QED a novel quaternary encoding to completely avoid re-labeling in XML updates
QED: A Novel Quaternary Encoding to Completely Avoid
Re-labeling in XML Updates
Changqing Li Tok Wang Ling
Department of Computer Science, National University of Singapore, Singapore, 117543
{lichangq, lingtw}@https://www.wendangku.net/doc/2116119826.html,.sg
ABSTRACT
The method of assigning labels to the nodes of the XML tree is called a labeling scheme. Based on the labels only, both ordered and un-ordered queries can be processed without accessing the original XML file. One more important point for the labeling scheme is the label update cost in inserting or deleting a node into or from the XML tree. All the current labeling schemes have high update cost, therefore in this paper we propose a novel quaternary encoding approach for the labeling schemes. Based on this encoding approach, we need not re-label any existing nodes when the update is performed. Extensive experimental results on the XML datasets illustrate that our QED works much better than the existing labeling schemes on the label updates when considering either the number of nodes or the time for re-labeling. Categories and Subject Descriptors
H.2.4 [Database Management]: Systems –Query processing General Terms
Algorithms, Performance.
Keywords
Dynamic XML, Labeling scheme, Update, Quaternary.
1.INTRODUCTION
As a standard to represent and exchange data on the web, XML [7] has gained a lot of attention from both research and enterprise areas. Presently there is a lot of interest in query processing over XML that conforms to an ordered tree-structured data model. There are two main techniques, viz. structural index and labeling (numbering) scheme, to facilitate the XML queries. The structural index approaches [10, 14, 15] can help to traverse the hierarchy of XML, but this traversal is costly. The labeling scheme approaches [1, 2, 21] require smaller storage space, yet they can efficiently determine the ancestor-descendant (A-D) and parent-child (P-C) relationships between any two elements of the XML. In this paper, we focus on the labeling schemes. If the XML is static, the current labelling schemes can efficiently process different queries. However if the XML is dynamically changed, how to efficiently update the labels of the labelling schemes becomes to an important issue.
As we know, the elements in the XML are intrinsically ordered, which is referred as document order (the element sequence in the XML). The relative order of two paragraphs in the XML is important because the order may influence the semantics, thus the standard XML query languages (e.g., XPath[5] and XQuery [6]) require the output of queries to be in document order by default. Hence it is very important to maintain the document order when the XML is updated.
Though some researches [3, 8, 17, 18, 19, 21] have been done to maintain the document order in updating, the update costs of these approaches are still expensive. Therefore in this paper we focus on how to efficiently update the XML.
The main contributions of this paper are summarized as follows: ·We propose a novel dynamic quaternary encoding (called QED) that can be applied to different labeling schemes.
·QED completely avoids the re-labeling when the XML is updated.
·We conduct comprehensive experiments to demonstrate the benefits of our QED over the previous approaches.
The rest of the paper is organized as follows. Section 2 reviews the related work. We propose our dynamic quaternary encoding in Section 3. The most important part of this paper is Section 4, in which we show that the approach proposed in this paper is much more efficient than the existing schemes in processing updates. The experimental results are illustrated in Section 5, and we conclude in Section 6.
2.RELATED WORK
In this section, we present three families of labeling schemes, viz. containment [2, 13, 24], prefix [1, 8, 17, 19] and prime [21].
2.1Containment Scheme
Zhang et al [24] use a labeling scheme in which every node is assigned three values: “start”, “end”and “level”. For any two nodes u and v, u is an ancestor of v iff u.start < v.start and v.end < u.end. In other words, the interval of v is contained in the interval of u. Node u is a parent of node v iff u is an ancestor of v and v.level – u.level = 1. For instance, in Figure 1, “5,6,3” is a child of “2,7,2” since interval [5, 6] is contained in interval [2, 7] and levels 3 – 2 = 1.
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Although the containment scheme is efficient to determine the ancestor-descendant relationship, the insertion of a node will lead to a re-labeling of all the ancestor nodes of this inserted node and all the nodes after this inserted node in document order (if we do not re-label, not only can not the document order be maintained, but the containment scheme can not work correctly to determine the A-D etc. relationships). This problem may be alleviated if the interval size is increased with values unused [13]. However, large interval size wastes a lot of numbers which causes the increase of storage, while small interval size is easy to lead to re-labeling.
To solve the re-labeling problem, [3] uses Float-point values for the “start”and “end”of the intervals. It seems that Float-point solves the re-labeling problem [19]. But in practice, the Float-point is represented in a computer with a fixed number of bits [3, 19]. As a result, only 18 nodes can be inserted at a fixed place [3] since [3] uses the consecutive integer values at the initial labeling. Even if [3] uses values with large gaps, it still can not avoid the re-labeling due to the float-point precision. In fact, Float-point is equivalent to the method in [13] that some values are unused at the beginning. Therefore, using real values instead of integers only provides limited benefit for the node updating [19, 21]. 2.2Prefix Scheme
In the prefix labeling scheme, the label of a node is that its parent’s label (prefix) concatenates its own (self) label. For any two nodes u and v, u is an ancestor of v iff label(u) is a prefix of label(v). Node u is a parent of node v iff label(v) has no prefix when removing label(u) from the left side of label(v).
DeweyID [19] labels the n th child of a node with an integer n, and this n should be concatenated to the prefix (its parent’s label) and delimiter (e.g. “.”) to form the complete label of this child node. In practice, DeweyID uses UTF8 [23] to process the delimiter. [8] uses Binary Strings (BinaryString) to label the XML tree. When a node is inserted, DeweyID [19] and BinaryString [8] both need to re-label the sibling nodes after this inserted node and the descendants of these siblings to maintain the document order. [11] is also based on binary strings, but it is dynamic. However [11] still can not completely avoid the re-labeling due to the overflow problem (see Example 3.2).
OrdPath [17] is similar to DeweyID, but it only uses the odd numbers at the initial labeling. When the XML tree is updated, it uses the even number between two odd numbers to concatenate another odd number. When the sizes of the OrdPath codes overflow (see Example 3.2 in Section 3 for more details about the overflow problem), it must re-label all the existing nodes.
[9] is similar to OrdPath [17] and [9] is not as compact as OrdPath. 2.3Prime Scheme
Wu et al [21] use Prime numbers to label XML trees. The root node is labeled with “1” (integer). Based on a top-down approach, each node is given a unique prime number (self_label) and the label of each node is the product of its parent node’s label (parent_label) and its own self_label. For any two nodes u and v, u is an ancestor of v iff label(v) mod label(u) = 0. Node u is a parent of node v iff label(v)/self_label(v) = label(u).
Prime [21] uses SC (Simultaneous Congruence) values in Chinese Remainder Theorem [4] to determine the document order, i.e. SC mod self_label = document order. When the order is changed, Prime needs to re-calculate the SC values instead of re-labeling. Although Prime supports order-sensitive updates without any re-labeling of the existing nodes, it needs to re-calculate the SC values based on the new ordering of nodes. The re-calculation is very time consuming.
2.4Motivation
All the current labeling schemes except Prime can not completely avoid the re-labeling in updates (OrdPath will encounter the overflow problem). Though Prime can completely avoid re-labeling, it needs to re-calculate the SC values which is much more expensive than re-labeling. Therefore the main objective of this paper is to dramatically decrease the update costs and completely avoid the re-labeling (see Sections 4.1, 5.1 and 5.2). 3.QED ENCODING
In this section, we elaborate our Quaternary Encoding for Dynamic XML data (QED) which supports label insertion without re-labeling or re-calculation (see Section 4.1). For the containment scheme shown in Figure 1 (Section 2.1), the “start”and “end” values are from 1 to 16. These decimal numbers can be encoded in binary codes with fixed length, called FixedLength1 (FixedLength column of Table 1) or with variable length, called VarLength (VarLength column of Table 1).
Definition 3.1 (Quaternary code)Four numbers “0”, “1”, “2”and “3” are used in the code and each number is stored with two bits, i.e. “00”, “01”, “10” and “11”.
Definition 3.2 (QED code)QED code is a quaternary code. The number “0” is used as the separator and only “1”, “2” and “3”are used in the QED code itself.
Now let us discuss how to encode the “start”s and “end”s (1-16) using our QED encoding (16 is only an example for the tree in Figure 1; our QED encoding can be applied to any other numbers; see the formal algorithm in Section 3.1). The following steps show the details of how to get the QED codes in Table 1 and these steps are examples for the formal algorithms in Section 3.1.
Step 1: In the encoding of the 16 numbers, we suppose there is one more number before number 1, say number 0, and one more number after number 16, say number 17.
