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1. Top results related to brute force algorithm

   • Brute force algorithm and string matching?

     Top Answer

     Answered Feb 19, 2017 · 0 votes

     So I think I've written up a pretty helpful program that might help you out. As @Scovetta mentioned, we can use Python's in operator. However, if you're adamant on using a brute force method and not using Python's nice syntactic sugar, then you can definitely do that.

     Here is an example program I wrote up. Let me know if this helps! Feel free to ask any questions that you might have.

     # This program is just going to find if a string contains a substringdef match(text, pattern): ''' Simple match program ''' if pattern in text: return True else: return Falsetest_cases = [['ab', 'bab'], ['cat', 'catdog'], ['banana', 'apple'], ['tree', 'tree']]def brute_match(text,pattern): ''' We just want to check that our text has our pattern So let's just iterate over our text ''' length_of_pattern = len(pattern) for i in range(len(text)): # Abuse python's splicing because we don't have to worry about end of the string nonsense if text[i:length_of_pattern+i] == pattern: return True return Falsefor case in test_cases: ans = match(case[1], case[0]) bans = brute_match(case[1], case[0]) print("{} is in {} (T/F): {}".format(case[0], case[1], ans)) print("{} is in {} (T/F): {}".format(case[0], case[1], bans))# Enter your main program # We want to loop infinitely unless we get qqqwhile True: text_string = raw_input("Please enter your text string: ") if text_string == 'qqq': break pattern_string = raw_input("Please enter your pattern string: ") print("Using 'in' operator:") print('-'*20) if match(text_string): print("IS a valid match") else: print("NOT a valid match") print("Using brute force:") print('-'*20) if brute_match(text_string): print("IS a valid match") else: print("NOT a valid match")

     I realize the code is a bit messy, but I just wanted to write it up quick for ya!

     Let me know if you have any questions.

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     1/5

   • Brute Force Algorithm

     Top Answer

     Answered Dec 20, 2012 · 0 votes

     Ok, so I just found an example what I wanted to do in C++ and I've managed to convert it to the following Visual Basic .NET code, which works perfectly. However the output seems a bit slow and I'd have thought the program would have used near 100% processing power, but it isn't. Could somebody tell me why this is and how I could change it please?

     Public Class Form1 Dim chars() As Char = "1234567890abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYX".ToCharArray Dim csize As Integer = chars.Length - 1 Dim upto As String Private Sub btnGo_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles btnGo.Click upto = " " Dim max_length As Integer = 50 For i = 1 To max_length bf_recursion(0, i) Update() Next End Sub Private Sub bf_recursion(ByVal index As Integer, ByVal depth As Integer) Dim current() As Char = upto.ToCharArray() For i = 0 To csize current(index) = chars(i) upto = CStr(current) Console.WriteLine(CStr(current)) '\\lblOutput.Text = CStr(current) If index <> (depth - 1) Then bf_recursion(index + 1, depth) End If Next End SubEnd Class

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     2/5

   • Brute force algorithm for the Traveling Salesman Problem in Java

     Top Answer

     Answered Jul 28, 2012 · 6 votes

     You should remove cities from the route after the recursive call returns. You do this:

     Route newRoute = r; newRoute.addToRoute(citiesNotInRoute.get(i)); BruteForceFindBestRoute(newRoute);-

     but never a newRoute.removeFromRoute or similar.

     Note that you are writing Java, and in Java, an assignment of an object is done by reference. Which means that r and newRoute are the same object. newRoute is not an independent copy of r as you might expect. So any modification to newRoute will change r as well. You might want to explicitely copy your route there. The Java term for this is clone. Make sure your clone is deep enough, i.e. you clone all relevant data structures, instead of sharing them between the original and its clone.

     Note: There are many places where you might make your code more efficient, but as brute force is far from efficient in any case, and you're only talking about few cities, perhaps you don't have to care. If you want to investigate alternatives, though, consider maintaining a single linked list of all unvisited cities. Each call would loop over that list, remove an element, recurse and put the element back. No need to build this list from scratch in each invocation. The idea of dancing links could be neatly applied to this task, as an alternative to premade linked list implementations.

