| http://www.w3.org/ns/prov#value | - As far as I know, it does not.here is an example worked out in full: // search for t in d[1]..d[n]: binary-search(t, d, n) { l := 1; h := n while h >= l do m := (l + h) / 2 if d[m] = t then return m else if d[m] > t then h := m - 1 else l = m + 1 done return false }unless I have got something wrong (I probably have[1]), the number ofcomparisons done for input size is bounded above by log2(n)+1.
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