Even though the problem is computationally difficult, many heuristics and exact algorithms are known, so that some cases with tens of thousands of city can be solved completely and even problems with millions of city can be estimated within a small fraction of 1%.The TSP has several applications even in its purest formulation, such as planning, logistics, and the manufacture of microchips. Thus, it is possible that the worst-case running time for any algorithm for the TSP increases superpolynomially (but no more than exponentially) with the number of city. For many other cases with millions of city, solutions can be found that are guaranteed to be within 2-3 % of an optimal tour. In the 1960s however a new approach was created, that instead of searching optimal solutions, one would produce a solution whose length is provably bounded by a multiple of the optimal length, and in making so make lower boundary for the problem; these may then be used with branch and bound approaches. While this paper make not give an algorithmic approach to TSP problems, the ideas that put within it were indispensable to later make exact solution methods for the TSP, though it would take 15 years to find an algorithmic approach in make these cuts.
""" The stock span problem is a financial problem where we have a series of n daily price quotes for a stock and we need to calculate span of stock's price for all n days. The span Si of the stock's price on a given day i is defined as the maximum number of consecutive days just before the given day, for which the price of the stock on the current day is less than or equal to its price on the given day. """ def calculateSpan(price, S): n = len(price) # Create a stack and push index of fist element to it st =  st.append(0) # Span value of first element is always 1 S = 1 # Calculate span values for rest of the elements for i in range(1, n): # Pop elements from stack while stack is not # empty and top of stack is smaller than price[i] while len(st) > 0 and price[st] <= price[i]: st.pop() # If stack becomes empty, then price[i] is greater # than all elements on left of it, i.e. price, # price, ..price[i-1]. Else the price[i] is # greater than elements after top of stack S[i] = i + 1 if len(st) <= 0 else (i - st) # Push this element to stack st.append(i) # A utility function to print elements of array def printArray(arr, n): for i in range(0, n): print(arr[i], end=" ") # Driver program to test above function price = [10, 4, 5, 90, 120, 80] S = [0 for i in range(len(price) + 1)] # Fill the span values in array S calculateSpan(price, S) # Print the calculated span values printArray(S, len(price))