1 Float-point, DeweyID, OrdPath, Prime, FixedLength and VarLength are all existing schemes. QED-PREFIX and QED (with all fonts capitalized) are schemes proposed in this paper.
Table 1. Binary and our QED encoding approaches
Decimal FixedLength VarLength QED
1 00001 1 112
2 00010 10 12
3 00011 11 122
4 00100 100 13
5 00101 101 132
6 00110 110 2
7 00111 111 212
8 01000 1000 22
9 01001 1001 23
10 01010 1010 232
11 01011 1011 3
12 01100 1100 312
13 01101 1101 32
14 01110 1110 322
15 01111 1111 33
16 10000 10000 332
Total size 80 54 76
Step 2:The (1/3)th number is encoded with “2”, and the (2/3)th number is encoded with “3”. The (1/3)th number is number 6, which is calculated in this way, 6 = round(0+(17–0)/3). The (2/3)th number is number 11 (11 = round(0+(17–0)′2/3)).
Step 3:The (1/3)th and (2/3)th numbers between number 0 and number 6 are number 2 (2 = round(0+(6–0)/3)) and number 4 (4 = round(0+(6–0)′2/3)). The QED code of number 0 (left code) is now empty with size 0 and the QED code of number 6 (right code) is now “2” with size 1 (here 1 refers to 2 bits). This is Case (a) where the left code size is smaller than the right code size. In this case, the (1/3)th code is that we change the last symbol of the right code to “1” and concatenate one more “2”, i.e. the code of number 2is “12”(“2”?“1”and “1”?“2”?“12”), and the (2/3)th code is that we change the last symbol of the right code to “1”and concatenate one more “3”, i.e. the code of number 4is “13” (“2”?“1” and “1”?“3”?“13”).
Step 4: The (1/3)th and (2/3)th numbers between numbers 6 and 11 are numbers 8 (8 = round(6+(11–6)/3)) and 9 (9 = round(6+(11–6)′2/3)). The QED code of number 6 (left code) is “2” with size 1 (here 1 refers to 2 bits) and the code of number 11 (right code) is “3” with size 1 (here 1 refers to 2 bits). This is Case (b) where the left code size is larger than or equal to the right code size. In this case, the (1/3)th code is that we directly concatenate one more “2”after the left code, i.e. the code of number 8is “22”(“2”?“2”?“22”), and the (2/3)th code is that we directly concatenate one more “3”after the left code, i.e. the code of number 9 is “23” (“2”?“3”?“23”).
Step 5: The (1/3)th and (2/3)th numbers between numbers 11 and 17 are numbers 13 (13 = round(11+(17–11)/3)) and 15 (15 = round(11+(17–11)′2/3)). The code of number 11 (left code) is “3” with size 1 and the code of number 17 (right code) is empty now with size 0. This is still Case (b). Therefore the QED code of number 13is “32”(“3”?“2”?“32”), and the code of number 15 is “33” (“3”?“3”?“33”). In this way, all the numbers will be encoded. Finally we need to discard the codes for numbers 0 and 17 since they do not exist actually. It should be noted that Step 1 is not compulsory, but with Step 1 the total code size is smaller. It should be noted also that if the (2/3)th number exactly refers to the (1/3)th number, the code for the (2/3)th number will not appear since this number has already been encoded with the (1/3)th code.
With the following example illustration for the total code size, the formal size analysis in Section 3.2 will be easier to understand. Example 3.1 Table 1 shows that VarLength has smaller total code size than FixedLength. However, we also need to store the size of each VarLength code, e.g., the size of “10000” is 5. We need to store this 5 using fixed length of bits (“101”; 3 bits). The sizes of other codes should also be stored using fixed length of bits (3 bits), therefore the total code size for VarLength is 3′16+54=102 bits which is larger than the bits required by FixedLength. Example 3.2The size of each VarLength code is stored with fixed length (e.g. 3), therefore if many nodes are inserted into the XML tree, the fixed length size (e.g. 3) is not enough for the new labels, then we have to re-label all the existing nodes. Even if we increase the size 3 to a larger number, it still can not completely avoid the re-labeling, and it will waste the storage space. This is called the overflow problem in this paper. Similarly FixedLength and OrdPath [17] will encounter the overflow problem. Example 3.3 On the other hand, our QED uses the separator “0”
(2 bits) to separate different codes instead of storing the sizes of the codes. For example, “1120120” will be separated to “112” and “12”. Therefore the size of our QED is 2′16+76=108 bits. Each size code of VarLength is stored with 3 bits, and this 3 will increase as the number of nodes increases, but the size 2 (bits) of our separator “0”will never increase. More important, we will never encounter the overflow problem in this way.
The most important feature of our QED is that it is based on the lexicographical order (for efficient updates in Section 4.1). Definition 3.3 (Lexicographical order )Given two Quaternary codes C A and C B, C A is lexicographically equal to C B iff they are exactly the same. C A is said to be lexicographically smaller than C B (C A C B) iff
(a)“0” “1” “2” “3”, or
(b)compare C A and C B symbol by symbol from left to right. If
the current symbol S A of C A and the current symbol S B of C B satisfy condition (a), then C A C B and stop the comparison, or
(c)C A is a prefix of C B.
Theorem 3.1Our QED codes are lexicographically ordered but not numerically ordered.
Example 3.4The QED codes in Table 1 are lexicographically ordered from top to bottom. For example, “132” “2”because the comparison is from left to right, and the 1st symbol of “132” is “1”, while the 1st symbol of “2” is “2”. Another example, “23” “232” because “23” is a prefix of “232”.
When we replace the “start”s and “end”s (1-16) in Figure 1 with our QED codes, and based on the lexicographical comparison, a QED containment scheme is formed.
3.1The Formal Encoding Algorithm
Algorithm 1: GetOneThirdAndTwoThirdCodes(Left_Code, Right_Code)
Input:QED codes Left_Code, Right_Code, and Left_Code Right_Code
Output:QED codes: Code_AtOneThirdPos, Code_AtTwoThirdPos, such that Left_Code Code_AtOneThirdPos Code_AtTwoThirdPos Right_Code
1: get the sizes of Left_Code and Right_Code
2: if Left_Code and Right_Code are both empty
3: then Code_AtOneThirdPos = “2”
4: Code_AtTwoThirdPos = “3”
5: else //the below? means concatenation
6: if size(Left_Code) < size(Right_Code) //case (a)
7: then Temp_Code = the Right_Code with the last
8: symbol changed to “1”
9: Code_AtOneThirdPos = Temp_Code ?“2”
10: Code_AtTwoThirdPos = Temp_Code ?“3”
11: else if size(Left_Code) 3 size(Right_Code) //case (b) 12: then Code_AtOneThirdPos = Left_Code ?“2”
13: Code_AtTwoThirdPos = Left_Code ?“3”
Figure 2. GetOneThirdAndTwoThirdCodes algorithm.
Algorithm 2: Encoding(TN)
Input: A positive integer TN
Output: The QED codes for numbers 1 to TN
1: suppose there is one more number before the first number,
called number 0, and one more number after the last number,
called number (TN+1)
2: SubEncoding(codeArr, 0, TN+1)
//here codeArr is an array with size TN+2
3: discard the 0th and (TN+1)th elements of the array codeArr
Procedure SubEncoding(codeArr, P L, P R)
/*SubEncoding is a recursive procedure; codeArr is an array, P L is the left position, and P R is the right position*/
4: P M1 = P L+round((P R-P L)/3) (P M1 is the (1/3)th position)
5: P M2 = P L+round((P R-P L)′2/3) (P M2 is the (2/3)th position)
6: if P L1P R
7: then GetOneThirdAndTwoThirdCodes(codeArr[P L],
codeArr[P R]) 8: if P M11P L and P M11P R
9: then codeArr[P M1]=Code_AtOneThirdPos //from line 7 10: if P M21P M1 and P M21 P R
11: then codeArr[P M2]=Code_AtTwoThirdPos //from line 7 12: if (P M11P L and P M11P R) or (P M21P L and P M21P R)
13: then SubEncoding(codeArr, P L, P M1)
14: SubEncoding(codeArr, P M1, P M2)
15: SubEncoding(codeArr, P M2, P R)
Figure 3. QED encoding algorithm.
Algorithm 1 and Algorithm 2 are a summary of Step 1 to Step 5 (in the previous page) which can be used to encode any total number (not only 16). Here we do not explain them in detail.
3.2Size Analysis
The size in this paper refers to bits. The log in this paper is used as the logarithm to base 2. The log3in this paper is used as the logarithm to base 3. In this section, we analyze the size required by Prime, Float-point, FixedLength, VarLength and our QED. The “D”and “N”are respectively used to denote the maximal depth and the number of nodes of an XML tree.
