加勒比久久综合,国产精品伦一区二区,66精品视频在线观看,一区二区电影

合肥生活安徽新聞合肥交通合肥房產(chǎn)生活服務(wù)合肥教育合肥招聘合肥旅游文化藝術(shù)合肥美食合肥地圖合肥社保合肥醫(yī)院企業(yè)服務(wù)合肥法律

代寫G6017 Program Analysis

時間:2024-01-14  來源:合肥網(wǎng)hfw.cc  作者:hfw.cc 我要糾錯



THE UNIVERSITY OF SUSSEX
BSc and MComp SECOND YEAR EXAMINATION January 2022 (A1)
Program Analysis Assessment Period: January 2022 (A1)
Candidate Number
G6017
 DO NOT TURN OVER UNTIL INSTRUCTED TO BY THE LEAD INVIGILATOR
Candidates should answer TWO questions out of THREE.
If all three questions are attempted only the first two answers will be marked.
The time allowed is TWO hours. Each question is worth 50 marks.
At the end of the examination the question paper and/or answer book, used or unused, will be collected from you before you leave the examination room.

G6017 Program Analysis
1.
(a) Precisely specify the conditions under which the following algorithm returns
true, and then discuss, in detail, the running time of the algorithm. If you think it has different best- and worst-case running times then these should be considered separately, and you should explain the conditions under which best and worst-cases arise.
You must fully explain your answer and use O,  and &#***3; appropriately to receive full marks.
  Algorithm Ex1 ((𝑎1, ... 𝑎𝑛), (𝑏1, ... , 𝑏𝑚))
𝑘←0
for 𝑖 ← 1 to 𝑛 do
𝑗←1
while 𝑗 ≤ 𝑚 do
If𝑎 ==𝑏 𝑖𝑗
𝑘←𝑘+1 𝑗←𝑗+1
return 𝑘 > 0
 (b) Precisely specify the conditions under which the following algorithm returns true, and then discuss, in detail, the running time of the algorithm. If you think it has different best- and worst-case running times then these should be considered separately, and you should explain the conditions under which best and worst-cases arise.
You must fully explain your answer and use O,  and &#***3; appropriately to receive full marks.
[10 marks]
 Algorithm Ex2 ((𝑎1, ... 𝑎𝑛), (𝑏1, ... , 𝑏𝑛)) w**2; ← w**5;w**3;w**6;Ү**;
for𝑖←1to𝑛 do 𝑗←𝑛
while𝑗>0andw**2; ==w**5;w**3;w**6;Ү**;do If𝑎 ==𝑏
𝑗←𝑗−1 return w**2;
𝑖𝑗
w**2; ← 𝑓𝑎𝑙w**4;Ү**;
 2
[10 marks]

(c) Precisely specify the conditions under which the following algorithm returns true, and then discuss, in detail, the running time of the algorithm. If you think it has different best- and worst-case running times then these should be considered separately, and you should explain the conditions under which best and worst-cases arise.
You must fully explain your answer and use O,  and &#***3; appropriately to receive full marks.
 Algorithm Ex3 ((𝑎1, ... 𝑎𝑛), 𝑏)
w**2;←0
w**8; ← 𝑓𝑎𝑙w**4;Ү**;
for 𝑖 ← 1 to 𝑛 − 1 do
𝑧 ← 𝑎𝑖 + 𝑎𝑖+1 If 𝑧 < 𝑏
w**2;←w**2;+1
else
If w**2; < 0 return w**8;
w**2;←w**2;−1 w**8; ← w**5;w**3;w**6;Ү**;
 (d) A data pattern analyser is to be built that can detect and count up the number of occurrences of two and three letter same letter sequences in a sequence (e.g. (a,a) or (b,b,b) ). The analyser should stop if it encounters * in the sequence and return the number of occurrences found up to that point in the form of a 2-tuple (#2𝐿Ү**;w**5;w**5;Ү**;w**3;𝑆Ү**;w**2;w**6;Ү**;𝑛𝑐Ү**;w**4;, #3𝐿Ү**;w**5;w**5;Ү**;w**3;𝑆Ү**;w**2;w**6;Ү**;𝑛𝑐Ү**;w**4;). A 3 letter sequence should only count as a 3 letter sequence, not an occurrence of two 2 letter sequences. No letter in the input sequence ever occurs more than 3 times in a row.
So, for example:
Input string (a,b,a,b,b,a,a,*)
(x,x,x,y,y) (p,q,p,z,z,*) (*,a,a)
(x,x,a,d,*, s,s,s,l)
2 letter sequences found
2 1 1 0 1
3 letter sequences found
0 1 0 0 0
[10 marks]
                     3

