Reputation: 71
I was wondering if there is an efficient premade algorithm for determining if the sum/difference of a group of numbers can equal a different number. Example:
5, 8, 10, 2, using + or -, to equal 9. 5 - 8 = -3 + 10 = 7 + 2 = 9
If there is a preexisting algorithm, what is it called. If not, I can figure out how to program it, though it may not be efficient.
Thank you!
Upvotes: 4
Views: 2701
Reputation: 2314
I actually wrote a simple java program, I was not actually aware of knapsack strategies. This is my own solution. Hope this helps
import java.util.ArrayList;
import java.util.List;
public class Puzzle {
public static void main(String[] args) {
int targetNumber = 0;
int min = 2147483647;
int[] numbers = {-10, -30, -20, -50};
//int[] numbers = {0,0,0,0};
//int[] numbers = {7, 2, 10};
//int[] numbers = {1, 2, 3, 4, 5};
//int[] numbers = {1000, 2, 3, 4, 100};
char set[] = {'+', '-'};
min = getNumberClosestToTarget(numbers, set, min, targetNumber);
System.out.println(String.format(" %d is closest to %d", min, targetNumber));
}
private static int getNumberClosestToTarget(int[] numbers, char[] set, int min, int targetNumber) {
List<String> operators = new ArrayList<>();
computeAllOperatorsCombination(set, "", set.length, numbers.length - 1, operators);
for (String operatorString : operators) {
String[] ops = operatorString.split("");
int sum = computeSum(numbers, ops, numbers.length - 1);
min = getClosestToTarget(min, targetNumber, sum);
}
return min;
}
static int computeSum(int[] numbers, String[] operators, int index) {
int result = numbers[index];
if (index == 0) {
return result;
} else {
switch (operators[index - 1]) {
case "+":
return computeSum(numbers, operators, index - 1) + result;
case "-":
return computeSum(numbers, operators, index - 1) - result;
}
return result;
}
}
static void computeAllOperatorsCombination(char set[], String prefix, int n, int k, List<String> result) {
if (k == 0) {
result.add(prefix);
return;
}
for (int i = 0; i < n; i++) {
String newPrefix;
newPrefix = prefix + set[i];
computeAllOperatorsCombination(set, newPrefix, n, k - 1, result);
}
}
private static int getClosestToTarget(int min, int targetNumber, int r) {
int distance = Math.abs(targetNumber - r) < Math.abs(r - targetNumber) ? Math.abs(targetNumber - r) : Math.abs(r - targetNumber);
if (distance < Math.abs(min)) {
min = distance;
if (r < 0) {
min = -distance;
}
}
return min;
}
}
Upvotes: 0
Reputation: 1
its not an NP problem, if the problem is to find a given number from adding or subtracting each elements of a list/array. if you think about AP. here is a sample code in C++
int Np( int mn, list<int>a, int c )
{
int size = a.size(), rst = 0, maxI = 0;
std::list<int>::iterator it;
while( size > c )
{
a.sort();
maxI += a.back();
a.pop_back();
rst = 0;
for( auto ele : a )
{
rst += ele;
cout << rst << endl;
}
if( (rst - maxI) == mn or (maxI - rst) == mn or (maxI + rst) == mn )
{
return mn;
}
size--;
}
return rst;
}
this should help. i think.
Upvotes: 0
Reputation: 23029
Yeah, this is basically knapsack problem, but it can be computed in pseudopolynomial time using dynamic programming.
I did it few month ago, so maybe this java code can help you, if you want to implement it :
public void solve() {
while (this.isEnd() == false) {
int priceSum = this.getItemsInstance().getTotalPrice()/divide;
int numOfItems = this.getItemsInstance().itemCount();
int maxWeight = this.getItemsInstance().getMaxWeight();
int[][] array = new int[numOfItems + 1][priceSum + 1];
boolean[][] arrayCounted = new boolean[numOfItems + 1][priceSum + 1];
for (int i = 0; i < numOfItems + 1; i++) {
array[i][0] = 0;
arrayCounted[i][0] = true;
}
int max = 0;
int price = 0;
for (int j = 1; j < priceSum + 1; j++) {
for (int i = 1; i < numOfItems + 1; i++) {
int temp = W(i, j, array, arrayCounted);
if (temp <= maxWeight) {
max = temp;
price = j;
}
}
}
}
}
private int W(int i, int c, int[][] array, boolean[][] arrayCounted) {
if (c < 0) {
return MAX_PRICE / divide;
}
if (i == 0) {
if (c == 0) {
return 0;
} else {
return MAX_PRICE / divide;
}
}
if (arrayCounted[i][c]) {
return array[i][c];
}
arrayCounted[i][c] = true;
array[i][c] = Math.min(W(i - 1, c, array, arrayCounted), W(i - 1, c - this.items[i - 1].price/divide, array, arrayCounted) + this.items[i - 1].weight);
return array[i][c];
}
Upvotes: 2