Im not at all sure how to word this question, but I will try my best. Im wanting to have examples of quantum algorithms that can complete logical parallel tasks. It extends beyond simply summation, for example multiplication, or finding the highest value, or a value closest to a given fixed value.
I know quantum algorithms can very easily "input" multiple states into a function/circuit and get a superposition of all answers, and even select specific desired outputs with grover's algorithm, but is it possible to incorporate multiple superposition into a final classical answer? Since there is no order to each "probability", obviously operations that depend on sequence are not possible.
Im trying to get into the mindset of how to make use of a quantum computer, and design circuits for it. Im not interested in theory or equations, just raw circuit/qasm diagrams.
Such examples that Im trying to refer to can be written as pseduo code like below
struct possibility {
float weight;
int value;
};
int summation(possibility[] input) {
int result = 0;
for (int i = 0; i < sizeof(input); i++) {
result += input[i].value * input[i].weight;
}
return result;
}
int multiplication(possibility[] input) {
int result = 1;
for (int i = 0; i < sizeof(input); i++) {
result *= input[i].value * input[i].weight;
}
return result;
}
int findClosest(possibility[] input, int toValue) {
int result = input[0].value;
int resultDistance = abs(toValue - result) * input[0].weight;
for (int i = 1; i < sizeof(input); i++) {
int distance = abs(toValue - input[i].value) * input[i].weight;
if (distance < resultDistance) {
result = input[i].value;
resultDistance = distance;
}
}
return result;
}
Sorry for my poor wording. Im not at all sure how to word this question better with my tiny knowledge in this subject. Any help at all is appreciated!
For accessing single point, I am using this line of code and it works
int intensity = gray_image.at<uchar>(Point(100, 100));
However when I use this code to access all the pixels in image, it gives memory error,
for (int i = 0; i < gray_image.rows;i++)
{
for (int j = 0; j < gray_image.cols; j++) {
intensity += gray_image.at<uchar>(Point(i, j));
}
}
When I run above code, it does not give compile time error but gives memory exception. Where am I going wrong?
You can just skip the use of Point and do the following.
for (int i = 0; i < gray_image.rows;i++)
{
for (int j = 0; j < gray_image.cols; j++) {
intensity += gray_image.at<uchar>(i, j);
}
}
You're requesting a pixel (j,i) that doesn't exist. This wouldn't have been an error in a square image (where the number of rows = number of columns), but you're using a rectangular image.
The Mat::at function has multiple prototypes, the two that you're concerned with are:
C++: template<typename T> T& Mat::at(int i, int j)
C++: template<typename T> T& Mat::at(Point pt)
The documentation for Mat::at states that Point pt is defined as the Element position specified as Point(j,i), so you've effectively swapped your rows and columns.
The reason this happens is because the image is stored in a 1D array of pixels, and to get a pixel Point (r,c) is translated to p = r * image.cols + c;
Consider the following code:
void foo(float* __restrict__ a)
{
int i; float val;
for (i = 0; i < 100; i++) {
val = 2 * i;
a[i] = val;
}
}
void bar(float* __restrict__ a)
{
int i; float val = 0.0;
for (i = 0; i < 100; i++) {
a[i] = val;
val += 2.0;
}
}
They're based on Examples 7.26a and 7.26b in Agner Fog's Optimizing software in C++ and should do the same thing; bar is more "efficient" as written in the sense that we don't do an integer-to-float conversion at every iteration, but rather a float addition which is cheaper (on x86_64).
Here are the clang and gcc results on these two functions (with no vectorization and unrolling).
Question: It seems to me that the optimization of replacing a multiplication by the loop index with an addition of a constant value - when this is beneficial - should be carried out by compilers, even if (or perhaps especially if) there's a type conversion involved. Why is this not happening for these two functions?
Note that if we use int's rather than float's:
void foo(int* __restrict__ a)
{
int i; int val = 0;
for (i = 0; i < 100; i++) {
val = 2 * i;
a[i] = val;
}
}
void bar(int* __restrict__ a)
{
int i; int val = 0;
for (i = 0; i < 100; i++) {
a[i] = val;
val += 2;
}
}
Both clang and gcc perform the expected optimization, albeit not quite in the same way (see this question).
You are looking for enabling induction variable optimization for floating point numbers. This optimization is generally unsafe in floating point land as it changes program semantics. In your example it'll work because both initial value (0.0) and step (2.0) can be precisely represented in IEEE format but this is a rare case in practice.
It could be enabled under -ffast-math but it seems this wasn't considered as important case in GCC as it rejects non-integral induction variables early on (see tree-scalar-evolution.c).
If you believe that this is an important usecase you might consider filing request at GCC Bugzilla.