Produce a formal statement of this problem, and then write an algorithm to solve the problem using a pseudo code style similar to the one shown in parts (a) to (c). State the bounds on the best and worst case performance of your algorithms using O,  and &#***3; appropriately to receive full marks.
[10 marks]
(e) A file is protected by a random password consisting of 𝑛 binary bits. All password combinations are equally probable. To access the file we need the correct password. The process of applying the password to the file takes 10ms regardless of the value of 𝑛. Brute force attack is always a viable basic strategy for guessing a password.
To ensure that the file remains sufficiently secure, we need to ensure that there is no more than a 1% chance over 30 days that the password is guessed by a hacker program utilizing brute force working 24 hours a day, 7 days a week. How many bits should be specified for the password?
2.
(a) A student has been asked to put some parcels on a shelf. The parcels all weigh different amounts, and the shelf has a maximum safe loading weight capacity of 100 Kg. The weight of parcels are as follows (in Kg):
𝒑𝒂𝒓𝒄𝒆𝒍 𝒘𝒆𝒊𝒈𝒉𝒕 (𝑲𝒈)
18 2 50 ** 4 15 54 65 7 20
The student has been asked to load the maximum weight possible parcels on the shelf subject to the maximum safe loading weight.
State two possible approaches for a greedy algorithm solution to solve this problem. In each case, state clearly the result you would get from applying that approach to this problem, stating whether the solution is optimal or not. If
[10 marks]
                  4

your answer does not produce an optimal solution, what algorithm could be employed to find one?
[10 marks]
(b) One example of a greedy algorithm is the Dijkstra algorithm for finding the lowest cost path through a weighted graph. The diagram below shows two weighted graphs that a student wants to investigate using Dijkstra’s algorithm. In each case the task it to find the lowest cost of reaching every node from v1. Each graph has a single negative weight in it.
Graph (a) Graph (b)
One of the graphs will yield a correct analysis of the lowest cost for all vertices, and the other will produce an incorrect analysis. Which of the two graphs will produce the incorrect analysis, and explain why the greedy nature of Dijkstra’s algorithm is responsible for the incorrect analysis. Your answer should include the key concept of an invariant.
[5 marks]
(c) The priority queue is a widely used data structure. Priority queues may be implemented using binary heaps and simple linear arrays. For the basic priority queue operations of:
• Building an initial queue
• Taking the highest priority item off the queue
• Adding a new item to the queue
Compare and contrast the running time complexities (best and worst cases) associated with implementations using binary heaps and simple linear arrays. You may find it helpful to use diagrams to support your answer.
[10 marks]
           5

(d) A recursive algorithm is applied to some data 𝐴 = (𝑎1, ... , 𝑎𝑚) where 𝑚 ≥ 2 and 𝑚 is even. The running time 𝑇 is characterised using the following recurrence equations:
𝑇(2) = 𝑐 when the size of 𝐴 is 2 𝑇(𝑚) = 𝑇(𝑚 − 2) + 𝑐 otherwise
Determine the running time complexity of this algorithm. Note that 𝑚 is even and the problem size reduces by 2 for each recursion.
[10 marks]
(e) Another recursive algorithm is applied to some data 𝐴 = (𝑎1, ... , 𝑎𝑚) where 𝑚 = 2w**9; (i.e. 2, 4, 8,16 ...) where w**9; is an integer ≥ 1. The running time T is characterised using the following recurrence equations:
𝑇(1) = 𝑐 when the size of 𝐴 is 1, and 𝑐 is a constant 𝑇(𝑚) = 2𝑇 (𝑚) + 𝑚 otherwise
2
Determine the running time complexity of this algorithm. You will find it helpful to recall that:
w**9;=∞ 1
∑2𝑘 →1 w**9;=1
And
2𝑙w**0;𝑔2(w**9;) = w**9;
3.
[15 marks]
(a) The subset sum problem can be reliably solved optimally using the dynamic programming algorithm shown below:
   SubsetSum(𝑛, 𝑊)
Let w**;(0,w**8;) = 0 for each w**8; ∈ {0,...,𝑊} for 𝑖 ← 1 w**5;w**0; 𝑛
for w**8; ← 0 w**5;w**0; 𝑊 ifw**8;<w**8;𝑖 then
w**;(𝑖, w**8;) ← w**;(𝑖 − 1, w**8;)
else
w**;(𝑖,w**8;)←max(w**8;𝑖 +w**;(𝑖−1,w**8;−w**8;𝑖),w**;(𝑖−1,w**8;))
 6

