Interval Subset Sum Problem (ISSP)

The Interval Subset Sum Problem (ISSP) is a generalization of the Subset Sum Problem. Given $n$ integer intervals $[l_i, u_i]$ $(0\leq i\leq n-1)$ and an upper bound $T$, the goal is to choose an integer value

\[\begin{aligned} v_i &\in \lbrace 0\rbrace \cup [l_i, u_i] && (i = 0,1,\dots,n-1), \end{aligned}\]

so as to satisfy the constraint

\[\begin{aligned} \sum_{i=0}^{n-1} v_i \leq T, \end{aligned}\]

and maximize the objective:

\[\begin{aligned} \sum_{i=0}^{n-1} v_i. \end{aligned}\]

HUBO formulation of the ISSP

An integer variable can be represented by multiple binary variables using a binary encoding. In PyQBPP, such integer variables can be defined easily using var_int.

Let $v_i$ $(0\leq i\leq n-1)$ be an integer variable that can take a value in $[l_i, u_i]$. We also introduce a binary variable $s_i$ $(0\leq i\leq n-1)$ such that $s_i=1$ if and only if interval $i$ is selected.

To model ISSP, we use the product $s_i v_i$ as the selected value:

\[\begin{aligned} s_iv_i &= 0 && \text{if } s_i= 0\\ &\in [l_i,u_i] && \text{if } s_i= 1 \end{aligned}\]

Let

\[\begin{aligned} \text{sum} &= \sum_{i=0}^{n-1} s_i v_i . \end{aligned}\]

In PyQBPP, we impose this inequality constraint via a penalty term:

\[\begin{aligned} \text{constraint} &= \sum_{i=0}^{n-1} \bigr(0\leq s_iv_i \leq T\bigl) \\ &= (T-\sum_{i=0}^{n-1} s_iv_i)^2 \end{aligned}\]

Since $s_i v_i$ is quadratic in binary variables, $\text{sum}$ is quadratic and $\text{constraint}$ becomes quartic.

Because the ISSP maximizes the sum under the upper bound $T$, we minimize the negative sum:

\[\begin{aligned} \text{objective} &= -\sum_{i=0}^{n-1} s_iv_i \end{aligned}\]

Finally, we combine the objective and the constraint penalty into a single HUBO function:

\[\begin{aligned} f &= \text{objective} + P\times\text{constraint}, \end{aligned}\]

where $P$ is a sufficiently large constant to prioritize feasibility.

PyQBPP program (HUBO formulation)

The following PyQBPP program solves an ISSP instance with 8 intervals. The lower and upper bounds $[l_i,u_i]$ are stored in the lists lower and upper, and $T=100$.

import pyqbpp as qbpp

lower = [18, 17, 21, 18, 20, 14, 14, 23]
upper = [19, 17, 22, 19, 20, 16, 15, 25]
T = 100
n = len(lower)

v = [qbpp.between(qbpp.var_int(f"v{i}"), lower[i], upper[i]) for i in range(n)]
s = qbpp.var("s", n)

total = qbpp.sum(v * s)
constraint = qbpp.between(total, 0, T)
f = -total + 1000 * constraint
f.simplify_as_binary()

solver = qbpp.EasySolver(f)
sol = solver.search({"target_energy": -T})
for i in range(n):
    if sol(s[i]) == 1:
        print(f"Interval {i}: val = {sol(v[i])}")
print(f"sum = {sol(total)}")

First, we define a list v of integer variables where each v[i] takes an integer value in [lower[i], upper[i]]. We also define an array s of binary variables, where s[i] = 1 means interval i is selected. The expression total represents $\sum_i v_i s_i$.

The inequality constraint between(total, 0, T) is stored in constraint. In PyQBPP, such a constraint is internally converted into a nonnegative penalty term that becomes zero when the constraint is satisfied.

Finally, we construct the HUBO objective function f as f = -total + P * constraint (with P = 1000 in this example). Minimizing f therefore maximizes total while heavily penalizing any violation of the constraint.

We set the target energy to -T because if the solver finds a feasible solution with total = T, then the penalty term is zero and the objective term becomes -T, i.e., the global minimum reaches -T.

For the obtained solution, the selected intervals and their values are printed. For example:

Interval 0: val = 18
Interval 1: val = 17
Interval 2: val = 22
Interval 4: val = 20
Interval 7: val = 23
sum = 100

This output confirms that a feasible solution achieving the maximum possible sum ($=T$) was obtained.