Prime According to the size analysis of Prime in [21], the size required to store all the nodes in the XML tree is:
))
log(
log(N
N
D
N′
′
′(1) Float-point According to [3], each float-point number is stored with 64 bits, thus the size of Float-point:
64
2′
N(2) FixedLength The size of FixedLength is:
))
2
log(log(
)
2
log(
2N
N
N+
)1
)
log(log(
2
)
log(
2+
+
+
=N
N
N
N(3) ))
2
log(log(N is the size of label size (stored only once). VarLength For VarLength, one number 1 is stored with one bit (see VarLength column of Table 1), two numbers 2 and 3 are stored with 2 bits, four numbers 4, 5, 6 and 7 are stored with 3 bits, ···, thus the total size of VarLength is:
)1
(
2
4
2
3
2
2
2
1
13
2+
′
+×××+
′
+
′
+
′
+
′n
n
1
21+
′
=+n
n(4) Since the number of nodes is N, the number of “start”s and “end”s is 2N which should be equal to 1
2
2
2
21
1
0-
=
+×××+
++n
n. However 2N is an even number and can not be equal to 1
21-
+
n. We assume that there are 2N+1 numbers. After getting the VarLength codes, we can discard one number, i.e. 2N numbers are left. There is only a constant difference between 2N and 2N+1, thus we assume that 2N+1=1
21-
+
n, then formula (4) becomes: 1
)1
log(
2
)1
log(
2+
+
+
+N
N
N(5) Note that formula (5) is a little larger than the actural size required by VarLength.
We need to store the size of each “start”and “end”. A fixed-length number is used to store the size of the VarLength code. The maximal size for a “start” or an “end” is )
2
log(N. To store this size, the bits required are ))
2
log(log(N, then the total bits required to store the sizes of VarLength codes are ))
2
log(log(
2N
N. When taking formula (5) into account, the total size of VarLength is:
1
)1
log(
2
)1
log(
2
))
2
log(log(
2+
+
+
+
+N
N
N
N
N(6) QED When considering our QED, it has two numbers 6 and 11 stored with size 1 (2 bits), 6 numbers 2, 4, 8, 9, 13 and 15 stored with size 2 (2 bits), ···, therefore its size is:
+
′
′
′
+
′
′
′
+
′
′
′)2
3(
)
3
2(
)2
2(
)
3
2(
)2
1(
)
3
2(2
1
)2
)1
((
)
3
2(′
+
′
′
+×××n
n (bits)
1
3
)1
2(1+
′
+
=+n
n(7) We assume +
′
+
′
=)
3
2(
)
3
2(
21
N1
3
)
3
2(1-
=
′
+×××+n
n (both sides are even numbers), then formula (7) becomes to:
N N N N 2)12(log 2)12(log 433-+++
(8)
When considering the separator (“0”) size 22′N , the total label size of our QED is:
N N N N 2)12(log 2)12(log 433++++
(9)
It can be seen that VarLength has larger label size than FixedLength. Prime has larger size than VarLength. To make the size of Float-point smaller than VarLength, N should be grossly larger than 264, but generally speaking, an XML file can not have so many nodes. That is to say, Float-point has larger label size than VarLength. If N>264, 64 bits are not enough to store the Float-point values.
When N=1, the size of our QED is 1.6 times of that of VarLength; when ]7,2[?N , the multiple is between 1.16 and 1.28; when ]38,8[?N , the multiple is between 1.12 and 1.15; when ]10000000,39[?N , the multiple is between 1.11 and 1.13. Thus the size of our QED is only a little larger than the size of VarLength.
Also we need to consider the “level ” size for the containment schemes which is ))log(log()log(D D N + and should be added into formulas (2), (3), (6) and (9) to form the total label sizes of these schemes.
Note that for simplicity, we omit the ceiling functions on the log functions in all the formulas.
3.3 Application Scope of QED
Property 3.1 Our QED is orthogonal to specific labeling schemes, thus it can be applied to all the labeling schemes or other applications which need to maintain the order.
Our QED encoding can be applied to the prefix scheme and prime scheme also to maintain the document order. When QED is applied to prefix, we call it QED-PREFIX.
Example 3.5 Figure 4 shows that we apply QED to the prefix scheme. The root has 4 children. To encode 4 numbers based on our QED, the codes will be “12”, “2”, “3” and “32”. Similarly if there are two siblings, their self_labels are “2” and “3”. If there is only one sibling, its self_label is “2”.
For the prefix scheme, the delimiter “.” can not be stored together with the numbers in the implementation to separate different components.
For our QED encoding, we use the following approach to process the delimiters. We use one separator “0” as the delimiter to separate different components of a label (e.g. separate “12” and “3” in “12.3”; the separator “0” is equivalent to the “.” in Figure 4), and use two consecutive separators “00” as the separator to separate different labels (e.g. separate “12.2” and “12.3”).
We can apply our QED to the prime labeling scheme also to record the document order. But because Prime employs the modular and division operations to determine the ancestor-descendant etc. relationships, its query efficiency is quite bad. Thus we do not discuss in detail how QED is applied to Prime. Similarly we can and it is better to apply our QED to the P-Containment scheme proposed in [12] to completely avoid re-labeling.
4. UPDATE
In Section 4.1, we show that our QED has much cheaper update cost. Section 4.2 analyzes the case for the frequent update.
4.1 Avoid Re-labeling in Updates
The deletion of a node will not affect the ordering of the nodes in the XML tree. Thus in this section, we only discuss the insertion. Algorithm 3 is similar to Algorithm 1, and their difference is marked in Figure 5 with italic fonts. Based on Algorithm 3, we can avoid the re-labeling. As the QED codes are long which can not be put in Figures 6, we still use decimal numbers in Figure 6 for the “start ” and “end ” values, but in practice, these numbers are stored using our QED codes. Refer to Table 1 for the mappings between the decimal numbers and our QED codes. We use an example to show how Algorithm 3 works.
Example 4.1 When inserting node “a ” (see Figure 6), we should insert a number between the “start ” of the parent “1” (Left_Code) and the “start ” of the first sibling “2” (Right_Code). If we use the traditional approach, we can not insert a number between “1” and “2”, and we must re-label the nodes. However, when referring to Table 1, our QED codes for “1” and “2” are “112” and “12”. Based on the GetInsertedCode algorithm, we insert a value between “112” and “12”, then the “start ” value of the inserted node “a ” is “113” (see lines 5-8 of the GetInsertedCode algorithm in Figure 5). The “end ” value of node “a ” is an insertion between the new “start ” “113” and the “start ” of the first sibling “12”, thus the “end ” value of “a ” is “1132” (see lines 5, 9 and 10 in Figure 5). Obviously, “112” “113” “1132” “12” lexicographically. We need not re-label any existing nodes, but we can keep the containment scheme working correctly.
Algorithm 3: GetInsertedCode Input: Left_Code, Right_Code
Output: Inserted_Code, such that Left_Code Inserted_Code Right_Code lexicographically.
2: if size(Left_Code) < size(Right_Code)
3: then Inserted_Code = (the Right_Code with the last
4: symbol changed to “1”) ? “2” 5: else if size(Left_Code) > size(Right_Code) 6: if the last symbol of Left_Code is “2”
7: then Inserted_Code = the Left_Code with the
8: last symbol changed from “2” to “3” 9: else if the last symbol of Left_Code is “3” 10: then Inserted_Code = Left_Code ? “2” 11: else if size(Left_Code) = size(Right_Code) 12: then Inserted_Code = Left_Code ? “2”
Figure 5. GetInsertedCode algorithm.
Theorem 4.1 Algorithm 3 guarantees that infinite number of QED codes can be inserted between any two consecutive QED codes with the orders kept and without any re-encoding of the existing numbers.
Based on Theorem 4.1, the insertions of nodes “b ”, “c ” and “d ” will not cause any re-labeling also. After insertion, the level values of “a ”, “b ” and “c ” are still 2 (2 is a decimal number for illustration), but the level value of node “d ” is 3. Now let us study the updates based on other schemes.
Example 4.2 For the containment schemes FixedLength and VarLength, when “a ” is inserted into the XML tree in Figure 6, all the “start ” and “end ” values except the “start ” value of the root need to be added with 2. The “end ” value of the root will be “18” after update. It is similar for “b ”, “c ” and “d ”.
Example 4.3 For the prefix schemes, our QED-PREFIX also need not re-label any existing nodes. DeweyID has to re-label all the sibling nodes after the inserted node and the descendants of these following sibling nodes. For Prime, the insertion of nodes will make the document order change (the orders of the nodes after this inserted node should all be added with 1), therefore Prime has to re-calculate the SC values.