Where 𝑛 is the number of requests, 𝑊 is the maximum weight constraint, w**8;𝑖 is the weight associated with request 𝑖, and w**; is the solution space.
You are given a set of requests and their corresponding weights. The maximum weight constraint 𝑊 is 12.
𝒊 𝒘𝒊
11 2** 10 46 53 62
Copy the following solution space table to your answer book (do not write your answer on the question paper) and complete the table to determine the optimal subset sum.
                       𝒊
6 5 4 3 2 1 0
𝒘
0 1 2 3 4 5 6 7 8 9 10 11 12
[10 marks]
     (b) The sequence alignment problem may be solved by the following dynamic programming algorithm:
   SequenceAlignment(X, Y):
Let w**;(𝑖, 0) ← 𝑖 × 𝛾 for each 1 ≤ 𝑖 ≤ 𝑛 Let w**;(0, 𝑗) ← 𝑗 × 𝛾 for each 1 ≤ 𝑗 ≤ 𝑚 For 𝑖 ← 1 w**5;w**0; 𝑛
For 𝑗 ← 1 w**5;w**0; 𝑚
w**;(𝑖, 𝑗) ← min [𝛿(w**9; , 𝑦 ) + w**;(𝑖 − 1, 𝑗 − 1),
𝑖𝑗
𝛾 + w**;(𝑖, 𝑗 − 1),
𝛾 + w**;(𝑖 − 1), 𝑗 ]
 7

Where 𝑋 = (w**9; ,...,w**9; ) and 𝑌 = (𝑦 ,...,𝑦 ) are two sequences to be aligned, 𝛿(w**1;,w**2;) 1𝑛 1𝑚
is a penalty associated with matching symbol w**1; to w**2;, and 𝛾 is a gap penalty. The sequence alignment algorithm is applied using the following data:
𝑋 =(𝑎,𝑏,𝑐)
𝑌 = (𝑎,𝑏,𝑎,𝑏,𝑏)
𝛾=4
The delta function is defined for symbols in the alphabet {𝑎, 𝑏, 𝑐} :
aBc a045 b407 c570
Generate the problem space matrix w**; and thus determine the optimal alignment between 𝑋 and 𝑌.
[15 marks] (c) Draw a Minimum Spanning Tree derived from the graph shown below.
[5 marks]
(d) The Ford-Fulkerson algorithm is used to determine network flow. The diagram below represents a data network that connects a Data Service Provider (DSP)
                         8

connected to w**7;1(w**4;) to a customer connected to w**7;6(w**5;). Each edge represents a single data transmission link.
The notation w**1;/w**2; indicates a current actual forwards flow w**1; measured in Gb/s in a pipe with a maximum capacity of w**2; also measured in Gb/s.
At the outset no data is being sent by the DSP to the customer.
i. Show the residual graph that will be created from the initial empty flow. When drawing the residual graph, show a forward edge with capacity w**9; and a backward edge with flow 𝑦 by annotating the edge w**9;⃗; 𝑦**; .
[2 marks]
ii. What is the bottleneck edge of the path (w**4;, w**7;3, w**7;4, w**5;) in the residual graph you have given in answer to part (a) ?
[2 marks]
iii. Show the residual graph after incorporating the simple path (w**4;, w**7;3, w**7;4, w**5;) that results from augmenting the flow based on the residual graph you have given in answer to part (a).
[4 marks]
iv. Repeat the process outlined above incorporating additionally the simple paths (w**4;, w**7;3, w**7;2, w**7;5, w**5;) , (w**4;, w**7;2, w**7;5, w**7;4, w**5;) and (w**4;, w**7;2, w**7;5, w**5;) showing each residual graph, to determine the maximum flow between w**4; and w**5;, and thus the maximum data bandwidth that can be achieved between the DSP and the customer.
請加QQ:99515681 或郵箱:99515681@qq.com   WX:codehelp

