Binary Search with nextafter() - c++

I input an double n and am trying to find the root by binary search. the break-condition with epsilon are kind of long and i was thinking about using nextafter() to compute the result with high precision.
double lower = 1, upper = n, mid, midSquared;
while (nextafter(lower,upper) < upper) {
mid = (l + u)/2.;
midSquared = mid * mid;
if (midSquared < n) lower = mid;
else upper = mid;
}
Am I guaranteed that that way my loop terminates? This actually is about mid. Will mid take a value between lower and upper bound, when there is a displayable one?
This seems to me as a nice approach because it matches the index comparison binary search on a sorted list. (Lower index + 1 < Upper index)

Initializing lower to 0 and upper to a minimum of 1 will allow your code to find square roots for values < 1.
I would do the error checking on the return value from nextafter if you aren't doing range validation on n. Check for NaN and HUGE_VAL.
double lower = 0;
double upper = (n < 1) ? 1 : n;
double next = nextafter(lower, upper);
while (!isnan(next) && (next < upper) && (next != HUGE_VAL))
{
double mid = (lower + upper) / 2.;
double midSquared = mid * mid;
if (midSquared < n)
{
lower = mid;
}
else
{
upper = mid;
}
next = nextafter(lower, upper);
}
And if you really want to be sure, you could put a counter on the loop as well and abort if it does too many loops.
int count = 0;
while (!isnan(next) && (next < upper) && (next != HUGE_VAL) && (count < 1000))
{
count++;
double mid = (lower + upper) / 2.;
...

Related

How to change function so that the difference between target and any element is minimum?

how can I change these two (or either of them) functions to get the minimum dist where dist=pow(target-v[i],2). I want to get the index of the element in the vector with the smallest distance from target considering that the vector is ordered and I want to find the element efficiently using binary search.
Thanks a lot.
int getClosest(int val1, int val2,int target, int i, int j)
{
if (pow(target - val1,2) >= pow(val2 - target,2))
return j;
else
return i;
}
// Returns element closest to target
int findClosest(vector<int> arr, int n, int target)
{
// Corner cases
if (target <= arr[0])
return 0;
if (target >= arr[n - 1])
return n - 1;
// Doing binary search
int i = 0, j = n, mid = 0;
while (i < j) {
mid = (i + j) / 2;
if (arr[mid] == target)
return mid;
/* If target is less than array element,
then search in left */
if (target < arr[mid]) {
// If target is greater than previous
// to mid, return closest of two
if (mid > 0 && target > arr[mid - 1])
{
return getClosest(arr[mid - 1],arr[mid], target, mid-1,mid);
}
/* Repeat for left half */
j = mid;
}
// If target is greater than mid
else {
if (mid < n - 1 && target < arr[mid + 1])
return getClosest(arr[mid], arr[mid + 1], target, mid,mid+1);
// update i
i = mid + 1;
}
}
// Only single element left after search
return mid;
}
enter code here
First of all, to use binary search algorithm, you must be sure about the given array or the vector is sorted or not. If the vector is sorted then only you can apply binary search to find the index of the closest distance.
Now considering the given vector is sorted, we can apply binary search.
Another thing that I see in your code, into the getClosest(int v1, int v2, int target, int i, int j) method that you are squaring both side to check with the positive integers. Rather you can consider mod operator or abs method in cpp to do so. abs will take less time rather than pow.
Modified Code:
int getClosest(int val1, int val2,int target, int i, int j)
{
if (abs(target - val1) >= abs(val2 - target))
return j;
else
return i;
}
// Returns element closest to target
int findClosest(vector<int> arr, int n, int target)
{
// Corner cases
if (target <= arr[0])
return 0;
if (target >= arr[n - 1])
return n - 1;
// Doing binary search
int i = 0, j = n, mid = 0;
while (i < j) {
mid = (i + j) / 2;
if (arr[mid] == target)
return mid;
/* If target is less than array element,
then search in left */
if (target < arr[mid]) {
// If target is greater than previous
// to mid, return closest of two
if (mid > 0 && target > arr[mid - 1])
{
return getClosest(arr[mid - 1],arr[mid], target, mid-1,mid);
}
/* Repeat for left half */
j = mid;
}
// If target is greater than mid
else {
if (mid < n - 1 && target < arr[mid + 1])
return getClosest(arr[mid], arr[mid + 1], target, mid,mid+1);
// update i
i = mid + 1;
}
}
// Only single element left after search
return mid;
}

