24 | April | 2008 | Yet Another Mathblog

This is a well-known trick but I’m posting it here since I often have a hard time tracking it down when I need it for an introductory coding theory talk or something. Hopefully someone else will find it amusing too!

Let $F = GF(2)$ and let $C$ be the set of all vectors in the third column below (for simplicity, we omit commas and parentheses, so $0000000$ is written instead of $(0,0,0,0,0,0,0)$ , for example).

The [7,4,3] Hamming code
decimal	binary	Hamming [7,4]
0	0000	0000000
1	0001	0001110
2	0010	0010101
3	0011	0011011
4	0100	0100011
5	0101	0101101
6	0110	0110110
7	0111	0111000
8	1000	1000111
9	1001	1001001
10	1010	1010010
11	1011	1011100
12	1100	1100100
13	1101	1101010
14	1110	1110001
15	1111	1111111

This is a linear code of length 7, dimension 4, and minimum distance 3. It is called the Hamming [7,4,3]-code.
In fact, there is a mapping from $F^4$ to $C$ given by $\phi(x_1,x_2,x_3,x_4)={\bf y}$ , where
${\bf y}= \left( \begin{array}{c} y_1 \\ y_2 \\ y_3 \\ y_4 \\ y_5 \\ y_6 \\ y_7 \end{array} \right) = \left( \begin{array}{cccc} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \\ 1&0&1&1 \\ 1&1&0&1 \\ 1&1&1&0 \end{array} \right) \left( \begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \end{array} \right) \ \pmod{2}$ . Moreover, the matrix $G=\left(\begin{array}{ccccccc} 1&0&0&0&1&1&1 \\ 0&1&0&0&0&1&1 \\ 0&0&1&0&1&0&1 \\ 0&0&0&1&1&1&0 \end{array}\right)$ is a generator matrix.

Now, suppose a codeword $c \in C$ is sent over a noisy channel and denote the received word by ${\bf w}=(w_1,w_2,w_3,w_4,w_5,w_6,w_7)$ . We assume at most 1 error was made.

Put $w_i$ in the region of the Venn diagram above associated to coordinate $i$ in the table below, $i=1,2,...,7$ .
Do parity checks on each of the circles $A$ , $B$ , and $C$ .

Decoding table
parity failure region(s)	error position
none	none
A, B, and C	1
B and C	2
A and C	3
A and B	4
A	5
B	6
C	7

Exercise: Decode ${\bf w} = (0,1,0,0,0,0,1)$ in the binary Hamming code using this algorithm.
(This is solved at the bottom of this column.)

In his 1976 autobiography Adventures of a Mathematician, Stanislaw Ulam [U] poses the following problem:

Someone thinks of a number between one and one million (which is just less than 220). Another person is allowed to ask up to twenty questions, to which the first person is supposed to answer only yes or no. Obviously, the number can be guessed by asking first: Is the number in the first half-million? and then again reduce the reservoir of numbers in the next question by one-half, and so on. Finally, the number is obtained in less than log2(1000000). Now suppose one were allowed to lie once or twice, then how many questions would one need to get the right answer?

We define Ulam’s Problem as follows: Fix integers $M\geq 1$ and $e\geq 0$ . Let Player 1 choose the number from the set of integers $0,1, ...,M$ and Player 2 ask the yes/no questions to deduce the number, to which Player 1 is allowed to lie at most e times. Player 2 must discern which number Player 1 picked with a minimum number of questions with no feedback (in other words, all the questions must be provided at once)

As far as I know, the solution to this version of Ulam’s problem without feedback is still unsolved if e=2. In the case e=1, see [N] and [M]. (Also, [H] is a good survey.)

