External sorting

number of passes: $1+ \lceil log_2 (N/M) \rceil$1+⌈log2​(N/M)⌉
seek time: O(number of passes)

a k-way merge
number of passes: $1+ \lceil log_k (N/M) \rceil$1+⌈logk​(N/M)⌉
require 2k tapes

polyphase merge
require k+1 tapes

Huffman tree
Total merge time = O ( the weighted external path length )

If the number of runs is a Fibonacci number $F_N$FN​, then the best way to distribute them is to split them into $F_{N–1}$FN–1​ and $F_{N–2}$FN–2​ .
For a k-way merge, $F_N^{(k)} = F_{N-1}^{(k)}+F_{N-2}^{(k)}$FN(k)​=FN−1(k)​+FN−2(k)​, where $F_N^{(k)}=0 \; (0 \leq N \leq k-2), F_{k-1}^{(k)}=1$FN(k)​=0(0≤N≤k−2),Fk−1(k)​=1

Exercises

hardware
外部排序主要开销在I/O上

$\lceil 1+log_2(100,000,000 \times 256 \div 128 \div 10^6) \rceil = 9$⌈1+log2​(100,000,000×256÷128÷106)⌉=9

Huffman tree，每次挑最短的两条链合并

https://blog.csdn.net/ailunlee/article/details/84548950