Some Unsolved Problems on Binary CodesAuthor(s): F. J. Budden and T. M. SportonSource: Mathematics in School, Vol. 11, No. 3 (May, 1982), pp. 26-28Published by: The Mathematical AssociationStable URL: http://www.jstor.org/stable/30213735 .

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Some Unsolved Problems

on Binary Codes

by F. J. Budden and T. M. Sporton

Progressive Sequences* We shall be concerned with strings, or lists, or sequences of N binary digits which have the property that, in proceeding down the sequences, only one binary digit changes at each line, while also there are no repetitions of binary groups. Such strings we shall call "progressive sequences" (PS). In the examples below (for which N = 3), (1), (2) and (3) are PS's, but (4) is not, because at the stages indicated, two or more bits are changed; (5) is not a PS because it contains the repetition marked. (6) is a PS for the first six rows only.

Examples - the columns are labelled A, B, C, and at each stage the bit changed is shown at the side of the binary group.

Row (1) (2) (3) (4) (5) (6)

0 000 000 011 000 000 00 1 A 1 0 0 C 0 0 1 Al 11 C0 B 0 1 0 A 1 0 0 2 B 1 1 0 A 1 0 1 B 1 0 1 BC A1 0 B 1 1 0 3 C111 111 C 1 0 0 C0i1 1 B B100 C 1 1 1 4 B 1 0 1 A01 1 B 1 1 0 AB 10 C101 A011 5 A001 C010 A010 C10

.1 B11 B001

6 B011 A11 0 B 0 0 0 BC 11 0Q C 1 01o COO 7 C010 8100 C001 C1 11

8 BOO0 A00O B011

A string containing all 2N possible binary groups will be called a Maximal Progressive Sequence (MPS). Thus (1), (2) and (3) are MPS's; the first six rows of (5) is a PS but not a MPS.

It will be necessary to consider whether or not PS's are distinguishable, or distinct, or whether there are senses in which they may be regarded as equivalent. For example, (2) may be derived from (1) by making a cyclic permutation of the three columns: (ACB), whereby column A is replaced by C, C by B and B by A, and so these two examples are really two versions of the "same", or rather "equivalent" sequence. Again, (3) is obtained by making the same successive changes as in (1), namely by changing digits A, B, C, B, A, B, C in turn, but whereas (1) starts with 0 0 0, (2) starts with 0 1 1. Thus (3) may be obtained by "complementation" of columns B and C in (1) (complementation means everywhere replacing 1 by 0 and 0 by 1). Again, in (1), a final change of bit B would bring us back to 0 0 0, the position from which we started; and a similar return is shown in (2) and (3). Suppose now in (3) we start in row 6 (0 0 0), and cycle round the sequence, the rows being numbered below as in the original sequence:

Row 6 000 7 C 001 0(8) B 0 11 1 A 111 2 B 101 3 C 100 4 B 110 5 A 010

6 B8000

*See for example, T. M. Sporton, A New Class of Error Correcting Codes for Displacement Transducers, Trans IMC, to be published.

Though we appear to have constructed a new MPS, closer inspection will show that the bit changes

CBABCBA (B)

are simply the original bit changes

ABCBABC(B)

only with the transposition of the columns A and C, and so we have nothing new - merely a permutation of the columns.

Finally, if (1) is written backwards, we obtain the bit changes: B C B A B C B (A)

and this is really the "same" as (1), only starting at a different point in the cycle. Thus all the MPS's so far listed may be considered to be equivalent.

Evidently the simplest way to describe a sequence is simply to write down the bit changes, such as A B C B A B C. However, while such a list does tell the whole story, it does not make it easy to detect any repetitions of binary groups. In example (5) above, with the sequence B A B C B C, it is easy to see that there has been a repetition from the successive changes B C B C, which restore the position. In general, a repetition occurs if between two lines every bit changes an even number of times. But examination of the parity of the bit changes between all 1/2N(N -1) pairs of lines would be tedious, and so repetitions are not readily revealed.

