Verify that the binary linear recurrence of the fourth order Un U(n in .NET Attach code 128 barcode in .NET Verify that the binary linear recurrence of the fourth order Un U(n

chapter 8 generate, create uss code 128 none on .net projects .NET Framework 2.0 To take an easily veri abl USS Code 128 for .NET e case rst: a sequence of order 4 might have a period of 15, which is 24 1. If we ignore the trivial case where all four multipliers are 0 there are 15 possible binary linear sequences of order 4.

Do any of these produce a binary sequence of maximum period 15 Just 2 of them do; they are Un U(n and Un U(n. 1) 3). Example 8.1 Verify that the binary linear recurrence of the fourth order Un U(n generates a sequence of maximum period 15. Veri cation We start with U0 U1 U2 U3 1 and generate each new term by adding together the terms three and four places earlier in the sequence and putting 0 or 1 according to whether the sum is even or odd. The sequence is then found to be 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 1, 1..

. and we see that the sequence begins to repeat from U15 onwards, but not before. (Note that if a binary sequence generated by a linear recurrence of order k has maximum period it must contain all possible 2k binary sequences of length k except the one consisting of all 0s.

We may therefore take any initial values except 00...

00 and the period will be seen to be maximal in every case. If the period is not maximal different starting values may produce different sequences.) Problem 8.

1 Verify that the binary recurrence Un U(n. also generates a sequence of period 15 but that the recurrence Un U(n does not. Producing random numbers and letters What about sequences of hi barcode standards 128 for .NET gher order A sequence of order 12, for example, might have a period as long as 212 1, which is 4095. Does any linear recurrence of order 12 generate such a maximal binary sequence As a result of some elegant and advanced mathematics a formula is known which tells us exactly how many binary linear recurrences of order k will produce a sequence of maximum period.

In the case of recurrences of order 12 the formula tells us that 144 binary linear recurrences of order 12 will produce binary sequences of period 4095. (The fact that 144 12 12 is a uke!) The remaining 3952 will fail to do so. The mathematical analysis does not lead directly to these 144, which must be found by a process somewhat akin to nding prime numbers.

Alternatively, using a computer, the 4095 possible recurrences can be tested and any which repeat before 4095 terms have been generated can be rejected. When this is done the rst successful sequence found is Un U(n. 12).. It is the rst in the se Code128 for .NET nse that, writing 0 and 1 for the multipliers of the 12 terms on the right-hand side of the linear recurrence, this sequence has the 12-bit representation 000001010011 which, interpreted as an integer written in binary form, is 64 16 2 1 83. No sequence with such an integer representation below 83 produces the maximum period of 4095.

For some of the mathematics behind all this see M11. By choosing a suf ciently high order and nding a linear recurrence that gives the maximum period we can produce a key stream that would seem to provide a pseudo-random binary stream which we could use as a key for encryption. It can be shown, for example, that 356 960 linear recurrences of order 23 will generate maximal key streams, which are more than 8 000 000 long (M11).

Since the initial starting values also provide over 8 000 000 possibilities one might think that such a key would present a formidable problem to the cryptanalyst and, initially, it would. Unfortunately for the cryptographers key which has been generated in this way has a fatal aw: given a fairly modest amount of the key the linear recurrence by which it has been generated and the initial values can be recovered This is a consequence of the following..

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