On the average time complexity of computation with random partition

Abstract

Some computations are based on structures of random partition. They take an n-size problem as input, then break this problem into sub-problems of randomized size, execute calculations on each sub-problems and combine results from these calculations at last. We propose a combinatorial method for analyzing such computations and prove that the averaged time complexity is in terms of Stirling numbers of the second kind. The result shows that the average time complexity is decreased about one order of magnitude compared to that of the original solution. We also show two application cases where random partition structures are applied to improve performance.

Appendices

Appendices

Appendix 1: Proof of Theorem 2

Proof

Among all solutions $[x_1,x_2, \ldots ,x_b ]$, for all i, there are the same occurrences for $x_i = k$. In other words, if there are X solutions where $x_i = k$ then there are also X solutions where $x_j = k,j \ne i$ because the chances for a value to occur anywhere are the same.

Suppose the value of $x_i$ is k, then all $x_j (j \ne i)$ constitutes an answer for$\sum \nolimits _{i = 1}^{b - 1} {x_i } = n - k$. With Proposition 1, the occurrences for $x_i = k$ are $\left( \begin{array}{l} n - k + b - 1 - 1 \\ b - 1 - 1 \\ \end{array} \right) = \left( \begin{array}{l} n - k + b - 2 \\ b - 2 \\ \end{array} \right)$, then Theorem 2 follows. $\square$

Appendix 2: Proof of Theorem 3

Proof

Let $\lambda _m (x) = \sum \nolimits _{k = 0}^\infty {\frac{{k^m }}{{(1 + r)^{k + 1} }}x^{k - 1} }$, computing the integral we have:

$$\begin{aligned} \frac{1}{x}\int _0^x {\lambda _m dx} = \sum \limits _{k = 0}^\infty \frac{{k^{m - 1} }}{{(1 + r)^{k + 1} }}x^{k - 1} = \lambda _{m - 1} \end{aligned}$$

So $\lambda _1 = \sum \nolimits _{k = 0}^\infty {\frac{k}{{(1 + r)^{k + 1} }}x^{k - 1} }$, computing the integral we get $\int _0^x {\lambda _1 dx} = \sum \nolimits _{k = 0}^\infty {\frac{1}{{(1 + r)^{k + 1} }}x^k } = \frac{1}{{1 + r - x}}$.

Based on the above analysis we have $\left\{ \begin{array}{l} \lambda _1 = \frac{1}{{(1 + r - x)^2 }} \\ \lambda _m = (x\lambda _{m - 1} )' \\ \end{array} \right.$.

With mathematical induction method, we now show

$$\begin{aligned} \lambda _m = \sum \limits _{k = 1}^m {S(m,k)x^{k - 1} \lambda _1 ^{(k - 1)} } \end{aligned}$$

(19)

Obviously, Eq. (19) holds for $m=1$, that is $\lambda _1 =\sum \nolimits _{k = 1}^1 {S(m = 1,k)x^{k - 1} \lambda _1 ^{(k - 1)}}$.

Suppose (19) holds for m, then:

$$\begin{aligned} \begin{array}{l} \lambda _{m + 1} = (x\lambda _m )' = \lambda _m + x\lambda _m ' \\ = \sum \limits _{k = 1}^m {S(m,k)x^{k - 1} \lambda _1 ^{(k - 1)} } + \sum \limits _{k = 1}^m {S(m,k)[(k - 1)x^{k - 1} \lambda _1 ^{(k - 1)} + } x^k \lambda _1 ^{(k)} ] \\ = \sum \limits _{k = 1}^m {kS(m,k)x^{k - 1} \lambda _1 ^{(k - 1)} } + \sum \limits _{k = 1}^m {S(m,k)x^k \lambda _1 ^{(k)} } \\ = \sum \limits _{k = 1}^m {kS(m,k)x^{k - 1} \lambda _1 ^{(k - 1)} } + \sum \limits _{k = 2}^{m + 1} {S(m,k - 1)x^{k - 1} \lambda _1 ^{(k - 1)} } \\ = \lambda _1 + \sum \limits _{k = 2}^m {S(m + 1,k)x^{k - 1} \lambda _1 ^{(k - 1)} + } x^m \lambda _1 ^{(m)} \\ = \sum \limits _{k = 1}^{m + 1} {S(m + 1,k)x^{k - 1} \lambda _1 ^{(k - 1)} } \\ \end{array} \end{aligned}$$

Thus (19) follows.

Let $x=1$, from (19) we have

$$\begin{aligned} \lambda _m (x = 1) = \sum \limits _{k = 1}^m {S(m,k) \lambda _1 ^{(k - 1)} } \vert _{x = 1}. \end{aligned}$$

(20)

We now can show that

$$\begin{aligned} \lambda _1 ^{(m)} = \frac{{(m + 1)!}}{{(1 + r - x)^{m + 2} }} \end{aligned}$$

(21)

and thus

$$\begin{aligned} \lambda _1 ^{(m)} (x = 1) = \frac{{(m + 1)!}}{{(1 + r - x)^{m + 2} }} = \frac{{(m + 1)!}}{{r^{m + 2} }}. \end{aligned}$$

(22)

Using the mathematical induction method as above, we know (21) holds for $m=0$, that is $\lambda _1 ^{(m = 0)} =\frac{{(0 + 1)!}}{{(1 + r - x)^{0 + 2} }}$.

Suppose $\lambda _1 ^{(m)} = \frac{{(m + 1)!}}{{(1 + r - x)^{m + 2} }}$, then $\lambda _1 ^{(m + 1)} = (\lambda _1 ^{(m)} )' = \frac{{(m + 1 + 1)!}}{{(1 + r - x)^{m + 1 + 2} }}$. Equation (21) follows and (22) also holds if we let $x=1$ in (21).

Therefore, based on (22), there is an induction:

$$\begin{aligned} \mathop {\lim }\limits _{n \rightarrow \infty } R(n,r,m)&= \mathop {\lim }\limits _{n \rightarrow \infty } r^2 \sum \limits _{k = 0}^n {\frac{{k^m }}{{(1 + r)^{k + 1} }}} \nonumber \\&= r^2 \sum \limits _{k = 1}^m {S(m,k)\lambda _1 ^{(k - 1)} } \vert _{x = 1} = \sum \limits _{k = 1}^m \frac{{S(m,k)k!}}{{r^{k - 1} }}. \end{aligned}$$

(23)

Theorem 3 follows. The proof is completed. $\square$