• 什么可以代替cellpadding，cellspacing，valign和HTML5表中的align？（什么能代替excel）
• Artificial Intelligence Pathfinding
• Cellpadding和Cellspacing之间的区别
• cellpadding和cellspacing有哪些区别
• cellpadding和cellspacing的区别

什么可以代替cellpadding，cellspacing，valign和HTML5表中的align？（什么能代替excel）

在 Visual Studio中 ，我看到以下警告：

• 验证（HTML 5）：属性“ cellpadding”不是元素“ table”的有效属性。
• 验证（HTML 5）：属性“ cellspacing”不是元素“ table”的有效属性。
• 验证（HTML 5）：属性“ valign”不是元素“ td”的有效属性。
• 验证（HTML 5）：属性“ align”不是元素“ td”的有效属性。
  

如果它们不是 HTML5中的 有效属性，那么用CSS替换它们是什么？

答案1

/* cellpadding */th, td { padding: 5px; }/* cellspacing */table { border-collapse: separate; border-spacing: 5px; } /* cellspacing="5" */table { border-collapse: collapse; border-spacing: 0; }   /* cellspacing="0" *//* valign */th, td { vertical-align: top; }/* align (center) */table { margin: 0 auto; }

Artificial Intelligence Pathfinding

School of Computer Science
The University of Adelaide
Artificial Intelligence
Assignment 1
Semester 1 2022
Due 11:59pm Wednesday 23 March 2022
1 Pathfinding
Pathfinding is the problem of finding a path between two points on a plane. It is a
fundamental task in robotics and AI. Perhaps the most obvious usage of pathfinding is
in computer games, when an object is instructed to move from its current position to a
goal position, while avoiding obstacles (e.g., walls, enemy fire) along the way.
Pathfinding in commercial games is frequently accomplished using search algorithms1.
We consider a simplified version in this assignment. The following shows a 2D map
drawn using ASCII characters:
1 1 1 1 1 1 4 7 8 X
1 1 1 1 1 1 1 5 8 8
1 1 1 1 1 1 1 4 6 7
1 1 1 1 1 X 1 1 3 6
1 1 1 1 1 X 1 1 1 1
1 1 1 1 1 1 1 1 1 1
6 1 1 1 1 X 1 1 1 1
7 7 1 X X X 1 1 1 1
8 8 1 1 1 1 1 1 1 1
X 8 7 1 1 1 1 1 1 1
Given a start position and an end position on the map, our aim is to find a path from the
start position to the end position. The character ‘X’ denotes an obstacle that cannot be
traversed by a path, while the digits represent the elevation at the respective positions.
Any position is indicated by the coordinates (i, j), where i is the row number (ordered
top to bottom) and j is the column number (ordered left to right). For example, the
1http://theory.stanford.edu/~a...
Semester 1 2022 Page 1 by Tat-Jun Chin
top left position is (1, 1), the bottom right is (10, 10), while the position with elevation
‘3’ is (4, 9). Given start position (1, 1) and end position (10, 10), a possible path is

