计算机科学外文翻译

Binomial heap

In computer science, a binomial heap is a heap similar to a binary heap but also supports quick merging of two heaps. This is achieved by using a special tree structure. It is important as an implementation of the mergeable heap abstract data type(also called meldable heap), which is a priority queue supporting merge operation.

Binomial tree

A binomial heap is implemented as a collection of binomial trees (compare with a binary heap, which has a shape of a single binary tree). A binomial tree is defined recursively:

∙ A binomial tree of order 0 is a single node

∙ A binomial tree of order k has a root node whose children are roots of binomial trees of orders k−1, k−2, ..., 2, 1, 0 (in this order).

Binomial trees of order 0 to 3: Each tree has a root node with subtrees of all lower ordered binomial trees, which have been highlighted. For example, the order 3 binomial tree is connected to an order 2, 1, and 0 (highlighted as blue, green and red respectively) binomial tree.

A binomial tree of order k has 2k nodes, height k.

Because of its unique structure, a binomial tree of order k can be constructed from two trees of order k−1 trivially by attaching one of them as the leftmost child of root of the

other one. This feature is central to the merge operation of a binomial heap, which is its major advantage over other conventional heaps.

The name comes from the shape: a binomial tree of order has nodes at depth

.

Structure of a binomial heap

A binomial heap is implemented as a set of binomial trees that satisfy the binomial heap properties:

∙Each binomial tree in a heap obeys the minimum-heap property: the key of a node is greater than or equal to the key of its parent.

∙There can only be either one or zero binomial trees for each order, including zero order.

The first property ensures that the root of each binomial tree contains the smallest key in the tree, which applies to the entire heap.

The second property implies that a binomial heap with n nodes consists of at most

log n + 1 binomial trees. In fact, the number and orders of these trees are uniquely determined by the number of nodes n: each binomial tree corresponds to one digit in the binary representation of number n. For example number 13 is 1101 in binary,

, and thus a binomial heap with 13 nodes will consist of three binomial trees of orders 3, 2, and 0 (see figure below).

Example of a binomial heap containing 13 nodes with distinct keys.

The heap consists of three binomial trees with orders 0, 2, and 3.

Implementation

Because no operation requires random access to the root nodes of the binomial trees, the roots of the binomial trees can be stored in a linked list, ordered by increasing order of the tree.

Merge

As mentioned above, the simplest and most important operation is the merging of two binomial trees of the same order within two binomial heaps. Due to the structure of binomial trees, they can be merged trivially. As their root node is the smallest element within the tree, by comparing the two keys, the smaller of them is the minimum key, and becomes the new root node. Then the other tree become a subtree of the combined tree. This operation is basic to the complete merging of two binomial heaps.

function mergeTree(p, q)

if p.root.key <= q.root.key

return p.addSubTree(q)

else

return q.addSubTree(p)

To merge two binomial trees of the same order, first compare the root key. Since 7>3, the black tree on the left(with root node 7) is attached to the grey tree on the

right(with root node 3) as a subtree. The result is a tree of order 3.

The operation of merging two heaps is perhaps the most interesting and can be used as a subroutine in most other operations. The lists of roots of both heaps are traversed simultaneously, similarly as in the merge algorithm

If only one of the heaps contains a tree of order j, this tree is moved to the merged heap. If both heaps contain a tree of order j, the two trees are merged to one tree of order j+1 so that the minimum-heap property is satisfied. Note that it may later be necessary to merge this tree with some other tree of order j+1 present in one of the heaps. In the course of the algorithm, we need to examine at most three trees of any order (two from the two heaps we merge and one composed of two smaller trees).

Because each binomial tree in a binomial heap corresponds to a bit in the binary representation of its size, there is an analogy between the merging of two heaps and the binary addition of the sizes of the two heaps, from right-to-left. Whenever a carry occurs during addition, this corresponds to a merging of two binomial trees during the merge.

Each tree has order at most log n and therefore the running time is O(log n).

function merge(p, q)

while not( p.end() and q.end() )

tree = mergeTree(p.currentTree(), q.currentTree())

if not heap.currentTree().empty()

tree = mergeTree(tree, heap.currentTree())

heap.addTree(tree)

else

heap.addTree(tree)

heap.next() p.next() q.next()

This shows the merger of two binomial heaps. This is accomplished by merging two binomial trees of the same order one by one. If the resulting merged tree has the same order as one binomial tree in one of the two heaps, then those two are merged again. Insert

Inserting a new element to a heap can be done by simply creating a new heap containing only this element and then merging it with the original heap. Due to the

merge, insert takes O(log n) time, however it has an amortized time of O(1) (i.e. constant).

