99爱在线视频这里只有精品_窝窝午夜看片成人精品_日韩精品久久久毛片一区二区_亚洲一区二区久久

合肥生活安徽新聞合肥交通合肥房產生活服務合肥教育合肥招聘合肥旅游文化藝術合肥美食合肥地圖合肥社保合肥醫院企業服務合肥法律

代寫CS1010S: Advanced Recursion

時間:2024-02-24  來源:合肥網hfw.cc  作者:hfw.cc 我要糾錯


CS1010S: Programming Methodology

Semester II, 2023/2024

Mission 4

Advanced Recursion

Release date: 16th February 2024

Due: 22nd February 2024, 23:59

Required Files

• mission04-template.py

Background

After demonstrating your abilities to Pharaoh Tyro, you were honored with the presti-gious role of bishop within his esteemed team. The anticipation was palpable as you entered his chambers, where Tyro’s eyes sparkled with expectation. With a grand ges-ture, he handed you three scrolls (Your mission tasks), each bearing the royal seal.

"These," he declared, his voice resonating with authority, "are your inaugural assign-ments as bishop. Execute them diligently and report to me during the upcoming CS1010S class."

This mission consists of three tasks.

Task 1: Number of ways to sum to an Integer (3 marks)

A positive integer n ≥ 2 can be expressed as the sum of a number of positive integers smaller than n. For example:

2 = 1 + 1

3 = 1 + 2

   = 1 + 1 + 1

4 = 1 + 3

   = 2 + 2

   = 1 + 1 + 2

   = 1 + 1 + 1 + 1

5 = 1 + 4

   = 1 + 1 + 3

   = 2 + 3

   = 1 + 2 + 2

   = 1 + 1 + 1 + 2

   = 1 + 1 + 1 + 1 + 1

The function num_sum returns the number of ways that an integer can be expressed as the sum of a number of positive integers. From the above examples, it should be clear that:

>>> num_sum ( 2 )

1

>>> num_sum ( 3 )

2

>>> num_sum ( 4 )

4

>>> num_sum ( 5 )

6

Hint: If you grasp the essence of the count change problem, you’ll recognize that this problem is a variation of it. You may want to consider implementing a helper function that model the count change process of this problem. Solving the problem using closed-form formulas are not allowed.

Task 2: Generalized Pathfinding: Enumerate All Paths (3 marks)

In Lecture Training 5, you faced a problem where you were required to assist Jon in im-plementing a function, num_of_possible_path(board). This function determined the num-ber of possible paths to move from the starting point "S" to the ending point "E" by either walking (covering 1 step) or jumping (covering 2 steps).

Now, you encountered a similar challenge. The game no longer restricts the steps to just 1 or 2; instead, it can be any arbitrary number of steps (i.e. 1, 2, 3, ..., n). Your task is to implement an iterative recursive function, num_of_possible_path(board), which calculates the number of possible paths to move from the starting point "S" to the ending point "E" given that there are n possible ways to move at each step.

You may assume substring(string, start, end, step) function is given.

Hint: Observe that this problem resembles a count change problem. At each step, you have the choice to move 1 step forward, or 2 steps forward, or 3 steps forward, and so on, up to n steps forward.

>>> num_of_possible_path ("S##E", 1 )

1

>>> num_of_possible_path ("S##E", 2 )

3

>>> num_of_possible_path ("S##E", 3 )

4

Task 3: Check valid brackets (5 marks)

Consider a string containing only brackets "(" and ")". A string of brackets is considered valid if:

• Every opening parenthesis has a corresponding closing parenthesis.

• Opening and closing parentheses are in the correct order.

• Each closing parenthesis has a matching opening parenthesis.

Implement a function, check_valid_brackets(s), that returns True if the string s is valid brackets, and False otherwise.

Hint: If a string of brackets is valid, it can repeatedly remove the innermost non-nested "()" until it becomes an empty string.

Subtask 3a: Illustrate Your Problem-Solving Approach

In Lecture 1, you have learnt the Polya’s Problem Solving Process:

1. Understand the Problem

2. Make a Plan (Create a Flowchart, as outlined in Lecture 1 slides)

3. Do the Plan

4. Review & Generalize

Apply the Polya problem-solving methodology, and demonstrate your problem-solving process for Task 3. You are tasked to write out each step, providing insights into your approach and decision-making. This exercise aims to reinforce your understanding and application of the problem-solving methodology.

Please submit your illustration to coursemology. Note that you must include Step 1 and Step 2 in your illustration; Step 3 and Step 4 are optional. (For an example, please refer to Coursemology -> Workbin -> PolyasProblemSolvingExample.pdf)

By using the idea of divide and conquer, here are the steps to solve Task 2

1. Implement an iterative function remove_bracket_pair(s) that takes in a string of brackets. This function iterates through the string from left to right, removing the first occurrence of the brackets pair "()" within the string s, and returns the modified string. You may assume substring(string, start, end, step) function is given.