Binary Search: how to determine half of the array

What's the difference between this two formulas
mid = low + (high - low) / 2;
mid = (high + low) / 2;
In the 2nd version, if high + low is greater than the maximum value of an int (assuming high is an int) then it can overflow, invoking undefined behavior. This particular bug is solved with the 1st version.
There are still issues with the 1st version, e.g. if low is a very large negative number, the difference can still overflow.
From c++20, you should use std::midpoint for this, which handles a whole bunch of corner cases, and does the right thing for all of them.
This seemingly simple function is actually surprisingly difficult to implement, and in fact, there's an hour long talk given by Marshall Clow at cppcon 2019, that covers the implementation of just this function.
The first one is superior (although still not perfect, see Binary Search: how to determine half of the array):
It works in cases where addition is not defined for high and low but is defined for adding an interval to low. Pointers are one such example, an object of a date type can be another.
high + low can overflow the type. For a signed integral type, the behaviour is undefined.
Both suffer from potential overflow. Signed integer overflow is undefined behavior (UB).
With unsigned math (often used in array indexing), then when low <= high, low + (high - low) / 2; does not overflow unlike potentially (high + low) / 2.
Same with signed math when low <= high and 0 <= low.
To avoid any overflow with signed math (or unsigned math with low > high) and still use only int/unsigned math, I thought the below would work.
mid = high/2 + low/2 + (high%2 + low%2)/2;
Yet that can fail when the sign of high/2 + low/2 differs from sign of (high%2 + low%2).
A more robust and tested version is below. Perhaps I'll simplify later.
#include <limits.h>
#include <stdio.h>
int midpoint(int a, int b) {
int avg = a/2 + b/2;
int small_sum = a%2 + b%2;
avg += small_sum/2;
small_sum %= 2;
if (avg < 0) {
if (small_sum > 0) avg++;
} else if (avg > 0) {
if (small_sum < 0) avg--;
}
return avg;
}
int midpoint_test(int a, int b) {
intmax_t lavg = ((intmax_t)a + (intmax_t)b)/2;
int avg = midpoint(a,b);
printf("a:%12d b:%12d avg_wide_math:%12jd avg_midpoint:%12d\n", a,b,lavg,avg);
return lavg == avg;
}
int main(void) {
int a[] = {INT_MIN, INT_MIN+1, -100, -99, -2, -1, 0, 1, 2, 99, 100, INT_MAX-1, INT_MAX};
int n = sizeof a/ sizeof a[0];
for (int i=0; i<n; i++) {
for (int j=0; j<n; j++) {
if (midpoint_test(a[i], a[j]) == 0) {
puts("Oops");
return 1;
}
}
}
puts("Success");
return 0;
}
The two formulae are different:
both may overflow depending on the values of low and high.
even when there is no overflow, they do not necessarily produce the same result: the first computes the midpoint and the second computes the average of 2 numbers.
For the rest of the discussion, we shall assume that low, mid and high have the same type. We are looking for a safe way to find the midpoint or average between low and high, which is always in the range of the type.
