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Can anyone give an example of 7 binary covering numbers that cover all 32 possible 5 digit binary numbers with an error of up to 1 place? I have proved that 6 binary numbers do not exist, but I am having trouble finding 7 that do. I have combed the interweaves but to no avail. Thank you so much!

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I think what you're asking for is a set of 7 5-bit numbers such that every 5-bit number agrees with at least one of your 7 numbers in at least 4 of its bits. Is that a correct understanding of your question? –  Gerry Myerson Feb 9 '12 at 2:56
    
Yes that is correct. And I have already proven there does not exist a set of 6 such numbers. –  Karl Feb 9 '12 at 2:58

1 Answer 1

Seems like a small enough problem to try to do by just doing it.

Write down the 32 5-bit numbers.

Circle one of them, then cross out all the ones that differ from it in exactly one place.

Now pick a number that hasn't been crossed out, circle it, and cross out all the ones that differ from it in exactly one place.

Repeat until everything is crossed out or circled. If only 7 numbers are circled, you win. If not, go back a couple of steps and see if you could have made some cleverer choices.

I think before asking for help, or searching the web, you should at least try this approach. It might work, or it might give you a better appreciation of the difficulties involved.

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