How many bit strings of length eight contain either three consecutive zeros or four consecutive ones?











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I have tried this question and I am able to get the answer as 149 by the principle of inclusion and exclusion but a standard book solutions gave it as 147. I want to confirm whether I am wrong or right. I am expecting a detailed answer as my question and clear and from a standard book.strong text










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    I have tried this question and I am able to get the answer as 149 by the principle of inclusion and exclusion but a standard book solutions gave it as 147. I want to confirm whether I am wrong or right. I am expecting a detailed answer as my question and clear and from a standard book.strong text










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      I have tried this question and I am able to get the answer as 149 by the principle of inclusion and exclusion but a standard book solutions gave it as 147. I want to confirm whether I am wrong or right. I am expecting a detailed answer as my question and clear and from a standard book.strong text










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      I have tried this question and I am able to get the answer as 149 by the principle of inclusion and exclusion but a standard book solutions gave it as 147. I want to confirm whether I am wrong or right. I am expecting a detailed answer as my question and clear and from a standard book.strong text







      discrete-mathematics






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      asked Nov 11 at 10:01









      YATIN DIXIT

      63




      63
























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          The correct answer seems to be 147.
          I did exhaustive search by constructing a binary tree with nodes that have as values 0 or 1 whose paths from the root to the leaves correspond to all possible 8-bit strings (that is, we start from the root with two branches that lead to two level-1 nodes with values 0 and 1. From each such node we draw another two branches that lead to two level-2 nodes with values 0 and 1. We go like this until we build all possible paths). Some pruning applies here: when one of the paths has three consecutive 0's or four consecutive 1's we stop right there. Likewise, if the path built so far does not allow for the possibility to contain a sequence of three 0's or four 1's we stop right there too.



          Below there is a simple piece of python code with an implementation (not optimized, only some pruning is performed).



          def pruning_pos(lst):
          '''Pruning function over a list ('lst')'''
          if len(lst) >= 3:
          # if there are three consecutive 0's...
          if lst[-3:] == [0, 0, 0]:
          return True
          else:
          if len(lst) >= 4:
          # if there are four consecutive 1's...
          if lst[-4:] == [1,1,1,1]:
          return True
          return False

          paths = [[0], [1]]
          count = 0
          for i in xrange(7):
          new_paths = # candidate paths
          for c in paths:
          for elem in [c+[1], c+[0]]:
          if pruning_pos(elem): # if it contains sequence of interest...
          count += 2**(8-len(elem)) # count all subpaths we are pruning
          else:
          new_paths.append(elem)
          paths = new_paths[:]

          print count





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            The correct answer seems to be 147.
            I did exhaustive search by constructing a binary tree with nodes that have as values 0 or 1 whose paths from the root to the leaves correspond to all possible 8-bit strings (that is, we start from the root with two branches that lead to two level-1 nodes with values 0 and 1. From each such node we draw another two branches that lead to two level-2 nodes with values 0 and 1. We go like this until we build all possible paths). Some pruning applies here: when one of the paths has three consecutive 0's or four consecutive 1's we stop right there. Likewise, if the path built so far does not allow for the possibility to contain a sequence of three 0's or four 1's we stop right there too.



            Below there is a simple piece of python code with an implementation (not optimized, only some pruning is performed).



            def pruning_pos(lst):
            '''Pruning function over a list ('lst')'''
            if len(lst) >= 3:
            # if there are three consecutive 0's...
            if lst[-3:] == [0, 0, 0]:
            return True
            else:
            if len(lst) >= 4:
            # if there are four consecutive 1's...
            if lst[-4:] == [1,1,1,1]:
            return True
            return False

            paths = [[0], [1]]
            count = 0
            for i in xrange(7):
            new_paths = # candidate paths
            for c in paths:
            for elem in [c+[1], c+[0]]:
            if pruning_pos(elem): # if it contains sequence of interest...
            count += 2**(8-len(elem)) # count all subpaths we are pruning
            else:
            new_paths.append(elem)
            paths = new_paths[:]

            print count





            share|improve this answer



























              up vote
              0
              down vote













              The correct answer seems to be 147.
              I did exhaustive search by constructing a binary tree with nodes that have as values 0 or 1 whose paths from the root to the leaves correspond to all possible 8-bit strings (that is, we start from the root with two branches that lead to two level-1 nodes with values 0 and 1. From each such node we draw another two branches that lead to two level-2 nodes with values 0 and 1. We go like this until we build all possible paths). Some pruning applies here: when one of the paths has three consecutive 0's or four consecutive 1's we stop right there. Likewise, if the path built so far does not allow for the possibility to contain a sequence of three 0's or four 1's we stop right there too.



