“斐波那契查找”真的比“二分查找”快么？
Is Fibonacci Search really "faster" than Binary Search?
申明：本文讨论的搜索对象为有序数组，不是数学上讨论的函数。
1. 介绍
对经过各种Sort算法排好序之后的有序数组进行检索的Search算法大致有以下三种：线性查找 O(n)，二分查找 O(log(n))，斐波那契查找 O(log(n))。
前两者用的比较多，对于 Fibonacci Search，应该蛮多人和我一样只闻其名，不见其人吧。
数学原理如下：
斐波那契数列：0、1、1、2、3、5、8、13、21、……（有人喜欢从1开始，随你~~~）
如果设F(n）为该数列的第n项（n∈N）。那么这句话可以写成如下形式：
F(0) = 0，F(1)=1，F(n)=F(n-1)+F(n-2) (n≥2)，显然这是一个线性递推数列,随着数列项数的增加，前一项与后一项之比越来越逼近黄金分割的数值0.6180339887..…
且看通项公式：
算法描述如下：
The Fibonacci Search Algorithm
Let Fk represent the k-th Fibonacci number where Fk+2=Fk+1 + Fk for k>=0 and F0 = 0, F1 = 1. To test whether an item is in a list of n = Fm ordered numbers, proceed as follows:
- Set k = m.
- If k = 0, finish - no match.
- Test item against entry in position Fk-1.
- If match, finish.
- If item is less than entry Fk-1, discard entries from positions Fk-1 + 1 to n. Set k = k - 1 and go to 2.
- If item is greater than entry Fk-1, discard entries from positions 1 to Fk-1. Renumber remaining entries from 1 to Fk-2, set k = k - 2 and go to 2.
If n is not a Fibonacci number, then let Fm be the smallest such number >n, augment the original array with Fm-n numbers larger than the sought item and apply the above algorithm for n'=Fm.
2.实验
实验数据：产生有序数组
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#define NUM
4
int *array = ( int *)calloc(NUM, sizeof( int)); for ( int i = 0; i < NUM; ++i) { array[i] = i; } |
实验方法：三种搜索算法函数原型pData-数组，n-数组个数，key-待搜索关键字
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int LineSearch(
int *pData,
int n,
int key);
int BinarySearch( int *pData, int n, int key); int FibonacciSearch( int *pData, int n, int key); |
方法实现：
线性查找
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int LineSearch(
int *pData,
int n,
int key)
{ int idx = 0; while(idx != n ) { if(pData[idx] != key) idx++; else return idx; } } |
二分查找
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int BinarySearch(
int *pData,
int n,
int key)
{ int center, left = 0, right = n - 1; while(left <= right) { center = (left + right) / 2; if (pData[center] > key) { right = center - 1; } else { if (pData[center] < key) left = center + 1; else return center; } } return - 1; } |
斐波那契查找
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/* Fibonaccian search for locating the index of "key" in an array "pData" of size "n" that is sorted in ascending order. See http://doi.acm.org/10.1145/367487.367496 Algorithm description ----------------------------------------------------------------------------- Let Fk represent the k-th Fibonacci number where Fk+2=Fk+1 + Fk for k>=0 and F0 = 0, F1 = 1. To test whether an item is in a list of n = Fm ordered numbers, proceed as follows: a) Set k = m. b) If k = 0, finish - no match. c) Test item against entry in position Fk-1. d) If match, finish. e) If item is less than entry Fk-1, discard entries from positions Fk-1 + 1 to n. Set k = k - 1 and go to b). f) If item is greater than entry Fk-1, discard entries from positions 1 to Fk-1. Renumber remaining entries from 1 to Fk-2, set k = k - 2 and go to b) If Fm>n then the original array is augmented with Fm-n numbers larger than key and the above algorithm is applied. */ int FibonacciSearch( int *pData, int n, int key) { register int k, idx, offs; static int prevn = - 1, prevk = - 1; /* Precomputed Fibonacci numbers F0 up to F46. This implementation assumes that the size n * of the input array fits in 4 bytes. Note that F46=1836311903 is the largest Fibonacci * number