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
discrete-mathematics
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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
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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
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
discrete-mathematics
asked Nov 11 at 10:01
YATIN DIXIT
63
63
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1 Answer
1
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up vote
0
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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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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
add a comment |
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
add a comment |
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
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
edited Nov 14 at 23:37
answered Nov 14 at 1:21
DavidPM
1216
1216
add a comment |
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