Sometimes Float-point [3] and OrdPath [17] also need not re-label the nodes. Sections 5.1 and 5.2 are the update performance comparisons among Float-point, OrdPath and our approach.
4.2 Frequent Update
The size analysis in Section 3.2 is based on the initial labeling of the XML. Our encoding algorithm (see Figure 3) is step by step insertions of nodes uniformly at different places. Therefore if a sequence of nodes are inserted randomly at different places of the XML, the size analysis in Section 3.2 is still valid.
For the case that nodes are always inserted at a fixed place of the XML, the size of our QED increases quickly. [8] proves that any deterministic labeling scheme which does not re-label nodes must in the worst case assign labels of size Omega(N). Our QED can not escape from this claim also, i.e. the label size of our QED increases linearly in the worst case. OrdPath [17] also has this skewed insertion problem. [18] uses B-tree to balance the update
and lookup performance which can be used to process this skewed insertion problem. The skewed insertion is not an emphasis of this paper. In this paper, we mainly focus on how to completely avoid the re-labeling.
5. PERFORMANCE STUDY
In this section, we evaluate and compare the performance of different labeling schemes. All the schemes are implemented in Java and all the experiments are carried out on a 3.0 GHz Pentium 4 processor with 1 GB RAM running Windows XP Professional. Table 2 shows the characteristics of the test datasets. D1, D2 and D3 are from [16], D5 and D6 are from [20], and all of them are real-world XML data. D4 is a benchmark generated by XMark [22].
Static XML is not the emphasis of this paper. However, we also test how our QED works on the static XML data which shows that our QED works not worse even in the static environment of XML (its label sizes are small in the six datasets shown in Table 2 and its query performance is not worse).
For the dynamic XML, in Section 5.1 we show that our QED works much better compared to the existing labeling schemes except OrdPath and Float-point. Although it seems that OrdPath and Float-point also work well in the intermittent insertions, the wide update performance difference between OrdPath and our QED-PREFIX, and between Float-point and our QED can be seen from Section 5.2, where frequent updates are performed.
OrdPath and our QED-PREFIX are prefix schemes. Float-point, FixedLength, VarLength, and our QED are containment schemes.
5.1 Intermittent Update
We select one XML file Hamlet in D2 to test the update performance (it is similar for other XML files). Hamlet has 5 act elements. We test the following six cases: inserting an act before act[1], inserting an act between act[1] and act[2], ···, inserting an act between act[4] and act[5], and inserting an act after act[5]. Figure 7(a) shows the number of nodes for re-labeling when applying different labeling schemes. FixedLength and VarLength have the same number of nodes to re-label in all the six cases. For case 1, 6596 (total 6636) nodes need to be re-labeled.
For Prime, the number of SC values that are required to re-calculate is counted in Figure 7(a). Because Prime uses each SC value for every five labels [21], the number of SC values required to re-calculate is 1/5 of the number of nodes required by FixedLength and VarLength to re-label. Note that it is impossible to use a single SC value for all the nodes in the XML since the SC value will be too large.
Table 2. Test datasets
Datasets Topics # of files
Max/average fan-out for a file
Max/average depth for a file
Total # of nodes for each dataset
D1 Company 24 529/135 5/3 161576 D2 Shakespeare ’s play
37
434/48 6/5 179689 D3 NASA 1882
1188/9 7/5 370292 D4 XMark 1 25500/3242 12/6 1666315 D5 Treebank 1 56384/1623 36/8 2437666 D6
DBLP
1
328858/65930
6/3
3332130
(b) Time to re-label
Figure 7. Performance study on intermittent update.
In all the six cases, OrdPath1 (without overflow here), OrdPath2 (without overflow here), our QED-PREFIX, Float-point (less than 18 nodes), and our QED need not re-label any existing nodes. Next we study the time required to re-label nodes or re-calculate SC values. Figure 7(b) shows that the time required by Prime to re-calculate the SC values is much larger (more than 202 times; sum time of Case 1 to Case 6) than the time required by FixedLength and VarLength to re-label the nodes. Prime theoretically is a good scheme in updating order-sensitive nodes, but it is not practicable. In contrast, our QED-PREFIX and QED need less than 0.001 second (processing time) for the insertion in all the 6 cases. The processing time of FixedLength or VarLength is at least 84 times of that of QED-PREFIX and QED even if we assume that our QED needs 0.001 second for the update (in fact, it will be much larger than 84 times; see Section 5.2).
It can be seen that OrdPath1, OrdPath2 and Float-point also need less than 0.001 second for the update. This is because only several nodes are inserted into the XML. The update performance difference among OrdPath, Float-point and our approach can be seen in Section 5.2 where frequent insertions are executed.
5.2Frequent Update
When intermittent nodes are inserted into the XML, Prime, FixedLength and VarLength have much larger update time, thus it will be a disaster for them to update the XML with frequent and tiny insertions, which makes them impossible to answer any queries in either the uniformly frequent or skewed frequent insertion environment. In this section, we mainly compare the update performance between OrdPath and our QED-PREFIX, and between Float-point and our QED. Section 5.2.1 discusses the case that frequent insertions are randomly at different places of the XML. Section 5.2.2 discusses the worst case that frequent insertions are always at a fixed place of the XML.
5.2.1Uniformly Frequent Update
In this section, we test the uniformly distributed frequent insertions. The Hamlet file has totally 6636 nodes. We insert 6635 nodes between every two consecutive nodes of the 6636 nodes. Based on the new file after insertion, we insert another 13270 nodes between any two consecutive nodes. We repeat this kind of insertion 6 times. After the 6th time insertion, the node number in Hamlet is 424641 that is 63.99 times of the original node number. The first two time insertions and the first four time insertions respectively will not cause OrdPath [17] (without overflow) and Float-point [3] (less than 18 nodes at a fixed place) to re-label the existing nodes. Even without re-labeling, Figure 8(a) shows that the update time of OrdPath is still at least 529 times of that of our QED-PREFIX and Figure 8(b) shows that the update time of Float-point is 386 times of that of our QED. OrdPath and Float-point have much larger update time because they need the addition and division operations to get the numbers between two numbers which are very expensive. On the other hand, our QED-PREFIX and QED only need to modify the last quaternary number (two bits) of the neighbor label to get the label of the inserted node which is much cheaper.
At the 3rd time insertion, OrdPath needs to re-label all the existing nodes, and at the 5th and 6th time insertions, Float-point needs to re-label all the existing nodes. The re-labeling time of OrdPath and Float-point is at least 2492 times of that of our QED-PREFIX and QED (see Figures 8(a) and 8(b)).
5.2.2Skewed Frequent Update
In this section, we test the case that nodes are always inserted at a fixed place of the Hamlet XML.
Figure 9 shows the re-labeling time (the update time without re-labeling in skewed frequent insertions is similar to the update time shown in Figure 8). For OrdPath (OrdPath1 and OrdPath2) [17], after inserting 163 nodes at the fixed place, it needs to re-label 5 times. The re-labeling time of OrdPath is at least 8126 times of that of our QED-PREFIX (see Figure 9(a)). When every 18 nodes are inserted at the fixed place of the XML, Float-point [3] needs to re-label. The update time of Float-point is 34383 times of that of our QED after 199 nodes are inserted (see Figure 9(b)).
The very large update time makes OrdPath and Float-point unsuitable to answer queries no matter in the uniformly or in the skewed frequent insertion environment. This means even if we do not use any skeweness processing technique, our QED still works the best to answer queries in the environment that frequent insertions are executed.
For the frequent update, our QED-PREFIX and QED have much cheaper update cost than OrdPath and Float-point.
(a) OrdPath1&2 vs QED-PREFIX
(b) Float-point vs QED (a) OrdPath1&2 vs QED-PREFIX
(b) Float-point vs QED
Figure 8. Performance study on uniformly frequent update. Figure 9. Performance study on skewed frequent update.
6.CONCLUSION
In this paper, we have proposed a novel Dynamic Quaternary Encoding (QED) approach for the labeling schemes. This encoding approach is orthogonal to specific labeling schemes, therefore it can be applied broadly to different labeling shemes, e.g. containment, prefix and prime schemes, to maintain the document order when the XML is updated.
The QED (or QED-PREFIX) proposed in this paper completely (no overflow) avoids the re-labeling in XML update. When a node is inserted, our QED (or QED-PREFIX) only needs to modify the last quaternary number (two bits) of the neighbor label to get the label of the inserted node which is very easy and cheap compared to Float-point (or OrdPath). The experimental results show that our QED (or QED-PREFIX) encoding is the only approach which supports frequent insertions efficiently.
