A special invertible k-bits to n-bits mapping

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The discussion centers on finding the minimum number of bits k needed to create an invertible mapping from k-bits to n-bits that excludes the all 1's sequence in the output. For linear mappings, it is established that k must be at least n-1, as this is the maximum capacity for such mappings. The inquiry extends to non-linear mappings, seeking a method that can achieve this with a lower overhead ratio than k/n > 2/3. However, it is concluded that no invertible mappings can meet these criteria, as they must maintain a one-to-one relationship, meaning the output range cannot have fewer sequences than the input domain. Thus, the search for a suitable non-linear mapping remains unresolved.
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I want a sequence of n bits without all 1's in the output. What is the minimum number of bits k given n that can be used in any linear or non-linear invertible mapping that will produce such a sequence at output. For example, consider n=3. I want to create a mapping that has all 2^3=8 minus 111 sequence. What is the minimum number of bits k required that will produce such a mapping? In this example, for a linear mapping, I can see that k=2. But with k=2, I am loosing 3 sequences at the output and the overhead of this mapping k/n is huge. Is there any other method or non-linear mapping that does this job with an overhead of k/n>2/3. I need a specific answer to n=4 or general answer with any n (reasonable range for me is n=2 to n=16. Please guide me to any keywords that may help in my search for the answer.
 
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ait.abd said:
I want a sequence of n bits without all 1's in the output. What is the minimum number of bits k given n that can be used in any linear or non-linear invertible mapping that will produce such a sequence at output.

You haven't defined the domain of mapping.
You haven't explained what it means to "use" bits in the mapping.
 
Domain of mapping is S=\{\text{All } 2^k \text{ sequences of k-bits}\}.
Range of the mapping is T=\{2^n-1 \text{ sequences of n-bits i.e. All sequences of n-bits excluding the all 1's sequence}\}
By using I mean defining a mapping S to T. In case of linear mapping, the answer is straightforward that the maximum number of bits k required are n-1. My question is whether there are any non-linear invertible mappings/transforms that take k bits and output a sequence of length k bits (not n) that doesn't contain all 1's sequence. I don't see any linear mapping fulfilling this criterion as k \le (n-1) is always true for linear mappings/transform.
*I may not be using mapping and transform words in the strict sense. By mapping or transform I mean is there any t = f(s); t \in T, s \in S; t=t_1,t_2,t_3,...,t_k; s=s_1,s_2,s_3,...,s_n; s_p,t_p \in \{0,1\}
 
ait.abd said:
My question is whether there are any non-linear invertible mappings/transforms that take k bits and output a sequence of length k bits (not n) that doesn't contain all 1's sequence.

There are no invertible mappings like that. An invertible mapping has to be 1-to-1. The range can't have a smaller cardinality than the domain.
 
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