[Rule] MinimumMultiwayCut to QUBO

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Source: MinimumMultiwayCut
Target: QUBO
Motivation: Enables solving minimum multiway cut on quantum annealers (D-Wave) and QUBO-based solvers via a direct Ising Hamiltonian with $kn$ binary variables.
Reference: Heidari, Dinneen & Delmas (2022), CDMTCS-565; Abbassi et al. (2026), arXiv:2601.00720

Reduction Algorithm

Notation:

• Source instance: graph $G=(V,E)$, $n=|V|$, edge weights $C({u,v})$, terminals $T={t_1,\ldots,t_k}$
• Target instance: QUBO matrix $Q \in \mathbb{R}^{kn \times kn}$
• $\alpha$ = penalty coefficient, $\alpha > \sum_{{u,v}\in E} C({u,v})$

Variable mapping:

Introduce $kn$ binary variables $x_{u,t}$ for each $u \in V$ and $t \in T$, where $x_{u,t} = 1$ means vertex $u$ is assigned to the component of terminal $t$. Variable index: $x_{u,t} \to u \cdot k + \mathrm{pos}(t)$.

Objective transformation (Heidari et al., eq. 2):

$$H_A = \alpha \left( \sum_{u \in V} \left(1 - \sum_{t \in T} x_{u,t}\right)^2 + \sum_{t \in T} \sum_{t' \neq t} x_{t,t'} \right)$$

$$H_B = \sum_{{u,v} \in E}\ \sum_{t \in T}\ \sum_{t' \neq t} C({u,v}), x_{u,t}, x_{v,t'}$$

$H_A$ enforces each vertex belongs to exactly one component (squared term) and each terminal $t$ is not misassigned to $t' \neq t$ (linear term), together forcing $x_{t,t}=1$. $H_B$ is the cut cost: for a feasible assignment each cut edge contributes $C({u,v})$ exactly once. Expanding $H = H_A + H_B$ in the binary variables yields the QUBO matrix $Q$.

Solution extraction: $E_m = {{u,v} \in E \mid x^_{u,t} = x^_{v,t'} = 1 \text{ for some } t \neq t'}$.

Size Overhead

Target metric (code name)	Polynomial (using symbols above)
num_vars	$kn$

Validation Method

• Solve small instances ($n \leq 6$, $k=3$) with BruteForce on both sides; verify $H(\mathbf{x}^*) = $ minimum cut cost with $H_A = 0$.
• Cross-check with the ILP reduction (issue [Rule] MinimumMultiwayCut to ILP #185) on the same instances.

Example

Source: $n=5$, $V={0,1,2,3,4}$, $T={0,2,4}$ ($k=3$), edges ${0,1}$:2, ${1,2}$:3, ${2,3}$:1, ${3,4}$:2, ${0,4}$:4, ${1,3}$:5. Set $\alpha=20$.

Variable index: $x_{u,t} \to u \cdot 3 + \mathrm{pos}(t)$ with $\mathrm{pos}(0)=0, \mathrm{pos}(2)=1, \mathrm{pos}(4)=2$. Total: $kn=15$ variables.

Optimal: $x_{0,0}=x_{1,2}=x_{2,2}=x_{3,2}=x_{4,4}=1$ (all others 0), giving components $\{0\}$, $\{1,2,3\}$, $\{4\}$.

• $H_A=0$ (valid partition, terminals correctly fixed)
• $H_B = C({0,1}) + C({0,4}) + C({3,4}) = 2+4+2 = \mathbf{8}$
• Ground-state energy $= 8 =$ minimum cut cost