• • • 1 1 1 4 7 8 X
• 1 * 1 1 1 1 5 8 8
• 1 * 7
• 1 1 1 1 X 1 1 * 6
• 1 1 1 1 X 1 1
• 1 1 1 1 1 1 * 1 1
• 1 1 1 1 X 1 *
• 7 1 X X X 1 1 1 *
• 8 1 1 1 1 1 1 1 *
  X 8 7 1 1 1 1 1 1 *
  Note that we use 4-connectedness for paths, which means any two points on the path
  are only connected if they are vertically or horizontally (but not diagonally!) adjacent.
  1.1 Problem formulation
  Following the lecture notes, we formulate the problem as follows:
  • States: Any obstacle-free position (i, j) on the map.
  • Initial state: A position (i0, j0) given by the user.
  • Actions: Since we consider 4-connectedness, only four actions are available: Up,
  down, left and right (your program must expand each node in this order).
  Available actions for positions adjacent to the map boundary or obstacles are
  reduced accordingly.
  • Transition model: Moving left moves the current position to the left, etc.
  • Goal test: Check if the current state is the end position (i∗, j∗) given by the user.
  • Path cost: Given a map M and a path P = {(i0, j0),(i1, j1), . . . ,(iN , jN )}, the
  cost of the path is calculated as
  g(P) = XNk=1
  c(ik∥1, jk∥1, ik, jk, M),
  Semester 1 2022 Page 2 by Tat-Jun Chin
  where
  c(a, b, c, d, M) = (
• • M(c, d) ⇒ M(a, b) if M(c, d) ⇒ M(a, b) > 0
• otherwise
  and M(a, b) is the elevation at position (a, b). In words, the cost of a path is the
  sum of the costs between two adjacent points of the path, and the cost between
  adjacent points is 1 plus the difference between the elevation of the two points if
  we climb “uphill”, or simply 1 if we stay “level” or slide “downhill”.
  This means shorter paths which avoid climbing cost less. As an example, the cost
  in the path in the previous page is 25. What is the optimal (cheapest) path?
  1.2 Your tasks
  Solve pathfinding using Breadth-First Search (BFS), Uniform-Cost Search (UCS) and
  A* Search. You should base your program on the pseudocode GRAPH-SEARCH in the
  lecture slides, and carefully think about the appropriate data structures to use. For A*
  Search, you must implement two heuristics:
  • Euclidean distance between current position and end position.
  • Manhattan distance between current position and end position.
  For the map in Page 1 with start position (1, 1) and end position (10, 10), your program
  should help you answer these questions:
• Are the paths returned by the three methods different?
• What about the optimality of the returned paths?
• Which method is the most computationally and memory efficient?
• Do the two heuristics for A* Search provide different solutions?
• Does checking for repeated states matter in this problem?
  Semester 1 2022 Page 3 by Tat-Jun Chin
  1.3 Deliverables
  Write your pathfinding program in Python 3 in a file called pathfinder.py. Your
  program must be able to be run as follows:
  $ python pathfinder.py [map] [algorithm] [heuristic]
  The inputs/options to the program are as follows.
  • [map] specifies the path to map, which is a text file formatted according to this
  example (see next page):
• 10
• 1
• 10
• 1 1 1 1 1 4 7 8 X
• 1 1 1 1 1 1 5 8 8
• 1 1 1 1 1 1 4 6 7
• 1 1 1 1 X 1 1 3 6
• 1 1 1 1 X 1 1 1 1
• 1 1 1 1 1 1 1 1 1
• 1 1 1 1 X 1 1 1 1
• 7 1 X X X 1 1 1 1
• 8 1 1 1 1 1 1 1 1
  X 8 7 1 1 1 1 1 1 1
  The first line indicates the size of the map (rows by columns), while the second
  and third line represent the start and end positions respectively. The map data
  then follows, where all elevation values are integers from 0 to 9 inclusive.
  • [algorithm] specifies the search algorithm to use, with the possible values of bfs,
  ucs, and astar. • [heuristic] specifies the heuristic to use for A* search, with the possible values
  of euclidean and manhattan. This input is ignored for BFS and UCS.
  Your program must then print to standard output the path returned by the
  search algorithm, in the following format:
  Semester 1 2022 Page 4 by Tat-Jun Chin
• • • 1 1 1 4 7 8 X
• 1 * 1 1 1 1 5 8 8
• 1 * 7
• 1 1 1 1 X 1 1 * 6
• 1 1 1 1 X 1 1
• 1 1 1 1 1 1 * 1 1
• 1 1 1 1 X 1 *
• 7 1 X X X 1 1 1 *
• 8 1 1 1 1 1 1 1 *
  X 8 7 1 1 1 1 1 1 *