I'm trying to learn concept about cuckoo search algorithm. Honestly, i'm learning cuckoo search using java source code. we know that cuckoo search using levy distribution for random walk. I have java source code impelements cuckoo search optimation , but i'm doubt that source code using levy distribution. Can anyone help me to inspect wheteher my source code using levy distribution or not?.
this is method that implements random walk..
public CSSolution randomWalk (OptimizationProblem prob, String distribution) {
int n = prob.getNumVar(); //2
// creates a neighborhood of size 1 times the scaling factor
double distanceSquared = Math.pow(rand.nextDouble() *
prob.getScalingFactor(),2);
System.out.println("distance Squared : "+distanceSquared);
// creates an ArrayList from 0 to n-1 (for indexing purposes only)
ArrayList<Integer> varIndices = new ArrayList<Integer>(n);
for (int i = 0; i < n; i++) {
varIndices.add(i, i);
}
ArrayList<Double> vars = this.getVars();
CSSolution newSol = new CSSolution(this.numVars);
newSol.initializeWithNull();
ArrayList<Double> newVars = newSol.getVars();
for (int i = 0; i < n; i++) {
/* Chooses a random variable index from the indices
* of the remaining/unwalked variables. */
int index = rand.nextInt(varIndices.size());
// Finds the variable value that this index corresponds to.
int varIndex = varIndices.get(index);
// System.out.println("varIndicesSize:"+varIndices.size()+" index: "+index+" varIndex: "+varIndex);
double curVar = vars.get(varIndex);
// use correct distribution to generate random double [0,1)
double r;
if (distribution == "weibull"){
r = weibull.random(1.5, 1, new uniform());
}
else if(distribution == "levy"){
r = 0.0;
}
else
r = rand.nextDouble();
// alters this variable coefficient by adding a random step between (-distance,distance)
double distance = Math.sqrt(distanceSquared);
System.out.println("distance : "+distance+" distance Squared : "+distanceSquared);
double varStep = r*distance*2-distance;
double newVar = curVar + varStep;
// System.out.println("x"+varIndex+" : "+curVar+" to "+newVar);
newVars.set(varIndex, newVar);
// removes the variable that has already been visited
varIndices.remove(index);
// updates distance for next for loop
distanceSquared -= Math.pow(varStep, 2);
}
// System.out.println("");
return newSol;
}
source code : https://github.com/cloudrave/Optimizer
It is a interview question. Given an array, e.g., [3,2,1,2,7], we want to make all elements in this array unique by incrementing duplicate elements and we require the sum of the refined array is minimal. For example the answer for [3,2,1,2,7] is [3,2,1,4,7] and its sum is 17. Any ideas?
It's not quite as simple as my earlier comment suggested, but it's not terrifically complicated.
First, sort the input array. If it matters to be able to recover the original order of the elements then record the permutation used for the sort.
Second, scan the sorted array from left to right (ie from low to high). If an element is less than or equal to the element to its left, set it to be one greater than that element.
Pseudocode
sar = sort(input_array)
for index = 2:size(sar) ! I count from 1
if sar(index)<=sar(index-1) sar(index) = sar(index-1)+1
forend
Is the sum of the result minimal ? I've convinced myself that it is through some head-scratching and trials but I haven't got a formal proof.
If you only need to find ONE of the best solution, here's the algorythm with some explainations.
The idea of this problem is to find an optimal solution, which can be found only by testing all existing solutions (well, they're infinite, let's stick with the reasonable ones).
I wrote a program in C, because I'm familiar with it, but you can port it to any language you want.
The program does this: it tries to increment one value to the max possible (I'll explain how to find it in the comments under the code sections), than if the solution is not found, decreases this value and goes on with the next one and so on.
It's an exponential algorythm, so it will be very slow on large values of duplicated data (yet, it assures you the best solution is found).
I tested this code with your example, and it worked; not sure if there's any bug left, but the code (in C) is this.
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
typedef int BOOL; //just to ease meanings of values
#define TRUE 1
#define FALSE 0
Just to ease comprehension, I did some typedefs. Don't worry.
typedef struct duplicate { //used to fasten the algorythm; it uses some more memory just to assure it's ok
int value;
BOOL duplicate;
} duplicate_t;
int maxInArrayExcept(int *array, int arraySize, int index); //find the max value in array except the value at the index given
//the result is the max value in the array, not counting th index
int *findDuplicateSum(int *array, int arraySize);
BOOL findDuplicateSum_R(duplicate_t *array, int arraySize, int *tempSolution, int *solution, int *totalSum, int currentSum); //resursive function used to find solution
BOOL check(int *array, int arraySize); //checks if there's any repeated value in the solution
These are all the functions we'll need. All split up for comprehension purpose.
First, we have a struct. This struct is used to avoid checking, for every iteration, if the value on a given index was originally duplicated. We don't want to modify any value not duplicated originally.
Then, we have a couple functions: first, we need to see the worst case scenario: every value after the duplicated ones is already occupied: then we need to increment the duplicated value up to the maximum value reached + 1.
Then, there are the main Function we'll discute later about.