The first formula, mid = low + (high - low) / 2; rounds toward low if the type is signed and may overflow if the type is signed and high and low are too far appart.
The second formula, mid = (high + low) / 2; rounds toward 0, but may overflow for large values of high and/or low for both signed and unsigned types.
In your particular application, computing the index of the middle element of a sorted array to perform binary search, the index values low and high are non-negative and low <= high. With this constraint, both formulas compute the same result, but the second can overflow whereas the first cannot.
Hence for your case, you should use mid = low + (high - low) / 2; as a safe replacement for mid = (high + low) / 2;.
In the general case, computing the average (second formula) without overflow is a tricky problem. Below is a set of solutions for the average formula, along with a test program inspired from chux' answer. They can be adapted for any signed integer type:
#include <limits.h>
#include <stdio.h>
#include <stdint.h>
int average_chqrlie(int a, int b) {
if (a <= b) {
if (a >= 0)
return a + ((b - a) >> 1);
if (b < 0)
return b - ((b - a) >> 1);
} else {
if (b >= 0)
return b + ((a - b) >> 1);
if (a < 0)
return a - ((a - b) >> 1);
}
return (a + b) / 2;
}
int average_chqrlie2(int a, int b) {
if (a > b) {
int tmp = a;
a = b;
b = tmp;
}
if (a >= 0)
return a + ((b - a) >> 1);
if (b < 0)
return b - ((b - a) >> 1);
return (a + b) / 2;
}
int average_chqrlie3(int a, int b) {
int half, mid;
if (a < b) {
half = (int)(((unsigned)b - (unsigned)a) / 2);
mid = a + half;
if (mid < 0)
mid = b - half;
} else {
half = (int)(((unsigned)a - (unsigned)b) / 2);
mid = b + half;
if (mid < 0)
mid = a - half;
}
return mid;
}
int average_chux(int a, int b) {
int avg = a / 2 + b / 2;
int small_sum = a % 2 + b % 2;
avg += small_sum / 2;
small_sum %= 2;
if (avg < 0) {
if (small_sum > 0)
avg++;
} else if (avg > 0) {
if (small_sum < 0)
avg--;
}
return avg;
}
int run_tests(const char *name, int (*fun)(int a, int b)) {
int array[] = { INT_MIN, INT_MIN+1, -100, -99, -2, -1, 0, 1, 2, 99, 100, INT_MAX-1, INT_MAX };
int n = sizeof(array) / sizeof(array[0]);
int status = 0;
printf("Testing %s:", name);
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
int a = array[i], b = array[j];
intmax_t lavg = ((intmax_t)a + (intmax_t)b) / 2; // assuming sizeof(intmax_t) > size(int)
int avg = fun(a, b);
if (lavg != avg) {
printf("\na:%12d b:%12d average_wide:%12jd average:%12d", a, b, lavg, avg);
status = 1;
}
}
}
puts(status ? "\nFailed" : " Success");
return status;
}
int main() {
run_tests("average_chqrlie", average_chqrlie);
run_tests("average_chqrlie2", average_chqrlie2);
run_tests("average_chqrlie3", average_chqrlie3);
run_tests("average_chux", average_chux);
return 0;
}
The first one will not result in overflow for a large value of low/high unlike second one. It's always preferred to use mid = low + (high - low) / 2.

