Question

Figure 1: An 8 x & Mario world and corresponding Mario graph. 2 Super Mario Run A Mario world M consists of a k xk gid. Each field in the grid is either empy or brick. Two empty fields are marked as start and goal (see Fig. 21a). The goal ot the game s to move the playe, called Mario, irom the start field to the goal field. When Mario is in field (x,y) he has the following options: Forward Mario moves to the hiekd (r +1,y). This move is possible if (x+ 1,y) is empty and (x + 1,y-1) s brick. Jump Mario moves to the field (r+1,y+2). This move is possible if the fields (x,y + 2), (r, y + 1), and (x +1,y+ 2) e emp % ana (x *1,y*1)15 brick. Fall Mario moves to the field (r +1,y) where y < y is the maximal value y' such that (x + 1, y'-1) is brick and x+ 1, y), (x+1,y-1),.., (x+ 1,y) are empty: For instance, Mario in field (4, 4) can move fonward to (5,4) or jump to field (5, 6). Amove is alid if the move can be done according to the above rules. All tields that can be reached via valid moves from the start field are the valid fields. For example, (4, 4) isa valid field, sinoe it can be reached from the start field in (1,2) doing the sequence of valid movesz forward, torward, jump. A Mario world M defines a directed graph G with n vertices and m edges (see Fig. 2(b)). All valid fields correspond vertex and the valid moves define the edges (there is an edge from field (r, y) to fie ld r,y)it Mario can move trom (x, y) to (x', y') with a valid move). The edges oorrespondin g to torward, junp and tall are denoted by forward edges, jump edges and fall edges. 2.1 Let M be a Mario world defined on a k xk grid and let G be the corresponding Mario graph. Indicate the maximum TALIImber ot nodes n that can appear an G as a function of k. Solution: ae(log k) De(vk) e(k) oeklogk) Detk) De(z) 2.2 Let M be a Mario worlid defined on a k Xxk grid and let G be the corresponding Mario graph. Indicate the maximum number of edges m that can appear in G as a function of k. Solution: e(log k) evk) ek) eklogk) 0ek) De(z

Answer 1

2.1- the answer is 6. i.e. o(2^k)

2.2- the answer is the first option i.e. o(log k)