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

COMP30026代做、C/C++設(shè)計(jì)程序代寫
COMP30026代做、C/C++設(shè)計(jì)程序代寫

時(shí)間:2024-08-30  來源:合肥網(wǎng)hfw.cc  作者:hfw.cc 我要糾錯(cuò)



Assignment 1
COMP30026 Models of Computation
School of Computing and Information Systems
Due: Friday 30 August at 8:00pm
Aims
To improve your understanding of propositional logic and first-order predicate
logic, including their use in mechanised reasoning; to develop your skills in
analysis and formal reasoning about complex concepts, and to practise writing
down formal arguments with clarity.
Marking
Each question is worth 2 marks, for a total of 12. We aim to ensure that
anyone with a basic comprehension of the subject matter receives a passing
mark. Getting full marks is intended to be considerably more difficult; the
harder questions provide an opportunity for students to distinguish themselves.
Your answers will be marked on correctness and clarity. Do not leave us
guessing! It is better to be clear and wrong; vague answers will attract few
if any marks. This also means you must show your working in mechanical
questions!
Finally, make sure your writing is legible! We cannot mark what we cannot
read. (Keep in mind that the exam will be on paper, so this will be even more
important later!)
Academic Integrity
In this assignment, individual work is called for. By submitting work for
assessment you declare that:
1. You understand the University  s policy on academic integrity.
2. The work submitted is your original work.
3. You have not been unduly assisted by any other person or third party.
4. You have not unduly assisted anyone else.
5. You have not used any unauthorized materials, including but not limited
to AI and translation software.
1
However, if you get stuck, you can use the discussion board to ask any ques-
tions you have. If your question reveals anything about your approach or work-
ing, please make sure that it is set to   private  .
You may only discuss the assignment in basic terms with your peers (e.g.
clarifying what the question is asking, or recommending useful exercises). You
may not directly help others in solving these problems, even by suggesting
strategies.
Soliciting or accepting further help from non-staff is cheating and will lead
to disciplinary action.
Q1 Propositional Logic: Island Puzzle
You come across three inhabitants of the Island of Knights and Knaves. Now, a
mimic has eaten one of them and stolen their appearance, as well as their status
as a knight or knave. (And is thus bound by the same rules. Remember that
knights always tell the truth, and knaves always lie!)
Each makes a statement:
1. A says:   C is either the mimic or a knight, or both.  
2. B says:   It is not the case that both A is the mimic and C is a knave.  
3. C says:   If B is a knight, then the mimic is a knave.  
Task A
Translate the information above into propositional formulas. Give an appropri-
ate interpretation of all propositional letters used. Use the same interpretation
throughout the question; do not give multiple interpretations.
Task B
Determine which of , , and is the mimic, and prove that it must be the
case using an informal argument.
Some advice: A good answer should not be much longer than about 250
words. But do not worry about the length of your first draft! Instead focus on
finding a proof in the first place. Once you have that, it is much easier to find a
shorter proof. Also, remember that clarity is key: write in complete sentences
with good grammar, but do not include irrelevant information or repeat yourself
unnecessarily.
Q2 Propositional Logic:
Validity and Satisfiability
For each of the following propositional formulas, determine whether it is valid,
unsatisfiable, or contingent. If it is valid or unsatisfiable, prove it by drawing
an appropriate resolution refutation. If it is contingent, demonstrate this with
two appropriate truth assignments.
1. ?    (    ?)