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中考语文课外阅读题小说题库
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古代晋灵公不君、齐晋鞌之战原文及译文
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农田水利学试题及答案
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3.渠道的输水损失包括以下四个部分：（） A、干渠、支渠、斗渠及农渠等四级渠道的水量损失。 B、渠床土质、地下水埋深、渠道的工作制度及输水时间。 C、自由渗流、顶托渗流、渠床土质、与渠道的工作制度等。 D、渠道水面蒸发损失、渠床渗漏损失、闸门漏水与渠道退水等。 4.什么叫渠道水的利用系数？（） A、灌溉渠系的净流量与毛流量的比值。 B、某一级渠道的净流量与毛流量的比值。 C、田间实际灌入的有效水量与末级渠道的供水量之比。 D、实际灌入农田的有效水量和渠首引入的水量之比。 5.在渠道规划设计中，渠道最小流量有何作用？（） A、用以校核对下一级渠道的水位控制条件。 B、用以校核渠道不淤条件。 C、用以确定修建节制闸的位置。 D、用以校核对下一级渠道的水位控制条件和确定修建节制闸的位置，并按最小流量验算渠道不淤条件。 6.什么叫雨水径流集蓄灌溉工程？（） A、导引、收集雨水径流，并把它蓄存起来加以有效灌溉利用的工程技术措施。 B、田面、坡面、路面及屋面庭院等各类集水工程。
如何翻译古文
如何翻译古文 学习古代汉语，需要经常把古文译成现代汉语。因为古文今译的过程是加深理解和全面运用古汉语知识解决实际问题的过程，也是综合考察古代汉语水平的过程。学习古代汉语，应该重视古文翻译的训练。 古文翻译的要求一般归纳为信、达、雅三项。“信”是指译文要准确地反映原作的含义，避免曲解原文内容。“达”是指译文应该通顺、晓畅，符合现代汉语语法规范。“信”和“达”是紧密相关的。脱离了“信”而求“达”，不能称为翻译；只求“信”而不顾“达”，也不是好的译文。因此“信”和“达”是文言文翻译的基本要求。“雅”是指译文不仅准确、通顺，而且生动、优美，能再现原作的风格神韵。这是很高的要求，在目前学习阶段，我们只要能做到“信”和“达”就可以了。 做好古文翻译，重要的问题是准确地理解古文，这是翻译的基础。但翻译方法也很重要。这里主要谈谈翻译方法方面的问题。 一、直译和意译 直译和意译是古文今译的两大类型，也是两种不同的今译方法。 1．关于直译。所谓直译，是指紧扣原文，按原文的字词和句子进行对等翻译的今译方法。它要求忠实于原文，一丝不苟，确切表达原意，保持原文的本来面貌。例如： 原文：樊迟请学稼，子曰：“吾不如老农。”请学为圃。子曰：“吾不如老圃。”（《论语?子路》） 译文：樊迟请求学种庄稼。孔子道：“我不如老农民。”又请求学种菜蔬。孔子道：“我不如老菜农。”（杨伯峻《论语译注》） 原文：齐宣王问曰：“汤放桀，武王伐纣，有诸？”（《孟子?梁惠王下》） 译文：齐宣王问道：“商汤流放夏桀，武王讨伐殷纣，真有这回事吗？（杨伯峻《孟子译注》） 上面两段译文紧扣原文，字词落实，句法结构基本上与原文对等，属于直译。 但对直译又不能作简单化理解。由于古今汉语在文字、词汇、语法等方面的差异，今译时对原文作一些适当的调整，是必要的，并不破坏直译。例如： 原文：逐之，三周华不注。（《齐晋鞌之战》） 译文：〔晋军〕追赶齐军，围着华不注山绕了三圈。
李文忠《摆渡老人》中考现代文阅读练习及答案
(三）课外阅读（13分） 摆渡老人 李文忠 ①读中学时，学校设在邻村，与我们村有一条河隔着，便认识了那摆渡老人。 ②当时，那老人一直是我们取乐的对象。每到放学，肚子饿得直叫的我们便一窝蜂向渡口跑去，挤在那儿。渡船只要一近岸，大家便争先恐后地向船上跃去。虽然老人很有经验，未到岸边便做好回撑的准备，但超载和落水的事仍然时有发生。超载时，船上的，岸上的，都望着老人手忙脚乱的样子大笑。夏天下雨时，大家总是把伞迎着风，看着渡船被吹到离渡口好远的地方，老人一番折腾，将船撑到渡口，我们便哄笑着上了岸。 ③最有意思的要数冬天（当时我是这样认为的）。冬天很冷，河面结了一层厚厚的冰。每天早晨，当我们赶到渡口时，总看见老人在那边晃着船，好长时间，才把船晃过来。望着老人跳舞似的笨拙姿势，我们总是哄笑，一直到船靠岸。我们往往因此耽误了早读，但我们都挺愿意。于是就希望天再冷些，便可迟点起来，反正去早了也过不了河。我们都为找到一个偷懒的理由而感到高兴。当时的我们，是“心忧读书愿天寒”。 ④那一天特别冷，我因有事起了个早，吃过早饭后，便不紧不慢地向学校走去。到渡口时我惊奇地发现，那老人不是我预料的那样正在“跳舞”，却已经到了这边。我和他攀谈起来。我问他：“你今天怎么这么早？”老人咳嗽了一声，缓缓地说：“许是人老了，早上总起不来。昨天听学校说耽误了学生上早读，今天我就起了早。谁知等了好长时间，才来了你一个人。”说完，便送我过河。 ⑤船到河心，老人忽然回头对我说道：“你们小孩子可要好好读书啊。”那双眼中竟满含着期待和爱意，我心中一震，看着老人瘦弱的身体，沾满冰碴的白胡子，一双手冻得发紫，猛地问道：“你干啥要吃这份苦呢？”好久，老人长叹一声说道：“干啥？哎，以前我也有一个可爱的儿子，他曾吵着要上学，但终究没上成。后来他得病去了，我一直感到对不起他，我在这儿摆渡，接送你们上学，心里好受些。”谈话间，船已到岸。待我上岸后，老人把船向河那边撑去。看着老人的背影，我不再感到笨拙可笑，鼻子有些酸酸的。 ⑥那天，我把老人的故事讲给同学们听了。以后，我们上船再也没有起哄。于是老人饱经风霜的脸上出现了笑容。 ⑦再后来，两岸之间架起了一座桥。最后一次过河时，老人对我们说：“以后不再送你们过河了，你们要记住，好好读书。” ⑧“好好读书”，这句话一直留在我记忆中。 ⑨真想再见见摆渡老人。 20、本文具体描述了摆渡老人接送“我们”过河上学的辛劳。先写老人因“我们”的___________而手忙脚乱，一番折腾；再写因_________，老人好不容易“才把船晃过来”；最后写老人不顾自己__________，特地起大早送“我们”过河上学。（3分） 21.第⑤段中摆渡老人说“我在这儿摆渡，接送你们上学，心里好受些”。请你用自己的语言简要说说为什么老人觉得这样做“心里好受些”。（3分） ________________ 22.第⑤段中划横线的语句表现了“我”当时___________和___________的心情。（2分） 23.结合全文内容回答以下问题。 原先，摆渡老人在孩子们的眼里是一个________________的形象；后来，摆渡老人在孩子们的心中却是一个________________的形象。（2分） 24.“摆渡老人”中的“摆渡”可以理解为老人对“我们”生活、思想上的帮助。其实生活中还有许多这样的“摆渡者”，请你展开联想，按照下面的示例仿写两个句子。
齐晋鞌之战原文和译文
