Editing Binomial heap
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Since a binomial heap is only allowed to contain at most one power-of-two tree of each size, the structure of a binomial heap is uniquely determined by the binary representation of its size. For example, if a binomial heap contains 13 elements, that is, {{Binary|1101}}, then we see that <math>13 = 2^0 + 2^2 + 2^3 = 1 + 4 + 8</math>, so a binomial heap of size 13 contains three power-of-two trees, of sizes 1, 4, and 8, respectively. | Since a binomial heap is only allowed to contain at most one power-of-two tree of each size, the structure of a binomial heap is uniquely determined by the binary representation of its size. For example, if a binomial heap contains 13 elements, that is, {{Binary|1101}}, then we see that <math>13 = 2^0 + 2^2 + 2^3 = 1 + 4 + 8</math>, so a binomial heap of size 13 contains three power-of-two trees, of sizes 1, 4, and 8, respectively. | ||
− | It follows that the number of power-of-two trees in a binomial heap is no more than <math>\lceil\ | + | It follows that the number of power-of-two trees in a binomial heap is no more than <math>\lceil\log(n+1)\rceil</math>, since that is the length of the number <math>n</math> expressed as a bit string. |
==Finding the maximum== | ==Finding the maximum== | ||
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==Merging== | ==Merging== | ||
− | Let two binomial heaps be denoted <math>A</math> and <math>B</math>, with sizes <math>n</math> and <math>m</math>, respectively. Let <math>A_k</math> denote <math>A</math>'s power-of-two tree of size <math>2^k</math>, if it has one, and <math>B_k</math> denote likewise a power-of-two tree of <math>B</math>. We want to create a new binomial | + | Let two binomial heaps be denoted <math>A</math> and <math>B</math>, with sizes <math>n</math> and <math>m</math>, respectively. Let <math>A_k</math> denote <math>A</math>'s power-of-two tree of size <math>2^k</math>, if it has one, and <math>B_k</math> denote likewise a power-of-two tree of <math>B</math>. We want to create a new binomial tree <math>S</math> that contains all the nodes from either <math>A</math> or <math>B</math> and has size <math>m+n</math>. After this operation, <math>A</math> and <math>B</math> will no longer exist as binomial heaps. |
(In theory, we could create a new binomial heap <math>S</math> that contained all the elements from both <math>A</math> and <math>B</math> ''without'' also destroying <math>A</math> and <math>B</math>. However, this would require copying over all the data from both heaps, which would take linear time (in the sum of their sizes). This would then be no more efficient than simply using the [[binary heap]]s. For this reason, we assume we usually do not want to do this.) | (In theory, we could create a new binomial heap <math>S</math> that contained all the elements from both <math>A</math> and <math>B</math> ''without'' also destroying <math>A</math> and <math>B</math>. However, this would require copying over all the data from both heaps, which would take linear time (in the sum of their sizes). This would then be no more efficient than simply using the [[binary heap]]s. For this reason, we assume we usually do not want to do this.) | ||
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</pre> | </pre> | ||
This procedure performs at most one merge (of a pair of power-of-two trees) at each iteration of the loop, which executes a logarithmic number of times; so it is <math>O(\log n)</math>, overall. | This procedure performs at most one merge (of a pair of power-of-two trees) at each iteration of the loop, which executes a logarithmic number of times; so it is <math>O(\log n)</math>, overall. | ||
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==Insertion== | ==Insertion== | ||
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==Deletion== | ==Deletion== | ||
Once we have identified the minimum element in the binomial heap, which takes logarithmic time, we can remove it, as follows. Suppose the minimum element is the root of the power-of-two tree with size <math>2^k</math>. Then, we remove this tree from the heap and split the tree into two trees of size <math>2^{k-1}</math>. Whichever one does not contain the minimum element, we split again, so that now we have a tree of size <math>2^{k-1}</math> and two trees of size <math>2^{k-2}</math>. One of these two new trees again does not contain the minimum element, so we split it again. We continue splitting until finally we end up with one power-of-two tree of size <math>2^{k-1}</math>, one of size <math>2^{k-2}</math>, and so on down; one of size 4, one of size 2, and two of size 1. One of the two singletons will be the minimum element, so we simply destroy that node. The rest of the power-of-two trees constitute an entire binomial heap in and of themselves, which is then merged back into the original heap from which we removed the tree of size <math>2^k</math>. This procedure uses a logarithmic number of splits, which takes logarithmic time, followed by a merge of two binomial heaps, which takes logarithmic time; so it is <math>O(\log n)</math> overall. | Once we have identified the minimum element in the binomial heap, which takes logarithmic time, we can remove it, as follows. Suppose the minimum element is the root of the power-of-two tree with size <math>2^k</math>. Then, we remove this tree from the heap and split the tree into two trees of size <math>2^{k-1}</math>. Whichever one does not contain the minimum element, we split again, so that now we have a tree of size <math>2^{k-1}</math> and two trees of size <math>2^{k-2}</math>. One of these two new trees again does not contain the minimum element, so we split it again. We continue splitting until finally we end up with one power-of-two tree of size <math>2^{k-1}</math>, one of size <math>2^{k-2}</math>, and so on down; one of size 4, one of size 2, and two of size 1. One of the two singletons will be the minimum element, so we simply destroy that node. The rest of the power-of-two trees constitute an entire binomial heap in and of themselves, which is then merged back into the original heap from which we removed the tree of size <math>2^k</math>. This procedure uses a logarithmic number of splits, which takes logarithmic time, followed by a merge of two binomial heaps, which takes logarithmic time; so it is <math>O(\log n)</math> overall. | ||
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==Binomial tree representation== | ==Binomial tree representation== | ||
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* A node in a binomial tree can have more than two children. | * A node in a binomial tree can have more than two children. | ||
* Binomial trees are fully max-heap-ordered, rather than simply left-max-heap-ordered like power-of-two trees. | * Binomial trees are fully max-heap-ordered, rather than simply left-max-heap-ordered like power-of-two trees. | ||
− | A binomial tree of height 0 is a single node. A binomial tree of height <math>k > 0</math> consists of a root node with <math>k</math> children. | + | A binomial tree of height 0 is a single node. A binomial tree of height <math>k > 0</math> consists of a root node with <math>k</math> children. The subtree rooted at each child is a tree of a different height. |
When we merge two binomial trees of the same size, we simply make the smaller root the child of the larger root. Observe that this instantly gives a binomial tree of twice the size (whose height is then one greater than the heights of the two original trees). When we split a binomial tree of height <math>k > 0</math>, we simply detach the subtree of height <math>k-1</math>, giving two binomial trees of size <math>k-1</math>. | When we merge two binomial trees of the same size, we simply make the smaller root the child of the larger root. Observe that this instantly gives a binomial tree of twice the size (whose height is then one greater than the heights of the two original trees). When we split a binomial tree of height <math>k > 0</math>, we simply detach the subtree of height <math>k-1</math>, giving two binomial trees of size <math>k-1</math>. | ||