While we have been discussing topological sort I'm the previous to previous post. I wanted to take a minute to discuss the advantages of breadth first traversal and depth first traversal.
As we know dfs uses an explicit agenda. It will recurse from the root to the leaves. It also uses less memory. A bfs on the other hand is exploratory. It covers nodes at a given level.
In the depth first search, the recursion termination is specified first. This takes care of covering the end case. If the end case is specified, then the progress can be specified in terms of smaller subspace.
Take the example of finding the factorial of a number for instance. This has a recursive solution.
but let us study its complexity. We can write that as a recursion as well.
If we take the complexity M(n) as the # of basic operations on input n,
M(n) = 1 + M(n-1), n > 0
and
M(0) = 1 which is the initial condition and base case.
We use backward substitution :
M(n) = M(n-1) + 1 = (M(n-2) + 1) + 1
= M(n-2) + 2 = (M(n-3) + 1) + 2
= M(n-3) + 3 = .... M(n-i) + i
= M(n-n) + n = M(0) + n
= 1 + n = O(n) multiplications
In general we can study the complexity of recursive algorithms such as DFS by following this:
Identify the input size
Identify the basic operations
Find the dependence on the input size ( the worst case, the best case )
Write the reurrence relation with initial conditions to compute total # of basic operations
and then solve the recurrence
In breadth first search, we rely on a collection or container such as a queue.
We determine the complexity based on the size and opeartions on this container.
As we know dfs uses an explicit agenda. It will recurse from the root to the leaves. It also uses less memory. A bfs on the other hand is exploratory. It covers nodes at a given level.
In the depth first search, the recursion termination is specified first. This takes care of covering the end case. If the end case is specified, then the progress can be specified in terms of smaller subspace.
Take the example of finding the factorial of a number for instance. This has a recursive solution.
but let us study its complexity. We can write that as a recursion as well.
If we take the complexity M(n) as the # of basic operations on input n,
M(n) = 1 + M(n-1), n > 0
and
M(0) = 1 which is the initial condition and base case.
We use backward substitution :
M(n) = M(n-1) + 1 = (M(n-2) + 1) + 1
= M(n-2) + 2 = (M(n-3) + 1) + 2
= M(n-3) + 3 = .... M(n-i) + i
= M(n-n) + n = M(0) + n
= 1 + n = O(n) multiplications
In general we can study the complexity of recursive algorithms such as DFS by following this:
Identify the input size
Identify the basic operations
Find the dependence on the input size ( the worst case, the best case )
Write the reurrence relation with initial conditions to compute total # of basic operations
and then solve the recurrence
In breadth first search, we rely on a collection or container such as a queue.
We determine the complexity based on the size and opeartions on this container.
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