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Assignment

Assignment
Assignment

Assignment

By 080835-Linda

080828-Naomi

080823-Claudia

080833-Sue

1. Nauseum (Repetition)

The continual repetition of the same idea over and over and over Explain: Because he said the reason about teachers’salary again and again.

2. Circulus in Demonstrando (Circular reasoning)

Explain: Because he explains the reason circularly.

3. Ad Logicam (Appeal to logic) [Straw Man]

An idea must be false because the opponent’s argument is not sound. Explain: Because the reason is not logical.

4. Red Herring

Explain: The introduction of an irrelevant idea in order to distract.

5. Circulus in Demonstrando (Circular reasoning)

Explain: Because he explains the reason circularly.

6. Ad Ignorantiam (Argument from Ignorance)

Explain: Never been proven false; therefore, must be true OR …Never been proven true; therefore, must be false

7. False dilemma

Explain: The artificial restricting of a choice to only two possibilities.

8. No True Scotsman

Changing the criteria to exclude any fact that may contradic ts one’s argument.

Explain: The fact is not reasonable.

9. Circumstantial

An irrelevant connection between the belief held and the circumstances of the person holding the belief.

10. Complex Question

An aggressive demand that a question be answered absol utely either “yes or no”

Explain: The answer is too absolutely.

11. Ad Populum / Ad Numerum

The mistaken belief that the majority believes something to be true alone makes something true automatically.

12. Ignoratio Elenchi (Irrelevant Conclusion)

The premisses “miss the point”: suddenly the premisses are re-directed to show an irrelevant conclusion.

13. Abusive

To consider a person’s integrity, not the argument

Explain: It is not an argument.

14. Circumstantial

An irrelevant connection between the belief held and the circumstances of the person holding the belief

chapter3-assignment

第三章 线性分组码 习题 1.证明[n ,k ]线性分组码的最大距离为n -k +1。 2.设一个[7,4]码的生成矩阵为 1000111010010100100110 00111 0G ????? ?=?????? (1)求出该码的全部码矢; (2)求出该码的一致校验矩阵; (3)作出该码的标准阵译码表。 3.证明定理3.1.3。 4.一个[8,4]系统码,它的一致校验方程为: c 0=m 1+m 2+m 3c 1=m 0+m 1+m 2c 2=m 0+m 1+m 2c 3=m 0+m 2+m 3 式中,m 0,m 1,m 2,m 3是信息位,c 0,c 1,c 2,c 3是校验位。找出该码的G 和H ,并证明该码的最小距离为4。 5.构造第4题中码的对偶码。 6.设H 1是[n ,k ]线性分组码C 1的校验矩阵,且有奇数最小距离为d 。作一个新的码C 2,它的校 验矩阵为 12000111 1H H ?? ????? ?=???????? M L (1)证明C 2是一个(n +1,k )分组码; (2)证明C 2中每一码字的重量为偶数; (3)证明C 2码的最小重量为d +1。 7.设C 1是一个有最小距离为d 1的[n 1,k ]线性系统码,生成矩阵为G 1=[P 1I k ]。C 2是一个有最小距离为d 2的[n ,k ]线性系统码,它的生成矩阵G 2=[P 2I k ]。对满足下述一致校验矩阵 212T k n n k T p H I P I ???? ? ?=??????

的[n 1+n 2,k]线性码,证明它有最小距离至少为d 1+d 2。 8.设一个二进制[n ,k ]码C 的G 矩阵不含全零列,将C 的所有码字排成2k ×n 的阵。 证明: (a )阵中不含有全零列; (b )阵中的每一列由2k -1个零和2k -1个1组成; (c )在一特定分量上为0的所有码字构成C 的一个子空间,问该子空间的维数是多少? 9.令是所有二进制[n ,k ]线性系统码的集合。证明非零二进制n 重V 或者恰巧含于的 ΓΓ(1)()2k n k ??个码中,或者不在的任一码中。 Γ10.证明二进制[23,12,7]Golay 码和三进制的[11,6,5]Golay 码是完备码。 11.若d 是码C 的最小重量,且为偶数,(1)/2t d =????? 。证明有两个重量均为t +1的矢量必在C 码的同一陪集中。 12.求出d =3,至多只有3个校验元的二进制码的码长n ;和d =5,至多只有8个校验元的二进 制码的码长n 。 13.计算二进制[24,12,8]扩张Golay 码的覆盖半径,及[8,4]RM 码的覆盖半径。 14.证明定理3.9.3。 15.构造三个二进制的[10,3,5]LUEP 码,其分离矢量分别为(8,2,2),(7,4,4), (6,4,4)。写出它们标准形式的G 和系统码形式的G 。 16.证明定理3.10.3。 17.构造一个具有最高码率的k =10,t =2的2-EC/AUED 码。 18.证明定理3.10.6。

财务分析

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中国海关的通关制度和海关业务制度改革

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solutions for assignment3

Chapter 3 Assignments P13.Consider a reliable data transfer protocol that uses only negative acknowledgments. Suppose the sender sends data infrequently. Would a NAK-only protocol be preferable to a protocol that uses ACKs? Why? Now suppose the sender has a lot of data to send and the end-to-end connection experiences few losses. In this second case, would a NAK-only protocol be preferable to a protocol that uses ACKs? Why? Answer: In a NAK only protocol, the loss of packet x is only detected by the receiver when packet x+1 is received. That is, the receivers receives x-1 and then x+1, only when x+1 is received does the receiver realize that x was missed. If there is a long delay between the transmission of x and the transmission of x+1, then it will be a long time until x can be recovered, under a NAK only protocol. On the other hand, if data is being sent often, then recovery under a NAK-only scheme could happen quickly. Moreover, if errors are infrequent, then NAKs are only occasionally sent (when needed), and ACK are never sent – a significant reduction in feedback in the NAK-only case over the ACK-only case. P23. We have said that an application may choose UDP for a transport protocol because UDP offers finer application control (than TCP) of what data is sent in a segment and when. a. Why does an application have more control of what data is sent in a segment? b. Why does an application have more control on when the segment is sent? Answer: a) Consider sending an application message over a transport protocol. With TCP, the application writes data to the connection send buffer and TCP will grab bytes without necessarily putting a single message in the TCP segment; TCP may put more or less than a singe message in a segment. UDP, on the other hand, encapsulates in a segment whatever the application gives it; so that, if the application gives UDP an application message, this message will be the payload of the UDP segment. Thus, with UDP, an application has more control of what data is sent in a segment. b) With TCP, due to flow control and congestion control, there may be significant delay from the time when an application writes data to its send buffer until when the data is given to the network layer. UDP does not have delays due to flow control and congestion control.

商务交际英语任务(assignment)

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通关作业无纸化背景下的随附单证“瘦身”方案

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assignment写作浅谈

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