I have to convert this recursive algorithm into an iterative one:
int alg(int A[], int x, int y, int k){
int val = 0;
if (x <= y){
int z = (x+y)/2;
if(A[z] == k){
val = 1;
}
int a = alg(A,x,z-1,k);
int b;
if(a > val){
b = alg(A,z+1,y,k);
}else{
b = a + val;
}
val += a + b;
}
return val;
}
I tried with a while loop, but I can't figure out how I can calculate "a" and "b" variables, so I did this:
int algIterative(int A[], int x, int y, int k){
int val = 0;
while(x <= y){
int z = (x+y)/2;
if(A[z] == k){
val = 1;
}
y = z-1;
}
}
But actually I couldn't figure out what this algorithm does.
My questions are:
What does this algorithm do?
How can I convert it to iterative?
Do I need to use stacks?
Any help will be appreciated.
I am not sure that alg computes anything useful.
It processes the part of the array A between the indexes x and y, and computes a kind of counter.
If the interval is empty, the returned value (val) is 0. Otherwise, if the middle element of this subarray equals k, val is set to 1. Then the values for the left and right subarrays are added and the total is returned. So in a way, it counts the number of k's in the array.
But, if the count on the left side is found to be not larger than val, i.e. 0 if val = 0 or 0 or 1 if val = 1, the value on the right is evaluated as the value on the left + val.
Derecursivation might be possible without a stack. If you look at the sequence of subintervals that are traversed, you can reconstruct it from the binary representation of N. Then the result of the function is the accumulation of partials results collected along a postorder process.
If the postorder can be turned to inorder, this will reduce to a single linear pass over A. This is a little technical.
A simple way could be smt like this with the aid of a two dimensional array:
int n = A.length;
int[][]dp = new int[n][n];
for(int i = n - 1;i >= 0; i--){
for(int j = i; j < n; j++){
// This part is almost similar to the recursive part.
int z = (i+j)/2;
int val = 0;
if(A[z] == k){
val = 1;
}
int a = z > i ? dp[i][z - 1] : 0;
int b;
if(a > val){
b = (z + 1 <= j) ? dp[z + 1][j] : 0;
}else{
b = a + val;
}
val += a + b;
dp[i][j] = val;
}
}
return dp[0][n - 1];
Explanation:
Notice that for i, it is decreasing, and j, it is increasing, so, when calculate dp[x][y], you need dp[x][z - 1] (with z - 1 < j) and dp[z + 1][j] (with z >= i), and those values should already be populated.
From "Introduction to algorithms" I am trying to implement the code as dividing n elements in n/5 groups, then recursively finding the median and then recursively finding the ith order statistics. Here is My code
bool isTrue(int *a, int *b)
{
if((*a) < (*b))
swap(*a, *b);
return *a < *b;
}
int select(int *A, int n ,int p)
{
int *B[(n / 5) + 2];
cout << (n / 5) + 2 << endl;
int i, j;
for(i = 1, j = 1; i <= (n - 5); i += 5, ++j)
{
sort(A + i, A + i + 5);
B[j] = A + i + 2;
}
sort(A + i, A + n + 1);
B[j] = A + i + (n - i) / 2;
sort(B + 1, B + (n / 5) + 2, isTrue);
}
this is as far as I can go. Now I am trying to find the median of B, then do B[median] - A as pivot. But It doesn't seems right. In the book it says to recursively find the median of medians.I can't catch that. Any help?
edit: I also note that in the wiki they didnt use any recursion!
I found a nice math problem, but I still can't solve it, I tried to find one solution using google and found that it can be solve using the Binary Indexed Tree data structure, but the solution is not clear to me.