QUBO formulation for the ISSP

The HUBO formulation above contains quartic terms because it uses products $s_i v_i$. We can avoid quartic terms by introducing auxiliary integer variables.

Let $a_i$ $(0\leq i\leq n-1)$ be an integer variable that can take a value in $[0,\, u_i-l_i]$. We also use a binary variable $s_i$ $(0\leq i\leq n-1$) such that $s_i=1$ if and only if interval $i$ is selected.

We define

\[\begin{aligned} v_i &= l_is_i + a_i && (0\leq i\leq n-1) \\ \end{aligned}\]

To ensure that $v_i$ becomes 0 when $s_i=0$, we add the following penalty term using the negated literal $\overline{s_i}$:

\[\begin{aligned} \text{constraint1} &= \sum_{i=0}^{n-1}\sum_j \overline{s_i}\,a_i \end{aligned}\]

Since $a_i \ge 0$ and $\overline{s_i} \ge 0$, we have $\text{constraint1}\ge 0$. Moreover, $\text{constraint1}=0$ holds if and only if $a_i=0$ whenever $s_i=0$. Therefore, the selected value $v_i$ satisfies

\[\begin{aligned} v_i & = 0 && \text{if } s_i=0,\\ & \in [l_i,u_i] &&\text{if } s_i=1. \end{aligned}\]

because $v_i = l_i + a_i$ and $a_i \in [0,u_i-l_i]$ when $s_i=1$.

Let

\[\begin{aligned} \text{sum} &= \sum_{i=0}^{n-1} v_i. \end{aligned}\]

The ISSP constraint is:

\[\begin{aligned} \text{constraint2} &= \sum_{i=0}^{n-1} \bigr(0\leq v_i \leq T\bigl) \\ &= (T-\sum_{i=0}^{n-1} v_i)^2 \end{aligned}\]

Finally, since ISSP maximizes $\text{sum}$ under the upper bound $T$, we minimize

\[\begin{aligned} \text{objective} &= -\sum_{i=0}^{n-1} v_i \end{aligned}\]

Combining the objective and the penalties, we obtain the QUBO expression:

\[\begin{aligned} f &= \text{objective} + P\times(\text{constraint1}+\text{constraint2}), \end{aligned}\]

where $P$ is a sufficiently large constant to prioritize feasibility.

PyQBPP program (QUBO formulation)

The following PyQBPP program solves the same ISSP instance using the QUBO formulation:

import pyqbpp as qbpp

lower = [18, 17, 21, 18, 20, 14, 14, 23]
upper = [19, 17, 22, 19, 20, 16, 15, 25]
T = 100
n = len(lower)

a = [qbpp.between(qbpp.var_int(f"a{i}"), 0, upper[i] - lower[i]) for i in range(n)]
s = qbpp.var("s", n)
v = [s[i] * lower[i] + a[i] for i in range(n)]

total = 0
for i in range(n):
    total += v[i]

constraint1 = 0
for i in range(n):
    constraint1 += ~s[i] * a[i]

constraint2 = qbpp.between(total, 0, T)
f = -total + 1000 * (constraint1 + constraint2)
f.simplify_as_binary()

solver = qbpp.EasySolver(f)
sol = solver.search({"target_energy": -T})
for i in range(n):
    if sol(s[i]) == 1:
        print(f"Interval {i}: val = {sol(v[i])}")
print(f"sum = {sol(total)}")

First, we define a list a of integer variables, where each a[i] takes an integer value in [0, upper[i] - lower[i]]. We also define an array s of binary variables. Using a and s, we construct v[i] = s[i] * lower[i] + a[i], which corresponds to $v_i = s_i l_i + a_i$. The expression constraint1 += ~s[i] * a[i] penalizes any solution with a[i] > 0 when s[i] = 0, thereby enforcing v[i] = 0 for unselected intervals. The inequality constraint constraint2 = between(total, 0, T) ensures that the total selected sum does not exceed T.

Finally, we minimize f = -total + P * (constraint1 + constraint2) with a sufficiently large penalty constant P. As in the previous example, passing {"target_energy": -T} to search() allows the solver to stop early if it finds a feasible solution achieving total = T (in which case the penalty terms are zero and the objective term becomes -T).

This produces the same result as the HUBO formulation.