  where the path is indicated by asterisks ‘*’ superimposed on the original map beginning
  from the start position and leading to the end position. Do not include extraneous
  spaces or other characters in the output.
  If the given map or problem does not have a feasible path, your program must print
  null
  Again, do not include extraneous spaces or other characters in the output.
  1.3.1 Python libraries
  You are allowed to use NumPy to write your pathfinding program. The marking program
  will not be able to run your program to completion if other Python libraries are used.
  1.4 Submission
  You must submit your program files on Gradescope. Instructions on accessing Grade￾scope and submitting assignments are provided at https://help.gradescope.com/
  article/5d3ifaeqi4-student-canvas. Please use the course code X3ZJZE to en￾rol into the course. For undergraduates, please submit your pathfinding program
  (pathfinder.py) to Assignment 1 - Undergraduates. If there are any questions or
  issues with Gradescope, please contact Andrew via email at andrew.du@adelaide.edu.au.
  1.5 Assessment
  I will compile and run your code on several test problems. If it passes all tests you will
  get 15% (undergrads) or 12% (postgrads) of the overall course mark.
  There will be no further manual inspection/grading of your program to award marks
  on the basis of coding style, commenting or “amount of code written.
  Semester 1 2022 Page 5 by Tat-Jun Chin
  1.6 Using other source code
  You may not use other source code for this assignment. You should personally and
  carefully implement the search algorithms to fully understand the concept.
  1.7 Due date and late submission policy
  This assignment is due by 11:59pm Wednesday 23 March 2022. If your submission is
  late, the maximum mark you can obtain will be reduced by 25% per day (or part thereof)
  past the due date or any extension you are granted.
  Continues next page for postgraduate section.
  Semester 1 2022 Page 6 by Tat-Jun Chin
• Pathfinding by direct optimisation
  For postgraduate students, completing this section successfully will give you the remain￾ing 3% of the marks.
  Here we shall attempt to directly optimise the path instead of step-by-step search￾ing. We consider the simulated annealing algorithm shown in Algorithm 1. For more
  background on simulated annealing, see Section 4.1 of Russell and Norvig (3rd ed.).
  Algorithm 1 Simulated annealing for path optimisation
  1: input Initial path P0
  , initial temperature Tini, final temperature Tf in, cooling rate
  α, segment length d.
  2: output Optimised path P.
  3: Initialise T ← Tini, P ← P0.
  4: while T > Tf in do
  5: Ph ← rand-local-adjust(P, d).
  6: ∆g ← g(P) ⇒ g(Ph)
  7: if ∆g > 0 then
  8: P ← Ph
  9: else
  10: With probability e∆g/T
  , P ← Ph.
  11: end if / Record T and g(P) here for bookkeeping./
  12: T ← αT
  13: end while
  14: return P
  The algorithm receives as input a feasible (but non-optimal) path joining a start
  position and an end position on a map. The core idea is to iteratively perform random
  local adjustments to the path, and accept the new path if the adjustments improve the
  path cost (defined in Sec. 1.1), or accept it probabilistically if the cost is not improved.
  The process is repeated until the annealing temperature T falls below a small value
  Tf in given by the user. The temperature reduction is conducted as T = αT , where
• < α < 1 is the cooling rate (also supplied by the user). See Section 4.1 of Russell and
  Norvig (3rd ed.) for more details.
  The main body of the algorithm is conceptually simple — the hardest part is the rou￾tine to perform the random adjustments. Fortunately we can rely on the BFS program
  written in the previous section. The method is shown in Algorithm 2.
  Note that the adjustments cannot make the path infeasible, i.e., any resulting path
  still joins the original start position and end position required by the user.
  Semester 1 2022 Page 7 by Tat-Jun Chin
  Algorithm 2 Make random local adjustment on path
  1: input Path P, segment length d.
  2: output Adjusted path Ph.
  3: Random pick a point (u, v) on P.
