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# Divide and Conquer Method using Stock Price Optimization Problem

##### December 15, 2020

Consider the scenario of the stocks and the stock markets. Everyday, millions of people trade equities and commodities on various exchanges across the globe. These processes are being technically equipped, and a majority of the functions are being automated. Algorithmic trading represents one such area of interest for brokerage firms.

### Understanding divide and conquer method using stock price optimization problem

Algorithmic trading involves coming up with new algorithms to predict the future price of a stock or commodity. However, before deploying it live, it is tested on historical data. This is called back-testing.

If the returns are comparable to the maximum possible profit during back-testing, then the algorithm is considered stable. On the other hand, if the returns are not close to the maximum potential profit, they are not yet ready to be deployed. This article will consider the algorithm to find the maximum possible profit given a list of historical prices.

### Background

We are given a list of historical prices over a specific period. Given the list, we will compute the maximum possible profit from the historical prices. One can perform either a BUY transaction or a SELL transaction at a given timestamp. The total number of buying and selling operations that can be executed is not limited.

### Objective

• We will use the divide and conquer paradigm to solve the maximization problem. The divide and conquer method recursively breaks down the problem into smaller sub-problems until they can be solved directly. The solutions to these sub-problems are then combined to give a solution to the original problem.

Divide and Conquer Paradigm is an algorithmic design approach used in problems that have a recursive structure to it. It is based on multi-branched recursion.

The algorithm operates by breaking down the problem into smaller sub-problems, making the problem small enough to be solved directly. In this article, we will observe the recursive approach taken to optimize the profit.

### Given list of prices

We are given a list of prices. The indices of the list indicate the day number. The values of the list indicate the stock value on a respective day.

### Given input

We are given a pythonic list of historical prices. Python’s list implements dynamic array under the hood. Dynamic arrays are resourceful as the size grows as and when required.

This saves space. Moreover, the amortized cost of inserting an element is O(1). The worst-case scenario is Θ(n) when the size of the array is doubled.

### Desired output

The output is the maximum potential profit rounded to the nearest ₹. We will output an integer, but depending on the accuracy of the output required, the data type can be chosen accordingly.

### Pseudocode

Let us look at the pseudocode for the problem above. The pseudocode explains the entire solution and is invariant to programming languages. The logic below is to split the problem into the smallest sub-problems and compute the profits. If the local profit is greater than the global profit, then the global profit value is updated.

``````function maximize_profit(list_of_prices, starting_index, ending_index):
{
IF length(list_of_prices)<2 THEN
raise an Error and exit the program

INITIALIZE maximum_profit to 0
# base condition for recursion
IF starting_index > ending_index THEN
return maximum_profit
FOR selling_counter in list_of_prices
UPDATE tentative_profit with the difference of the current two prices at hand
RECURSIVELY call maximize_profit on the following two slices of the list:
2. sell: end
ADD the values obtained to the variable tentative_profit
COMPUTE the maximum of maximum_profit and tentative profit
& UPDATE the variable maximum_profit with the maximum computed.
ENDFOR
ENDFOR

RETURN maximum_profit
}

function main():
{
STORE the list of prices in LIST_OF_PRICES
CALL maximize_profit() with argument LIST_OF_PRICES
STORE the value returned by the function in variable PROFIT
print PROFIT on the console.
}
``````

### Python code

We convert the pseudocode given above into Python code. Try coding it on your own. The solution is given below.

``````def maximize_profit(price_list, start, end):
if(len(price_list)<2):
raise ValueError('At least 2 prices required')

maximum_profit = 0
if start >= end:
return maximum_profit
tentative_profit = price_list[sell] - price_list[buy] + maximize_profit(price_list, start, buy - 1) + maximize_profit(price_list, sell, end)
maximum_profit = max(maximum_profit, tentative_profit)

return maximum_profit

# inputs
price1 = [1, 5, 3, 2]
price2 = [7, 2, 8, 9]
price3 = [1, 6, 7, 9]
price4 = [9, 7, 4, 1]
price5 = [10, 5, 1, 0]
price6 = [100, 180, 260, 310, 40, 535, 695]

print("     List", "     --> Profit(₹)")
print("1.", price1, "--> ₹", maximize_profit(price1,0,len(price1)-1))
print("2.", price2, "--> ₹", maximize_profit(price2,0,len(price2)-1))
print("3.", price3, "--> ₹", maximize_profit(price3,0,len(price3)-1))
print("4.", price4, "--> ₹", maximize_profit(price4,0,len(price4)-1))
print("5.", price5, "--> ₹", maximize_profit(price5,0,len(price5)-1))
print("6.", price6, "--> ₹", maximize_profit(price6,0,len(price6)-1))
``````

### Output

The program outputs the following profits:

``````     List      --> Profit(₹)
1. [1, 5, 3, 2] --> ₹ 4
2. [7, 2, 8, 9] --> ₹ 7
3. [1, 6, 7, 9] --> ₹ 8
4. [9, 7, 4, 1] --> ₹ 0
5. [10, 5, 1, 0] --> ₹ 0
6. [100, 180, 260, 310, 40, 535, 695] --> ₹ 865
``````

### Output analysis

Let’s analyze the edge cases. This helps us identify the advantages and disadvantages of the divide and conquer approach.

#### Case 1:

Input List: [1,5,3,2]

Output: Maximum Profit: 4: Buy at 1: Sell at 5

Best Profit Possible: 4

#### Case 2:

Input List: [9,7,4,1]

Output: Maximum Profit: 0 :Buy at 9 :Sell at 7

Best Profit Possible: -2

In this case, the output remains at 0 and does not output negative numbers. Therefore, this approach can be used to find only the greatest profit possible.

#### Case 3:

Input List: [100,180,260,310,40,535,695]

Output: Maximum Profit: 865

Optimal Solution

Buy at 100, Sell at 310 Buy at 40, Sell at 695

Total Profit = 865

Compared to the greedy approach, this method gives the optimal solution to the problem. We observed that the maximum profit in the greedy method was 655.

In this case, the total profit is 865. We solved the above problem using the greedy method in the earlier article. You may go through the solution at this link.

• The program can work with many buy and sell operations due to its recurring property. Multiple buy and sell operations take place, and the best profits are returned.
• Divide and Conquer algorithm’s solutions are always optimal. The comparison of code output: scenario - 3 shows the same. The greedy algorithm outputs 655, whereas the divide and conquer algorithm outputs 865.

• The worst-case time complexity of the function `maximize_profit`() is Θ(n^2*log(n)). The time complexity is arrived at through computational methods. A good exercise would be to come up with recurrence relations for the above algorithm.
• Space complexity of the function is Θ(log(n)). This is computed using the maximum memory utilization by the program at a given time. At any given time, the algorithm has a maximum of `log(n)` function calls. Each function call represents the function solving a subproblem.
• The minimum profit is 0. Losses are not accounted for and displayed by the program. The base case drives DNC methods, and in this case, the base case is when `starting_index`> `ending_index.` When the following happens, 0 is returned.