Here the Problem called Finding Magic Triplets, it can be found in Uva online judge:
(a + b^2) mod k = c^3 mod k, where a<=b<=c and 1 <= a, b, c <= n.
given n and k (1 <= n, k <= 10^5), how many different magic triplets exist for known values of n and k. A triplet is different from another if any of the three values is not same in both triplets.
and here the solution that I found:
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long int64;
const int MAX_K = (int)(1e5);
int N, K;
struct BinaryIndexedTree{
typedef int64 bit_t;
static const int MAX_BIT = 3*MAX_K + 1;
bit_t data[MAX_BIT+1];
int SIZE;
void init(int size){
memset(data, 0, sizeof(data));
SIZE = size;
}
bit_t sum(int n){
bit_t ret = 0;
for(;n;n-=n&-n){
ret += data[n];
}
return ret;
}
bit_t sum(int from, int to){
return sum(to)-sum(from);
}
void add(int n, bit_t x){
for(n++;n<=SIZE;n+=n&-n){
data[n]+=x;
}
}
};
BinaryIndexedTree bitree;
void init(){
scanf("%d%d", &N, &K);
}
int64 solve(){
bitree.init(2*K+1);
int64 ans = 0;
for(int64 i=N; i>=1; i--){
int64 b = i * i % K, c = i * i * i % K;
bitree.add(c, 1);
bitree.add(c+K, 1);
bitree.add(c+2*K, 1);
int64 len = i;
if(len >= K){
ans += (len / K) * bitree.sum(K);
len %= K;
}
if(len > 0){
ans += bitree.sum(b + 1, b + len + 1);
}
}
return ans;
}
int main(){
int T;
scanf("%d", &T);
for(int i=0; i<T; i++){
init();
printf("Case %d: %lld\n", i+1, solve());
}
return 0;
}
Are you determined to use BITs? I would have thought ordinary arrays would do. I would start by creating three arrays of size k, where arrayA[i] = the number of values of a in range equal to i mod k, arrayB[i] = the number of values of b in range where b^2 = i mod k, and arrayC[i] = the number of values of c in range where c^3 = i mod k. N and k are both <= 10^5 so you could just consider each value of a in turn, b in turn, and c in turn, though you can be cleverer if k is much smaller than n, because will be some sort of fiddly fence-post counting expression that allows you to work out how many numbers in the range 0..n are equal to i mod k for each i.
Given those three arrays then consider each possible pair of numbers i, j where 0<=i,j < k and work out that there are arrayA[i] * arrayB[j] pairs which have those values mod k. Sum these up in arrayAB[i + j mod k] to find the number of ways that you can chose a + b^2 mod k = x for 0<=x < k. Now you have two arrays arrayAB and arrayC, where arrayAB[i] * arrayC[i] is the number of ways of finding a triple where a + b^2 = c^3] = i, so sum this over all 0<=i < k to get your answer.
Answering to another question, I wrote the program below to compare different search methods in a sorted array. Basically I compared two implementations of Interpolation search and one of binary search. I compared performance by counting cycles spent (with the same set of data) by the different variants.
However I'm sure there is ways to optimize these functions to make them even faster. Does anyone have any ideas on how can I make this search function faster? A solution in C or C++ is acceptable, but I need it to process an array with 100000 elements.
