The most basic function of firewalls is to

What is the worst-case complexity of the maxrepeats function? assume that the longest string in the names array is at most 25 characters wide (i.e., string comparison can be treated as o( class namecounter { private: int* counts; int nc; string* names; int nn; public: namecounter (int ncounts, int nnames); int maxrepeats() const; }; int namecounter: : maxrepeats () { int maxcount = 0; for (int i = 0; i < nc; ++i) { int count = 1; for (int j = i+1; j < nc; ++j) { if (names[i] == names[j]) ++count; } maxcount = max(count, maxcount); } return maxcount; }

Into what form does the barcode reader convert individual bar patterns?

The idea that, for each pair of devices v and w, there’s a strict dichotomy between being “in range” or “out of range” is a simplified abstraction. more accurately, there’s a power decay function f (·) that specifies, for a pair of devices at distance δ, the signal strength f(δ) that they’ll be able to achieve on their wireless connection. (we’ll assume that f (δ) decreases with increasing δ.) we might want to build this into our notion of back-up sets as follows: among the k devices in the back-up set of v, there should be at least one that can be reached with very high signal strength, at least one other that can be reached with moderately high signal strength, and so forth. more concretely, we have values p1 ≥ p2 ≥ . . ≥ pk, so that if the back-up set for v consists of devices at distances d1≤d2≤≤dk,thenweshouldhavef(dj)≥pj foreachj. give an algorithm that determines whether it is possible to choose a back-up set for each device subject to this more detailed condition, still requiring that no device should appear in the back-up set of more than b other devices. again, the algorithm should output the back-up sets themselves, provided they can be found.\