     EDIT:

     The following variation of your code works for me:

     import java.util.*;-class SO11703827 {- private static ArrayList<Integer> bestRoute; public static void bruteForceFindBestRoute (ArrayList<Integer> r, ArrayList<Integer> citiesNotInRoute) { if(!citiesNotInRoute.isEmpty()) { for(int i = 0; i<citiesNotInRoute.size(); i++) { Integer justRemoved = (Integer) citiesNotInRoute.remove(0); ArrayList<Integer> newRoute = (ArrayList<Integer>) r.clone(); newRoute.add(justRemoved); bruteForceFindBestRoute(newRoute, citiesNotInRoute); citiesNotInRoute.add(justRemoved); } } else //if(citiesNotInRoute.isEmpty()) { if(isBestRoute(r)) bestRoute = r; } } private static boolean isBestRoute(ArrayList<Integer> r) { System.out.println(r.toString()); return false; } public static void main(String[] args) { ArrayList<Integer> lst = new ArrayList<Integer>(); for (int i = 0; i < 6; ++i) lst.add(i); ArrayList<Integer> route = new ArrayList<Integer>(); bruteForceFindBestRoute(route, lst); }}

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     3/5

   • Brute force Sudoku algorithm

     Top Answer

     Answered Apr 17, 2012 · 2 votes

     I think your problem may lie here

     for(int i = 1; i <= board.getDimension(); i++) { if(legalValue(i)) { setValue(i); getNext().setNumberMeAndTheRest(board); } }

     If legalValue(i) returns true independent of the current state of i, then you're back tracking, if not, you're not backtracking

     What most backtracking looks like is osmething like htis

     for(int i = 1; i <= board.getDimension(); i++) { if(legalValue(i)) { setValue(i); // boolean indicating whether solution was found if(getNext().setNumberMeAndTheRest(board)) return true; else unsetValue(i) } }

     We need more code to know if legalValue returns false when square i is already set

     Try this to see if I'm on the right track or post all of your code

     System.out.println("STARTING ITERATION") for(int i = 1; i <= board.getDimension(); i++) {- if(legalValue(i)) { System.out.println("GOING " + i) setValue(i); getNext().setNumberMeAndTheRest(board); } } System.out.println("ENDING ITERATION")

     If it fills out the grid and then stops without backtracking, your problem is that you calling setValue(i) and then calling legalValue(i+1) and it is return false because the value is alraedy set, not because it's not legal. If this is so, you need an equivalent 'unset' after the reucrsion

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     4/5

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     5/5

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2. www.freecodecamp.org › news › brute-force-algorithmsBrute Force Algorithms Explained - freeCodeCamp.org

   www.freecodecamp.org › news › brute-force-algorithms

   • Cached

   Jan 6, 2020 · Brute Force Algorithms are exactly what they sound like – straightforward methods of solving a problem that rely on sheer computing power and trying every possibility rather than advanced techniques to improve efficiency. For example, imagine you have a small padlock with 4 digits, each from 0-9.

3. www.geeksforgeeks.org › brute-force-approach-andBrute Force Approach and its pros and cons - GeeksforGeeks

   www.geeksforgeeks.org › brute-force-approach-and

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   Jan 18, 2024 · A brute force algorithm is a simple, comprehensive search strategy that systematically explores every option until a problem’s answer is discovered. It’s a generic approach to problem-solving that’s employed when the issue is small enough to make an in-depth investigation possible.

4. en.wikipedia.org › wiki › Brute-force_searchBrute-force search - Wikipedia

   en.wikipedia.org › wiki › Brute-force_search

   • Cached

   In computer science, brute-force search or exhaustive search, also known as generate and test, is a very general problem-solving technique and algorithmic paradigm that consists of systematically checking all possible candidates for whether or not each candidate satisfies the problem's statement.

5. www.scholarhat.com › tutorial › datastructuresBrute Force Algorithm in Data Structures: Types, Advantages ...

   www.scholarhat.com › tutorial › datastructures

   • Cached

   Jun 6, 2024 · Here's a general outline of the brute force algorithm: Define the problem: Clearly understand the problem you are trying to solve and the constraints involved. Enumerate all possible solutions: Generate or iterate through all possible candidates for the solution.

6. www.codecademy.com › learn › learn-data-structuresLearn Data Structures and Algorithms with Python: Brute Force ...

   www.codecademy.com › learn › learn-data-structures

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   Brute Force Algorithms. Cheatsheet. Searching for smallest or largest value using linear search. Linear search can be used to search for the smallest or largest value in an unsorted list rather than searching for a match.

7. jsinkers.github.io › 05_brute_forceBrute Force and Exhaustive Search · Notes

   jsinkers.github.io › 05_brute_force

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   Brute Force. brute force: straightforward approach to solving problem, “just do it”. shouldn’t be overlooked: brute force is applicable to wide variety of problems. for some problems, produces reasonable algorithms of practical value with no limit on instance size.

8. math.uaa.alaska.edu › ~afkjm › cs351Brute Force Algorithms CS 351, Chapter 3

   math.uaa.alaska.edu › ~afkjm › cs351

   For most of the algorithms portion of the class we’ll focus on specific design strategies to solve problems. One of the simplest is brute force, which can be defined as: Brute force is a straightforward approach to solving a problem, usually directly based on the problem’s statement and definitions of the concepts involved.