#include <stdlib.h>
#include <stdio.h>
#include <time.h>
#include <stdint.h>
#include <assert.h>
static __inline__ unsigned long long rdtsc(void)
{
unsigned long long int x;
__asm__ volatile (".byte 0x0f, 0x31" : "=A" (x));
return x;
}
int interpolationSearch(int sortedArray[], int toFind, int len) {
// Returns index of toFind in sortedArray, or -1 if not found
int64_t low = 0;
int64_t high = len - 1;
int64_t mid;
int l = sortedArray[low];
int h = sortedArray[high];
while (l <= toFind && h >= toFind) {
mid = low + (int64_t)((int64_t)(high - low)*(int64_t)(toFind - l))/((int64_t)(h-l));
int m = sortedArray[mid];
if (m < toFind) {
l = sortedArray[low = mid + 1];
} else if (m > toFind) {
h = sortedArray[high = mid - 1];
} else {
return mid;
}
}
if (sortedArray[low] == toFind)
return low;
else
return -1; // Not found
}
int interpolationSearch2(int sortedArray[], int toFind, int len) {
// Returns index of toFind in sortedArray, or -1 if not found
int low = 0;
int high = len - 1;
int mid;
int l = sortedArray[low];
int h = sortedArray[high];
while (l <= toFind && h >= toFind) {
mid = low + ((float)(high - low)*(float)(toFind - l))/(1+(float)(h-l));
int m = sortedArray[mid];
if (m < toFind) {
l = sortedArray[low = mid + 1];
} else if (m > toFind) {
h = sortedArray[high = mid - 1];
} else {
return mid;
}
}
if (sortedArray[low] == toFind)
return low;
else
return -1; // Not found
}
int binarySearch(int sortedArray[], int toFind, int len)
{
// Returns index of toFind in sortedArray, or -1 if not found
int low = 0;
int high = len - 1;
int mid;
int l = sortedArray[low];
int h = sortedArray[high];
while (l <= toFind && h >= toFind) {
mid = (low + high)/2;
int m = sortedArray[mid];
if (m < toFind) {
l = sortedArray[low = mid + 1];
} else if (m > toFind) {
h = sortedArray[high = mid - 1];
} else {
return mid;
}
}
if (sortedArray[low] == toFind)
return low;
else
return -1; // Not found
}
int order(const void *p1, const void *p2) { return *(int*)p1-*(int*)p2; }
int main(void) {
int i = 0, j = 0, size = 100000, trials = 10000;
int searched[trials];
srand(-time(0));
for (j=0; j<trials; j++) { searched[j] = rand()%size; }
while (size > 10){
int arr[size];
for (i=0; i<size; i++) { arr[i] = rand()%size; }
qsort(arr,size,sizeof(int),order);
unsigned long long totalcycles_bs = 0;
unsigned long long totalcycles_is_64 = 0;
unsigned long long totalcycles_is_float = 0;
unsigned long long totalcycles_new = 0;
int res_bs, res_is_64, res_is_float, res_new;
for (j=0; j<trials; j++) {
unsigned long long tmp, cycles = rdtsc();
res_bs = binarySearch(arr,searched[j],size);
tmp = rdtsc(); totalcycles_bs += tmp - cycles; cycles = tmp;
res_is_64 = interpolationSearch(arr,searched[j],size);
assert(res_is_64 == res_bs || arr[res_is_64] == searched[j]);
tmp = rdtsc(); totalcycles_is_64 += tmp - cycles; cycles = tmp;
res_is_float = interpolationSearch2(arr,searched[j],size);
assert(res_is_float == res_bs || arr[res_is_float] == searched[j]);
tmp = rdtsc(); totalcycles_is_float += tmp - cycles; cycles = tmp;
}
printf("----------------- size = %10d\n", size);
printf("binary search = %10llu\n", totalcycles_bs);
printf("interpolation uint64_t = %10llu\n", totalcycles_is_64);
printf("interpolation float = %10llu\n", totalcycles_is_float);
printf("new = %10llu\n", totalcycles_new);
printf("\n");
size >>= 1;
}
}
If you have some control over the in-memory layout of the data, you might want to look at Judy arrays.
Or to put a simpler idea out there: a binary search always cuts the search space in half. An optimal cut point can be found with interpolation (the cut point should NOT be the place where the key is expected to be, but the point which minimizes the statistical expectation of the search space for the next step). This minimizes the number of steps but... not all steps have equal cost. Hierarchical memories allow executing a number of tests in the same time as a single test, if locality can be maintained. Since a binary search's first M steps only touch a maximum of 2**M unique elements, storing these together can yield a much better reduction of search space per-cacheline fetch (not per comparison), which is higher performance in the real world.
n-ary trees work on that basis, and then Judy arrays add a few less important optimizations.
Bottom line: even "Random Access Memory" (RAM) is faster when accessed sequentially than randomly. A search algorithm should use that fact to its advantage.
Benchmarked on Win32 Core2 Quad Q6600, gcc v4.3 msys. Compiling with g++ -O3, nothing fancy.
Observation - the asserts, timing and loop overhead is about 40%, so any gains listed below should be divided by 0.6 to get the actual improvement in the algorithms under test.