鞌之战选自《左传》又名《鞍之战》原文：楚癸酉，师陈于鞌(1)。邴夏御侯，逢丑父为右②。晋解张御克，郑丘缓为右(3)。侯日：“余姑翦灭此而朝食(4)”。不介马而驰之⑤。克伤于矢，流血及屦2 未尽∧6)，曰：“余病矣(7)！”张侯曰：“自始合(8)，而矢贯余手及肘(9)，余折以御，左轮朱殷(10)，岂敢言病吾子忍之！”缓曰：“自始合，苟有险，余必下推车，子岂＿识之(11)然子病矣！”张侯曰：“师之耳目，在吾旗鼓，进退从之。此车一人殿之(12)，可以集事(13)，若之何其以病败君之大事也擐甲执兵(14)，固即死也(15)；病未及死，吾子勉之(16)！”左并辔(17) ，右援拐鼓(18)。马逸不能止(19)，师从之，师败绩。逐之，三周华不注(20) 韩厥梦子舆谓己曰：“旦辟左右！”故中御而从齐侯。邴夏曰：“射其御者，君子也。”公曰：“谓之君子而射之，非礼也。”射其左，越于车下；射其右，毙于车中。綦毋张丧车，从韩厥，曰：“请寓乘。”从左右，皆肘之，使立于后。韩厥俛，定其右。逢丑父与公易位。将及华泉，骖絓于木而止。丑父寝于轏中，蛇出于其下，以肱击之，伤而匿之，故不能推车而及。韩厥执絷马前，再拜稽首，奉觞加璧以进，曰：“寡君使群臣为鲁、卫请，曰：‘无令舆师陷入君地。’下臣不幸，属当戎行，无所逃隐。且惧奔辟而忝两君，臣辱戎士，敢告不敏，摄官承乏。” 丑父使公下，如华泉取饮。郑周父御佐车，宛茷为右，载齐侯以免。韩厥献丑父，郤献子将戮之。呼曰：“自今无有代其君任患者，有一于此，将为戮乎”郤子曰：“人不难以死免其君，我戮之不祥。赦之，以劝事君者。”乃免之。译文1：在癸酉这天，双方的军队在鞌这个地方摆开了阵势。齐国一方是邴夏为齐侯赶车，逢丑父当车右。晋军一方是解张为主帅郤克赶车，郑丘缓当车右。齐侯说:“我姑且消灭了这些人再吃早饭。”不给马披甲就冲向了晋军。郤克被箭射伤，血流到了鞋上，但是仍不停止擂鼓继续指挥战斗。他说:“我受重伤了。”解张说:“从一开始接战，一只箭就射穿了我的手和肘，左边的车轮都被我的血染成了黑红色，我哪敢说受伤您忍着点吧！”郑丘缓说:“从一开始接战，如果遇到道路不平的地方，我必定（冒着生命危险）下去推车，您难道了解这些吗不过，您真是受重伤了。”daier 解张说:“军队的耳朵和眼睛，都集中在我们的战旗和鼓声，前进后退都要听从它。这辆车上还有一个人镇守住它，战事就可以成功。为什么为了伤痛而败坏国君的大事呢身披盔甲，手执武器，本来就是去走向死亡，伤痛还没到死的地步，您还是尽力而为吧。”一边说，一边用左手把右手的缰绳攥在一起，用空出的右手抓过郤克手中的鼓棰就擂起鼓来。（由于一手控马，）马飞快奔跑而不能停止，晋军队伍跟着指挥车冲上去，把齐军打得打败。晋军随即追赶齐军，三次围绕着华不注山奔跑。韩厥梦见他去世的父亲对他说:“明天早晨作战时要避开战车左边和右边的位置。”因此韩厥就站在中间担任赶车的来追赶齐侯的战车。邴夏说:“射那个赶车的，他是个君子。”齐侯说: “称他为君子却又去射他，这不合于礼。”daier 于是射车左，车左中箭掉下了车。又射右边的，车右也中箭倒在了车里。（晋军的）将军綦毋张损坏了自己的战车，跟在韩厥的车后说: “请允许我搭乗你的战车。”他上车后，无论是站在车的左边，还是站在车的右边，韩厥都用肘推他，让他站在自己身后——战车的中间。韩厥又低下头安定了一下受伤倒在车中的那位自己的车右。于是逢丑父和齐侯（乘韩厥低头之机）互相调换了位置。将要到达华泉时，齐侯战车的骖马被树木绊住而不能继续逃跑而停了下来。（头天晚上）逢丑父睡在栈车里，有一条蛇从他身子底下爬出来，他用小臂去打蛇，小臂受伤，但他（为了能当车右）隐瞒了这件事。由于这样，他不能用臂推车前进，因而被韩厥追上了。韩厥拿着拴马绳走到齐侯的马前，两次下拜并行稽首礼，捧着一杯酒并加上一块玉璧给齐侯送上去，
农田水利学试题及答案资料
农田水利学课程考试试题及答案 姓名 年级 专业 学号 一、名词解释（每小题 2 分 共 10 分） 1. 灌水率： 2. 排涝模数： 3. 平均排除法： 4. （排涝计算中的）设计内水位： 5. 容泄区： 二．选择题（共 10 分） 1. 灌溉设计标准是反映灌区效益达到某一水平的一个重要技术指标，一般以 （ ）与 （ ）表示？ A 、灌溉设计保证率、抗旱天数。 B 、水文年型、降水量。 C 、设计灌溉用水量全部获得满足的年数、抗旱天数。 D 、水源来水量、灌区需水量。 2. 什么叫田间渠系的灌排相邻布置？ （ ） 3. 渠道的输水损失包括以下四个部分： （ ） A 、干渠、支渠、斗渠及农渠等四级渠道的水量损失。 B 、渠床土质、地下水埋深、渠道的工作制度及输水时间。 C 、自由渗流、顶托渗流、渠床土质、与渠道的工作制度等。 D 、渠道水面蒸发损失、渠床渗漏损失、闸门漏水与渠道退水等。 4. 什么叫渠道水的利用系数？ （ ） A 、灌溉渠系的净流量与毛流量的比值。 B 、某一级渠道的净流量与毛流量的比值。 C 、田间实际灌入的有效水量与末级渠道的供水量之比。 D 、实际灌入农田的有效水量和渠首引入的水量之比。 5. 在渠道规划设计中，渠道最小流量有何作用？ （ ） A 、用以校核对下一级渠道的水位控制条件。 A 、灌溉渠道与排水沟道的规划布置。 B 、田间各级渠道规划布置的形 式。
B、用以校核渠道不淤条件。 C、用以确定修建节制闸的位置。 D、用以校核对下一级渠道的水位控制条件和确定修建节制闸的位置，并按最小流量验算渠道不淤条件。 6.什么叫雨水径流集蓄灌溉工程？（） A 、导引、收集雨水径流，并把它蓄存起来加以有效灌溉利用的工程技术措施。 B、田面、坡面、路面及屋面庭院等各类集水工程。 C、各类形式的水窖、水窑窖等蓄水工程。 D、各类最为有效节水的灌溉方式。 7.集流面的处理主要包括哪三类方法？（） A 、采用混凝土、水泥土、三七灰土进行表面处理。 B、采用塑料薄膜、或塑膜复沥青、复草泥。 C、植被管理；地表处理；化学处理。 D、采用钠盐、硅有机树脂及粗石蜡等化学处理方法。 8.蓄水工程有哪几种主要的类型？（） A 、引水渠沟或管道、入水口、拦污栅、沉沙槽、蓄水设施以及放水装置等。 B、涝池、旱井、田间蓄水池、水窖、水窑窖等。 C、引水渠、蓄水窑洞与放水暗管与放水暗渠。 D 、沉沙池、进水管、水窖等。 9.什么叫续灌方式？（） A 、类似于自来水管道可随机用水的供水方式。 B、输配水管道进行输水、配水和灌水的方式。 C、是指上一级管道按预先划分好的轮灌组分组向下一级管道配水的方式。 D 、是指上一级管道向所有的下一级管道同时配水的方式。 10.什么叫集水效率？（） A 、降水特征（次降雨量、降雨强度）和集水面质地、坡度、前期含水量与集水面尺寸。 B、集水面的处理材料、集水面积、集流路径和汇流时间。 C、随降水强度的增大而提高。 D、某时段内或某次降雨集水面的集水量占同一时期内的降雨量的比值。 三、简答题（每题6分，共30分） 1.四种地表取水方式的使用条件
摆渡老人的阅读答案
摆渡老人的阅读答案 摆渡老人 ①读中学时，学校设在邻村，与我们村有一条河隔着，便认识了那摆渡老人。 ②当时，那老人一直是我们取乐的对象。每到放学，肚子饿得直叫的我们便一窝蜂向渡口跑去，挤在那儿。渡船只要一近岸，大家便争先恐后地向船上跃去。虽然老人很有经验，未到岸边做好回撑的准备，但超载和落水的事仍然时有发生。超载时，船上的，岸上的，都望着老人手忙脚乱的样子大笑。夏天下雨时，大家总是把伞迎着风，看着渡船被吹淌到离渡口好远的地方，老人一番折腾，将船撑到渡口，我们便哄笑着上了岸。 ③最有意思的要数冬天。冬天很冷，河面结了一层厚厚的冰。每天早晨，当我们赶到渡口时，总看见老人在那边晃着船，好长时间，才把船晃过来。望着老人跳舞似的笨拙姿势，我们总是哄笑，一直到船靠岸。我们往往因此耽误了早读，但我们都挺愿意。于是就希望天再冷些，便可迟点起来，反正去早了也过不了河。我们都为找到一个偷懒的理由而感到高兴。当时的我们，是心忧读书愿天寒。 ④那一天特别冷，我因有事起了个早。吃过早饭后，便不紧不慢地向学校走去。到渡口时我惊奇地发现，那老人不