              Below there is a simple piece of python code with an implementation (not optimized, only some pruning is performed).



              def pruning_pos(lst):
              '''Pruning function over a list ('lst')'''
              if len(lst) >= 3:
              # if there are three consecutive 0's...
              if lst[-3:] == [0, 0, 0]:
              return True
              else:
              if len(lst) >= 4:
              # if there are four consecutive 1's...
              if lst[-4:] == [1,1,1,1]:
              return True
              return False

              paths = [[0], [1]]
              count = 0
              for i in xrange(7):
              new_paths = # candidate paths
              for c in paths:
              for elem in [c+[1], c+[0]]:
              if pruning_pos(elem): # if it contains sequence of interest...
              count += 2**(8-len(elem)) # count all subpaths we are pruning
              else:
              new_paths.append(elem)
              paths = new_paths[:]

              print count





              share|improve this answer

























                up vote
                0
                down vote










                up vote
                0
                down vote









                The correct answer seems to be 147.
                I did exhaustive search by constructing a binary tree with nodes that have as values 0 or 1 whose paths from the root to the leaves correspond to all possible 8-bit strings (that is, we start from the root with two branches that lead to two level-1 nodes with values 0 and 1. From each such node we draw another two branches that lead to two level-2 nodes with values 0 and 1. We go like this until we build all possible paths). Some pruning applies here: when one of the paths has three consecutive 0's or four consecutive 1's we stop right there. Likewise, if the path built so far does not allow for the possibility to contain a sequence of three 0's or four 1's we stop right there too.



                Below there is a simple piece of python code with an implementation (not optimized, only some pruning is performed).



                def pruning_pos(lst):
                '''Pruning function over a list ('lst')'''
                if len(lst) >= 3:
                # if there are three consecutive 0's...
                if lst[-3:] == [0, 0, 0]:
                return True
                else:
                if len(lst) >= 4:
                # if there are four consecutive 1's...
                if lst[-4:] == [1,1,1,1]:
                return True
                return False

                paths = [[0], [1]]
                count = 0
                for i in xrange(7):
                new_paths = # candidate paths
                for c in paths:
                for elem in [c+[1], c+[0]]:
                if pruning_pos(elem): # if it contains sequence of interest...
                count += 2**(8-len(elem)) # count all subpaths we are pruning
                else:
                new_paths.append(elem)
                paths = new_paths[:]

                print count





                share|improve this answer














                The correct answer seems to be 147.
                I did exhaustive search by constructing a binary tree with nodes that have as values 0 or 1 whose paths from the root to the leaves correspond to all possible 8-bit strings (that is, we start from the root with two branches that lead to two level-1 nodes with values 0 and 1. From each such node we draw another two branches that lead to two level-2 nodes with values 0 and 1. We go like this until we build all possible paths). Some pruning applies here: when one of the paths has three consecutive 0's or four consecutive 1's we stop right there. Likewise, if the path built so far does not allow for the possibility to contain a sequence of three 0's or four 1's we stop right there too.



                Below there is a simple piece of python code with an implementation (not optimized, only some pruning is performed).



                def pruning_pos(lst):
                '''Pruning function over a list ('lst')'''
                if len(lst) >= 3:
                # if there are three consecutive 0's...
                if lst[-3:] == [0, 0, 0]:
                return True
                else:
                if len(lst) >= 4:
                # if there are four consecutive 1's...
                if lst[-4:] == [1,1,1,1]:
                return True
                return False

                paths = [[0], [1]]
                count = 0
                for i in xrange(7):
                new_paths = # candidate paths
                for c in paths:
                for elem in [c+[1], c+[0]]:
                if pruning_pos(elem): # if it contains sequence of interest...
                count += 2**(8-len(elem)) # count all subpaths we are pruning
                else:
                new_paths.append(elem)
                paths = new_paths[:]

                print count






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                edited Nov 14 at 23:37

























                answered Nov 14 at 1:21









                DavidPM

                1216




                1216






























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