segment tree space complexity

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Автор arjun95, история, 7 лет назад, По-английски

how to prove space complexity in segment tree is O(4*n).

arjun95
7 лет назад
14

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gongy

7 лет назад, # |

← Rev. 2 →

+13

Segment tree with 2^x leaf nodes will have 2^(x+1) — 1 total nodes because it is a perfect binary tree.

Now, you will have useless leaf nodes if you are using general n. Therefore at worst you will have almost 2*n leaf nodes because you must round up to the next power of two. If n is 2^j + 1, then you must have 2^(j+1) leaf nodes = O(2*n).

And as we proved before total nodes is ~2 * leaf nodes so we have 2*2*n = O(4n) total space.

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arjun95

7 лет назад, # ^ |

-28

thanks for explanation. is there any other mathematical proof to prove this?

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Azret

7 лет назад, # ^ |

← Rev. 2 →

If you're not satisfied by the above proof here is another:

Last layer: $\text{[math]}$ nodes
Pre-last layer: $\text{[math]}$ nodes
$\text{[math]}$

So, we have $\text{[math]}$ nodes.
You can see the infinite geometric progression on the right. Its sum is equal to $\text{[math]}$ , where $\text{[math]}$ is first element of progression ( $\text{[math]}$ here) and $\text{[math]}$ is ratio between two adjacent elements ( $\text{[math]}$ here).
So, there is a total of $\text{[math]}$ nodes in a segment tree.

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Devised_positive

3 года назад, # ^ |

-11

why would it be infinite G.P. on RHS? I mean it could be finite as well right?

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Arpa

7 лет назад, # |

← Rev. 3 →

-13

This problem was one of the INOI 2012-2013 problems.

At first note that we need exactly 2·n - 1 nodes, but if we use arrays and set id of left child of node with id Id to 2·Id and 2·Id + 1 id to right one, we need an array with size 4·n, let's prove.

Let f(n) the maximum id that we need in segment tree. Let's prove f(2ⁿ + 1) = 4·2ⁿ - 1. We can prove it using induction, f(2¹ + 1) = 7, f(2ⁿ + 1) = 2·f(2^n - 1 + 1) + 1 = 2·(4·2^n - 1 - 1) + 1 = 4·2ⁿ - 1.

Edit: Definition of f(n) has changed.

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depevlad

7 лет назад, # |

← Rev. 2 →

Let k be the smallest natural number such that 2^k ≥ n. Note that 2^k < 2 × n. We will find the answer for 2^k. The segment tree, and indeed any other binary tree formed will have exactly k + 1 levels, the i-th containing 2ⁱ nodes. Then, the total number of nodes will be a geometric progression of the form 2⁰ + 2¹ + 2² + ... + 2^k, which is precisely equal to 2^k + 1 - 1. Since 2^k < 2 * n, it follows immediately that 2^k + 1 - 1 < 4 × n, so the number of nodes of the new tree — greater than our answer — is still less than 4 × n.

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Al.Cash

7 лет назад, # |

Non-recursive segment trees use exactly 2n - 1 nodes.

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xsc

7 лет назад, # ^ |

@AI.Cash: I've read u non-recursive segment tree. It's very easy, powerful as general segment-tree and required less memory space. But, in non-recursive segment tree how to find lower bound of position for given sum ?? // for perfect binary tree (i.e. n = 2^k):

   int find(int sum){
     int result = 1; 
     while(result < n){ 
          result <<= 1; 
          if(tree[result]< sum)sum-=tree[result++]; } 
     }
     return result;
}

when n = 2^k, this works fine, but n != 2^k not.

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Al.Cash

7 лет назад, # ^ |

Indeed, for n ≠ 2^k we basically get not one tree but O(logn) separate perfect trees. So, it doesn't support the techniques where we need to start from the root and move to the leaves, like binary search and fractional cascading.

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