>>> remove_bracket_pair (" ()()() ")

" ()() "

>>> remove_bracket_pair (" (()()) ")

" (()) "

>>> remove_bracket_pair (" ((())) ")

" (()) "

>>> remove_bracket_pair (")()")

")"

>>> remove_bracket_pair ("()")

""

>>> remove_bracket_pair (" (())((())) ")

" ()((())) "

2. Using the above iterative remove_bracket_pair(s) function, implement a recursive check_valid_brackets(s) that takes in a string of brackets and returns True if the string s is valid brackets, and False otherwise.

>>> check_valid_brackets ("()")

True

>>> check_valid_brackets (" (()) ")

True

>>> check_valid_brackets (" ()() ")

True

>>> check_valid_brackets (" (()")

False

>>> check_valid_brackets (" ())")

False

>>> check_valid_brackets (" ())( ")

False

Subtask 3b: Execute Your Plan

1. Implement the iterative function remove_bracket_pair(s).

2. Implement the recursive function check_valid_brackets(s).

You may assume substring(string, start, end, step) function is given.

You are highly encouraged to test your functions with additional test cases.

Optional: Spiral Maze Iterative Recursively

Write an iterative recursive function num_of_steps that takes in 4 arguments, the x and y coordinates of ending point, x and y, width of the maze, W and height of the maze, H. The function returns the number of steps to navigate from the bottom-left corner (origin) of the maze to the specified ending point. Please follow the question requirements any closed form formula or pure iterative solution will not be accepted.

Hint: You will need to iterate until the boundary, then recursively call the function with the new boundary and updated x & y.



Figure 1: A spiral maze with height 3 and width 3. The number of steps from the origin to the ending point (1, 1) is 8.

num_of_steps (1 , 1 , 3 , 3 )

>>> 8

num_of_steps (0 , 0 , 3 , 3 )

>>> 0

num_of_steps (1 , 1 , 3 , 2 )

>>> 4

num_of_steps (1 , 3 , 5 , 7 )

>>>

Optional: Alternative approach of Task 2

There are many ways to solve the problem in Task 2. You are encouraged to explore alternative approaches to solve the problem.

You may assume substring(string, start, end, step) function is given in this task.

Implement a function, check_valid_brackets_alt(s), that returns True if the string s is valid brackets, and False otherwise.

Completely Iterative Approach (Easy)

You can implement the function purely iterative. Please confine your implementation to what you’ve learned from CS1010S thus far.

Completely Recursive Approach (Challenging)

You may also implement the function purely recursively.

Warning: This is a challenging task.

請加QQ:99515681  郵箱:99515681@qq.com   WX:codehelp 

掃一掃在手機打開當前頁
  • 上一篇:代寫ELEC-4840 編程
  • 下一篇:代寫 Financial Derivatives and Financial
  • 無相關信息
    合肥生活資訊

    合肥圖文信息
    急尋熱仿真分析?代做熱仿真服務+熱設計優化
    急尋熱仿真分析?代做熱仿真服務+熱設計優化
    出評 開團工具
    出評 開團工具
    挖掘機濾芯提升發動機性能
    挖掘機濾芯提升發動機性能
    海信羅馬假日洗衣機亮相AWE  復古美學與現代科技完美結合
    海信羅馬假日洗衣機亮相AWE 復古美學與現代
    合肥機場巴士4號線
    合肥機場巴士4號線
    合肥機場巴士3號線
    合肥機場巴士3號線
    合肥機場巴士2號線
    合肥機場巴士2號線
    合肥機場巴士1號線
    合肥機場巴士1號線
  • 短信驗證碼 豆包 幣安下載 AI生圖 目錄網

    關于我們 | 打賞支持 | 廣告服務 | 聯系我們 | 網站地圖 | 免責聲明 | 幫助中心 | 友情鏈接 |

    Copyright © 2025 hfw.cc Inc. All Rights Reserved. 合肥網 版權所有
    ICP備06013414號-3 公安備 42010502001045