区間部分和問題 (ISSP)

区間部分和問題 (Interval Subset Sum Problem, ISSP) は部分和問題の一般化です。 $n$ 個の整数区間 $[l_i, u_i]$ $(0\leq i\leq n-1)$ と上限 $T$ が与えられたとき、整数値

\[\begin{aligned} v_i &\in \lbrace 0\rbrace \cup [l_i, u_i] && (i = 0,1,\dots,n-1) \end{aligned}\]

を選び、制約

\[\begin{aligned} \sum_{i=0}^{n-1} v_i \leq T \end{aligned}\]

を満たしつつ、目的関数

\[\begin{aligned} \sum_{i=0}^{n-1} v_i \end{aligned}\]

を最大化することが目標です。

ISSPのHUBO定式化

整数変数はバイナリ符号化を用いて複数のバイナリ変数で表現できます。 PyQBPPでは、var_int を使って整数変数を簡単に定義できます。

$v_i$ $(0\leq i\leq n-1)$ を $[l_i, u_i]$ の値をとる整数変数とします。 また、区間 $i$ が選択されるときかつそのときに限り $s_i=1$ となるバイナリ変数 $s_i$ $(0\leq i\leq n-1)$ を導入します。

ISSPをモデル化するために、選択された値として積 $s_i v_i$ を使用します：

\[\begin{aligned} s_iv_i &= 0 && \text{if } s_i= 0\\ &\in [l_i,u_i] && \text{if } s_i= 1 \end{aligned}\]

次を定義します：

\[\begin{aligned} \text{sum} &= \sum_{i=0}^{n-1} s_i v_i . \end{aligned}\]

PyQBPPでは、この不等式制約をペナルティ項で課します：

\[\begin{aligned} \text{constraint} &= \sum_{i=0}^{n-1} \bigr(0\leq s_iv_i \leq T\bigl) \\ &= (T-\sum_{i=0}^{n-1} s_iv_i)^2 \end{aligned}\]

$s_i v_i$ はバイナリ変数の2次式であるため、$\text{sum}$ は2次、$\text{constraint}$ は4次になります。

ISSPは上限 $T$ のもとで和を最大化するため、負の和を最小化します：

\[\begin{aligned} \text{objective} &= -\sum_{i=0}^{n-1} s_iv_i \end{aligned}\]

最後に、目的関数と制約ペナルティを単一のHUBO関数にまとめます：

\[\begin{aligned} f &= \text{objective} + P\times\text{constraint}, \end{aligned}\]

ここで $P$ は実行可能性を優先するための十分大きな定数です。

HUBO定式化のPyQBPPプログラム

以下のPyQBPPプログラムは、8個の区間を持つISSPインスタンスを解きます。 下限と上限 $[l_i,u_i]$ はリスト lower と upper に格納され、$T=100$ です。

import pyqbpp as qbpp

lower = [18, 17, 21, 18, 20, 14, 14, 23]
upper = [19, 17, 22, 19, 20, 16, 15, 25]
T = 100
n = len(lower)

v = [qbpp.between(qbpp.var_int(f"v{i}"), lower[i], upper[i]) for i in range(n)]
s = qbpp.var("s", n)

total = qbpp.sum(v * s)
constraint = qbpp.between(total, 0, T)
f = -total + 1000 * constraint
f.simplify_as_binary()

solver = qbpp.EasySolver(f)
sol = solver.search({"target_energy": -T})
for i in range(n):
    if sol(s[i]) == 1:
        print(f"Interval {i}: val = {sol(v[i])}")
print(f"sum = {sol(total)}")

まず、各 v[i] が [lower[i], upper[i]] の整数値をとる整数変数のリスト v を定義します。 また、s[i] = 1 が区間 i が選択されることを意味するバイナリ変数の配列 s を定義します。 式 total は $\sum_i v_i s_i$ を表します。

不等式制約 between(total, 0, T) は constraint に格納されます。PyQBPPでは、このような制約は内部的に非負のペナルティ項に変換され、制約が満たされると0になります。

最後に、HUBO目的関数 f を f = -total + P * constraint（この例では P = 1000）として構築します。 f を最小化することで、制約違反に大きなペナルティを課しつつ total を最大化します。

ターゲットエネルギーを -T に設定するのは、ソルバーが total = T の実行可能解を見つけた場合、ペナルティ項が0になり目的関数項が -T になる、すなわち大域最小値が -T に達するためです。

得られた解について、選択された区間とその値が表示されます。例えば：

Interval 0: val = 18
Interval 1: val = 17
Interval 2: val = 22
Interval 4: val = 20
Interval 7: val = 23
sum = 100