  4: Pick as (x, y) the point of d positions away from (u, v) along P towards the end
  position. If such a point does not exist, use the end position for (x, y).
  5: Find a random path S joining (u, v) and (x, y) using randomised BFS (see text
  below).
  6: Replace path segment in P between (u, v) and (x, y) with S. Store new path as Ph.
  7: return Ph.
  To perform randomised BFS, only a minor tweak to the original BFS algorithm is
  required — the order of actions for expanding each node in the search tree is randomised
  every time. For example, while in Sec. 1.1 the order is fixed as UDLR (up, down, left,
  right), we randomise this at every instance to be LURD, DLUR, etc. The following
  shows randomised adjustments with d = 5, and (u, v) = (8, 1) and (x, y) = (10, 4).
• 1 8 1 1 2 4 7 8 X 1 8 1 1 2 4 7 8 X 1 8 1 1 2 4 7 8 X
• 1 1 5 1 5 1 5 8 8 1 1 5 1 5 1 5 8 8 1 1 5 1 5 1 5 8 8
• 4 2 2 1 6 1 4 6 7 4 2 2 1 6 1 4 6 7 4 2 2 1 6 1 4 6 7
• 5 1 7 0 3 5 1 1 6 5 1 7 0 3 5 1 1 6 5 1 7 0 3 5 1 1 6
• 7 8 1 2 6 8 1 5 1 7 8 1 2 6 8 1 5 1 7 8 1 2 6 8 1 5 1
• 7 4 1 1 4 2 2 4 2 7 4 1 1 4 2 2 4 2 7 4 1 1 4 2 2 4 2
• 5 1 2 1 2 7 5 1 6 5 1 2 1 2 7 5 1 6 5 1 2 1 2 7 5 1 6
• 7 1 3 4 2 0 4 2 1 1 3 4 2 0 4 2 1 * 7 1 3 4 2 0 4 2 1
• • 1 1 1 5 1 1 9 1 8 1 1 1 5 1 1 9 1 * 1 5 1 1 9 1
    X X X 8 7 *
    2.1 Your tasks
    Implement simulated annealing for path optimisation. As a sanity check, test your
    program on the following map with start position (1, 1) and end position (10, 10), with
    the initial path given by your (deterministic) BFS method from the previous section.
• 1 8 1 1 2 4 7 8 X
• 1 1 5 1 5 1 5 8 8
• 4 2 2 1 6 1 4 6 7
• 5 1 7 0 3 5 1 1 6
  Semester 1 2022 Page 8 by Tat-Jun Chin
• 7 8 1 2 6 8 1 5 1
• 7 4 1 1 4 2 2 4 2
• 5 1 2 1 2 7 5 1 6
• 7 1 3 4 2 0 4 2 1
• 8 1 1 1 5 1 1 9 1
  X 8 7 1 3 1 7 1 0 0
  Use parameter values Tini = 10, Tf in = 0.001, α = 0.99, and d = 5. Your program
  should help you answer the following questions:
• Does simulated annealing find the optimal path every time?
• How important is it to be able to accept an inferior path? Investigate by disabling
  Step 10 in Algorithm 1.
• How sensitive is the performance to the parameter settings? Investigate by chang￾ing the values of Tini, Tf in, α and d.
  2.2 Deliverables
  Write your simulated annealing pathfinder program in Python 3 in a file called sapathfinder.py.
  Your program must be able to be run as follows:
  $ python sapathfinder.py [map] [init] [tini] [tfin] [alpha] [d]
  The test program will assume that you would use the same programming language as in
  Sec. 1, and that you have a working program (pathfinder.py) for the tasks in Sec. 1.
  The inputs/options to the program are as follows.
  • [map] specifies the path to a map, formatted according to Sec. 1.3.
  • [init] specifies the path to an initial path, encoded according to the output of
  the program in Sec. 1.3.
  • [tini] and [tfin] specifies the initial and final temperature respectively.
  • [alpha] specifies the cooling rate.
  • [d] specifies the segment length for random local path adjustments.
  Your program must then print to standard output the optimised path, as well as
  the evolution of the temperature and path cost, in the manner of this example:
  Semester 1 2022 Page 9 by Tat-Jun Chin
• 1 8 1 1 2 4 7 8 X
• 1 1 5 1 5 1 5 8 8
• 4 2 2 1 6 1 4 6 7
• 5 1 7 0 3 5 1 1 6
• 7 8 1 2 6 8 1 5 1
• 7 4 1 1 4 2 2 4 2
• • • • 1 2 7 5 1 6
• 7 1 * 4 2 0 4 2 1
• 8 1 * 1 5 1 1 9 1
  X 8 7 *
  T = 10.000000, cost = 38
  T = 9.900000, cost = 44
  T = 9.801000, cost = 42
  ...
  T = 5.151371, cost = 40
  T = 5.099857, cost = 40
  T = 5.048859, cost = 41
  ...
  T = 0.001014, cost = 23
  T = 0.001004, cost = 23
  Do not include extraneous spaces or other characters in the output.
  Submit your program in the same way as the submission for Sec. 1. For postgradu￾ates, please submit your pathfinding programs (pathfinder.py and sapathfinder.py)
  to Assignment 1 - Postgraduates. The due date, late submission policy and code
  reuse policy are also the same as in Sec. 1.
  ∼∼∼ The End ∼∼∼
  Semester 1 2022 Page 10 by Tat-Jun Chin
  WX：codehelp