是我预料的那样正在跳舞，却已经到了这边。我和他攀谈起来。我问他：你今天怎么这么早?老人咳嗽了一声，缓缓地说：许是人老了，早上总起不来。昨天听学校说耽误了学生上早读，今天我就起了早。谁知等了好长时间，才来了你一个人。说完，便送我过河。 ⑤船到河心，老人忽然回头对我说道：你们小孩子可要好好读书啊。那双眼中竟满含着期待和爱意。我心中一辱，看着老人瘦弱的身体，沾满冰碴的白胡子，一双手冻得发紫，猛地问道：你干啥要吃这份苦呢?好久，老人长叹一声说道：干啥？哎，以前我也有一个可爱的儿子，他曾吵着要上学，但终究没上成。后来他得病去了，我一直感到对不起他，我在这儿摆渡，接送你们上学，心里好受些。谈话间，船已到岸。待我上岸后，老人又把船向河那边撑去。看着老人的背影，我不再感到笨拙可爱，鼻子有些酸酸的。 ⑥那天，我把老人的故事讲给同学们听了。以后，我们上船再也没有起哄。于是老人饱经风霜的脸上出现了笑容。 ⑦再后来，两岸之间架起了一座桥。最后一次过河时，老人对我们说：以后不再送你们过河了，你们要记住，好好读书。 ⑧好好读书，这句话一直留在我记忆中。 ⑨真想再见见摆渡老人。 13、本文具体描述了摆渡老人接送我们过河上学的辛劳。
《鞌之战》阅读答案(附翻译)原文及翻译
《鞌之战》阅读答案（附翻译）原文及翻 译 鞌之战[1] 选自《左传成公二年（即公元前589年）》 【原文】 癸酉，师陈于鞌[2]。邴夏御齐侯[3]，逢丑父为右[4]。晋解张御郤克，郑丘缓为右[5]。齐侯曰：余姑翦灭此而朝食[6]。不介马而驰之[7]。郤克伤于矢，流血及屦，未绝鼓音[8]，曰：余病[9]矣！张侯[10]曰：自始合，而矢贯余手及肘[11]，余折以御，左轮朱殷[12]，岂敢言病。吾子[13]忍之！缓曰：自始合，苟有险[14]，余必下推车，子岂识之[15]？然子病矣！张侯曰：师之耳目，在吾旗鼓，进退从之[16]。此车一人殿之[17]，可以集事[18]，若之何其以病败君之大事也[19]？擐甲执兵，固即死也[20]。病未及死，吾子勉之[21]！左并辔[22]，右援枹而鼓[23]，马逸不能止[24]，师从之。齐师败绩[25]。逐之，三周华不注[26]。 【注释】 [1]鞌之战：春秋时期的著名战役之一。战争的实质是齐、晋争霸。由于齐侯骄傲轻敌，而晋军同仇敌忾、士气旺盛，战役以齐败晋胜而告终。鞌：通鞍，齐国地名，在今山东济南西北。 [2]癸酉：成公二年的六月十七日。师，指齐晋两国军队。陈，
列阵，摆开阵势。 [3]邴夏：齐国大夫。御，动词，驾车。御齐侯，给齐侯驾车。齐侯，齐国国君，指齐顷公。 [4]逢丑父：齐国大夫。右：车右。 [5]解张、郑丘缓：都是晋臣，郑丘是复姓。郤（x ）克，晋国大夫，是这次战争中晋军的主帅。又称郤献子、郤子等。 [6]姑：副词，姑且。翦灭：消灭，灭掉。朝食：早饭。这里是吃早饭的意思。这句话是成语灭此朝食的出处。 [7]不介马：不给马披甲。介：甲。这里用作动词，披甲。驰之：驱马追击敌人。之：代词，指晋军。 [8] 未绝鼓音：鼓声不断。古代车战，主帅居中，亲掌旗鼓，指挥军队。兵以鼓进，击鼓是进军的号令。 [9] 病：负伤。 [10]张侯，即解张。张是字，侯是名，人名、字连用，先字后名。 [11]合：交战。贯：穿。肘：胳膊。 [12]朱：大红色。殷：深红色、黑红色。 [13]吾子：您，尊敬。比说子更亲切。 [14]苟：连词，表示假设。险：险阻，指难走的路。 [15]识：知道。之，代词，代苟有险，余必下推车这件事，可不译。 [16]师之耳目：军队的耳、目（指注意力）。在吾旗鼓：在我们
农田水利学试题六
农田水利学课程考试试题 姓名年级专业学号 一、名词解释（每小题2分共10分） 1、渠道设计流量： 2、灌溉水利用系数： 3、最小流量： 4、田间净流量:： 5、不冲流速： 二．单向选择题（共10分） 1.地下水临界深度是指? （） A、地下水埋藏深度。 B、在一定的自然条件和农业技术措施条件下，为了保证土壤不产生渍害，所要求保持的地下水最小埋深。 C、在一定的自然条件和农业技术措施条件下，为了保证土壤不产生盐碱化和作物不受盐害，所要求保持的地下水最小埋深。 D、在一定的自然条件和农业技术措施条件下，为了保证土壤不产生盐碱化和作物不受盐害，所要求保持的地下水最大埋深。 2.对于控制一定地下水位要求的农田排水系统，下列哪种说法是正确的？（） A、在同一排水沟深度的情况下，排水沟的间距愈大，地下水位下降速度愈快，在一定时间内地下水位的下降值愈大，在规定时间内地下水位的下降值也愈大。 B、在允许的时间内要求达到的地下水埋藏深度ΔH一定时，排水沟的间距愈大，需要的深度也愈大。 C、在允许的时间内要求达到的地下水埋藏深度ΔH一定时，排水沟的间距愈小，需要的深度也愈大。 D、在同一排水沟间距的情况下，排水沟的深度愈小，地下水位下降速度愈快，在一定时间内地下水位的下降值愈大，在规定时间内地下水位的下降值也愈大。
3.设计排涝标准时，需选择发生一定重现期的暴雨，一般选择标准是？（） A、1-5年。 B、5-10年。 C、10-15年。 D、15-20年。 4.对渍害最不敏感的作物是？（） A、小麦； B、玉米； C、高粱； D、水稻。 5.特别适宜防治土壤次生盐碱化的农田排水方式是？（） A、明沟排水； B、竖井排水； C、暗管排水； D、暗沟排水。 6.在进行排水沟设计时，用来校核排水沟的最小流速的设计流量是？（） A、排涝设计流量； B、排渍设计流量； C、日常排水设计流量； D、排涝模数。 7.农田长期渍水不会造成下列后果？ A、土壤的透气性很差。 B、土层都处于强烈的氧化状态。 C、利于硫化氢等硫化物的形成，对作物根系产生永久性伤害。 D、有机质矿化程度低，分解释放的有效养分少，不能满足作物生长的需要。 8.防治土壤盐碱化的水利技术不包括？（） A、明沟排水 B、井灌井排 C、灌水冲洗 D、放淤改良 9.什么叫计划用水？（） A、灌溉水量的分配方法。 B、就是按作物的需水要求与灌溉水源的供水情况，结合渠系工程状况，有计划地蓄水、引水、配水与灌水。 C、是指灌溉水在灌区各级渠系调配、管理的方式。 D、是指灌溉水通过各级渠道流入田间的方法。 10.灌区用水计划一般来说有哪四种主要类型？（） A、水源引水计划、渠系配水计划与田间的用水计划等。 B、水权集中、统筹兼顾、分级管理、均衡受益。 C、年度轮廓用水计划、某灌季全渠系用水计划、干支渠段用水计划及用水单位的用水计划。 D、上下结合、分级编制，统一调度、联合运用。
摆渡老人阅读题答案