この出力は、最大可能な sum ($=T$) を達成する実行可能解が得られたことを確認しています。

ISSPのQUBO定式化

上記のHUBO定式化は積 $s_i v_i$ を使用するため4次項を含みます。 補助整数変数を導入することで4次項を避けることができます。

$a_i$ $(0\leq i\leq n-1)$ を $[0,\, u_i-l_i]$ の値をとる整数変数とします。 また、区間 $i$ が選択されるときかつそのときに限り $s_i=1$ となるバイナリ変数 $s_i$ $(0\leq i\leq n-1)$ を使用します。

次を定義します：

\[\begin{aligned} v_i &= l_is_i + a_i && (0\leq i\leq n-1) \\ \end{aligned}\]

$s_i=0$ のときに $v_i$ が0になることを保証するため、否定リテラル $\overline{s_i}$ を用いた次のペナルティ項を追加します：

\[\begin{aligned} \text{constraint1} &= \sum_{i=0}^{n-1}\sum_j \overline{s_i}\,a_i \end{aligned}\]

$a_i \ge 0$ かつ $\overline{s_i} \ge 0$ であるため、$\text{constraint1}\ge 0$ が成り立ちます。 さらに、$s_i=0$ のときに $a_i=0$ であるときかつそのときに限り $\text{constraint1}=0$ が成り立ちます。 したがって、選択された値 $v_i$ は次を満たします：

\[\begin{aligned} v_i & = 0 && \text{if } s_i=0,\\ & \in [l_i,u_i] &&\text{if } s_i=1. \end{aligned}\]

$s_i=1$ のとき $v_i = l_i + a_i$ かつ $a_i \in [0,u_i-l_i]$ であるためです。

次を定義します：

\[\begin{aligned} \text{sum} &= \sum_{i=0}^{n-1} v_i. \end{aligned}\]

ISSPの制約は：

\[\begin{aligned} \text{constraint2} &= \sum_{i=0}^{n-1} \bigr(0\leq v_i \leq T\bigl) \\ &= (T-\sum_{i=0}^{n-1} v_i)^2 \end{aligned}\]

最後に、ISSPは上限 $T$ のもとで $\text{sum}$ を最大化するため、次を最小化します：

\[\begin{aligned} \text{objective} &= -\sum_{i=0}^{n-1} v_i \end{aligned}\]

目的関数とペナルティを組み合わせて、QUBO式を得ます：

\[\begin{aligned} f &= \text{objective} + P\times(\text{constraint1}+\text{constraint2}), \end{aligned}\]

ここで $P$ は実行可能性を優先するための十分大きな定数です。

QUBO定式化のPyQBPPプログラム

以下のPyQBPPプログラムは、QUBO定式化を用いて同じISSPインスタンスを解きます：

import pyqbpp as qbpp

lower = [18, 17, 21, 18, 20, 14, 14, 23]
upper = [19, 17, 22, 19, 20, 16, 15, 25]
T = 100
n = len(lower)

a = [qbpp.between(qbpp.var_int(f"a{i}"), 0, upper[i] - lower[i]) for i in range(n)]
s = qbpp.var("s", n)
v = [s[i] * lower[i] + a[i] for i in range(n)]

total = 0
for i in range(n):
    total += v[i]

constraint1 = 0
for i in range(n):
    constraint1 += ~s[i] * a[i]

constraint2 = qbpp.between(total, 0, T)
f = -total + 1000 * (constraint1 + constraint2)
f.simplify_as_binary()

solver = qbpp.EasySolver(f)
sol = solver.search({"target_energy": -T})
for i in range(n):
    if sol(s[i]) == 1:
        print(f"Interval {i}: val = {sol(v[i])}")
print(f"sum = {sol(total)}")

まず、各 a[i] が [0, upper[i] - lower[i]] の整数値をとる整数変数のリスト a を定義します。 また、バイナリ変数の配列 s を定義します。 a と s を用いて v[i] = s[i] * lower[i] + a[i] を構築し、これは $v_i = s_i l_i + a_i$ に対応します。 式 constraint1 += ~s[i] * a[i] は、s[i] = 0 のときに a[i] > 0 となる解にペナルティを課し、選択されていない区間に対して v[i] = 0 を強制します。 不等式制約 constraint2 = between(total, 0, T) は、選択された合計が T を超えないことを保証します。

最後に、十分大きなペナルティ定数 P で f = -total + P * (constraint1 + constraint2) を最小化します。 前の例と同様に、search() に {"target_energy": -T} を渡すことで、total = T を達成する実行可能解が見つかった場合にソルバーを早期停止させることができます（この場合、ペナルティ項は0になり目的関数項は -T になります）。

HUBO定式化と同じ結果が得られます。