When we have four sets, there is no longer a unique solution to the resulting linear system.
There are now six sorts of edges, see the attached figure. The linear system is:
\(p_1+p_4+p_5=k_1\)
\(p_1+p_2+p_6=k_2\)
\(p_2+p_3+p_5=k_3\)
\(p_3+p_4+p_6=k_4\)
This is four equations in six unknowns, and so, assuming there are perfect matchings to count, there is a two-parameter family of them.
Donovan Young /
npub16e…2pu5f
2024-12-30 18:03:15
in reply to nevent1q…2nuf
Author Public Key
npub16evzp8e2ra26hyyh96ut6ld5hvz3830nqhdn3dx06cn0tyej724sp2pu5fPublished at
2024-12-30 18:03:15Event JSON
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"content": "When we have four sets, there is no longer a unique solution to the resulting linear system. \n\nThere are now six sorts of edges, see the attached figure. The linear system is:\n\n\\(p_1+p_4+p_5=k_1\\)\n\\(p_1+p_2+p_6=k_2\\)\n\\(p_2+p_3+p_5=k_3\\)\n\\(p_3+p_4+p_6=k_4\\)\n\nThis is four equations in six unknowns, and so, assuming there are perfect matchings to count, there is a two-parameter family of them.",
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