On Nov 22, 2012, at 9:44 AM, Thomas Colthurst <thomaswc@gmail.com> wrote:
The n=6 code reminds me that I had a question about the product construction of unique union codes: how does it work exactly? For example, how do you put two copies of the n=3 code (000, 001, 010, 100) together to get an n=6 code?
-Thomas C
I was wrong about Grossman's construction being a product code in the original sense of a "unique union code", but it works in Michael's generalization, where there is a parameter k which bounds the equivocation, i.e. number of codewords that can produce a given pairwise union. Say we have code 1 with parameters (n1,k1) and code 2 with parameters (n2,k2), then the product code has parameters (n1+n2,k1*k2). Let M(n,k) be the maximum size code with parameters (n,k); we then have the product-code bounds M(10,4) >= M(2,2)M(8,2) = 3*13 = 39 M(10,4) >= M(3,2)M(7,2) = 4*10 = 40 M(10,4) >= M(4,2)M(6,2) = 5*8 = 40 M(10,4) >= M(5,2)M(5,2) = 6*6 = 36 The best gives a bottle-elimination ratio 4/40 = 1/10, which is beat by your size 21 (10,2)-code (ratio 2/21) and Michael's size 46 (10,4)-code (ratio 4/46). -Veit