substring), but for storing that I need additional information, but by this additional bits I do not have $$ C(u) \le C(u_i) + (n - |u_i|) $$ from which the above would follow. So any hints?
This is Exercise 2.2.2 (c) from An Introduction to Kolmogorov Complexity and Its Applications.