Simple answers:
On my machine replacing the int64_t with int for "low", "high" and "mid" in interpolationSearch gives a 20% to 40% speed up. This is the fastest easy method I could find. It is taking about 150 cycles per look-up on my machine (for the array size of 100000). That's roughly the same number of cycles as a cache miss. So in real applications, looking after your cache is probably going to be the biggest factor.
Replacing binarySearch's "/2" with a ">>1" gives a 4% speed up.
Using STL's binary_search algorithm, on a vector containing the same data as "arr", is about the same speed as the hand coded binarySearch. Although on the smaller "size"s STL is much slower - around 40%.
I have an excessively complicated solution, which requires a specialized sorting function. The sort is slightly slower than a good quicksort, but all of my tests show that the search function is much faster than a binary or interpolation search. I called it a regression sort before I found out that the name was already taken, but didn't bother to think of a new name (ideas?).
There are three files to demonstrate.
The regression sort/search code:
#include <sstream>
#include <math.h>
#include <ctime>
#include "limits.h"
void insertionSort(int array[], int length) {
int key, j;
for(int i = 1; i < length; i++) {
key = array[i];
j = i - 1;
while (j >= 0 && array[j] > key) {
array[j + 1] = array[j];
--j;
}
array[j + 1] = key;
}
}
class RegressionTable {
public:
RegressionTable(int arr[], int s, int lower, int upper, double mult, int divs);
RegressionTable(int arr[], int s);
void sort(void);
int find(int key);
void printTable(void);
void showSize(void);
private:
void createTable(void);
inline unsigned int resolve(int n);
int * array;
int * table;
int * tableSize;
int size;
int lowerBound;
int upperBound;
int divisions;
int divisionSize;
int newSize;
double multiplier;
};
RegressionTable::RegressionTable(int arr[], int s) {
array = arr;
size = s;
multiplier = 1.35;
divisions = sqrt(size);
upperBound = INT_MIN;
lowerBound = INT_MAX;
for (int i = 0; i < size; ++i) {
if (array[i] > upperBound)
upperBound = array[i];
if (array[i] < lowerBound)
lowerBound = array[i];
}
createTable();
}
RegressionTable::RegressionTable(int arr[], int s, int lower, int upper, double mult, int divs) {
array = arr;
size = s;
lowerBound = lower;
upperBound = upper;
multiplier = mult;
divisions = divs;
createTable();
}
void RegressionTable::showSize(void) {
int bytes = sizeof(*this);
bytes = bytes + sizeof(int) * 2 * (divisions + 1);
}
void RegressionTable::createTable(void) {
divisionSize = size / divisions;
newSize = multiplier * double(size);
table = new int[divisions + 1];
tableSize = new int[divisions + 1];
for (int i = 0; i < divisions; ++i) {
table[i] = 0;
tableSize[i] = 0;
}
for (int i = 0; i < size; ++i) {
++table[((array[i] - lowerBound) / divisionSize) + 1];
}
for (int i = 1; i <= divisions; ++i) {
table[i] += table[i - 1];
}
table[0] = 0;
for (int i = 0; i < divisions; ++i) {
tableSize[i] = table[i + 1] - table[i];
}
}
int RegressionTable::find(int key) {
double temp = multiplier;
multiplier = 1;
int minIndex = table[(key - lowerBound) / divisionSize];
int maxIndex = minIndex + tableSize[key / divisionSize];
int guess = resolve(key);
double t;
while (array[guess] != key) {
// uncomment this line if you want to see where it is searching.