Binary Search for a range

I was making binary search program to find the number of elements between the Left and Right values in a range .
I code it :
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
vector<int> arr(20);
int search(int value,int low,int high)
{
if (high <= low)
return low;
int mid = (low + high) / 2;
if (arr[mid] > value)
return search(value,low,mid-1);
else
return search(value,mid+1,high);
}
int main(){
int n;
cin>>n;
//ENTER SORTED ARRAY
for(int i=0;i<n;i++){
cin>>arr[i];
}
int left;
cin>>left;
//RIGHT IS GREATER OR EQUAL TO LEFT
int right;
cin>>right;
cout<<search(right,0,n-1)-search(left,0,n-1)+1<<"\n";
}
It's giving right answer for some ranges.
But for some its giving wrong like If N=6 and array be [1 3 5 8 10 13] and say the range be [5,9] then it's giving 1 as the answer but it should be 2 as 5 and 8 both are in the range.
try this
int search(int value,int low,int high)
{
if (high <= low)
return low;
int mid = (low + high) / 2;
if(arr[mid]==value){ // add this line it would be work for you
return mid;
}
if (arr[mid] > value)
return search(value,low,mid-1);
else
return search(value,mid+1,high);
}
and make correction in main()
cout<<search(right,0,n-1)-search(left,0,n-1)<<"\n";
int search(int value,int low,int high)
{
if (high <= low + 1)
return low;
int mid = (low + high) / 2;
if (arr[mid] > value)
return search(value,low,mid);
else
return search(value,mid,high);
}
And in your main function
cout<<search(right+1,0,n-1)-search(left,0,n-1)<<"\n";
One problem is that when arr[mid] == value, you just ignore it and recurse to the right.
You'll need to either include mid in your right range, or return mid if arr[mid] == value.
I also see duplicate values (if these are possible) being a problem - when recursing to find the left-most position, you need to find the first duplicate value, when recursing to find the right-most position, you need to find the last duplicate value, so a single function without a flag to indicate which one we're doing isn't going to work. To illustrate the problem:
If the range is [5,5] and the input is [1,2,5,5,5,6,8], the same recursive call finding the position of 5 will always return the position of the same 5, where-as you need to return index 2 for the left range and index 4 for the right, as to get 3 as your output.
There is no check that arr[mid] can be == value. In your example, first iteration for left == 5 gives mid == ( 0 + (6-1) )/2 = 5/2 = 2 and arr[2] is exactly 5. We should stop, but your code goes to the branch search(5, 3, 5);
The logic of your program seems wrong, if you want to find the number of elements in the arr that are in the range of [left,right], try this:
int i;
int count = 0;
for(i = 0; i < n; i++) {
if (arr[i] >= left && arr[i] <= right)
count++;
}
If you insist on using binary search try this :
static int search(int value,int low,int high)
{
if (high <= low)
return low;
int mid = (low + high) / 2;
if (arr[mid] == value)
return mid;
int idx;
if (arr[mid] > value)
idx = search(value,low,mid-1);
else
idx = search(value,mid+1,high);
if (value == arr[idx]) {
return idx;
}
else {
if(value > arr[idx])
return mid +1;
else
return mid;
}
}

C++ Binary Search Algorithm not working

So I have a vector of ints named bList that has information already in it. I have it sorted before running the binary search.
//I have already inserted random ints into the vector
//Sort it
bubbleSort();
//Empty Line for formatting
cout << "\n";
//Print out sorted array.
print();
cout << "It will now search for a value using binary search\n";
int val = binSearch(54354);
cout<<val;
My bubble sort algorithm does work.
I have it returning an int which is the location of the searched value in the list.
//Its one argument is the value you are searching for.
int binSearch(int isbn) {
int lower = 0;
int upper = 19;//Vector size is 20.
int middle = (lower + upper) / 2;
while (lower < upper) {
middle = (lower + upper) / 2;
int midVal = bList[middle];
if (midVal == isbn) {
return middle;
break;
} else if (isbn > midVal) {
lower = midVal + 1;
} else if (isbn < midVal) {
upper - midVal - 1;
}
}
}
But for some reason, when I run it, it just keeps running and doesn't return anything.
Here the bug is:
// ...
} else if (isbn > midVal) {
lower = midVal + 1;
} else if (isbn < midVal) {
upper - midVal - 1;
}
You may want
lower = middle + 1;
and
upper = middle - 1;
instead.
You also need to explicitly return something when the required number cannot be found.
You still have a slight logic problem with your while condition:
int binary_search(int i, const std::vector<int>& vec) // you really should pass in the vector, if not convert it to use iterators
{
int result = -1; // default return value if not found
int lower = 0;
int upper = vec.size() - 1;
while (lower <= upper) // this will let the search run when lower == upper (meaning the result is one of the ends)
{
int middle = (lower + upper) / 2;
int val = vec[middle];
if (val == i)
{
result = middle;
break;
}
else if (i > val)
{
lower = middle + 1; // you were setting it to the value instead of the index
}
else if (i < val)
{
upper = middle - 1; // same here
}
}
return result; // moved your return down here to always return something (avoids a compiler error)
}
Alternatively, you could switch it to use iterators instead:
template<class RandomIterator>
RandomIterator binary_search(int i, RandomIterator start, RandomIterator end)
{
RandomIterator result = end;
while (start <= end) // this will let the search run when start == end (meaning the result is one of the ends)
{
RandomIterator middle = start + ((end - start) / 2);
if (*middle == i)
{
result = middle;
break;
}
else if (i > *middle)
{
start = middle + 1;
}
else if (i < *middle)
{
end = middle - 1;
}
}
return result;
}

What is the fastest search method for a sorted array?

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