掃一掃在手機(jī)打開當(dāng)前頁
  • 上一篇:代做SIPA U6500、代寫 java,python 程序設(shè)計(jì)
  • 下一篇:TEMU FUND海外拼多多是什么?一篇文章告訴你
  • 無相關(guān)信息
    合肥生活資訊

    合肥圖文信息
    2025年10月份更新拼多多改銷助手小象助手多多出評軟件
    2025年10月份更新拼多多改銷助手小象助手多
    有限元分析 CAE仿真分析服務(wù)-企業(yè)/產(chǎn)品研發(fā)/客戶要求/設(shè)計(jì)優(yōu)化
    有限元分析 CAE仿真分析服務(wù)-企業(yè)/產(chǎn)品研發(fā)
    急尋熱仿真分析?代做熱仿真服務(wù)+熱設(shè)計(jì)優(yōu)化
    急尋熱仿真分析?代做熱仿真服務(wù)+熱設(shè)計(jì)優(yōu)化
    出評 開團(tuán)工具
    出評 開團(tuán)工具
    挖掘機(jī)濾芯提升發(fā)動機(jī)性能
    挖掘機(jī)濾芯提升發(fā)動機(jī)性能
    海信羅馬假日洗衣機(jī)亮相AWE  復(fù)古美學(xué)與現(xiàn)代科技完美結(jié)合
    海信羅馬假日洗衣機(jī)亮相AWE 復(fù)古美學(xué)與現(xiàn)代
    合肥機(jī)場巴士4號線
    合肥機(jī)場巴士4號線
    合肥機(jī)場巴士3號線
    合肥機(jī)場巴士3號線
  • 短信驗(yàn)證碼 目錄網(wǎng) 排行網(wǎng)

    關(guān)于我們 | 打賞支持 | 廣告服務(wù) | 聯(lián)系我們 | 網(wǎng)站地圖 | 免責(zé)聲明 | 幫助中心 | 友情鏈接 |