2
2. (    (    (    )))    (?    ?(?    ?))
3. ?((    )    )    ( ? )    (    ?)
4. ( ? )    ((    ) ? ( ? ))
Hint: If you are unsure, you can use a truth table to help you decide!
Q3 Predicate Logic: Translation and Seman-
tics
Task A
Translate the following English sentences into formulas of predicate logic. Give
an appropriate interpretation of any non-logical symbols used. Use the same
interpretation throughout this question; do not give multiple interpretations.
1. Iron is heavier than oxygen.
2. All actinides are radioactive.
3. Some, but not all, lanthanides are radioactive.
4. Actinides are heavier than lanthanides.
5. Both lanthanides and actinides are heavier than iron and oxygen.
6. At least three isotopes of lanthanides are radioactive, but the only lan-
thanide without any non-radioactive isotopes is promethium.
Task B
By arguing from the semantics of predicate logic, prove that the universe of
every model of following formula has at least 3 distinct elements. (Resolution
refutations will receive 0 marks.)
??((, )    ?(, ))    ??((, ))
Q4 Predicate Logic: Red-Black Trees
The use of function symbols in our notation for predicate logic allows us to
create a simple representation of binary trees. Namely, let the constant symbol
represent the root node of the tree, and the unary functions and represent
the left and right children of a node. The idea is that () is the left child of the
root node, (()) is the right child of the left child of the root node, and so on.
With this representation defined, we can now prove statements about trees.
A red-black tree is a special type of binary tree that can be searched faster,
in which each node is assigned a colour, either red or black. Let the predicates
and denote whether a node is red or black respectively. A red-black tree is
faster to search because it must satisfy some constraints, two of which are:
3
1. Every node is red or black, but not both:
?((()    ?())    (()    ?())) (1)
2. A red node does not have a red child:
?(()    (?(())    ?(()))) (2)
Task
Use resolution to prove that these two conditions entail that a tree consisting
of a non-black root with a red left child is not a red-black tree.
Q5 Informal Proof: Palindromes
Assume the following definitions:
1. A string is a finite sequence of symbols.
2. Given a symbol , we write the string consisting of just also as .
3. Given strings and , we write their concatenation as .
4. Given a collection of symbols 1,   , , we have the following:
(a) The expression 1   stands for the string of symbols whose th
symbol is equal to for all integers from 1 to .
(b) The reverse of the empty string is the empty string.
(c) The reverse of a nonempty string 1   of length is the string
1   where = ?+1 for all positive integers    .
(d) The expression   1 stands for the reverse of 1  .
5. A string is a palindrome if and only if it is equal to its reverse.
Task
The proof attempt below has problems. In particular, it does not carefully
argue from these definitions. Identify and describe the problems with the proof.
Then, give a corrected proof.
Theorem. Let be a palindrome. Then is also a palindrome.
Proof (attempt). We have = 1   for some symbols 1,   , where is
the length of . Since is a palindrome, it is by definition equal to itself under
reversal, so =   1 and = ?+1 for all positive integers    .
Therefore = 1    1, and hence there exist symbols 1,   , 2
such that = 1  2. Since the reverse of 1    1 is itself, it follows
that is a palindrome, as desired.
4
d f g h ie
b
a
c
Figure 1: Diagram of our 9-segment display. Colour key: horizontal segments
are blue, vertical segments are green, and diagonal segments are orange.
Q6 Propositional Logic: Logic on Display
One common practical application of propositional logic is in representing logic
circuits. Consider a 9-segment LED display with the segments labelled a through
i, like the one shown on Figure 1. To display the letter   E  , for example, you