//cout << "Regression Guessing " << guess << ", not there." << endl;
if (array[guess] < key) {
minIndex = guess + 1;
}
if (array[guess] > key) {
maxIndex = guess - 1;
}
if (array[minIndex] > key || array[maxIndex] < key) {
return -1;
}
t = ((double)key - array[minIndex]) / ((double)array[maxIndex] - array[minIndex]);
guess = minIndex + t * (maxIndex - minIndex);
}
multiplier = temp;
return guess;
}
inline unsigned int RegressionTable::resolve(int n) {
float temp;
int subDomain = (n - lowerBound) / divisionSize;
temp = n % divisionSize;
temp /= divisionSize;
temp *= tableSize[subDomain];
temp += table[subDomain];
temp *= multiplier;
return (unsigned int)temp;
}
void RegressionTable::sort(void) {
int * out = new int[int(size * multiplier)];
bool * used = new bool[int(size * multiplier)];
int higher, lower;
bool placed;
for (int i = 0; i < size; ++i) {
/* Figure out where to put the darn thing */
higher = resolve(array[i]);
lower = higher - 1;
if (higher > newSize) {
higher = size;
lower = size - 1;
} else if (lower < 0) {
higher = 0;
lower = 0;
}
placed = false;
while (!placed) {
if (higher < size && !used[higher]) {
out[higher] = array[i];
used[higher] = true;
placed = true;
} else if (lower >= 0 && !used[lower]) {
out[lower] = array[i];
used[lower] = true;
placed = true;
}
--lower;
++higher;
}
}
int index = 0;
for (int i = 0; i < size * multiplier; ++i) {
if (used[i]) {
array[index] = out[i];
++index;
}
}
insertionSort(array, size);
}
And then there is the regular search functions:
#include <iostream>
using namespace std;
int binarySearch(int array[], int start, int end, int key) {
// Determine the search point.
int searchPos = (start + end) / 2;
// If we crossed over our bounds or met in the middle, then it is not here.
if (start >= end)
return -1;
// Search the bottom half of the array if the query is smaller.
if (array[searchPos] > key)
return binarySearch (array, start, searchPos - 1, key);
// Search the top half of the array if the query is larger.
if (array[searchPos] < key)
return binarySearch (array, searchPos + 1, end, key);
// If we found it then we are done.
if (array[searchPos] == key)
return searchPos;
}
int binarySearch(int array[], int size, int key) {
return binarySearch(array, 0, size - 1, key);
}
int interpolationSearch(int array[], int size, int key) {
int guess = 0;
double t;
int minIndex = 0;
int maxIndex = size - 1;
while (array[guess] != key) {
t = ((double)key - array[minIndex]) / ((double)array[maxIndex] - array[minIndex]);
guess = minIndex + t * (maxIndex - minIndex);
if (array[guess] < key) {
minIndex = guess + 1;
}
if (array[guess] > key) {
maxIndex = guess - 1;
}
if (array[minIndex] > key || array[maxIndex] < key) {
return -1;
}
}
return guess;
}
And then I wrote a simple main to test out the different sorts.
#include <iostream>
#include <iomanip>
#include <cstdlib>
#include <ctime>
#include "regression.h"
#include "search.h"
using namespace std;
void randomizeArray(int array[], int size) {
for (int i = 0; i < size; ++i) {
array[i] = rand() % size;
}
}
int main(int argc, char * argv[]) {
int size = 100000;
string arg;
if (argc > 1) {
arg = argv[1];
size = atoi(arg.c_str());
}
srand(time(NULL));
int * array;
cout << "Creating Array Of Size " << size << "...\n";
array = new int[size];
randomizeArray(array, size);
cout << "Sorting Array...\n";
RegressionTable t(array, size, 0, size*2.5, 1.5, size);
//RegressionTable t(array, size);
t.sort();
int trials = 10000000;
int start;
cout << "Binary Search...\n";
start = clock();
for (int i = 0; i < trials; ++i) {
binarySearch(array, size, i % size);
}
cout << clock() - start << endl;
cout << "Interpolation Search...\n";
start = clock();
for (int i = 0; i < trials; ++i) {
interpolationSearch(array, size, i % size);
}
cout << clock() - start << endl;
cout << "Regression Search...\n";
start = clock();
for (int i = 0; i < trials; ++i) {
t.find(i % size);
}
cout << clock() - start << endl;
return 0;
}
Give it a try and tell me if it's faster for you. It's super complicated, so it's really easy to break it if you don't know what you are doing. Be careful about modifying it.
I compiled the main with g++ on ubuntu.
