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SOLUTIONS TO CHAPTER 9 PROBLEMS 9.1 in .NET Drawer PDF 417 in .NET SOLUTIONS TO CHAPTER 9 PROBLEMS 9.1

SOLUTIONS TO CHAPTER 9 PROBLEMS 9.1 using barcode writer for visual .net control to generate, create pdf417 2d barcode image in visual .net applications. Beaware of Malicious QR Codes Hamming distance = 3. SOLUTIONS TO PROBLEMS ASCII Q = 1010001 1 0 11 10 1 9 0 0 8 7 C8 0 6 0 5 0 1 4 3 C4 0 1 2 1 C2 C1 (a) k + r + 1 2r for k = 6. Simplifying yields: 7 + r 2r for which r = 4 is the smallest value that satis es the relation. (b).

1 10 0 9 1 1 8 7 C8 1 6 0 5 0 0 4 3 C4 1 1 2 1 C2 C1 (c) 11 (d) 101111011001. k + r + 1 2r .net framework pdf417 2d barcode for k = 1024. Simplifying yields 1025 + r 2r for which r = 11 is the smallest value that satis es the relation.

. Code 11100101 11100110 11100111 11101000 11101001 11100101 Character V W X Y Z Checksum (a) 4096 (b) 8. First, we look for errors in the check bit computations for the SEC code. If all of the check bit computations are correct, then there are no single bit errors nor any double bit errors. If any of the SEC check bit computations are wrong, then there is either a single bit error or a double bit error.

The DED parity bit creates even parity over the entire word when there are no errors. If the DED parity computation is in error, then there is an odd number of errors, which we can assume is a single bit. 602 SOLUTIONS TO PROBLEMS error for this problem. If the DED parity computation is even, and the SEC computation indicates there is an error, then there must be at least a double bit error..

9.8 9.9 9.

10 9.11 9.12.

CRC = 101. The entire frame to be transmitted is 101100110 101. 32 bits Class B 27 + 214 + 221 63 ns for bit- PDF-417 2d barcode for .NET serial transfer, 32 ns for word-parallel. The word-parallel version requires 32 times as much hardware, but only halves the transmission time, as a result of the time-of- ight delay through the connection.

[Placeholder for solution.]. SOLUTIONS TO CHAPTER 10 PROBLEMS 10.1 CPIAVG = 1 + 5(.25)(.5) = 1.625 10 = 16.25 ns Execution ef c iency = 1/1.625 = 62%. No, for the AR barcode pdf417 for .NET C architecture, we cannot use %r0 immediately after the st because st needs two cycles to nish execution. If we did reuse %r0 in the second line, then a nop (or some other instruction that does not produce a register con ict) would have to be inserted between the st and sethi lines.

However, for the SPARC architecture, the pipeline stalls when a con ict is detected, so nops are never actually needed for delayed loads in the SPARC. The nop instruction is needed for delayed branches in the SPARC, however..

10.3 10.4.

[Placeholder f PDF417 for .NET or solution.] [Placeholder for solution.

]. SOLUTIONS TO PROBLEMS * x2 + x2 + y2 * xy * y2 2 2xy x2 + 2xy + y2 1 S = -------- .net vs 2010 PDF417 --------------------------- = 16.8 1 0.

05 0.05 + -----------------100. 16.8 --------- = .168, or 16.

8% 100. n=p=6: complex visual .net barcode pdf417 ity = 306. n=p=2: complexity = 180.

n=p=4: complexity = 231. n=p=3: complexity = 200. 15.

SOLUTIONS TO APPENDIX A PROBLEMS A.1 A B AB 604 SOLUTIONS TO PROBLEMS AND A AB B A NOT OR A B A+B A B C 0 0 0 0 1 1 1 1 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 F 0 1 0 1 0 1 1 1 G 1 0 1 0 0 0 1 0 A B C 0 0 0 0 1 1 1 1 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 XOR 0 1 1 0 1 0 0 1 SOLUTIONS TO PROBLEMS A B C A BC f(A,B,C). B C E D A g(A,B,C,D,E). Not equivalent:. A B C 0 0 0 0 1 1 1 1 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 F 0 0 1 0 0 0 0 1 G 0 0 0 1 0 0 1 0 606 SOLUTIONS TO PROBLEMS Alternatively, we can also disprove equivalence algebraically:. g ( A, B, C ) VS .NET PDF-417 2d barcode = ( A C )B a = ( AC + AC )B a = ABC + ABC a f ( A, B, C ). A B C 0 0 0 0 1 1 1 1 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 F 0 1 0 0 0 0 0 0 F(A,B,C) = ABC A.10. b3 a3 b2 a2 b1 a1 b0 a0 A = B A.11. x0 x1 x2 x3 x4 F SOLUTIONS TO PROBLEMS A.12. 0 1 1 0 00 01 visual .net barcode pdf417 10 11 0 0 0 1 00 01 10 11. A.13. 0 C 1 00 01 10 11. A.14. A B 00 01 10 11 A.15. 000 001 010 01 PDF417 for .NET 1 100 101 110 111 F(A,B,C). A B C G(A,B,C). 608 SOLUTIONS TO PROBLEMS A.16. d0 d1 d2 d3 d4 d5 d6 d7 0 1 0 1 0 1 0 1 0 1 0 1 0 1 C B A F(A,B,C). A.17. B A 0 1 1 0 1 AB A.18. A B C D 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 b 1 0 0 0 1 1 1 0 1 1 1 1 1 0 1 1 1 0 0 0 1 1 1 0 1 1 1 1 1 0 1 1 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 ABCD.
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