PYTHON 3.6

def on_ridge(map, r, c): Given a map and a location, is that spot on a ridge? We define a ridge as a place that has two neighbors on opposite side that are both land and both lower than the spot itself. We consider sideto-side, above-and-below, as well as opposing diagonal neighbors as ways to show we are on a ridge. When the adjacent spot is off the map, it can't be used to claim we're on a ridge. If (r,c) isn't on the map, the answer is False..

on_ridge(map1,2,2) True

on_ridge(map6,1,3) False

on_ridge(map6,1,4) True

SEE PIC FOR REFERENCE FUNCTIONS

def show (map): mystr'' substrings[] max_len max ([len (str(j)) for i in map for j in i]) for i in map: i [' '"(int (max-len)-len(str(j)))+str(j) for j substrings.append( ".join(i)) in i] return " ".join(substrings) +'n' #map 1 is coming from above function def highest_point (map): max=0 row-0 column-0 for r in range(1en (map)): for c in range(1en (map[r])) if map[r][c]>max: max map[r][c] column-c if max 0: return None else: return (row, column) def on_map (map, r, c): if(r)s0 r=0 and and else return False def is_map (map): if(len (map)>0): ssetO for x in range(len(map)): s.add (len (map[x])) for y in map[x]: if (not isinstance(y, int)): return False elif(yc0): return False if(len (s) 1): return True else: return False else return False def show (map): mystr'' substrings[] max_len max ([len (str(j)) for i in map for j in i]) for i in map: i [' '"(int (max-len)-len(str(j)))+str(j) for j substrings.append( ".join(i)) in i] return " ".join(substrings) +'n' #map 1 is coming from above function def highest_point (map): max=0 row-0 column-0 for r in range(1en (map)): for c in range(1en (map[r])) if map[r][c]>max: max map[r][c] column-c if max 0: return None else: return (row, column) def on_map (map, r, c): if(r)s0 r=0 and and else return False def is_map (map): if(len (map)>0): ssetO for x in range(len(map)): s.add (len (map[x])) for y in map[x]: if (not isinstance(y, int)): return False elif(yc0): return False if(len (s) 1): return True else: return False else return False