Find minimum

To find the minimum element of the heap, find the minimum among the roots of the binomial trees. This can again be done easily in O(log n) time, as there are just O(log n) trees and hence roots to examine.

By using a pointer to the binomial tree that contains the minimum element, the time for this operation can be reduced to O(1). The pointer must be updated when performing any operation other than Find minimum. This can be done in O(log n) without raising the running time of any operation.

Delete minimum

To delete the minimum element from the heap, first find this element, remove it from its binomial tree, and obtain a list of its subtrees. Then transform this list of subtrees into a separate binomial heap by reordering them from smallest to largest order. Then merge this heap with the original heap. Since each tree has at most log n children, creating this new heap is O(log n). Merging heaps is O(log n), so the entire delete minimum operation is O(log n).

function deleteMin(heap)

min = heap.trees().first()

for each current in heap.trees()

if current.root < min then min = current

for each tree in min.subTrees()

tmp.addTree(tree)

heap.removeTree(min)

merge(heap, tmp)

Decrease key

After decreasing the key of an element, it may become smaller than the key of its parent, violating the minimum-heap property. If this is the case, exchange the element with its parent, and possibly also with its grandparent, and so on, until the

minimum-heap property is no longer violated. Each binomial tree has height at most log n, so this takes O(log n) time.

Delete

To delete an element from the heap, decrease its key to negative infinity (that is, some value lower than any element in the heap) and then delete the minimum in the heap.

Performance

All of the following operations work in O(log n) time on a binomial heap with n elements:

∙Insert a new element to the heap

∙Find the element with minimum key

∙Delete the element with minimum key from the heap

∙Decrease key of a given element

∙Delete given element from the heap

∙Merge two given heaps to one heap

Finding the element with minimum key can also be done in O(1) by using an additional pointer to the minimum.

二项堆

在计算机科学中，二项堆是一个二叉堆类似的堆结构，但是支持两个二项堆快速合并。这是通过使用一个特殊的树结构。实现可合并堆的抽象数据类型（也称为meldable堆）是重要的，这个抽象数据结构就是支持合并操作的优先级队列。

二项树

二项堆是二项树的集合（与二项堆相比二叉堆则只有一颗二项树）。二项式树的递归定义：

(1)0阶二项树是一个单一的节点

(2)k阶二项树有根订单K-1，K-2，...，2，1，0（这个顺序）二叉树根节点的孩子。

上图是0阶到3阶的二项树：每一棵树都有一个根节。连接在根节点上的是所有低阶的子树，这在图中被高亮显示。例如，3阶二叉树连接到2阶，1和0（蓝色，绿色和红色高亮显示）二叉树。

k阶二项树高度为K，共有2k个节点。

由于其独特的结构，k阶的二叉树可以由两棵k-1阶的二叉树构成，通过将其中一棵二叉树的根节点作为另外一棵二叉树的最左子节点。这个特点对二叉堆的合并操作来说至关重要，这也是二项堆相对于传统堆拥有的主要优势。

二项树的名字来源于形状：n阶二叉树在深度n处共有个节点。

二项堆结构

二项式堆实现为一组满足堆性质的二项树集合：

(1)在堆的每二项式服从的最小堆属性：一个节点的关键是大于或等于它的父键。

(2)任意阶的二项树只能有1个或者0个。包括0阶二叉树。

第一属性确保每个二叉树的根包含树中最小的键，它适用于整个堆中的二叉树。第二个属确保n个节点的二项堆最多包含lgn + 1棵二叉树。事实上，二叉树的数目和阶数是由二叉堆的节点个数决定的：每颗二项树和n的二进制表示中的一位对应。例如13的二进制表示是1101，因此包含13个节点的二项堆包括三棵二叉树，阶数分别为3，2和0（见下图）。

