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Is Lovasz theta not equal to Form of the Shannon capacity for Heptagon?

Lovasz theta number for $n$-cyle is $\frac{ncos(\frac{\pi}{n})}{1+cos(\frac{\pi}{n})}$.

Is the $0$-error capacity of $7$-cycle:

$(1)$ known not to be of form $\frac{7cos(\frac{\pi}{7})}{1+cos(\frac{\pi}{7})}$?

$(2)$ known to be of form $7^q$ for some $q\in \mathbb Q$?

Is Lovasz theta not equal to Shannon capacity for Heptagon?

Lovasz theta number for $n$-cyle is $\frac{ncos(\frac{\pi}{n})}{1+cos(\frac{\pi}{n})}$.

Is the $0$-error capacity of $7$-cycle:

$(1)$ known not to be of form $\frac{7cos(\frac{\pi}{7})}{1+cos(\frac{\pi}{7})}$?

$(2)$ known to be of form $7^q$ for some $q\in \mathbb Q$?

Form of the Shannon capacity for Heptagon?

Is the $0$-error capacity of $7$-cycle:

$(1)$ known to be of form $7^q$ for some $q\in \mathbb Q$?

Source Link
Turbo
  • 13.9k
  • 1
  • 27
  • 76

Is Lovasz theta not equal to Shannon capacity for Heptagon?

Lovasz theta number for $n$-cyle is $\frac{ncos(\frac{\pi}{n})}{1+cos(\frac{\pi}{n})}$.

Is the $0$-error capacity of $7$-cycle:

$(1)$ known not to be of form $\frac{7cos(\frac{\pi}{7})}{1+cos(\frac{\pi}{7})}$?

$(2)$ known to be of form $7^q$ for some $q\in \mathbb Q$?