Algorithm for Adding/Subtracting numbers to find if number can be made

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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

It is somewhere nearer to Knapsack problem.

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The sum/diff of any two numbers in the group, or the sum/diff of any number of numbers in the group? Are all numbers in the group unique? Can there be negative numbers? Is there a known boundary to the numbers? Is the special number guaranteed to be between the highest and the lowest number in the group?

Roses are FF0000, violets are 0000FF... All my base are belong to you.

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```Example data:

151.15   --> is the number we need ti get by doing arithmetic operations(only add or subtract) on below set of numbers (each number can be used any number of times).

Set of numbers:

2.77

8.24

2.77

3.48

3.48

3.99

3.48

3.48

3.48

17.4

3.48

3.48

3.48

2.77

4.85

5.67

17.67

2.77

3.48

2.77

3.58

2.77

3.48

2.77

2.77

3.48

3.48

2.77

3.48

3.48

3.48

3.48

3.48

2.77

2.77

3.48

3.48

2.77

3.48

3.48

3.48

3.48

4.8

5

6.76

3.48

3.48

3.48

14.4

17.94

4.85

3.48

3.48

3.48

3.48

8.39

4.58

4.85

3.48

2.77

3.48

2.77

2.77

3.48

3.48

3.48

4.03

3.48

2.77

21.76

3.48

3.48

10.03

3.48

2.77

9.21

2.77

3.65

1.23

1.23

1.11

0.37

0.47

0.47

0.54

0.47

0.47

0.47

2.34

0.47

0.47

0.47

0.37

0.65

0.76

2.37

0.37

0.47

0.37

0.48

1.23

1.33

1.23

0.37

0.47

0.47

0.37

0.47

0.47

1.33

1.33

0.47

0.37

0.37

0.47

0.47

0.37

0.47

0.47

0.47

0.47

0.64

0.67

0.91

0.47

0.47

0.47

1.94

2.41

1.51

1.33

0.47

0.47

0.47

1.13

1.48

0.65

0.47

0.37

0.47

0.37

0.37

1.33

0.47

0.47

0.54

0.47

0.37

2.92

0.47

0.47

1.35

0.47

0.37

1.24

0.37

0.46```

Edited by Melba23
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There's nothing "premade" you're going to find here. In all case, scale up all values (including target) to integers to avoid floating-point issues (here, * 100), then feed the baby to a classical knapsack variant. Of course you know it's NP-Complete.

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This is what I did meanwhile for your initial post:

```;Brute force method

Global \$aIntegers[4] = [5, 8, 10, 2]
ConsoleWrite(BruteForce_Calc(\$aIntegers, 1) & "=" & 1 & @CRLF)
ConsoleWrite(BruteForce_Calc(\$aIntegers, 9) & "=" & 9 & @CRLF)
ConsoleWrite(BruteForce_Calc(\$aIntegers, 15) & "=" & 15 & @CRLF)
ConsoleWrite(BruteForce_Calc(\$aIntegers, -11) & "=" & -11 & @CRLF)

Func BruteForce_Calc(ByRef \$aIntegers, \$iTarget)
Local \$aOp[2] = ["+", "-"], \$i = 1, \$j, \$z = UBound(\$aIntegers), \$iCalc, \$k, \$l, \$sTerm
Do
For \$j = 0 To UBound(\$aIntegers) - 1
\$l = 2^(\$z - \$j)
\$k = (Mod(Mod(\$i - 1, \$l), \$i)) < (\$l / 2) ? 0 : 1
\$sTerm &= \$aOp[\$k] & \$aIntegers[\$j]
Next
\$iCalc = Execute(\$sTerm)
If \$iCalc = \$iTarget Then Return (StringLeft(\$sTerm, 1) = "+" ? StringTrimLeft(\$sTerm, 1) : \$sTerm)
\$sTerm = ""
\$i += 1
Until \$i > 2 ^ \$z
EndFunc```

I didn't test it properly and thus I don't know whether it will work also for other variations.

Edited by UEZ

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