The check Function only checks if there's any duplicated value in a temporary solution.
int main() { //testing purpose
int i;
int testArray[] = { 3,2,1,2,7 }; //test array
int nTestArraySize = 5; //test array size
int *solutionArray; //needed if you want to use the solution later
solutionArray = findDuplicateSum(testArray, nTestArraySize);
for (i = 0; i < nTestArraySize; ++i) {
printf("%d ", solutionArray[i]);
}
return 0;
}
This is the main Function: I used it to test everything.
int * findDuplicateSum(int * array, int arraySize)
{
int *solution = malloc(sizeof(int) * arraySize);
int *tempSolution = malloc(sizeof(int) * arraySize);
duplicate_t *duplicate = calloc(arraySize, sizeof(duplicate_t));
int i, j, currentSum = 0, totalSum = INT_MAX;
for (i = 0; i < arraySize; ++i) {
tempSolution[i] = solution[i] = duplicate[i].value = array[i];
currentSum += array[i];
for (j = 0; j < i; ++j) { //to find ALL the best solutions, we should also put the first found value as true; it's just a line more
//yet, it saves the algorythm half of the duplicated numbers (best/this case scenario)
if (array[j] == duplicate[i].value) {
duplicate[i].duplicate = TRUE;
}
}
}
if (findDuplicateSum_R(duplicate, arraySize, tempSolution, solution, &totalSum, currentSum));
else {
printf("No solution found\n");
}
free(tempSolution);
free(duplicate);
return solution;
}
This Function does a lot of things: first, it sets up the solution array, then it initializes both the solution values and the duplicate array, that is the one used to check for duplicated values at startup. Then, we find the current sum and we set the maximum available sum to the maximum integer possible.
Then, the recursive Function is called; this one gives us the info about having found the solution (that should be Always), then we return the solution as an array.
int findDuplicateSum_R(duplicate_t * array, int arraySize, int * tempSolution, int * solution, int * totalSum, int currentSum)
{
int i;
if (check(tempSolution, arraySize)) {
if (currentSum < *totalSum) { //optimal solution checking
for (i = 0; i < arraySize; ++i) {
solution[i] = tempSolution[i];
}
*totalSum = currentSum;
}
return TRUE; //just to ensure a solution is found
}
for (i = 0; i < arraySize; ++i) {
if (array[i].duplicate == TRUE) {
if (array[i].duplicate <= maxInArrayExcept(solution, arraySize, i)) { //worst case scenario, you need it to stop the recursion on that value
tempSolution[i]++;
return findDuplicateSum_R(array, arraySize, tempSolution, solution, totalSum, currentSum + 1);
tempSolution[i]--; //backtracking
}
}
}
return FALSE; //just in case the solution is not found, but we won't need it
}
This is the recursive Function. It first checks if the solution is ok and if it is the best one found until now. Then, if everything is correct, it updates the actual solution with the temporary values, and updates the optimal condition.
Then, we iterate on every repeated value (the if excludes other indexes) and we progress in the recursion until (if unlucky) we reach the worst case scenario: the check condition not satisfied above the maximum value.
Then we have to backtrack and continue with the iteration, that will go on with other values.
PS: an optimization is possible here, if we move the optimal condition from the check into the for: if the solution is already not optimal, we can't expect to find a better one just adding things.
The hard code has ended, and there are the supporting functions:
int maxInArrayExcept(int *array, int arraySize, int index) {
int i, max = 0;
for (i = 0; i < arraySize; ++i) {
if (i != index) {
if (array[i] > max) {
max = array[i];
}
}
}
return max;
}
BOOL check(int *array, int arraySize) {
int i, j;
for (i = 0; i < arraySize; ++i) {
for (j = 0; j < i; ++j) {
if (array[i] == array[j]) return FALSE;
}
}
return TRUE;
}
I hope this was useful.
Write if anything is unclear.
Well, I got the same question in one of my interviews.
Not sure if you still need it. But here's how I did it. And it worked well.
num_list1 = [2,8,3,6,3,5,3,5,9,4]
def UniqueMinSumArray(num_list):
max=min(num_list)
for i,V in enumerate(num_list):
while (num_list.count(num_list[i])>1):
if (max > num_list[i]+1) :
num_list[i] = max + 1
else:
num_list[i]+=1
max = num_list[i]
i+=1
return num_list
print (sum(UniqueMinSumArray(num_list1)))
You can try with your list of numbers and I am sure it will give you the correct unique minimum sum.
I got the same interview question too. But my answer is in JS in case anyone is interested.
For sure it can be improved to get rid of for loop.
function getMinimumUniqueSum(arr) {
// [1,1,2] => [1,2,3] = 6
// [1,2,2,3,3] = [1,2,3,4,5] = 15
if (arr.length > 1) {
var sortedArr = [...arr].sort((a, b) => a - b);
var current = sortedArr[0];
var res = [current];
for (var i = 1; i + 1 <= arr.length; i++) {
// check current equals to the rest array starting from index 1.
if (sortedArr[i] > current) {
res.push(sortedArr[i]);
current = sortedArr[i];
} else if (sortedArr[i] == current) {
current = sortedArr[i] + 1;
// sortedArr[i]++;
res.push(current);
} else {
current++;
res.push(current);
}
}
return res.reduce((a,b) => a + b, 0);
} else {
return 0;
}
}