实现

因为没有操作需要随机访问二叉树的根节点，因此二叉树的根节点可以按照对应阶数的增序存储在一个链表中。

合并

正如上面所提到的，最简单的和最重要的操作是在两个二叉堆中合并两个阶数相同的二叉树。由于二叉树的结构，它们能够被轻易的合并。由于二叉树的根节点具有最小关键字，通过比较两个根节点的关键字大小，其中较小的成为最小关键字，并成为新的根节点。另外一棵树成为合并后的树的一颗子树。这个操作对于合并两个二项堆而言是基础的。

function mergeTree(p, q)

if p.root.key <= q.root.key

return p.addSubTree(q)

else

return q.addSubTree(p)

对应上图，为了合并两棵阶数相同的二项树，首先比较根节点的关键字。因为7> 3，在左边的黑色树（根节点7）连接到灰树（与根结点3）右边的子树。结果是棵3阶树。

合并两个堆的操作也许是最有趣的，可以用来作为在大多数其他操作的子程序。和合并算法相似的是两个二项堆的根节点链表同时被遍历。

如果只有一个堆包含j阶的树，这棵树被移动到合并后的堆。如果两个堆都包含j阶的树，两棵树被合并成一颗满足最小堆性质的阶数为j+1的二叉树。注意这棵树以后也可能要和某个堆中j+1阶的二叉树合并。在算法的过程中，我们需要研究最三棵树的任何顺序的情况。

因为二项堆中的每棵二项树与表示二项堆大小的二进制表示中的一位对应，合并两个二项堆与两个二进制数相加是相似的。由右至左，每当一个进位产生时都意味着两棵二项树的合并。

每棵树的阶数不超过logn，因此运行时间是O（log n）的。

function merge(p, q)

while not( p.end() and q.end() )

tree = mergeTree(p.currentTree(), q.currentTree())

if not heap.currentTree().empty()

tree = mergeTree(tree, heap.currentTree())

heap.addTree(tree)

else

heap.addTree(tree)

heap.next() p.next() q.next()

上图表明合并两个二项堆的过程。这是连续的合并阶数相同的二叉树来实现的。如果合并后的二叉树又和两个二项堆中某棵二叉树阶数相同，那么这两棵二叉树再次被合并。

插入

可以通过简单地创建一个只包含此元素的新堆，然后将它与原来的堆合并。由于合并，插入需要O（log n）的时间，但它有一个摊销时间为O（1）。

查找最小节点

查找堆的最小节点可以通过查找最小二项树的根部。这可以再次轻松完成在O （log n）的时间，因为有只有O（log n）的二叉树，因此要检查的根节点不超过O(log n)。

通过使用指向最小节点的指针，此操作的时间可以降低到O（1）。在执行Find 操作以外的任何操作必须更新指针的值。这些操作的时间复杂度依然是O(log n)。

删除最小节点

要删除堆的最小元素，首先找到这个元素然后从二叉树中删除它，获得它的子树的列表。通过将子树按照阶数从小到大重新排列形成一个新堆。然后合并这堆与原来的堆。由于每棵树至多有 log n个孩子，创建这个新堆的时间复杂度是O （log n）。合并堆的时间复杂度是O(log n)，所以整个删除最小节点操作的时间复杂度是O(log n)。

function deleteMin(heap)

min = heap.trees().first()

for each current in heap.trees()

if current.root < min then min = current

for each tree in min.subTrees()

tmp.addTree(tree)

heap.removeTree(min)

merge(heap, tmp)

减小节点值

减小某个节点关键字之后，它可能会比其父节点的关键字来的小，从而违背了最小堆的性质。如果这种情况发生，交换其与父节点的关键字，也可能交互其与祖父节点的关键字，重复这个操作直到最小堆的性质不再被违背。每棵二叉树高度至多为log n，所以这需要O(log n)的时间。