Player 1 has to convert his number to binary and encode it as a Hamming [7,4,3] codeword. A table containing all 16 codewords is included above for reference. Player 2 then asks seven questions of the form, “Is the i-th bit of your 7-tuple a 1?” Player 1’s answers form the new 7-tuple, $(c_1,c_2,...,c_7)$ , and each coordinate is placed in the corresponding region of the circle. If Player 1 was completely truthful, then the codeword’s parity conditions hold. This means that the 7-tuple generated by Player 2’s questions will match a 7-tuple from the codeword table above. At this point, Player 2 just has to decode his 7-tuple into an integer using the codeword table above to win the game. A single lie, however, will yield a 7-tuple unlike any in the codeword table. If this is the case, Player 2 must error-check the received codeword using the three parity check bits and the decoding table above. In other words, once Player 2 determines the position of the erroneous bit, he corrects it by “flipping its bit”. Decoding the corrected codeword will yield Player 1’s original number.

References

[H] R. Hill, “Searching with lies,” in Surveys in Combinatorics, ed. by P. Rowlinson, London Math Soc, Lecture Notes Series # 218.

[M] J. Montague, “A Solution to Ulam’s Problem with Error-correcting Codes”

[N] I. Niven, “Coding theory applied to a problem of Ulam,” Math Mag 61(1988)275-281.

[U] S. Ulam, Adventures of a mathematician, Scribner and Sons, New York, 1976 .

Solution to exercise using SAGE:

sage: MS = MatrixSpace(GF(2), 4, 7) sage: G = MS([[1,0,0,0,1,1,1],[0,1,0,0,0,1,1],[0,0,1,0,1,0,1],[0,0,0,1,1,1,0]]) sage: C = LinearCode(G) sage: V = VectorSpace(GF(2), 7) sage: w = V([0,1,0,0,0,0,1]) sage: C.decode(w) (0, 1, 0, 0, 0, 1, 1)

In other words, Player picked 4 and lied on the 6th question.

Here is a Sage subtlety:
sage: C = HammingCode(3, GF(2)) sage: C Linear code of length 7, dimension 4 over Finite Field of size 2 sage: V = VectorSpace(GF(2), 7) sage: w = V([0,1,0,0,0,0,1]) sage: C.decode(w) (1, 1, 0, 0, 0, 0, 1)
Why is this decoded word different? Because the Sage Hamming code is distinct (though “equivalent” to) the Hamming code we used in the example above.

Unfortunately, SAGE does not yet do code isomorphisms (at least not as easily as GUAVA), so we use GUAVA’s CodeIsomorphism function, which calls J. Leon’s GPL’s desauto C code function:

gap> C1 := GeneratorMatCode([[1,0,0,1,0,1,0],[0,1,0,1,0,1,1],[0,0,1,1,0,0,1],[0,0,0,0,1,1,1]],GF(2)); a linear [7,4,1..3]1 code defined by generator matrix over GF(2) gap> C2 := GeneratorMatCode([[1,0,0,0,1,1,1],[0,1,0,0,0,1,1],[0,0,1,0,1,0,1],[0,0,0,1,1,1,0]],GF(2)); a linear [7,4,1..3]1 code defined by generator matrix over GF(2) gap> CodeIsomorphism(C1,C2); (2,6,7,3) gap> Display(GeneratorMat(C1)); 1 . . 1 . 1 . . 1 . 1 . 1 1 . . 1 1 . . 1 . . . . 1 1 1 gap> Display(GeneratorMat(C2)); 1 . . . 1 1 1 . 1 . . . 1 1 . . 1 . 1 . 1 . . . 1 1 1 .

Now we see that the permutation (2,6,7,3) sends the “SAGE Hamming code” to the “circle Hamming code”:

sage: H = HammingCode(3, GF(2)) sage: G = MS([[1,0,0,0,1,1,1],[0,1,0,0,0,1,1],[0,0,1,0,1,0,1],[0,0,0,1,1,1,0]]) sage: C = LinearCode(G) sage: C.gen_mat() [1 0 0 0 1 1 1] [0 1 0 0 0 1 1] [0 0 1 0 1 0 1] [0 0 0 1 1 1 0] sage: S7 = SymmetricGroup(7) sage: g = S7([(2,6,7,3)]) sage: H.permuted_code(g) == C True

Yet Another Mathblog

Daily Archives: 2008/04/24

“Circle decoding” of the [7,4,3] Hamming code