# Subset sum problem with an extra condition [closed]

A sub-set sum problem is described that giving a set of integers $$S$$ and a constant $$C$$, find a subset $$s$$ of $$S$$ so that the sum of integers in $$s$$ is maximum but not greater than $$C$$. $$maximize\sum_{i=0}^{|S|} v_ix_i \ \ \ subject\ to \sum_{i=0}^{|S|} v_ix_i \le C$$ where $$v_i \in S$$ and $$x_i\in \{0,1\}$$

We have an extra problem here. Give two sets of numbers $$A = \{a_1, a_2, a_3, a_4, a_5\}$$ and $$K = \{k_1, k_2, k_3, k_4\}$$ and a number of lists in order as below: \begin{align*} L_1 &= \{a_1, a_4\} \\ L_2 &= \{a_1, a_2, a_3, a_4\} \\ L_3 &= \{a_3, a_4, a_5\} \\ L_4 &= \{a_5\} \end{align*} Assign $$S_i$$ is the expected result for sub-set sum problem of $$L_i$$. Find a subset of $$A$$ so that if an element of this subset appears both $$S_i$$ and $$L_{i+1}$$, it must also be in $$S_{i+1}$$

For each list, find a subset sum with a respective constraint in $$K$$ ($$L_1 \leftrightarrow k_1$$, $$L_2 \leftrightarrow k_2$$, and so on). The procedure is described as following:

• First, we can see that $$L_1$$ contains the subsets like $$\{a_1\}, \{a_4\}$$ and $$\{a_1, a_4\}$$, we then assume a subset $$S_1=\{a_1,a_4\}$$, with $$a_1$$ and $$a_4$$ sum up to $$k_1$$, is the expected result for $$L_1$$
• Next, we have to solve the subset sum problem with $$L_2$$ by finding a subset $$S_2$$ with a restriction (because $$a_1,a_4$$ also appear in $$L_2$$, so $$S_2$$ must contain $$a_1,a_4$$). Therefore all possible states of $$S_2$$ are $$\{a_1, a_4, a_3\}, \{a_1, a_4, a_2\}, \{a_1, a_4, a_2, a_3\}$$. Assume that $$S_2 = \{a_1, a_4, a_3\}$$ is the expected subset for $$L_2$$
• We then continue playing with $$L_3$$. Similarly, we can see $$a_3, a_4$$ also appear in $$L_3$$, thus $$S_3$$ must contain $$a_3, a_4$$, i.e., $$S_3=\{a_1,a_4,a_3\}$$.
• Finally, the subset $$B = \{a_1, a_4, a_3\}$$ is final result what I expect.

For example $$A = \{1, 2, 3, 4, 5\}$$ and $$K = \{7, 8, 9, 10\}$$
$$L_1 = \{1, 4\}, k_1 = 7$$, due to $$1 < 4 < (1+4) < 7$$ so $$S_1=\{1, 4\}$$
$$L_2 = \{1, 2, 3, 4\}, k_2 = 8$$, due to $$(1+4+2) < (1+4+3) <= 8 < (1+4+2+3)$$ so $$S_2 = \{1, 4, 3\}$$ $$L_3 = \{3, 4, 5\}, k_3 = 9$$, due to $$(4+3) < 9 < (4+3+5)$$ then $$S_3$$ is same with $$S_2$$
$$L_4 = \{5\}, k_4 = 10$$, due to 5 is not in $$S_3$$ so we do not consider it, therefore $$S_4$$ is empty
The final result is $$B = S_3 = \{1, 4, 3\}$$

Currently, I can solve this problem with the brute-force algorithm. However, I'm finding any other faster ways. Does anyone have any suggestion for me?

## closed as unclear what you're asking by Yuval Filmus, Evil, GillesDec 16 '18 at 19:37

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• Please credit the original source of the problem. – Apass.Jack Dec 16 '18 at 10:32
• I don't understand which source I have to credit, this is my own problem. – hdng Dec 16 '18 at 13:49
• Could you please share a bit of background how you come up with the problem? – Apass.Jack Dec 16 '18 at 14:03
• This problem I came up with by expanding the special case of knapsack problem. – hdng Dec 16 '18 at 15:03
• The problem is still extremely hard to understand, since you're not explaining it. – Yuval Filmus Dec 16 '18 at 18:56