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

    色综合综合网| 亚洲女同中文字幕| 久久精品国产免费| 在线亚洲成人| 第九色区aⅴ天堂久久香| 另类小说视频一区二区| 欧美a级成人淫片免费看| 久久精品亚洲成在人线av网址| 国内久久视频| 日本一区二区三区视频在线| 老司机精品福利视频| 欧美日韩在线网站| 日韩欧美久久| 一区二区三区日本视频| 丁香婷婷久久| av综合电影网站| 在线视频免费在线观看一区二区| 色爱av综合网| 亚洲一区二区免费在线观看| av日韩久久| 久久精品欧洲| avav成人| 香蕉伊大人中文在线观看| 国产毛片一区| 午夜久久免费观看| 成人三级视频| av不卡一区| 经典三级久久| 亚洲最好看的视频| 99综合久久| 久久久久观看| 久久精品一区二区国产| 亚洲aⅴ网站| 日韩精选视频| 伊人久久综合一区二区| 色天天久久综合婷婷女18| 亚洲一区二区伦理| 亚洲欧美日韩高清在线| 欧美中文一区二区| 伊人春色精品| 亚洲二区精品| 欧洲激情视频| 先锋资源久久| 婷婷综合久久| 黄色在线成人| 婷婷综合视频| 欧洲杯半决赛直播| 欧美日韩国产免费观看视频| 欧美+亚洲+精品+三区| 欧美午夜精品一区二区三区电影| 久久久久国内| 欧美va天堂| 日韩午夜一区| 首页亚洲欧美制服丝腿| 男男成人高潮片免费网站| 日韩在线a电影| segui88久久综合9999| 97精品一区| 女人高潮被爽到呻吟在线观看| 日韩亚洲一区在线| 日韩一区二区三区在线免费观看| 波多视频一区| 日日夜夜精品| 麻豆视频一区二区| 在线精品在线| 国产精品片aa在线观看| 日韩精品视频在线看| 在线视频亚洲欧美中文| 久久97久久97精品免视看秋霞| 欧美性感美女一区二区| 激情91久久| 91欧美在线| 先锋影音网一区二区| 亚洲三级网站| 日韩三级毛片| 欧美综合自拍| 99成人在线| av资源在线播放| 成人国产一区| www久久久| 亚洲精品不卡在线观看| 欧美69视频| 首页国产欧美久久| 高清av一区| 成人久久精品| 禁果av一区二区三区| 欧美特黄一级| 欧美亚洲日本精品| 老司机免费视频一区二区| 亚洲精品进入| 极品裸体白嫩激情啪啪国产精品| 六月婷婷一区| 国产亚洲欧美日韩精品一区二区三区 | 日本欧美一区二区三区乱码 | 成人动漫视频在线观看| 国产欧美88| 尤物精品在线| 日本国产亚洲| 国产免费av一区二区三区| 风间由美中文字幕在线看视频国产欧美 | 国产精品一区2区3区| 国内精品免费| 老牛嫩草一区二区三区日本 | 免费成人在线观看视频| 高清欧美日韩| 国产情侣一区在线| 亚洲男女av一区二区| 深夜成人福利| 亚洲图片久久| 亚洲美女视频在线免费观看| 日本美女久久| 日韩精品一级| 久久动漫亚洲| 日本不卡免费在线视频| 久久国产精品免费精品3p| 蜜桃久久久久久| 中文字幕亚洲精品乱码| 久久久久国产一区二区| 九色porny视频在线观看| 欧美精品播放| 极品少妇一区二区三区| 欧洲精品一区二区三区| 久久av免费| 男人的天堂亚洲| 中文成人在线| 亚州av乱码久久精品蜜桃 | 伊人久久精品| 亚洲精品网址| 日本中文字幕一区| 久久久久91| 69堂免费精品视频在线播放| 久久久久毛片免费观看| 国产精品视区| 国产视频网站一区二区三区 | 国产精品一线| 日韩国产激情| 日韩免费一级| 久草在线中文最新视频| 日韩在线黄色| 欧美国产一区二区三区激情无套| 国内精品久久久久久99蜜桃| 国产精品毛片一区二区三区| 亚洲精品影院在线观看| 成人aaaa| 国内自拍一区| 欧美网站在线| 电影一区中文字幕| 性欧美xxxx大乳国产app| 亚洲最大黄网| 久久av一区| 偷拍一区二区| 亚洲国产福利| 99精品在免费线中文字幕网站一区 | 欧美精品自拍| 欧美高清一级片| 欧美hentaied在线观看| 日韩高清在线免费观看| 福利在线免费视频| 91欧美极品| 欧美成人免费全部网站| 精品日本12videosex| 国产精品va| 人人精品人人爱| av日韩精品| 99精品国产在热久久| 欧美午夜a级限制福利片| 国产精品一区二区精品视频观看| 老司机午夜精品视频| 精品国产18久久久久久二百| 全球中文成人在线| 免费观看久久av| 电影一区二区三区久久免费观看| 色琪琪久久se色| 亲子伦视频一区二区三区| 老司机午夜精品| 色综合天天爱| 久久久综合网| 精品一区二区三区视频在线播放| 欧美oldwomenvideos| 久久久久久影院| 欧美日韩 国产精品| 日韩综合一区| 亚洲一级二级| 欧美男男gaytwinkfreevideos| 日韩在线观看电影完整版高清免费悬疑悬疑| 欧美三级午夜理伦三级在线观看 | 日韩激情av在线| 久久精品黄色| 成人亚洲一区| 亚洲手机在线| 日韩最新在线| 日本vs亚洲vs韩国一区三区二区| 蜜臀a∨国产成人精品| 日本久久成人网| 国产精一区二区| 久久国产麻豆精品| 蜜桃传媒麻豆第一区在线观看| 欧洲精品一区| 五月天亚洲色图| 日本不卡的三区四区五区|