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(b) k-SPANNING TREE is NP-complete (Hint: Start with k = 2 and consider the relation between this problem and R UDRATA PATH) 813 Determine which of the following problems are NP-complete and which are solvable in polynomial time In each problem you are given an undirected graph G = (V, E), along with: (a) A set of nodes L V , and you must nd a spanning tree such that its set of leaves includes the set L (b) A set of nodes L V , and you must nd a spanning tree such that its set of leaves is precisely the set L (c) A set of nodes L V , and you must nd a spanning tree such that its set of leaves is included in the set L (d) An integer k, and you must nd a spanning tree with k or fewer leaves (e) An integer k, and you must nd a spanning tree with k or more leaves (f) An integer k, and you must nd a spanning tree with exactly k leaves (Hint: All the NP-completeness proofs are by generalization, except for one) 814 Prove that the following problem is NP-complete: given an undirected graph G = (V, E) and an integer k, return a clique of size k as well as an independent set of size k, provided both exist 815 Show that the following problem is NP-complete M AXIMUM COMMON SUBGRAPH Input: Two graphs G1 = (V1 , E1 ) and G2 = (V2 , E2 ); a budget b Output: Two set of nodes V1 V1 and V2 V2 whose deletion leaves at least b nodes in each graph, and makes the two graphs identical 816 We are feeling experimental and want to create a new dish There are various ingredients we can choose from and we d like to use as many of them as possible, but some ingredients don t go well with others If there are n possible ingredients (numbered 1 to n), we write down an n n matrix giving the discord between any pair of ingredients This discord is a real number between 00 and 10, where 00 means they go together perfectly and 10 means they really don t go together Here s an example matrix when there are ve possible ingredients 1 2 3 4 5 1 00 04 02 09 10 2 04 00 01 10 02 3 02 01 00 08 05 4 09 10 08 00 02 5 10 02 05 02 00.

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However, geostationary satellites have certain limitations A geostationary orbit requires constant adjustment, because a tiny change in altitude will cause the satellite to get out of sync with the earth s rotation Geostationary satellites are expensive to launch and maintain When communicating through them, there is always a delay because of the path length It takes high transmitter power, and a sophisticated, precisely aimed antenna, to communicate reliably These problems with geostationary satellites have brought about a revival of the LEO scheme Instead of one single satellite, the new concept is to have a large fleet of them A good LEO satellite system is launched and maintained in such a way that, for any point on the earth, there is always at least one satellite in direct line-of-sight range The satellites can relay messages throughout the fleet Thus, any two points on the surface can always make, and maintain, contact through the satellites The satellites are placed in polar orbits (routes that pass over or near the earth s geographic poles) to optimize the geographical coverage Even if you re at or near the north geographic pole or the south geographic pole, you can use a LEO satellite system This is not true of geostationary satellite networks, where the regions immediately around the geographic poles are not seen by the satellites A LEO satellite wireless communications link is easier to access and use than a geostationary satellite link A small, simple antenna will suffice, and it doesn t have to be aimed in any particular direction The transmitter can reach the network using only a few watts of power The latency is less than 100 milliseconds (ms), compared with as much as 400 ms for geostationary satellite links

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In this case, ingredients 2 and 3 go together pretty well whereas 1 and 5 clash badly Notice that this matrix is necessarily symmetric; and that the diagonal entries are always 00 Any set of ingredients incurs a penalty which is the sum of all discord values between pairs of ingredients For instance, the set of ingredients {1, 3, 5} incurs a penalty of 02 + 10 + 05 = 17 We want this penalty to be small

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