Math Project 1 Determination of Magnetic Codes

Transcription

1 Math Project 1 Determination of Magnetic Codes 1. Gist of the Problem This project concerns the storage of data on magnetic tape. The first thing to understand is the basic mechanism by which a tape recorder/player stores data on magnetic tape. This works as follows: Magnetic tape is typically mylar tape which is has been uniformly coated with millions of specks of a magnetic material like ferric oxide. The magnetic coating can be thought of as millions of tiny bar magnets whose North-South pole orientations can be forced into alignment by the application of a strong external magnetic field. When magnetic tape is run under a recording head (an electromagnet whose polarity can be arbtrarily changed by reversing the flow of current through it), different portions of the tape can be assigned particular magnetic polarities - by changing the polarity of the recording head s magnetic field as the tape runs under the recording head. After recording then, the magnetic tape is essentially a long linear medium populated by contiguous regions of locally fixed, but globally varying, magnetic polarity. Fig. 1 : Physical Tape Fig. 2 : Regions of Fixed Polarity Next one can (mentally) subdivide the tape into subregions of uniform length and map the tapes magnetic arrangement to a sequence of 0 s and 1 s ½ 0 if magnetic polarity is postive in element of sequence subinterval 1 if magnetic polarity is negative in subinterval Thus, we can map a length of recorded to tape to an -digit binary number. For example, the binary sequence associated with an arrangement like would be Fig

2 2 Figure 1 Alternatively, rather than attach 1 s and 0 s to the individual subregions of the tape, one could focus instead on the boundaries between regions and set ( 1 if polarity changes between the element of sequence and ( +1) subregion 0 if and ( +1) subregion have same polarity Thus, thus arrangement in Fig. 3 would yield or Now comes the basic caveats to this method of storing binary data on tape. (i) There is a limitation as to how short a region of constant polarity can be (as the actual edges between of fixed polarity are not sharp on the physical tape) (ii) There is a limitation as to how long a region of fixed polarity can be. This is because the tape can get misaligned with the recording head unless there is a realignment mechanism built into the way the tape is recorded. The realignment mechanism is just the stipulation that polarity transitions occur at prescribed places along the tape. However, because of tape speed inaccuracies, the precise lengths of long segments of the constant polarity can not be accurately obtained. 2. AMathematicalProblem The mechanism described above is just one step (albeit a crucial one) in general mechanism for recording and reading data on magnetic tape. A more complete outline of a recording process would be (1) Human recognizable data is converted into numerical data (ascii code or integers). (2) Numerical data is converted into strings of binary numbers of fixed length. (3) Strings of binary numbers of fixed length are mapped to recording instructions to a recording head. (4) The recording head writes the instructions it receives to the physical tape. Note that each stage is essentially a (literal) translation of some sort. The reading process by which data is retrieved is essentially be the reverse of the process outlined above. Thisprojectfocusesonstep3anditsinverse. Weshallabstract the situation as follows. A binary string of length is to be recorded somehow on a segment of magnetic tape. Via the mechanisms described above, a segment of recorded magnetic tape can also be mapped to a sequence of binary numbers. Let be an input data string; an ordered list of 0 s and 1 s and let be the element of.

3 3 Then an obvious way to record on magnetic tape would be to divide a segment of tape up into subintervals and then use the recording head polarize the subintervals according to the rule ½ ¾ ½ ¾ positive polarity 0 region has whenever negative polarity is 1 However, this naive method has two drawbacks. The length of tape need to record will be where is the minimal length for a region of fixed polarity. (We can do better than this.) Not all possible data strings can be accurately recorded and retrived this way. We will run into problems if we have long substrings of consecutive 0 s (or 1 s) as the reading/recording head and the tape can get misaligned when there are long gaps between changes of polarity. To get around these limitations we will first abstract the problem mathematically, and then look for a mathematical solution. First, let s separate completely the data string to be recorded from the binary string representing the instructions to recording head (the instructions telling the recording head when it should be polarized one way or the other). We shall continue to call the input data string and its individual digits, 1. Let me use to denote a list of recording instructions to the recording head. We will refer to as a code string. We will think of also as an ordered list of 0 s and 1 s but its length need not be. To be precise, we will regard a code string to be a binary string of length. However, rather than regarding the 0 s and 1 s in as being instructions for writing corresponding polarities onto the tape, we shall think of the 1 s in as being an instruction to reverse the polarity of the recording head, and a 0 as an instruction to leave the polarity of the recording head unchanged. Thus, as in Fig. 4 above, rather than attaching 1 s and 0 s to subintervals of the tape, we attach 1 s and 0 s to the boundaries between subintervals of the tape. With this way of implementing a code string, we can frame the restrictions on the recording strings as follows. (i ) Theremustbeatleast 0 s between consecutive 1 s in (so that changes in polarity do not occur too quickly) (ii ) The maximal number of consecutive 0 s is (so that the gaps between changes in polarity are not so long that the tape and recording head get mis-aligned). Question 2.1. Let ( ) #of binary strings of length such that ½ there are at least 0 s between consecutive 1 s there are at most consecutive 0 s There are 2 possible data strings of length. How might one choose,,, so that each possible data string can be mapped to a distinct code string meeting the above requirements; viz., for what values of, and do we have 2 ( )? Next we shall explore how different choices of and affect the efficiency of storage. But for now let s concentrate on developing means for computing ( ) 3. A Recursive Approach to Computing ( ) Rather than compute ( ) correctly, we first compute a simpler related number. Let ( ) {binary strings in ( ) that start and end with 1 s} ( ) # ( ) We shall write down a recursive formula for computing ( ). Each string in ( ) startsand ends with a 1. Let s divide up the strings in ( ) according to where the second 1 (starting from the

