I was reading page 147 of Goodrich and Tamassia, Data Structures and Algorithms in Java, 3rd Ed. (Google books). It gives example of linear sum algorithm which uses linear recursion to calculate sum of all elements of the array:
Algorithm linearSum (arr , n)
if (n == 1)
return arr[0]
else
return linearSum (arr , n-1) + arr[n-1]
end linearSum
And the binary sum algorithm which uses binary recursion to calculate sum of all elements of the array:
Algorithm binarySum (arr, i, n)
if (n == 1)
return arr[i]
return binarySum (arr, i, ⌈n/2⌉) + binarySum (arr, i+⌈n/2⌉, ⌊n/2⌋)
end binarySum
It further says:
The value of parameter $n$ is halved at each recursive call
binarySum()
. Thus, the depth of the recursion, that is, the maximum number of method instances that are active at the same time, is 1ドル + \log_2 n$. Thus the algorithmbinarySum()
uses $O(\log n)$ additional space. This is big improvement over $O(n)$ needed by thelinearSum()
algorithm.
I did not understood how the maximum number of method instances that are active at the same time, is 1ドル + \log_2n$.
For example consider the below calls to method with method parameters given in rounded box:
enter image description here
Then in two recursive calls of second row from top, $n = 8$. So, 1ドル + \log_2 8 = 4$. Now I dont get what maximum limit this 4 represent?
1 Answer 1
- In a given tree, all the vertices of this tree correspond to
binarySum()
calls. - The value of parameter
n
tobinarySum()
is halved at each recursive call. - Also, each recursive call finishes after all its children finish. Thus at each recursive call, number of active calls include all the ancestor calls in call sequence.
- Thus when any
binarySum()
call corresponding to leaves in above call tree is active, its parent, grandparent and so on are still active as well. - Thus, the depth of the recursion, that is, the maximum number of method instances that are active at the same time, which is always equal to the height of the recursive calls tree is 1 + log$_2$n.
- For example in
binarySum()
recursive calls tree above, with n = 8, at any of the calls correspnding to leaves,
enter image description here
Why does the depth of recursion affect the space, required by an algorithm? Each recursive function call usually needs to allocate some additional memory (for temporary data) to process its arguments. At least, each such a call has to store some information about its parent call - just to know where to return after finishing. Let's imagine you are performing a task, and you need to perform a sub-task inside this first task - so you need to remember (or write down on a paper) where you stopped in the first task to be able to continue it after you finish the sub-task. And so on, sub-sub-task inside a sub-task... So, a recursive algorithm will require space O(depth of recursion)
.
@randomA mentioned the Call Stack, which is normally used when a function invokes another function (including itself). The call stack is the part of the computer memory, where a recursive algorithm allocates its temporary data.
-
$\begingroup$ So if you have two recursive functions in binary recursion with the same parent function, does that mean that they will use the same memory to store information about their parent function which is more efficient than if each of them was storing that information in a different location like they would be in linear recursion? $\endgroup$GilbertS– GilbertS2020年11月19日 20:29:02 +00:00Commented Nov 19, 2020 at 20:29
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binarySum
calls. Each abinarySum
call finishes after all its children finish. So, whenbinarySum
on the leaves level is active, its parent, grandparent and so on are still active as well. It's called a depth of recursion, which is equal to the height of this tree. $\endgroup$binarySum
calls on the same level of the tree can use the same piece of space. $\endgroup$