8:24 Note that ${n+1 \choose 2}$ is just the summation equation of 1 + ... + n. 9:20 This inequality comes from flipping the previous inequality. Instead of requiring that the elements in S are greater than the summation, we require that the "remaining" elements of [n] - S (denoted as T) is less than the sum of all values in n minus the sum of values in S ( |S| = n - |T|)
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8:24 Note that ${n+1 \choose 2}$ is just the summation equation of 1 + ... + n.
9:20 This inequality comes from flipping the previous inequality. Instead of requiring that the elements in S are greater than the summation, we require that the "remaining" elements of [n] - S (denoted as T) is less than the sum of all values in n minus the sum of values in S ( |S| = n - |T|)
21:18 So we've basically converted the permutahedron formulation using an exponential number of sigmas, into one that uses an nxn matrix Y.