Question

A 3-dimensional surface contains a saddle point when the point is a global minimum along a line parallel to either the x or the y axis, but a global maximum along the other axis. You'll usually see a "U" shape meeting an upside-down "U" shape. The classic horseback riding saddle has one such saddle point, and, not coincidentally, it is a rider's most stable position The input will be a 2-dimensional array AInIn] of numbers, representing a lattice approximation of a surface. That is, given an (x,) pair of integers, the array entry A[x]y contains the height (z-coordinate) of the surface In the sample surface shown (thank yo to Wikipedial), the array might be 0.50 0.62 0.72 0.80 0.88 0.93 0.97 0.99 1.00 0.99 0.97 0.93 0.88 0.80 0.72 0.62 0.50 0.38 0.50 0.60 0.69 0.76 0.81 0.85 0.88 0.88 0.88 0.85 0.81 0.76 0.69 0.60 Ο.50 0.38 0.28 0.40 0.50 0.59 0.66 0,71 0.75 0.770.78 0.77 0.75 0.71 0.66 0.59 0.50 0.40 0.28 0.20 0.31 0.41 0.50 0.57 0.62 0.66 0.69 0.70 0.69 0.66 0.62 0.57 0.50 0.41 0.31 0.20 0.12 0.24 0.34 0.43 0.50 0.55 0.59 0.62 0.62 0.62 0.59 0.55 0.50 0.43 0.34 0.24 0.12 0.07 0.19 0.29 0.38 0.45 0.50 0.54 0.56 0.57 0.56 0.54 0.50 0,45 0.38 0.29 0.19 0.07 0.03 0.15 0.25 0.34 0.41 0.46 0.50 0,52 0.53 0,52 0.50 0,46 0.41 0.34 0,25 0.15 0.03 0.01 0.12 0.23 0,31 0.38 0,44 0.48 0,50 0.51 0.50 0.48 0.44 0,38 0.31 0.23 0.12 0.01 0.00 0.12 0.22 0.30 0.38 0.43 0.47 0.49 0.50 0,49 0.47 0.43 0.38 0.30 0.22 0.12 0.00 0.01 0.12 0.23 0.31 0.38 0.44 0.48 0.50 0.51 0.50 0.48 0.44 0.38 0.31 0.23 0.12 0.01 0.03 0.15 0.25 0.34 0.41 0.46 0.50 0.52 0.53 0.52 0.50 0.46 0.41 0.34 0.25 o.15 0.03 0.07 0.19 0.29 0.38 0.45 0.50 0.54 0.56 0.57 0.56 0.54 0.50 0.45 0.38 0.29 0.19 0.07 0.12 0.24 0.34 0.43 0.50 0.55 0.59 0.62 0.62 0.62 0.59 0.55 0.50 0.43 0.34 0.24 0.12 0.20 0.31 0.41 0.50 0.57 0,62 0.66 0,69 0.70 0.69 0.66 0,62 0.57 0.50 0,41 0.31 0.20 0.28 0.40 0.50 0.59 0.66 0.71 0.75 0.770.78 0.77 0.75 0.71 0.66 0.59 0.50 0.40 0.28 0.38 0.50 0.60 0.69 0.76 0.81 0.85 0.88 0.88 0.88 0.85 0.81 0.76 0.69 0.60 0.50 0.38 0.50 0.62 0.72 0.80 0.88 0.93 0.97 0,99 1.00 0.99 0.97 0,93 0.88 0.80 0.72 0.62 0.50 There are 5 saddle points in the sample. One is at A[81[8] and has the value 0.50, which is the maximum value of its row, but the minimum value of its column. The other 4 are at the corners of the matrix A[OI[O], AIO][16], A [16][0] and A[16][16]) each of which is the minimum of its row, and the maximum of its column. Your algorithm will locate and output all saddle points, i.e., all entries A[xJlyl for which the entry is the max in its row and min in its column, or vice versa. There might be 1 saddle point, several saddle points or perhaps no saddle points. Write your algorithm using pseudocode and document it with assertions and comments. You may assume that max and min are functions that work in the expected fashion, i.e., your algorithm can freely make use of max and min. That doesn't mean that max and min are "free" in terms of cost! Each of them will still cost time in proportion to the number of elements it needs to scan. Determine the running time of your algorithm using O-notation, as a function of n. Remember that time is a precious resource: the faster your algorithm, the better your grade

Answer 1

algorithm wil be flky "The PSedodo耘 len do den do index-min-d pxint i ithm い! This is keex than the notnal naive o a う a Saddk Pointes it would hav token O()