Summary of MPS's for N= 3 ABC B A B C CABACAB BCAC BCA BACA B AC C BABC B C A AC BC B C B

A BACA BA CAC B CAC BC B A B C B BABC BAB BABC BAB C B CAC B C

by permutations of columns

by starting each of the above six at a different point; or by other mean

All the above 12 are equivalent in the sense that we have dis- cussed. Are there any other distinguishable MPS's for N=3? The only possible one is:

Example (7) 000 A 100 B 110 A 010 C 011 B 001 A 101 B 11 1

This is different from the previous examples in that we finish up with 1 1 1, and so cannot reach our original group 0 0 0 by a single bit change. A MPS which "cycles" like Exs (1), (2) and (3) we shall call a CMPS (cycling maximal progressive sequence). Thus example (7) is a MPS but not a CMPS; while example

26 Mathematics in School, May 1982

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(6) is a CPS but not a CMPS. From the industrial applications point of view, the cycling property is not important. Example (7) may be used to construct a MPS for N= 4, as follows:

Example (8)

Row 0 Bit changed 0000 1 A 1000 2 B 1100 3 A 0100 4 C 0110 5 B 0010 6 A 1010 7 B 1110 8 D 1111 9 B 1011

10 A 001 1 11 B 01 11 12 C 0101 13 A 1101 14 B 1001 15 A 0001

16 D 0000

(It may be thought that there is a large "imbalance" about this in the sense that whereas A and B are changed six times, C and D have been changed only twice.)

Restrictions to the Construction of Progressive Sequences In the sequence defined by the bit changes:

2 5 A

1

we note that bit A is switched in rows 1, 3 and 8. The "distance" between successive changes of bit A are 2 and 5 respectively. Similarly the "distance" between the changes of bit B made at lines 4 and 7 is 3.

We are now able to introduce the concept of "Minimum Distance". If in a PS the distance between a successive pair of changes of any of the bits is never less than (say) 4, but is sometimes equal to 4, we say that the sequence has a Minimum Distance of 4, or MD = 4. Thus for each of the examples quoted so far, MD=2, with the single exception of example (6), where MD = 3.

In order to produce a PS which has the most efficient error- correcting properties, it is desirable to have MD as large as possible. This restriction does of course make it more difficult to construct a MPS. For example, it is easy (as we have seen in example 8) to produce a MPS with N= 4, MD= 2, but curiously enough it is impossible to produce a MPS with N = 4, MD= 3. The longest PS which is possible in this case is of length 14, not 16. An example of this is given below, written horizontally (transposed) to save space: Example (9)

Row 0 1 2 3 4 5 6 7 8 9 10 11 12 13

Bit A B C D A C B A D A B C changed

A 01111000111000 B 0 0 1 1 1 1 1 0 0 0 0 0 1 1 C 0 0 0 1 1 1 0 0 0 1 1 1 1 0 D 0 0 0 0 1 1 1 1 1 1 0 0 0 0

Any change of bit will now lead to a repetition

It may be verified by laboriously drawing tree diagrams, tracing through the various possibilities and keeping a record of the binary groups so as to avoid repetitions, that this is in fact the only (essentially different) possible solution, and this can be done with paper and pencil in an hour or so. Note that the minimum distance (= 3) is attained four times, as indicated by the brackets. However, if we were to "close" the cycle by changing bit B at the end to restore the group 0 0 0 0, we should then contravene the MD= 3 requirement. This means that we cannot in this case generate any variations by starting the cycle at a different point.

It is of course immediately obvious that 2 _MD !N. We may easily dispose of the case MD=N. One example of this was given in (6), with N = 3. In a similar way, if N = 6, we have the only possible set of bit changes:

ABCDEFABCDEF

giving a CPS of length 12. In general, for N =MD, the maxi- mum possible length of a PS is 2N. Such a sequence is called a "walking code".

If MD=N-1, there will be two possible choices for each stage reached - each node of the tree will have two branches emerging from it, e.g. if N = 6; MD = 5, we could have:

FB< F

F

A-B-C-D--E and so on. SB< SB<

F

In view of what was said at the outset, it is clear that we can, without loss d6fgenerality, always begin our PS with the sequence A B C D E (when MD= 5, say). This simply means that we choose to assign the label "A" to the first bit to be changed, "B" to the second, and so on.

Though it is impossible to construct a MPS with N=4, MD=3, it is possible to do so in the case N= 5, MD=4. One possible such sequence of bit changes is as follows:

Example (1 0) ABCDEBADCBADEBCDABCDEBADCBADEBCD

This sequence containing six changes of each of A and C, and eight of each of B and D and four of E, and it returns to the original group, so is a CMPS. Note that the sequence began A B C D E, though A B C D A would have been a possible start. This CMPS could easily be derived from the above by starting at the second letter and cycling round, and then re- labelling by the cyclic permutation which replaces A by E, B by A, C by B, D by C and E by D to give:

ABCDAECBAECDABCEABCDAECBAECDABCE

Moreover, it would be of interest to investigate the equivalence or otherwise of other MPS's with N = 5, MD = 4.

Sequences of this size and beyond need to be handled by com- puter. The case N = 6, MD = 5 has been thoroughly investigated by computer, and the result is that it appears to be impossible to construct a MPS, and that the longest PS extends to only 50 of the possible 64; while in the case N = 7, MD = 6, only 114 of the 128 can be obtained. The latter investigation required many hundreds of hours of CPU time!