would turn on LEDs , , , and , and turn the rest off.
Arrays of similar displays are commonly used to show numbers on digital
clocks, dishwashers, and other devices. Each LED segment can be turned on or
off, but in most applications, only a small number of on/off combinations are of
interest (e.g. displaying a digit in the range 0 C9 only uses 10 combinations). In
that case, the display can be controlled through a small number of input wires.
For this question, we are interested in creating a display for eight symbols
from the proto-science of alchemy. Since we only want eight different symbols
(see Figure 2), we only need three input wires: , , and .
Figure 2: Table of symbols, their encodings in terms of , and , and the
corresponding on/off state of the segments  C.
So, for example, is represented by = = = 0, and so when all three
wires are unpowered, we should turn on segments , and ? and turn off the
other segments. Similarly, is represented by = = 0 and = 1, so when
wires and are off and the wire is on, we should turn on , and , and
turn off the other segments.
5
Note that each of the display segments  C can be considered a propositional
function of the variables , , and . For example, segment e is on when the
input is one of 101, 110, or 111, and is off otherwise. That is, we can capture
its behavior as the following propositional formula:
(    ?    )    (       ?)    (       ).
The logic display must be implemented with logic circuitry. Here we assume
that only three types of logic gates are available:
1. An and-gate takes two inputs and produces, as output, the conjunction
(  ) of the inputs.
2. An or-gate implements disjunction (  ).
3. An inverter takes a single input and negates (?) it.
Task
Design a logic circuit for each of  C using as few gates as possible. Your answer
does not need to be optimal1 to receive full marks, but it must improve upon
the trivial answer. (Incorrect answers will receive 0 marks.)
We can specify the circuit by writing down the Boolean equations for each
of the outputs  C. For example, from what we just saw, we can define
= (    ?    )    (       ?)    (       )
and thus implement using 10 gates. But the formula (    ?  )    (   )
is equivalent, so we can in fact implement using 5 gates.
Moreover, the nine functions might be able to share some circuitry. For
example, if we have a sub-circuit defined by = ?    , then we can define
=    (    ?    ?), and also possibly reuse in other definitions. That is,
we can share sub-circuits among multiple functions. This can allow us to reduce
the total number of gates. You can define as many   helper   sub-circuits as you
please, to create the smallest possible solution.
Submission
Go to   Assignment 1 (Q6)   on Gradescope, and submit a text file named q6.txt
consisting of one line per definition. This file will be tested automatically, so it
is important that you follow the syntax exactly.
We write ? as - and    as +. We write    as ., or, simpler, we just leave it
out, so that concatenation of expressions denotes their conjunction. Here is an
example set of equations (for a different problem):
# An example of a set of equations in the correct format:
a = -Q R + Q -R + P -Q -R
b = u + P (Q + R)
c = P + -(Q R)
d = u + P a
u = -P -Q
# u is an auxiliary function introduced to simplify b and d
1Indeed, computing an optimal solution to this problem is extremely difficult!
6
Empty lines, and lines that start with   #  , are ignored. Input variables are
in upper case. Negation binds tighter than conjunction, which in turn binds
tighter than disjunction. So the equation for says that = (?    )    (   
?)    (    ?    ?). Note the use of a helper function , allowing and
to share some circuitry. Also note that we do not allow any feedback loops
in the circuit. In the example above, depends on , so is not allowed to
depend, directly or indirectly, on (and indeed it does not).