4 4 left) occurs. According to the rules for constructing strins in ( ), the second 1 has to be preceded by at least and at most 0 s. So the beginning of ( ) musthaveoneofthefollowingforms with 0 s between the first two 1 s, or with ( + 1) 0 s between the first two 1 s, or with ( + 2) 0 s between the first two 1 s, or with ( 1) 0 s between the first two 1 s, or with ( 1) 0 s between the first two 1 s Note that each of the cases listed above is distinct, so we ll get the total number of possibilities for by summing up the cardinalities for each subcase. Next, tail end of such an in ( ) (the substring after the initial 1 and 0 s) will still have to satisfy the requirements that the number of 0 s between consecutive 1 s must always lie between and. Sothetailendofsuchan will be another string of type ( 0 ), but for a smaller value of. In the first subcase, we stipulate that beginswitha1followedby 0 s. The length of the tail 1 1 will then be 1. And so the number possibilities for completing the string in this subcase will be ( 1). The second subcase corresponds to beginning with a 1 followed by +1 0 s. The tail end in this case will be of length 2, and so we have ( 2) ways of completing this sort of head to a string in ( ). In general: thenumberofstringsin ( ) that start with a 1 followed by 0 s is ( 1) This will be true for all between and. Since each between and corresponds to a particular subcase, we get the following recursive formula for ( ) (1) ( ) ( 1) We now are in a position to derive a formula for ( ). The basic idea will be to build the strings in ( ) fromstringin ( 0 ) by adding 0 s to either side. Suppose you have a string ( ) that begins with consecutive 0 s and ends with consecutive 0 s The number of ways of filling in the interior will be ( ); because between the initial 0 s and the final 0 s is a string of type ( ). Since we can have as many as initial 0 s and as many as final 0 0,weseethat (2) ( ) ( ) 0 0 This last formula gives us a means of computing ( ) so long as we know enough initial values for ( ). In fact, we can readily derive a recursive formula for ( ) thatavoidstheexplicit computation of the numbers ( ). This is obtained as follows. Replacing the summation index in (1) by and the substituting for in (1), we have (3) ( ) ( 1)

5 5 If we now replace ( ) ontherighthandsideof(2)withthelefthandsideof(3)weget or ( ) ( 1) ( 1) 0 0 ( 1) ( 1) (4) ( ) ( 1) after changing the order of summation Note the amazing resemblance to the formula (1) for ( ). Despite the similarity their formulas, we do not have ( ) ( ). For the values for ( ) and ( ) dependasmuch upon their the initial values (small ) as the recursive formula used to compute the values for ( ) and ( ) for higher values of 4. Compression Recall that we have been looking at ( ) {binary strings of length with at least and at most zero s between consecutive one s} ( ) # ( ) (the cardinality of ( )) Each string in ( ) corresponds to a list of instructions to the write head of our magnetic tape recorder: let ( ) andlet instuction 1 Think of as the instruction given to the write head at time ( being the period elasped between successive instructions). If 0, the instruction to the write head at time is to do nothing, if 1, the instruction to the write head at time is to reverse its magnetic polarity. Let minimal length of tape allowed between successive changes in polarity velocity of tape as it moves under the write head The parameter in ( ) is the minimal number of 0 s between successive 1 s of a string ( ). It corresponds to the minimal time min that can elapse between successive reversals in magnetic polarity: min ( +1) E.g. if 5,ifwehadasubstringlike and the time that would elaspe between the two 1 instuctions would actually be 6. Now during the time min, the length of tape that has passed under the write head will be min min

6 6 If we set this min equal to the physical limitation of the tape we get ( +1) min implies ( +1) Thus, in order to implement an instruction set using strings in ( ), we pass instructions to the write head every (+1) seconds. This means that each instruction in a string in ( ) uses µ ( +1) +1 units of tape (whatever the units used to prescribe are). But since each ( ) contains instructions, the strings in ( ) requirealength ( ) +1 of tape. Here you see for fixed, we can improve storage efficiency by increasing. But perhaps a better measure of storage efficiency would be (5) ( ) log 2 ( ( )) # of raw data bits capable of being mapped 1:1 to strings in ( ) ( ) length of tape required by strings in ( ) 5. Specific Tasks To Be Carried Out Write an Abstract for the report Write an Introductory section for the report, explaining the magnetic code problem. Write a section that transports the application problem to a mathematical problem and develop a mathematical solution. Include in this discussion the derivations for the formulas (1), (2), (4) and (5) above. Carry out some sample computations. For this you can restrict 25. Look for choices of that would be optimal in the following sense: log 2 ( ( )) 32 so that the instruction code would be capable of handling raw data consisting of 32 bit datums. is as small as possible; ( ) is as large as possible (larger means you can squeeze more raw data onto thesameamountoftape). Summarize your findings. Write a reference section (even though it may be minimal for this project).