The known results are tabulated, giving the maximum known lengths:

N 3 4 5 6 7

MD= 3 6 14 X X? 4 - 8 32 X? 5 - - 10 50 6 - - - 12 114 7 - - - - 14

X= of no interest

For a given value of N, as MD is decreased (i.e. the restric- tion relaxed), the length of the PS will increase till the upper limit of 2N is reached, and there is no point in reducing the MD further. We need not therefore consider the cases marked X in the table. The sort of questions which arise and which need to be answered are as follows: with N= 6, can we achieve a MPS with MD= 4, or might it be necessary to reduce to MD = 3? Can we construct a MPS with N = 7, MD = 5; and of special interest (from the industrial application point of view) is the case N= 8, MD= 6 - what is the maximum length of PS in this case?

Mathematics in School, May 1982 27

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011 111 110 .

111 011 111

001 001 /

011 10 lo

-1- 1o I ,

- - 010/10010010 i 0

B 110

000 100 000 001 0 EXo(1) EO(a0 - -00 EX b 00

A B 011 011 111 011 111

011 001 --- io 001 101 010 1110 - 110010

11 110 110

000 100 000 100 000 100

EX(3) c EX (b) EX (7)

0110 1110 0110 1110

0010 10 B 0010 1010

1100 D

10- -- 1100 \ \

S \ 0111 1111 11 1111 ooo \ 1oo \ \ 0000 1000 /

00111 1011 \ 001----- 1011 I

0101 0101 1101 - - - -)1101

0001 1001 0001 1001

N z4 EX (8) N 4 EX (9) MD 2 CMPS MPD3 CPS

Another question that needs to be answered is as follows: The incomplete PS's N=4, MD=3 (length 14) and N=6, MD=5 (length 50) were found by using laborious exhaustive processes; in the first case by paper and pencil, in the second by computer. Is there a short (mathematical) proof that these maximum lengths of PS are 14 and 50 respectively? If there is, then such methods might be applied to the cases N = 6, MD = 4 and N=7, MD=5 and N=8, MD=6 to give answers to the questions posed above without the expenditure of hundreds of hours of CPU time.

Many other mathematical problems exist, related to the design of decoding algorithms that will cope with as many errors as possible. Some of these may be the subject of a future article.

Graphical Treatment One approach to the problem of constructing PS's is to regard

01100 11100

S 00100 10100 /

I / o 110 0/110

00101 011oo

01001 - 11001 0011

0000100 - -0 10 - 11

00011 10011

AE EX(10) N=5, MD=

C.M.P.S.

B

D

each binary numeral in the sequence as the co-ordinates of a vertex of an N-dimensional cube. The problem is then one of visiting each vertex once only by moving along the edges of the cube. The minimum distance criterion then requires that after each step in a particular dimension, MD-1 steps must be made in different dimensions before repeating a step in that direction. The diagrams below are intended to show this graphical repre- sentation corresponding to examples (1), (2), (3), (6), (8), (9) and (10) of the text. It will be seen for example that the cyclic permutation (ACB) of the bits referred to in paragraph 2 above corresponds to a rotation of the cube about the diagonal joining 000 to 111 in Figure (2) a and b. In the figure for example (9) where the PS falls short of the 16, the vertices have been parti- tional into two sets, the two which fail to come within the orbit having been ringed in the diagram.

The above discussion seems very much to border on such areas as Graph Theory, Finite Geometries, and even Campa- nology, but the connections are at the time of writing, obscure. Elucidation from knowledgeable readers would be most welcome.

Short Notices

Essentials of Pure Mathematics by J. R. Irwin Edward Arnold, 352 pp., paper, a5.95

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Activities for Junior High School and Middle School Mathematics edited by K. E. Easterday, L. L. Henry and F. M. Simpson NCTM, 218 pp., paper, $8.25

A series of readings, taken from the Arith-

28

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Mathematical Circus by Martin Gardner Penguin, 274 pp., paper, a1.95

A further collection of mathematical games, puzzles, paradoxes and other mathematical activities taken from the columns of Scientific American. Some of the activities are simple and requirellittle development, but the major- ity, although starting from a simple idea are developed into significant, and sometimes, unexpected depths. Those teachers who, like Martin Gardner, believe that a good mathematical puzzle can stimulate a child's imagination faster than a practical applica-

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A series of eight booklets based upon the BBC radio series of the same name, for chil- dren in the age range 8-11 years. Each booklet begins with a simple narrative version

Mathematics in School, May 1982

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