Fig. 3: Second-order versus higher-order networks when solving kSAT problems.
From: Efficient optimization with higher-order Ising machines
a The mean higher-order energy at the end of the simulation is plotted against the number of problem variables for hard instances of 3SAT problems. As the problem size increases, the difference in energy between the second-order oscillator Ising machines and the higher-order oscillator Ising machines increases. b The mean percent of constraints satisfied at the end of simulation versus the problem size for 3SAT problems. c The probability of satisfying all constraints for different problem sizes and models for 3SAT problems. d The mean percent of constraints satisfied at the end of simulation versus the problem size for higher-order oscillator Ising machines for 3SAT problems. e The mean time to satisfy 95% of constraints for higher-order Ising machines for 3SAT problems. d, e Lines indicate different linear annealing schedules for the sub-harmonic injection locking coefficients, qi. In all plots, error bars represent the sample standard deviation computed over problem instances and trial simulations. f Comparing resources and solutions of 5SAT and 7SAT problems to their 3SAT reductions. Reducing kSAT problems to 3SAT for k > 3 increases the number of variables and connections (left two columns). The 5th-order and 7th-order Ising machines find lower energy states corresponding to a greater fraction of constraints satisfied compared to the 3rd-order Ising machine (right two columns).
