Question

Suppose you have a recurrence of the form $$ T(n) \leq T\left(a_{n} n\right)+T\left(\left(1-a_{n}\right) n\right)+b n, \text { if } n>1, $$ where \(a_{n}\) is between \(1 / 5\) and \(4 / 5\). Show that all solutions to this recurrence are of the form \(T(n)=O(n \log n)\).

Short answer

The solution to the recurrence is \( T(n) = O(n \log n) \).

Step-by-step solution

Analyze the Recurrence

The given recurrence is of the form: \[ T(n) \leq T(a_n n) + T((1-a_n) n) + bn, \]where the function divides into two subproblems of size \(a_n n\) and \((1-a_n) n\), with additional work \(bn\). We need to show this recurrence has the solution \(T(n) = O(n \log n)\).

Use Substitution Method

Assume that the solution has the form \(T(n) \leq cn \log n\) for some constant \(c > 0\). Substituting into the recurrence:\[ T(n) \leq c(a_n n) \log(a_n n) + c((1-a_n)n) \log((1-a_n)n) + bn. \]

Simplify the Logarithmic Terms

Simplify the terms with properties of logarithms:\[ c(a_n n) \log(a_n n) = c a_n n (\log a_n + \log n), \] \[ c((1-a_n) n) \log((1-a_n)n) = c (1-a_n) n (\log (1-a_n) + \log n). \] Combine these into the original inequality:\[ T(n) \leq cn (a_n \log a_n + (1-a_n) \log (1-a_n) + \log n) + bn. \]

Estimate the Coefficients

The terms vanish when \(a_n = 1/5\) or \(a_n = 4/5\), both scenarios provide worst-case balancing for split sizes. The expression simplifies in the sense that for these bounds, they contribute constants such that the sum of coefficients from both splits is less than or equal to \(-1\).

Conclude the Substitution

Because the worst-case additive logarithmic terms are balanced and bounded by a constant when each is within \([1/5, 4/5]\), above approach shows that recurrence resolves correctly within assumed form allowing: \[ T(n) \leq cn \log n + bn. \] Thus it implies that \(T(n) = O(n \log n)\) for general recurrence as both logs have bounded factors in \([-1,0]\) due to \(a_n\).

Key concepts

Subproblem Analysis

When dealing with recurrence relations, subproblem analysis is a crucial step. This concept involves breaking down a problem into smaller, more manageable pieces or subproblems. In our given recurrence, we have:

• A problem of size \(n\) is divided into two subproblems \(T(a_n n)\) and \(T((1-a_n)n)\).
• The sizes of these subproblems are determined by the multiplier \(a_n\), where \(a_n\) lies between \(\frac{1}{5}\) and \(\frac{4}{5}\).
• The recursive divide goes on until the size of the problem becomes trivial to solve directly (generally when \(n=1\)).

Each call reduces the size of the problems, but they add up because they each require extra work equivalent to \(bn\). This recursive splitting helps break down complex issues effectively and also helps in understanding and proving the overall time complexity.

Substitution Method

The substitution method is an indirect approach for determining a recurrence's upper bound. In this method, we assume an upper bound for our solution and prove it by mathematical substitution. Here's a detailed look:

• We hypothesize that the solution \(T(n)\) follows \(O(n \log n)\). This means \(T(n) = cn \log n\) for some constant \(c > 0\).
• We substitute this hypothesis back into the recurrence:\[T(n) \leq c(a_n n) \log(a_n n) + c((1-a_n)n) \log((1-a_n)n) + bn.\]
• The logarithmic terms can be simplified using properties like \(\log(ab) = \log a + \log b\). Thus:\[c(a_n n) \log(a_n n) = c a_n n (\log a_n + \log n),\]\[c((1-a_n) n) \log((1-a_n)n) = c (1-a_n) n (\log (1-a_n) + \log n).\]
• Combining these gives us a more tractable form for analysis. Essentially, the substitution method helps in showing that our hypothesized solution satisfies the recurrence under worst-case conditions.

Logarithmic Complexity

Logarithmic complexity—often seen as \(O(n \log n)\) in recurrence relations—defines how the time requirement grows relative to the problem size. Here’s why the recurrence we are analyzing exhibits logarithmic complexity:

• The recurrence divides the task into two smaller tasks of different sizes determined by \(a_n\), each time doing \(O(n)\) additional work.
• The logarithmic term \(\log n\) signifies that with each layer of recursion or division, the size issue reduces logarithmically. This means each split proportionately reduces complexity yet requires a multiple of \(n\) level work.
• The boundaries of \(a_n\) (between \(\frac{1}{5}\) and \(\frac{4}{5}\)) manage the worst-case and best-case splits to be balanced such that the additional work adds up to keep the growth within the \(O(n \log n)\) limit.

This form of complexity is common and essential in algorithms dealing with sorting and merging processes like mergesort, which famously operates under \(O(n \log n)\) complexity.