删除

要从堆中删除一个元素，减少其值至负无穷大（即低于堆中任意节点的关键字），然后删除堆中的最小节点。

性能

对于有n个节点的二项堆，以下操作的时间复杂度为O(log n)：

(1)插入一个新的节点

(2)查找最小节点

(3)删除最小节点

(4)减小给定节点的关键字

(5)删除指定节点

(6)合并两个二项堆

寻找最小节点也可以通过使用一个额外的指针在O(1)时间内完成。

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7.1 Enter ActionMappings The Model 2 architecture (see chapter 1) encourages us to use servlets and Java- Server Pages in the same application. Under Model 2, we start by calling a servlet. The servlet handles the business logic and directs control to the appropriate page to complete the response. The web application deployment descriptor (web.xml) lets us map a URL pattern to a servlet. This can be a general pattern, like *.do, or a specific path, like saveRecord.do. Some applications implement Model 2 by mapping a servlet to each business operation. This approach works, but many applications involve dozens or hundreds of business operations. Since servlets are multithreaded, instantiating so manyservlets is not the best use of server resources. Servlets are designed to handle any number of parallel requests. There is no performance benefit in simply creating more and more servlets. The servlet’s primary job is to interact with the container and HTTP. Handling a business operation is something that a servlet could delegate to another component. Struts does this by having the ActionServlet delegate the business operation to an object. Using a servlet to receive a request and route it to a handler is known as the Front Controller pattern [Go3]. Of course, simply delegating the business operation to another component does not solve the problem of mapping URIs [W3C, URI] to business operations. Our only way of communicating with a web browser is through HTTP requests and URIs. Arranging for a URI to trigger a business operation is an essential part of developing a web application. Meanwhile, in practice many business operations are handled in similar ways. Since Java is multithreaded, we could get better use of our server resources if we could use the same Action object to handle similar operations. But for this to work, we might need to pass the object a set of configuration parameters to use with each operation. So what’s the bottom line? To implement Model 2 in an efficient and flexible way, we need to: Enter ActionMappings 195

计算机科学外文翻译

Binomial heap In computer science, a binomial heap is a heap similar to a binary heap but also supports quick merging of two heaps. This is achieved by using a special tree structure. It is important as an implementation of the mergeable heap abstract data type(also called meldable heap), which is a priority queue supporting merge operation. Binomial tree A binomial heap is implemented as a collection of binomial trees (compare with a binary heap, which has a shape of a single binary tree). A binomial tree is defined recursively: ∙ A binomial tree of order 0 is a single node ∙ A binomial tree of order k has a root node whose children are roots of binomial trees of orders k−1, k−2, ..., 2, 1, 0 (in this order). Binomial trees of order 0 to 3: Each tree has a root node with subtrees of all lower ordered binomial trees, which have been highlighted. For example, the order 3 binomial tree is connected to an order 2, 1, and 0 (highlighted as blue, green and red respectively) binomial tree. A binomial tree of order k has 2k nodes, height k. Because of its unique structure, a binomial tree of order k can be constructed from two trees of order k−1 trivially by attaching one of them as the leftmost child of root of the

计算机专业英语翻译参考

计算机专业英语翻译参考

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计算机专业中英文翻译外文翻译文献翻译

英文参考文献及翻译 Linux - Operating system of cybertimes Though for a lot of people , regard Linux as the main operating system to make up huge work station group, finish special effects of " Titanic " make , already can be regarded as and show talent fully. But for Linux, this only numerous news one of. Recently, the manufacturers concerned have announced that support the news of Linux to increase day by day, users' enthusiasm to Linux runs high unprecedentedly too. Then, Linux only have operating system not free more than on earth on 7 year this piece what glamour, get the favors of such numerous important software and hardware manufacturers as the masses of users and Orac le , Informix , HP , Sybase , Corel , Intel , Netscape , Dell ,etc. , OK? 1.The background of Linux and characteristic Linux is a kind of " free (Free ) software ": What is called free, mean users can obtain the procedure and source code freely , and can use them freely , including revise or copy etc.. It is a result of cybertimes, numerous technical staff finish its research and development together through Inte rnet, countless user is it test and except fault , can add user expansion function that oneself make conveniently to participate in. As the most outstanding one in free software, Linux has characteristic of the following: (1)Totally follow POSLX standard, expand the network operating system of supporting all AT&T and BSD Unix characteristic. Because of inheritting Unix outstanding design philosophy , and there are clean , stalwart , high-efficient and steady kernels, their all key codes are finished by Li nus Torvalds and other outstanding programmers, without any Unix code of AT&T or Berkeley, so Linu x is not Unix, but Linux and Unix are totally compatible. (2)Real many tasks, multi-user's system, the built-in network supports, can be with such seamless links as NetWare , Windows NT , OS/2 ,

合肥工业大学各学院、专业名称及其英文翻译

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