that is less or equal to the 4-byte INT_MAX (=2147483647). The next Fibonacci * number, i.e. F47, is 2971215073 and is larger than INT_MAX, implying that it does not * fit in a 4 byte integer. Note also that the last array element is INT_MAX rather than * F47. This ensures correct operation for n>F46. */ const static int Fib[ 47 + 1] = { 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986, 102334155, 165580141, 267914296, 433494437, 701408733, 1134903170, 1836311903, INT_MAX }; /* find the smallest fibonacci number that is greater or equal to n. Store this * number to avoid recomputing it in the case of repetitive searches with identical n. */ if(n != prevn) { register int f0, f1, t; for(f0 = 0, f1 = 1, k = 1; f1 < n; t = f1, f1 += f0, f0 = t, ++k); prevk = k; prevn = n; } else k = prevk; /* If the sought value is larger than the largest Fibonacci number less than n, * care must be taken top ensure that we do not attempt to read beyond the end * of the array. If we do need to do this, we pretend that the array is padded * with elements larger than the sought value. */ for(offs = 0; k > 0; ) { idx = offs + Fib[--k]; /* note that at this point k has already been decremented once */ if(idx >= n || key < pData[idx]) // index out of bounds or key in 1st part { continue; } else if (key > pData[idx]) { // key in 2nd part offs = idx; --k; } else // key==pData[idx], found return idx; } return - 1; // not found } |
算法正确性验证：
C++ Code
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printf(
"LineSearch Index : %d\n", LineSearch(array, NUM, array[
3]));
printf( "BinarySearch Index : %d\n", BinarySearch(array, NUM, array[ 3])); printf( "FibonacciSearch Index : %d\n", FibonacciSearch(array, NUM, array[ 3])); |
分析：
| （NUM=4） | 0 | 1 | 2 | 3 | total |
| Line | 0 | 1 | 2 | 3 | 7 |
| Binary | 1 | 0 | 1 | 2 | 4 |
| Fibonacci | 4 | 2 | 1 | 0 | 7 |
ps:第一行为待查找的关键值，本文将遍历序数组中的每个关键值的查找次数并相加（如果要表征平均，除以NUM即可，因为一样，所以就不除了），来表征平均搜索效率，因为当假设每个值被查找的概率相同，即符合均匀分布，还是有一定的合理性的。即total值越小，对单个关键值的搜索次数约有效率。
算法性能实验：
| NUM | 4 | 4E1 | 4E2 | 4E3 | 4E4 | 4E5 | 4E6 | 4E7 |
| Line | 6 | 780 | 79800 | 7998000 | 79998000 | - | - | - |
| Binary | 4 | 143 | 2687 | 39917 | 534481 | 6675732 (0.07s) |
79805719 (0.81s) |
932891163 (20.34s) |
| Fibonacci | 7 | 179 | 3189 | 42428 | 572324 | 7206170 (0.06s) |
86747879 (0.77s) |
1012299070 (12.31s) |
得到次数需要改写一些代码，改写的整个代码放在最后附录，需要的自己去验证下。
结论： 与二分查找相比，斐波那契查找的明显优点在于它只涉及加法和减法运算，而不用除法（可能用“>>2”要好点）。因为除法比加减法要占去更多的机时，因此，斐波那契查找的运行时间比二分查找短。 但前者的O(log(n))还是后者的大点，对某一个对象的平均搜索次数要小。所以，要取决于你如何看待“faster”了。
Reference:
附录：
Search.h
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#include <stdio.h>
#include <stdlib.h> #include <limits.h> #include <time.h> #define NUM 4 int LineSearch( int* pData, int n, int key); int BinarySearch( int* pData, int n, int key); int FibonacciSearch( int* pData, int n, int key); |
Search.c
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#include
"search.h"
//四种搜索算法 搜索遍历 有序数组所用的总次数 int main( int argc, char const *argv[]) { clock_t start, finish; double time_duration; //产生有序数组 int *array = ( int *)calloc(NUM, sizeof( int)); for ( int i = 0; i < NUM; ++i) { array[i] = i; } start = clock(); int tTmp; //线性查找 tTmp = 0; for ( int i = 0; i < NUM; ++i) { tTmp += LineSearch(array, NUM, array[i]); } printf( "LineSearch Index : %d\n", tTmp); //二分查找 tTmp = 0; for ( int i = 0; i < NUM; ++i) { tTmp += BinarySearch(array, NUM, array[i]); } printf( "BinarySearch Index : %d\n", tTmp); //斐波那契查找 tTmp = 0; for ( int i = 0; i < NUM; ++i) { tTmp += FibonacciSearch(array, NUM, array[i]); } printf( "FibonacciSearch Index : %d\n", tTmp); free(array); finish = clock(); time_duration = ( double)(finish - start) / CLOCKS_PER_SEC; printf( "Times Cost : %.2f \n", time_duration ); return 0; } int LineSearch( int *pData, int n, int key) { int idx = 0; int times = 0; while(idx != n ) { if(pData[idx] != key) times++, idx++; else return times; } return - 1; } int BinarySearch( int *pData, int n, int key) { int center, left = 0, right = n - 1; int times = 0; while(left <= right) { center = (left + right) / 2; if (pData[center] > key) { times++, right = center - 1; } else { if (pData[center] < key) times++, left = center + 1; else return times; } } return - 1; } /* Fibonaccian search for locating the index of "key" in an array "pData" of size "n" that is sorted in ascending order. See http://doi.acm.org/10.1145/367487.367496 Algorithm description ----------------------------------------------------------------------------- Let Fk represent the k-th Fibonacci number where Fk+2=Fk+1 + Fk for k>=0 and F0 = 0, F1 = 1. To test whether an item is in a list of n = Fm ordered numbers, proceed as follows: a) Set k = m. b) If k = 0, finish - no match. c) Test item against entry in position Fk-1. d) If match, finish. e) If item is less than entry Fk-1, discard entries from positions Fk-1 + 1 to n. Set k = k - 1 and go to b). f) If item is greater than entry Fk-1, discard entries from positions 1 to Fk-1. Renumber remaining entries from 1 to Fk-2, set k = k - 2 and go to b) If Fm>n then the original array is augmented with Fm-n numbers larger than key and the above algorithm is applied. */ int FibonacciSearch( int *pData, int n, int key) { register int k, idx, offs; static int prevn = - 1, prevk = - 1; int times = 0; /* Precomputed Fibonacci numbers F0 up to F46. This implementation assumes that the size n * of the input array fits in 4 bytes. Note that F46=1836311903 is the largest Fibonacci * number that is less or equal to the 4-byte INT_MAX (=2147483647). The next Fibonacci * number, i.e. F47, is 2971215073 and is larger than INT_MAX, implying that it does not * fit in a 4 byte integer. Note also that the last array element is INT_MAX rather than * F47. This ensures correct operation for n>F46. */ const static int Fib[ 47 + 1] = { 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986, 102334155, 165580141, 267914296, 433494437, 701408733, 1134903170, 1836311903, INT_MAX }; /* find the smallest fibonacci number that is greater or equal to n. Store this * number to avoid recomputing it in the case of repetitive searches with identical n. */ if(n != prevn) { register int f0, f1, t; for(f0 = 0, f1 = 1, k = 1; f1 < n; t = f1, f1 += f0, f0 = t, ++k); prevk = k; prevn = n; } else k = prevk; /* If the sought value is larger than the largest Fibonacci number less than n, * care must be taken top ensure that we do not attempt to read beyond the end * of the array. If we do need to do this, we pretend that the array is padded * with elements larger than the sought value. */ for(offs = 0; k > 0; ) { idx = offs + Fib[--k]; /* note that at this point k has already been decremented once */ if(idx >= n || key < pData[idx]) // index out of bounds or key in 1st part { times++; continue; } else if (key > pData[idx]) { // key in 2nd part times++; offs = idx; --k; } else // key==pData[idx], found return times; } return - 1; // not found } |