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





 

掃一掃在手機(jī)打開當(dāng)前頁
  • 上一篇:MAST30027代做、Java/C++設(shè)計(jì)程序代寫
  • 下一篇:ECOS3010 代做、代寫 c/c++,java 語言程序
  • 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(dòng)機(jī)性能
    挖掘機(jī)濾芯提升發(fā)動(dòng)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

    午夜影院欧美| 少妇精品久久久一区二区| 天堂av在线一区| 精品视频在线你懂得| www.成人在线.com| 成人片免费看| 黄色一区二区三区四区| av日韩精品| 国产麻豆一区二区三区| 欧美激情福利| 久久uomeier| 中文一区在线| 久久久999| 日韩av网址大全| 裸体一区二区三区| 麻豆精品蜜桃| 午夜亚洲视频| 亚洲特色特黄| 欧美色资源站| 久久丁香四色| 久久综合另类图片小说| 成人精品国产亚洲| 日韩中文在线电影| 免费在线一区观看| 在线综合亚洲| 免费视频亚洲| 激情久久久久| 久久九九国产| 精品国产一区二区三区香蕉沈先生| 国产精品一级在线观看| 国产一区二区三区成人欧美日韩在线观看 | 久久国产生活片100| 色资源二区在线视频| 免费成人在线观看| 中文国产一区| 天天射综合网视频| 成人av动漫在线观看| 亚洲国产影院| 国精品一区二区| 欧美一级精品片在线看| 国产一二三在线| 欧美视频导航| 999久久精品| 国产色99精品9i| 青草伊人久久| 日韩欧美久久| 日韩精品一级| 久久精品九色| 国产极品模特精品一二| 亚洲精品在线国产| 精品久久久久久久久久久aⅴ| 亚洲视频一起| 精品理论电影| 国产91久久精品一区二区| 久久神马影院| 欧美特黄一级| 日韩中文欧美在线| 色天天久久综合婷婷女18| 日韩1区在线| 天堂av中文在线观看| 日韩欧美一区二区三区免费看| 在线天堂新版最新版在线8| 日韩免费高清| 国产一区二区高清在线| 日韩欧美三区| 久久男人av| 日韩在线成人| 久久免费黄色| 99成人精品| 综合一区二区三区| 欧美黄免费看| 日韩精品1区2区3区| 精品国产精品久久一区免费式 | 国产精品一国产精品| 久久久久久亚洲精品美女| 国产另类在线| 日韩亚洲精品在线| 亚洲日本天堂| 麻豆国产欧美日韩综合精品二区| 欧美激情一区| avtt综合网| 亚洲精品国产偷自在线观看| 美女精品在线观看| 欧洲av不卡| 99综合久久| 国内一区二区三区| 国产欧洲在线| 日韩久久99| 亚洲小说图片| 久久九九免费| 男人的j进女人的j一区| 日韩美女在线| 久久不见久久见国语| 91精品啪在线观看国产18| 亚洲欧美bt| 欧美亚洲福利| 国产真实有声精品录音| 欧美三级午夜理伦三级小说| 亚洲一卡久久| 国产一区影院| 日韩区欧美区| 日韩天天综合| 日韩综合久久| 日本一区二区三区视频在线看| 国产字幕视频一区二区| 欧美韩日一区| 中文字幕成人| 久久免费黄色| 成人欧美一区二区三区的电影| 欧美日本一区| 久久在线免费| 精品免费av在线| 日韩黄色网络| 99成人免费视频| 久久精品国产网站| 超碰在线亚洲| 国产中文在线播放| 99精品美女视频在线观看热舞| 91精品国产福利在线观看麻豆| 日av在线不卡| 91精品麻豆| 欧美日韩国产传媒| 美日韩一区二区| 精品色999| 日韩理论电影院| 精品视频在线观看免费观看| 亚洲一卡久久| 久久中文字幕导航| 91久久电影| 国产日韩欧美一区在线| 99精品网站| 久久精品资源| 久久国产成人午夜av影院宅| 三级成人在线| 欧美精品国产白浆久久久久| a国产在线视频| 日韩av一二三| 日本一区二区三区视频| 亚洲图片久久| 国产精品原创| 亚洲精品不卡在线观看| av女在线播放| 成人性生交大片免费看96| 少妇淫片在线影院| 日韩中文字幕无砖| 涩涩av在线| 精品国产中文字幕第一页 | 国产成人视屏| 夜久久久久久| 99久久999| 免费精品视频最新在线| 亚洲盗摄视频| 色资源二区在线视频| 激情亚洲另类图片区小说区| 日韩一区二区三区免费视频| 北条麻妃一区二区三区在线| 97成人超碰| 国产一区欧美| 综合一区在线| 国产精品久久天天影视| 免费一区二区三区在线视频| 久久99久久99精品免观看软件| 999成人精品视频线3| 日本视频一区二区| 噜噜噜躁狠狠躁狠狠精品视频| 国产中文字幕一区二区三区| 爱啪啪综合导航| 成人羞羞视频在线看网址| 久久尤物视频| 首页国产欧美日韩丝袜| 99这里只有精品视频| 久久精品国产99国产| 久久一级大片| 91综合国产| 狠狠入ady亚洲精品| 日韩av不卡一区| 国产福利亚洲| 国产亚洲亚洲| 国产成人tv| 国产精品v一区二区三区| 日韩www.| 久久精选视频| 亚洲婷婷影院| 国产精品久久久久久模特| 亚洲在线一区| 色哟哟精品丝袜一区二区| 欧美精品99| 91精品韩国| 日韩午夜av在线| 色天天色综合| 欧美区一区二区| 久久久久久久性潮| 免费观看日韩av| 欧美一区二区三区高清视频| 成人豆花视频| av成人在线播放| 免费精品视频最新在线| 欧美 日韩 国产一区二区在线视频| 亚洲国产精品嫩草影院久久av| 亚洲国产1区|