Unless your data is known to have special properties, pure interpolation search has the risk of taking linear time. If you expect interpolation to help with most data but don't want it to hurt in the case of pathological data, I would use a (possibly weighted) average of the interpolated guess and the midpoint, ensuring a logarithmic bound on the run time.
One way of approaching this is to use a space versus time trade-off. There are any number of ways that could be done. The extreme way would be to simply make an array with the max size being the max value of the sorted array. Initialize each position with the index into sortedArray. Then the search would simply be O(1).
The following version, however, might be a little more realistic and possibly be useful in the real world. It uses a "helper" structure that is initialized on the first call. It maps the search space down to a smaller space by dividing by a number that I pulled out of the air without much testing. It stores the index of the lower bound for a group of values in sortedArray into the helper map. The actual search divides the toFind number by the chosen divisor and extracts the narrowed bounds of sortedArray for a normal binary search.
For example, if the sorted values range from 1 to 1000 and the divisor is 100, then the lookup array might contain 10 "sections". To search for value 250, it would divide it by 100 to yield integer index position 250/100=2. map[2] would contain the sortedArray index for values 200 and larger. map[3] would have the index position of values 300 and larger thus providing a smaller bounding position for a normal binary search. The rest of the function is then an exact copy of your binary search function.
The initialization of the helper map might be more efficient by using a binary search to fill in the positions rather than a simple scan, but it is a one time cost so I didn't bother testing that. This mechanism works well for the given test numbers which are evenly distributed. As written, it would not be as good if the distribution was not even. I think this method could be used with floating point search values too. However, extrapolating it to generic search keys might be harder. For example, I am unsure what the method would be for character data keys. It would need some kind of O(1) lookup/hash that mapped to a specific array position to find the index bounds. It's unclear to me at the moment what that function would be or if it exists.
I kludged the setup of the helper map in the following implementation pretty quickly. It is not pretty and I'm not 100% sure it is correct in all cases but it does show the idea. I ran it with a debug test to compare the results against your existing binarySearch function to be somewhat sure it works correctly.
The following are example numbers:
100000 * 10000 : cycles binary search = 10197811
100000 * 10000 : cycles interpolation uint64_t = 9007939
100000 * 10000 : cycles interpolation float = 8386879
100000 * 10000 : cycles binary w/helper = 6462534
Here is the quick-and-dirty implementation:
#define REDUCTION 100 // pulled out of the air
typedef struct {
int init; // have we initialized it?
int numSections;
int *map;
int divisor;
} binhelp;
int binarySearchHelp( binhelp *phelp, int sortedArray[], int toFind, int len)
{
// Returns index of toFind in sortedArray, or -1 if not found
int low;
int high;
int mid;
if ( !phelp->init && len > REDUCTION ) {
int i;
int numSections = len / REDUCTION;
int divisor = (( sortedArray[len-1] - 1 ) / numSections ) + 1;
int threshold;
int arrayPos;
phelp->init = 1;
phelp->divisor = divisor;
phelp->numSections = numSections;
phelp->map = (int*)malloc((numSections+2) * sizeof(int));
phelp->map[0] = 0;
phelp->map[numSections+1] = len-1;
arrayPos = 0;
// Scan through the array and set up the mapping positions. Simple linear
// scan but it is a one-time cost.
for ( i = 1; i <= numSections; i++ ) {
threshold = i * divisor;
while ( arrayPos < len && sortedArray[arrayPos] < threshold )
arrayPos++;
if ( arrayPos < len )
phelp->map[i] = arrayPos;
else
// kludge to take care of aliasing
phelp->map[i] = len - 1;
}
}
if ( phelp->init ) {
int section = toFind / phelp->divisor;
if ( section > phelp->numSections )
// it is bigger than all values
return -1;
low = phelp->map[section];
if ( section == phelp->numSections )
high = len - 1;
else
high = phelp->map[section+1];
} else {
// use normal start points
low = 0;
high = len - 1;
}
// the following is a direct copy of the Kriss' binarySearch
int l = sortedArray[low];
int h = sortedArray[high];
while (l <= toFind && h >= toFind) {
mid = (low + high)/2;
int m = sortedArray[mid];
if (m < toFind) {
l = sortedArray[low = mid + 1];
} else if (m > toFind) {
h = sortedArray[high = mid - 1];
} else {
return mid;
}
}
if (sortedArray[low] == toFind)
return low;
else
return -1; // Not found
}
The helper structure needs to be initialized (and memory freed):
help.init = 0;
unsigned long long totalcycles4 = 0;
... make the calls same as for the other ones but pass the structure ...
