## 2 Materials and Methods
### Hardness of Decoding Payload in Absence of Disambiguation Information  
The non-UD coding process represents the input data with two elements as the payload and the disambiguation information. We have shown in Lemma 2 that the disambiguation information occupies \(\approx\frac{2}{d}\) of the input size. Thus, if one encrypts the disambiguation information and leaves the payload plain, the retrieval of the original sequence will require decoding the payload without the disambiguation information. We provide a combinatorial analysis on the hardness of this decoding below by first proving that disambiguation information of a payload is specific per se, and then use this fact to count the number of possible distinct raw sequences that map to a given payload with the same code-word mapping of \(\Sigma\to W\). **Lemma 3**. : _Each distinct disambiguation information on a given NPF stream decodes into distinct input sequence._  
Proof. : Let \(S=s_{1}s_{2}\ldots s_{r}\) and \(S^{\prime}=s^{\prime}_{1}s^{\prime}_{2}\ldots s^{\prime}_{r}\) denote two different disambiguation information, which are actually the code-word length sequences such that \(\sum_{i=1}^{r}s_{i}=\sum_{i=1}^{r}s^{\prime}_{i}=|NPF(\mathcal{A})|\), where \(s_{i}\) and \(s^{\prime}_{i}\) are the bit lengths of the \(i\)th code-word in the corresponding decoding of the \(NPF(\mathcal{A})\). Since \(S\) and \(S^{\prime}\) are distinct, for some \(1\leq j\leq r\), \(s_{j}\) and \(s^{\prime}_{j}\) are not equal, which causes the corresponding code-words extracted from the NPF code-stream to be of different length. Although there is a chance that different code-words of length \(d\) may map onto the same symbol in \(W\) according to Definition 2, it is for sure  
\begin{table}
\begin{tabular}{c c c c c c c} \hline \hline
**File Name** & **Payload Size** & **Dis.Info. Size** & **Overhead Bits** & **Payload Size** & **Dis.Info. Size** & **Overhead Bits** \\ \hline chr22 & 1555 & 500 & 7 & 40,961 & 8213 & 22 \\ etex99 & 1515 & 533 & 0 & 41,103 & 8060 & 11 \\ gcc & 1491 & 578 & 21 & 40,764 & 8410 & 22 \\ howtobwt & 1510 & 538 & 0 & 41,058 & 8127 & 33 \\ howto & 1551 & 511 & 14 & 41,079 & 8095 & 22 \\ jdk & 1522 & 540 & 14 & 41,074 & 8122 & 44 \\ rtail & 1535 & 527 & 14 & 41,075 & 8088 & 11 \\ rfc & 1522 & 526 & 0 & 40,769 & 8394 & 11 \\ sprot34 & 1538 & 524 & 14 & 41,049 & 8125 & 22 \\ w3c2 & 1529 & 519 & 0 & 41,016 & 8158 & 22 \\ mp3 & 1470 & 585 & 7 & 41,021 & 8131 & 0 \\ jpg & 1426 & 636 & 14 & 40,819 & 8344 & 11 \\ mp4 & 1415 & 654 & 21 & 41,373 & 7779 & 0 \\ webm & 1496 & 566 & 14 & 40,835 & 8317 & 0 \\ ogv & 1456 & 592 & 0 & 40,985 & 8189 & 22 \\ fv & 1571 & 484 & 7 & 40,796 & 8367 & 11 \\
**Expected** & **1554** & **508** & **14** & **40,986** & **8188** & **22** \\ \hline \multicolumn{6}{c}{\(d=8\)} & \multicolumn{6}{c}{\(d=12\)} \\ \hline chr22 & 917,579 & 131,027 & 30 & 18,873,040 & 2,098,575 & 95 \\ etex99 & 917,289 & 131,302 & 15 & 18,872,686 & 2,098,853 & 19 \\ gcc & 917,397 & 131,209 & 30 & 18,879,975 & 2,091,602 & 57 \\ howtobwt & 917,518 & 131,088 & 30 & 18,875,332 & 2,096,207 & 19 \\ howto & 917,812 & 130,794 & 30 & 18,875,502 & 2,096,037 & 19 \\ jdk & 917,412 & 131,179 & 15 & 18,873,190 & 2,098,406 & 76 \\ rtail & 917,139 & 131,437 & 0 & 18,876,497 & 2,095,042 & 19 \\ rfc & 917,346 & 131,275 & 45 & 18,873,175 & 2,098,364 & 19 \\ sprot34 & 918,158 & 130,433 & 15 & 18,878,705 & 2,092,891 & 76 \\ w3c2 & 917,077 & 130,899 & 30 & 18,876,613 & 2,094,945 & 38 \\ mp3 & 914,926 & 133,695 & 45 & 18,863,837 & 2,107,721 & 38 \\ jpg & 915,821 & 132,770 & 15 & 18,878,914 & 2,092,625 that when the two code-word lengths are different they cannot map onto the same symbol. Thus, it is not possible to find two distinct \(S\) and \(S^{\prime}\) that generate equal \(\mathcal{A}\) sequences. **Lemma 4**.