Cellpadding和Cellspacing之间的区别

cellpadding和cellspacing是table表格中使用的属性，可以设置table表格单元格中的空白，是两个有助于控制网页中表格的样式和布局的属性。那么它们之间有什么区别？下面本篇文章就来带大家简单比较一下，介绍cellpadding和cellspacing之间的区别，希望对大家有所帮助。

Cellpadding属性

在内容和外围边界之间设置单元格空间的属性称为cellpadding。换句话说，通过cellpadding属性管理单元格所围绕的内容与单元格边界之间的距离。此距离的单位可以隐含在HTML中的像素或百分比。

根据用户的需要，cellpadding属性值也可以是0值。它用于将文本彼此分开，从而改善外观。

简单示例：

<!DOCTYPE html>
<html>
<head>
    <meta charset="UTF-8">
    <title>Cellpadding属性的简单示例</title>
    <style>
    table{
    font-size: 20px;
    width: 100%;
    text-align: center;
    }
    td {background-color:lavender;}
    </style>
</head>
<body>
<table border = "2" cellpadding = "20">
<tbody>
<tr><td>内容</td><td>内容</td><td>内容</td></tr>
<tr><td>内容</td><td>内容</td><td>内容</td></tr>
<tr><td>内容</td><td>内容</td><td>内容</td></tr>
</tbody>
</table>
</body>
</html>

输出：

注：较旧版本的HTML支持cellpadding命令，但更高版本的HTML5不允许使用cellpadding，因此替代CSS用于在需要时提供相同的格式。

Cellspacing属性

Cellspacing属性可以控制表格中单个单元格之间的距离。通过使用此属性，设计人员可以轻松地改变不同相邻单元格边缘之间的空间。

Cellspacing属性主要用于网页设计。不同的语言对于cellspacing具有完全不同的语法，例如在CSS中使用“border-spacing”属性，而在HTML中使用“cellspacing”。

简单示例：

<!DOCTYPE html>
<html>
<head>
    <meta charset="UTF-8">
    <title>Cellspacing属性的简单示例</title>
    <style>
    table{
    font-size: 20px;
    width: 100%;
    text-align: center;
    }
    td {background-color:lavender;}
    </style>
</head>
<body>
<table border = "2" cellspacing = "20">
<tbody>
<tr><td>内容</td><td>内容</td><td>内容</td></tr>
<tr><td>内容</td><td>内容</td><td>内容</td></tr>
<tr><td>内容</td><td>内容</td><td>内容</td></tr>
</tbody>
</table>
</body>
</html>

输出：

总结：

cellpadding和cellspacing之间的主要区别在于cellpadding用于固定单元格边缘与其内容之间的宽度，即在文本和单元格边框之间的单元格内创建空白。而，cellspacing可以用于管理表格中单个单元格之间的空间。

以上就是本篇文章的全部内容，希望能对大家的学习有所帮助。更多精彩内容大家可以关注php中文网相关教程栏目！！！

以上就是Cellpadding和Cellspacing之间的区别的详细内容，更多请关注php中文网其它相关文章！

cellpadding和cellspacing有哪些区别

cellpadding和cellspacing的区别：1、影响范围；2、作用对象；3、属性值。详细介绍：1、影响范围，cellpadding影响单元格内部内容与单元格边框之间的距离，即单元格内的空白空间，cellspacing影响相邻单元格之间的间距，即单元格之间的空白空间；2、作用对象，cellpadding作用于单元格内部，用于控制单元格内部内容与边框的距离等等。

`cellpadding`和`cellspacing`是HTML表格元素中的两个属性，它们用于控制表格中的单元格之间的间距和单元格内部内容与单元格边框之间的距离。虽然它们在过去的HTML版本中经常使用，但在HTML5中已经不再建议使用，因为更现代的CSS样式可以更灵活地控制表格的外观和布局。以下是关于这两个属性的详细解释和区别：

一、`cellpadding`（单元格内边距）：

`cellpadding`属性用于定义单元格内部内容与单元格边框之间的距离，也就是单元格内的空白区域。这个属性通常用于调整单元格内部的文本或内容与单元格边框之间的距离。以下是有关`cellpadding`的一些关键点：

1.属性值：`cellpadding`的属性值可以是像素（px）或百分比（%）。如果你设置`cellpadding="10"`，那么单元格内的内容将在其边框内部保留10像素的空白空间。

2.影响范围：`cellpadding`只会影响单元格内部的空白区域，不会影响单元格之间的间距。

3.用途：`cellpadding`通常用于控制单元格内部文本或内容与单元格边框之间的间距，以使表格看起来更美观。这可用于增加单元格内的空白，以改善内容的可读性或美观性。

二、`cellspacing`（单元格间距）：

`cellspacing`属性用于定义表格中相邻单元格之间的间距，也就是控制单元格之间的空白空间。这个属性通常用于调整表格的整体外观，以使表格行和列之间具有一定的间距。以下是有关`cellspacing`的一些关键点：