binarySearchHelp(&help, arr,searched[j],length);
if ( help.init )
free( help.map );
help.init = 0;
Look first at the data and whether a big gain can be got by data specific method over a general method.
For large static sorted datasets, you can create an additional index to provide partial pigeon holing, based on the amount of memory you're willing to use. e.g. say we create a 256x256 two dimensional array of ranges, which we populate with the start and end positions in the search array of elements with corresponding high order bytes. When we come to search, we then use the high order bytes on the key to find the range / subset of the array we need to search. If we did have ~ 20 comparisons on our binary search of 100,000 elements O(log2(n)) we're now down to ~4 comarisons for 16 elements, or O(log2 (n/15)). The memory cost here is about 512k
Another method, again suited to data that doesn't change much, is to divide the data into arrays of commonly sought items and rarely sought items. For example, if you leave your existing search in place running a wide number of real world cases over a protracted testing period, and log the details of the item being sought, you may well find that the distribution is very uneven, i.e. some values are sought far more regularly than others. If this is the case, break your array into a much smaller array of commonly sought values and a larger remaining array, and search the smaller array first. If the data is right (big if!), you can often achieve broadly similar improvements to the first solution without the memory cost.
There are many other data specific optimizations which score far better than trying to improve on tried, tested and far more widely used general solutions.
Posting my current version before the question is closed (hopefully I will thus be able to ehance it later). For now it is worse than every other versions (if someone understand why my changes to the end of loop has this effect, comments are welcome).
int newSearch(int sortedArray[], int toFind, int len)
{
// Returns index of toFind in sortedArray, or -1 if not found
int low = 0;
int high = len - 1;
int mid;
int l = sortedArray[low];
int h = sortedArray[high];
while (l < toFind && h > toFind) {
mid = low + ((float)(high - low)*(float)(toFind - l))/(1+(float)(h-l));
int m = sortedArray[mid];
if (m < toFind) {
l = sortedArray[low = mid + 1];
} else if (m > toFind) {
h = sortedArray[high = mid - 1];
} else {
return mid;
}
}
if (l == toFind)
return low;
else if (h == toFind)
return high;
else
return -1; // Not found
}
The implementation of the binary search that was used for comparisons can be improved. The key idea is to "normalize" the range initially so that the target is always > a minimum and < than a maximum after the first step. This increases the termination delta size. It also has the effect of special casing targets that are less than the first element of the sorted array or greater than the last element of the sorted array. Expect approximately a 15% improvement in search time. Here is what the code might look like in C++.
int binarySearch(int * &array, int target, int min, int max)
{ // binarySearch
// normalize min and max so that we know the target is > min and < max
if (target <= array[min]) // if min not normalized
{ // target <= array[min]
if (target == array[min]) return min;
return -1;
} // end target <= array[min]
// min is now normalized
if (target >= array[max]) // if max not normalized
{ // target >= array[max]
if (target == array[max]) return max;
return -1;
} // end target >= array[max]
// max is now normalized
while (min + 1 < max)
{ // delta >=2
int tempi = min + ((max - min) >> 1); // point to index approximately in the middle between min and max
int atempi = array[tempi]; // just in case the compiler does not optimize this
if (atempi > target)max = tempi; // if the target is smaller, we can decrease max and it is still normalized
else if (atempi < target)min = tempi; // the target is bigger, so we can increase min and it is still normalized
else return tempi; // if we found the target, return with the index
// Note that it is important that this test for equality is last because it rarely occurs.
} // end delta >=2
return -1; // nothing in between normalized min and max
} // end binarySearch