摆渡老人阅读题答案 读中学时学校与我们村隔着一条河由此认识了那摆渡老人。当时那老人一直是我们取乐的对象。每到放学肚子饿得直叫的我们便一窝蜂向渡口跑去挤在那儿。渡船只要一靠近岸大家便争先恐后地向船上跃去。虽然老人很有经验来到岸边便做好回撑的准备但超载和落水的事仍然时有发生。超载时船上的岸上的都望着老人手忙脚乱的样子大笑。夏天下雨时大家总是把伞迎着风看着渡船被吹到离渡口好远的地方老人一番折腾将船撑到渡口我们便哄笑着上了岸。 最有意思的要数冬天。冬天很冷河面结了一层厚厚的冰。每天早晨当我们赶到渡口时总看见老人在那边摇着船好长时间才把船摇过来。一直到船靠岸。我们往往因此耽误了早读但我们都挺愿意。于是就希望天再冷些便可迟点起来反正去早了也过不了河。我们都为找到一个偷懒的理由而感到高兴。当时的我们是“心忧读书愿天寒”。 那一天特别冷我因有事起了个早。吃过早饭后便不紧不慢地向学校走去。到渡口时我惊奇地发现那老人不是我预料的那样正在“跳舞”却已经到了这边。我和他攀谈起来。我问他“你今天怎么这么早”老人咳嗽了一声缓缓地说“许是人老了早上总起不来。昨天听学校说耽误了学生上早读今天我就起了早。谁知等了好长时间才来了你一个人。”说完便送我过河。 船到河心老人忽然回头对我说“你们小孩子可要好好读书啊。”那双眼睛竟满含着期待和爱意。我心中一震看着老人疲弱的身体沾满水碴?6?7碴á?6?8小碎块。的白胡子一双冻得发紫的手猛地问到“你干啥要受这份苦呢”好久老人长叹一声说道“干啥哎以前我也有一个可爱的儿子他曾吵着要上学但终究没上成。后来他得病去了我一直感到对不起他。我在这儿摆渡接送你们上学心里好受些。”谈话间船已到岸。待我上岸后老人又把船向河那边撑去。看着老人的背影我不再感到笨拙可笑鼻子有些酸酸的。 那天我把老人的故事讲给同学们听了。以后我们乘船再也没有起哄。于是老人饱经风霜的脸上现出了笑容。 再后来两岸之间架起了一座桥。最后一次过河时老人对我们说“以后不再送你们过河了你们要记住好好读书啊。” “好好读书”这句话一直留在我的记忆中。 真想再见见摆渡老人。 1本文具体描述了摆渡老人接送“我们”过河上学的辛劳。先写老人因“我们”的______而手忙脚乱一番折腾再写因 ______老人好不容易“才把船晃过来”最后写老人不顾自己________特地起大早送“我们”过河上学。用概括的语言填空 2第⑤段中摆渡老人说“我在这儿摆渡接送你们上学心里好受些”。请你用自己的语言简要说说为什么老人觉得这样做“心里好受些”。分析摆渡老人的心理答______________________________________________________ 3第⑤段中加粗的词句表现了“我”当时_______和_______的心情。 4原先摆渡老人在孩子们的眼里是一个______的形象后来摆渡老人在孩子们的心中却是一个______的形象。用自己的话填空5“摆渡老人”中的“摆渡”可以理解为老人对“我们”生活、思想上的帮助。其实生活中还有许多这样的“摆渡者”请你展开联想按照下面的示例仿写三个句子。新题型示例1作家把读者摆渡到精神的彼岸。 训练联想和表达能力 2 “希望工程”把贫穷的孩子摆渡到求知的校园。 仿句1______________________________________________________
《鞌之战》阅读答案附翻译
《鞌之战》阅读答案（附翻译） 《鞌之战》阅读答案（附翻译） 鞌之战[1] 选自《左传·成公二年（即公元前589年）》 【原文】 癸酉，师陈于鞌[2]。邴夏御齐侯[3]，逢丑父为右[4]。晋解张御郤克，郑丘缓为右[5]。齐侯曰：“余姑 翦灭此而朝食[6]。”不介马而驰之[7]。郤克伤于矢， 流血及屦，未绝鼓音[8]，曰：“余病[9]矣！”张侯[10]曰：“自始合，而矢贯余手及肘[11]，余折以御，左轮 朱殷[12]，岂敢言病。吾子[13]忍之！”缓曰：“自始合，苟有险[14]，余必下推车，子岂识之[15]？——然 子病矣！”张侯曰：“师之耳目，在吾旗鼓，进退从之[16]。此车一人殿之[17]，可以集事[18]，若之何其以 病败君之大事也[19]？擐甲执兵，固即死也[20]。病未 及死，吾子勉之[21]！”左并辔[22]，右援枹而鼓[23]，马逸不能止[24]，师从之。齐师败绩[25]。逐之，三周 华不注[26]。 【注释】 [1]鞌之战：春秋时期的著名战役之一。战争的实质是齐、晋争霸。由于齐侯骄傲轻敌，而晋军同仇敌忾、 士气旺盛，战役以齐败晋胜而告终。鞌：通“鞍”，齐
国地名，在今山东济南西北。 [2]癸酉：成公二年的六月十七日。师，指齐晋两国军队。陈，列阵，摆开阵势。 [3]邴夏：齐国大夫。御，动词，驾车。御齐侯，给齐侯驾车。齐侯，齐国国君，指齐顷公。 [4]逢丑父：齐国大夫。右：车右。 [5]解张、郑丘缓：都是晋臣，“郑丘”是复姓。郤（xì）克，晋国大夫，是这次战争中晋军的主帅。又称郤献子、郤子等。 [6]姑：副词，姑且。翦灭：消灭，灭掉。朝食：早饭。这里是“吃早饭”的意思。这句话是成语“灭此朝食”的出处。 [7]不介马：不给马披甲。介：甲。这里用作动词，披甲。驰之：驱马追击敌人。之：代词，指晋军。 [8]未绝鼓音：鼓声不断。古代车战，主帅居中，亲掌旗鼓，指挥军队。“兵以鼓进”，击鼓是进军的号令。 [9]病：负伤。 [10]张侯，即解张。“张”是字，“侯”是名，人名、字连用，先字后名。 [11]合：交战。贯：穿。肘：胳膊。 [12]朱：大红色。殷：深红色、黑红色。 [13]吾子：您，尊敬。比说“子”更亲切。
扬州大学农科农田水利学计算题过程及答案
农田水利学计算题 1、某小型灌区作物单一为葡萄，某次灌水有600亩需灌水，灌水定额为25m 3/亩，灌区灌溉水利用系数为0.75，试计算该次灌水的净灌溉用水量和毛灌溉用水量。 解：W 毛=MA/η水=25*600/0.75=20000（m3） W 净=MA=25*600=15000（m3） 2、某灌区A =0.2万亩，A 蔬菜=0.16万亩，A 花卉=0.04万亩，m 蔬菜=20m 3/亩，m 花卉=15m 3/亩。求综合净灌水定额m 综及净灌溉用水量。 解：m 综=α1m 1+α2m 2=0.8*20+0.2*15=16+3=19（m3/亩） W=m 综*A=19*2000=38000（m3） 3、某小型提水灌区，作物均为果树，面积1000亩，用水高峰期最大灌水定额为25m3/亩，灌溉水利用系数为0.75，灌水延续4天，每天灌水20小时。试计算水泵设计流量。 解：Q 设= =25*1000/(3600*20*4*0.75)=0.12（m3/s ） 4、已知苏南某圩区，F=3.8Km2，其中旱地占20%，水田占80%。水田日耗水e=5mm/d ，水田滞蓄30mm ，旱地径流系数为0.6 。排涝标准采用1日暴雨200mm ，2天排除，水泵每天工作时间22小时。试求泵站设计排涝流量Q 和综合设计排涝模数q 。 解：R 水田=P-h 田蓄-eT=200-30-5*2=160（mm ） 水ηTt A m t T W Q k j j ij i i i 360036001∑==??=
R旱田=αP=0.6*200=120（mm） ∴Q=(R水田F水田+R旱田F旱田)/3.6Tt =(160*3.8*0.8+120*3.8*0.2)/3.6*2*22 =3.65(m3/s) ∴q =Q/F=3.65/3.8=0.96(m3/km2) 5、冬小麦播前土壤最大计划湿润层深度为0.6m，土壤平均孔隙率42.5%（占土壤体积百分比），土壤田间持水率为70%（孔隙百分比）。播前土壤含水率为45.0%（孔隙百分比）。计算冬小麦的播前灌水定额。解：M=667Hn(θmax-θ0)=667*0.6*0.425*(0.7-0.45) =42.5(m3/亩) 6、已知某渠系如图1-4-3所示，干、支渠采用续灌，设计灌水率q=0.78m3/（s·万亩），一支灌溉面积为2万亩，二支灌溉面积为2.4万亩，三支灌溉面积2.5万亩，支渠的灌溉水利用系数为0.82，干渠渠道水利用系数0.9。 【要求】 （1）计算各支渠设计流量； （2）计算干渠设计流量和灌区灌溉水利用系数。 解：（1）Q1=qA1/η支水=0.78*2/0.82=1.90(m3/s) Q2=qA2/η支水=0.78*2.4/0.82=2.28(m3/s) Q3=qA3/η支水=0.78*2.5/0.82=2.38(m3/s) （2）η水=η支水*η干=0.82*0.9=0.74 7、1）下图渠系干、支渠续灌
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