# Binomial Number System
## 2 Theory of BNS
\begin{table}
\begin{tabular}{c c c} \hline \hline
**No** & **Binomial Codeword** & **Quantitative Equivalent** \\ \hline
0 & 00 & \(0C_{4}^{4}+0C_{4}^{4}\) \\
1 & 010 & \(0C_{5}^{4}+1C_{4}^{4}+0C_{3}^{3}\) \\
2 & 0110 & \(0C_{5}^{4}+1C_{4}^{4}+1C_{3}^{3}+0C_{2}^{2}\) \\
3 & 01110 & \(0C_{5}^{4}+1C_{4}^{4}+1C_{3}^{3}+1C_{2}^{2}+0C_{1}^{1}\) \\
4 & 01111 & \(0C_{5}^{4}+1C_{4}^{4}+1C_{3}^{3}+1C_{2}^{2}+1C_{1}^{1}\) \\
5 & 100 & \(1C_{5}^{4}+0C_{4}^{4}+0C_{3}^{3}\) \\
6 & 1010 & \(1C_{5}^{4}+0C_{4}^{4}+1C_{3}^{3}+0C_{2}^{2}\) \\
7 & 10110 & \(1C_{5}^{4}+0C_{3}^{4}+1C_{3}^{3}+1C_{2}^{2}+0C_{1}^{1}\) \\
8 & 10111 & \(1C_{5}^{4}+0C_{4}^{4}+1C_{3}^{3}+1C_{2}^{2}+1C_{1}^{1}\) \\
9 & 1100 & \(1C_{5}^{4}+1C_{3}^{4}+0C_{3}^{3}+0C_{2}^{2}\) \\
10 & 11010 & \(1C_{5}^{4}+0C_{3}^{4}+0C_{3}^{3}+1C_{2}^{2}+0C_{1}^{1}\) \\
11 & 11011 & \(1C_{5}^{4}+1C_{3}^{4}+0C_{3}^{3}+1C_{2}^{2}+1C_{1}^{1}\) \\
12 & 11100 & \(1C_{5}^{4}+1C_{3}^{4}+1C_{3}^{2}+0C_{2}^{2}+0C_{1}^{1}\) \\
13 & 11101 & \(1C_{5}^{4}+1C_{4}^{4}+1C_{3}^{2}+0+1C_{1}^{1}\) \\
14 & 1111 & \(1C_{5}^{4}+1C_{4}^{4}+1C_{3}^{2}+1C_{2}^{1}\) \\ \hline \hline \end{tabular}
\end{table} 
Table 1: Binomial non-uniform codewords with parameters \(n=6\), \(k=4\) and quantitative equivalents. (LSB). The codewords in the considered example, containing \(k=4\) ones and \(1\) in the LSB, are determined by the binomial coefficient \(C_{5}^{4}=5\). These are the codewords--01111, 10111, 11011, 11101, 11111. The codeword \(1111=1C_{5}^{4}+1C_{4}^{3}+1C_{3}^{2}+1C_{2}^{1}=14\) in this example is the largest. For the remaining codewords, based on the system of Equations (4) and (5), with the number of units \(q=0\), code length \(r=n-k+q=6-4+0=2\), for \(q=k-1=4-1=3\), \(r=n-k+q=n-k+k-1=n-1=6-1=5\). Moreover, all these codewords must contain \(0\) in the LSB. Consequently, the code length \(r\) of the binomial code for equalities (4) and (5) varies from \(n-k\) to \(n-1\) value. For the number of ones \(q=0\), as follows from equality (4), codeword obtained encoding \(0\) and consisting only of \(n-k=6-4=2\) zeros. The presence of such a zero-code combination for positional numbers of any class is required. The codewords with \(0\) in the LSB, for the example under consideration in Table 1, are determined by the binomial coefficient \(C_{5}^{2}=10\). These are the codewords--00, 010, 0110, 01110, 100, 1010, 10110, 1100, 11100, 11100. The code combination \(00=0C_{5}^{4}+0C_{4}^{4}=0\) in this example is the smallest. Summing the numbers of codewords, the LSB containing \(0\) and \(1\), we obtain the range of binomial codes for the parameters \(n=6\) and \(k=4-C_{5}^{2}=C_{5}^{3}+C_{5}^{4}=C_{6}^{4}=15\). In Table 1, the \(r\) of binomial codes in classes containing \(0\) and \(1\) in the LSB does not exceed the value \(n-1=5\). If the codewords in the LSB contain \(1\), then it must contain \(q=k=4\) ones, if the LSB contains \(0\), then the number of \(q\) units changes from \(0\) to \(k-1=4-1=3\), wherein the number of zeros constant and equals \(n-k=6-4=2\). Violation of these conditions is a sign of errors in binomial codes. For example, if binary code combinations represented by binomial codes have the form \(11111\), \(000\), \(0001\), they contain errors since they belong to the code combinations prohibited for binomial codes. In the first case, the sequence has more than five ones, and in the second and third, there are more than two zeros in the code combination. The above conditions for binomial codes forbid both cases. However, the indicated errors in binomial codes detect if markers of the MSB and LSB of the binomial code combinations. These markers indicate using zeros, which contain the LSB that transforms non-uniform binomial code into uniform code, with length \(r=n-1\), shown in Table 2.  
Thus, non-uniform binomial codewords are binomial codewords. Binomial uniforms codewords are converted binomial code combinations with added redundancy.