    99爱在线视频这里只有精品_窝窝午夜看片成人精品_日韩精品久久久毛片一区二区_亚洲一区二区久久

          香蕉乱码成人久久天堂爱免费| 亚洲国产经典视频| 欧美在线关看| 激情自拍一区| 欧美日韩999| 午夜精品偷拍| 激情丁香综合| 欧美人与禽猛交乱配| 午夜精品999| 亚洲第一天堂av| 欧美日韩国产精品| 久久国产毛片| 亚洲美女中文字幕| 国产女精品视频网站免费| 久久久综合香蕉尹人综合网| 亚洲精品麻豆| 国产日韩高清一区二区三区在线| 久久男女视频| 亚洲视频中文字幕| 在线观看成人一级片| 欧美特黄一级大片| 鲁鲁狠狠狠7777一区二区| 亚洲天堂网在线观看| 在线观看一区视频| 国产精品资源在线观看| 欧美精品v日韩精品v国产精品| 午夜精品久久久久久久99黑人| 亚洲国产精品电影| 国产日韩欧美高清免费| 欧美精品一区二区在线观看| 久久高清福利视频| 亚洲视频高清| 亚洲激情午夜| 国产亚洲精品aa午夜观看| 欧美日韩国产免费观看| 老司机精品视频网站| 亚洲欧美日韩综合aⅴ视频| 亚洲人成在线观看| 狠狠色丁香婷婷综合| 国产精品视频99| 欧美日韩国产一区二区| 嫩草成人www欧美| 久久精品国产综合| 亚洲欧美日韩精品久久奇米色影视 | 欧美精品久久久久久久免费观看 | 亚洲精品中文字幕有码专区| 国产一区二区三区高清播放| 国产精品久久久久高潮| 欧美欧美天天天天操| 免费高清在线一区| 久久久无码精品亚洲日韩按摩| 午夜精品福利在线观看| 亚洲视频一二区| 一二三四社区欧美黄| 亚洲免费av观看| 亚洲免费大片| 99在线精品视频| 99天天综合性| 99国内精品| 亚洲裸体视频| 日韩视频不卡| 一区二区三区高清视频在线观看| 亚洲精品视频一区| 亚洲三级视频| av成人激情| 亚洲视频一区二区在线观看| 一区二区三区精品视频| 日韩一级成人av| 一本色道久久加勒比88综合| 一区二区三区欧美激情| 一区二区高清| 亚洲一区免费视频| 性欧美大战久久久久久久久| 性做久久久久久久久| 欧美在线观看视频一区二区三区 | 国产精品久久久久久av下载红粉| 欧美三级免费| 国产乱码精品一区二区三区忘忧草| 国产精品亚洲综合久久| 国产亚洲精品v| 尤物精品在线| 日韩亚洲不卡在线| 亚洲一区日韩在线| 久久精品国产精品亚洲精品| 久久一二三四| 欧美日韩精品欧美日韩精品一| 国产精品h在线观看| 国产手机视频一区二区| 影音先锋久久精品| 夜夜夜久久久| 久久精品理论片| 欧美黑人国产人伦爽爽爽| 欧美三级第一页| 国产一区二区欧美| 亚洲精品久久7777| 亚洲一区二区视频在线观看| 久久精品电影| 欧美区二区三区| 国产欧美日韩视频一区二区三区| 影音先锋亚洲精品| 一本大道久久精品懂色aⅴ| 欧美在线观看视频| 欧美乱大交xxxxx| 国产日韩在线一区二区三区| 亚洲人成欧美中文字幕| 亚洲欧美在线高清| 欧美激情1区2区3区| 国产精品亚洲欧美| 亚洲欧洲综合| 午夜精品久久久久久99热| 欧美成人r级一区二区三区| 国产精品毛片大码女人| 精品成人在线观看| 亚洲午夜精品久久久久久浪潮| 六十路精品视频| 国产日韩精品一区二区三区| 日韩视频二区| 久久精品视频va| 国产精品久久久久免费a∨| 在线成人av网站| 亚洲欧美不卡| 欧美另类专区| 亚洲国产电影| 久久精品亚洲| 国产欧美日韩| 亚洲婷婷综合久久一本伊一区| 另类综合日韩欧美亚洲| 国产欧美视频一区二区三区| 在线亚洲电影| 欧美激情一区在线观看| 伊人久久婷婷色综合98网| 亚洲愉拍自拍另类高清精品| 欧美激情精品久久久久久大尺度 | 欧美亚洲综合网| 国产精品啊啊啊| 亚洲毛片在线观看.| 毛片一区二区| 激情一区二区| 欧美在线免费观看亚洲| 国产精品二区在线| 在线一区二区三区四区五区| 欧美成人国产| 亚洲国产一二三| 猛男gaygay欧美视频| 韩国av一区| 久久九九国产精品怡红院| 国产精品天天摸av网| 亚洲男人的天堂在线aⅴ视频| 欧美日韩在线播放三区四区| 亚洲美女黄网| 欧美日韩黄色一区二区| 一本色道久久综合亚洲91| 欧美欧美全黄| 中日韩男男gay无套| 欧美日韩国产欧| 一区二区三区免费网站| 国产精品v欧美精品v日韩精品| 99re8这里有精品热视频免费| 欧美日韩伦理在线| 亚洲一区二区不卡免费| 国产精品一区在线观看你懂的| 亚洲自拍三区| 国产日韩欧美一区二区| 久久久久国产一区二区| 在线观看日韩一区| 欧美99在线视频观看| 亚洲毛片在线看| 国产精品成人一区二区| 亚洲欧美三级伦理| 国产一区二区三区的电影 | 欧美大胆a视频| 99国产精品99久久久久久粉嫩| 欧美视频中文字幕在线| 午夜欧美精品久久久久久久| 韩国精品在线观看| 欧美精品高清视频| 中文精品视频| 国产亚洲一区二区三区在线观看| 久久九九国产| 亚洲精品一级| 国产九色精品成人porny| 久久一区视频| 日韩一区二区精品| 国产欧美日韩综合| 欧美成人一品| 亚洲综合日韩| 亚洲国产精品久久久久秋霞影院| 欧美日本高清一区| 久久精品一区| 亚洲精品1区2区| 国产精品香蕉在线观看| 久久免费的精品国产v∧| 99re热精品| 尤物九九久久国产精品的分类| 欧美日本在线观看| 久久精品99国产精品| 日韩亚洲综合在线| 激情另类综合| 欧美性视频网站| 久久久久久精|