1.属性值：`cellspacing`的属性值也可以是像素（px）或百分比（%）。如果你设置`cellspacing="5"`，那么相邻单元格之间将有5像素的间距，这将使表格看起来更加分隔。

2.影响范围：`cellspacing`只会影响相邻单元格之间的间距，不会影响单元格内部的内容和边框。

3.用途：`cellspacing`通常用于控制整个表格的外观，以增加表格行和列之间的间距，使表格看起来更整洁和易读。

三、区别总结：

`cellpadding`和`cellspacing`之间的主要区别可以总结如下：

1.影响范围：

-`cellpadding`影响单元格内部内容与单元格边框之间的距离，即单元格内的空白空间。

-`cellspacing`影响相邻单元格之间的间距，即单元格之间的空白空间。

2.作用对象：

-`cellpadding`作用于单元格内部，用于控制单元格内部内容与边框的距离。

-`cellspacing`作用于整个表格，用于控制相邻单元格之间的间距。

3.属性值：

-两者都可以接受像素（px）或百分比（%）作为属性值，以定义距离或间距的大小。

需要注意的是，虽然`cellpadding`和`cellspacing`在过去的HTML中非常常见，但在现代web开发中，使用CSS更为常见和灵活，因为CSS允许更精细地控制表格的外观和布局。所以，建议尽量避免使用这两个属性，而是使用CSS样式来定制表格的样式和间距，以实现更好的可维护性和可扩展性。

以上就是cellpadding和cellspacing有哪些区别的详细内容，更多请关注php中文网其它相关文章！

cellpadding和cellspacing的区别

cellpadding和cellspacing的区别：1、cellpadding属性用于设置单元格内容与单元格边框之间的距离，而cellspacing属性用于设置相邻单元格之间的距离；2、cellpadding可以为单元格内的内容提供一定的空白区域，使内容与边框之间有一定的间隔，而cellspacing控制了相邻单元格之间的空白区域大小，使表格看起来更加整齐等等。

cellpadding和cellspacing是HTML表格中的两个属性，用于控制表格中单元格之间的间距。它们的区别如下：

cellpadding（单元格内边距）：

cellpadding属性用于设置单元格内容与单元格边框之间的距离。

它可以为单元格内的内容提供一定的空白区域，使内容与边框之间有一定的间隔。

通过设置cellpadding属性的值，可以调整单元格内部内容与单元格边框之间的距离。

cellspacing（单元格间距）：

cellspacing属性用于设置相邻单元格之间的距离。

它控制了相邻单元格之间的空白区域大小，使表格看起来更加整齐。

通过设置cellspacing属性的值，可以调整相邻单元格之间的间距大小。

综上所述，cellpadding属性主要用于控制单元格内部内容与单元格边框之间的距离，而cellspacing属性主要用于控制相邻单元格之间的间距大小。

在HTML中，可以通过在table标签中设置这两个属性的值来实现相应的效果。例如：

<table cellpadding="10" cellspacing="5">
  <tr>
    <td>单元格1</td>
    <td>单元格2</td>
  </tr>
  <tr>
    <td>单元格3</td>
    <td>单元格4</td>
  </tr>
</table>

在上述示例中，设置了cellpadding为10和cellspacing为5，这意味着单元格内部内容与边框之间的距离为10个像素，相邻单元格之间的间距为5个像素。

需要注意的是，HTML5中已经不推荐使用cellpadding和cellspacing属性，而是通过CSS来控制表格的样式。可以使用CSS的padding属性和border-spacing属性来实现类似的效果。例如：

<style>
  table {
    border-spacing: 5px;
  }
  td {
    padding: 10px;
  }
</style>
<table>
  <tr>
    <td>单元格1</td>
    <td>单元格2</td>
  </tr>
  <tr>
    <td>单元格3</td>
    <td>单元格4</td>
  </tr>
</table>

通过CSS样式表中的border-spacing属性和padding属性，可以实现与cellpadding和cellspacing相同的效果。

总之，cellpadding和cellspacing是HTML表格中用于控制单元格内部内容与边框之间的距离以及相邻单元格之间的间距的属性。它们的使用可以使表格更加美观和易读。但是在HTML5中，推荐使用CSS来控制表格样式，而不是依赖这两个属性。

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