hi, Okay- I need 5 bits to represent 32 coins.I count as coin 0,coin 1,... coin 31. If it is a perfectly random fair coin throwing experiment,then 50 percent of them will be heads. So I know that 16 of them will be heads. What we do is i simply place all the 32 coins on the table in a row or column. I look at the first coin and determine if it is a head or a tail. I repeat the same proccess till i count 16 heads. If I count 15 heads at coin 31, then I cant reduce the entropy. How ever, if i count 16 heads at coin 30,then I dont have to check that coin 31,I already know its a tail,so I have less than 5 bits of entropy. So if it is a perfectly random experiment,I wouldn't get 16 heads before i look at coin 31,which is the last coin and thats what you said-isn't it? So how did chaitin get to compress the information from k instances of the turing machine in http://www.cs.umaine.edu/~chaitin/summer.html under the sub-section redundant? he says- "Is this K bits of mathematical information? K instances of the halting problem will give us K bits of Turing's number. Are these K bits independent pieces of information? Well, the answer is no, they never are. Why not? Because you don't really need to know K yes/no answers, it's not really K full bits of information. There's a lot less information. It can be compressed. Why? " If the input programs are truely random-there is no redundancy and thats a contradiction to the claim in the paper. Thanks. Regards Sarath.
It's simple, if I am correct. The redundancy simply makes you care less about the specific instance you are looking at.
To represent 32 coins-i need 5 bits of information. Since the experiment is truely random-i know half of them will be heads,so in this case using 5 bits of information,i can determine all the coins that are heads and that are tails.
Same deal, unless you are counting pairs, in which case you cannot distinguish between the members of a pair. You need an extra bit to tell a head from a tail.
So-the question is what is the minimum number of bits or entropy required to determine which all coins are heads and which all coins are tails,is it 5 bits or 6 bits of information?
With 5 bits, you can count to 31, so you need 6.
Just my two tails.
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