## 3 Proposed CS-DDSG Scheme
### Does the Measurement Matrices Satisfy RIP?  
The structure of measurement matrix \(\Phi\) is random and relies on the receiving nodes. Thus, CS-DDSG avoids measuring the lost nodes and avoids the packet loss. The question is: Does \(\Phi\) obey RIP to utilize the CS theory? Unfortunately, it is an NP-hard problem to prove the RIP property of a matrix. However, [PERSON] et al. [21] reported that recovery performance can be guaranteed with high probability when the rows of the measurement matrix are linearly independent. We investigate this proposition below.  
The rows of \(\Phi\) are linearly dependent when one of the following two situations occurs.  
**Case 1.**_Any row \(\varphi_{k}\) can be expressed as a linear combination of other rows._  
**Proof.** The measurement coefficient is 1; thus, if \(\mathbf{\varphi}_{k}\) can be expressed as a linear combination of rows \(\mathbf{\varphi}_{k_{1}},\cdots,\mathbf{\varphi}_{k_{q}}\), \(\mathbf{q=2,\cdots,N-1}\), they satisfy the following:  
\[\mathbf{\varphi}_{k}=\mathbf{\varphi}_{k_{1}}+\cdots+\mathbf{\varphi}_{k_{q}}. \tag{17}\]  
Suppose \(\mathbf{I}_{k}=\{j|\mathbf{\varphi}_{k,j}\
eq\mathbf{0}\}\) and \(\mathbf{I}_{k_{i}}=\{j|\mathbf{\varphi}_{k,ij}\
eq\mathbf{0}\}\); if the condition of Equation (17) is satisfied, then \(\mathbf{I}_{k}=\cup_{i=1}^{q}\mathbf{I}_{k_{i}}\) and we can obtain  
\[|\mathbf{I}_{k}|>|\mathbf{I}_{k_{i}}|,\mathbf{i}\in[\mathbf{1},\mathbf{q}], \tag{18}\]  
where \(|\cdot|\) denotes the number of elements in the set. Thus, Equation (17) can be satisfied when one of the following two situations occurs. The first situation would occur if node \(\mathbf{k}\) were to receive packets from nodes \(\mathbf{k_{1}},\cdots,\mathbf{k_{q}}\) and merge their packets. However, this situation contradicts the reception condition under which each node receives one packet. Thus, the condition of Equation (17) cannot occur.  
The second situation would occurs when node \(\mathbf{k_{2}}\) receives a packet from node \(\mathbf{k_{1}}\) and node \(\mathbf{k_{3}}\) receives a packet from node \(\mathbf{k_{2}}\). It follows that node \(\mathbf{k_{q}}\) receives a packet from node \(\mathbf{k_{q-1}}\). Finally, node \(\mathbf{k}\) receives the packet from node \(\mathbf{k_{q}}\). According to the condition in Equation (10), \(\mathbf{I}_{\mathbf{k_{q}}}\) satisfies the following:  
\[\mathbf{I}_{\mathbf{k_{q}}}=\{\mathbf{k_{q}}\}\cup\Big{(}\cup_{i=1}^{q-1}\mathbf{I}_{k_{i}} \Big{)}. \tag{19}\]  
After node \(\mathbf{k}\) updates its packet, \(\mathbf{I}_{k}\) satisfies:  
\[\mathbf{I}_{\mathbf{k}}=\{\mathbf{k}\}\cup\mathbf{I}_{\mathbf{k_{q}}}. \tag{20}\]  
Obviously, \(\mathbf{k}\in\mathbf{I}_{\mathbf{k}}\) but \(\mathbf{k}\
otin\mathbf{I}_{\mathbf{k_{q}}}\). Thus, \(\mathbf{I}_{\mathbf{k}}=\{\mathbf{k}\}\cup\Big{(}\cup_{i=1}^{q}\mathbf{I}_{k_{i}}\Big{)}\) and Equation (17) is false.  
Consequently, it can be concluded that no rows can be linearly expressed by other rows. \(\Box\)  
**Case 2.**_Any two rows \(\mathbf{\varphi}_{i}\) and \(\mathbf{\varphi}_{j}\) are linearly dependent_.  
**Proof.**\(\mathbf{\varphi}_{i}\) and \(\mathbf{\varphi}_{j}\) are linearly dependent if and only if they are precisely the same. However, according to the reception condition, each node receives only one packet and merges with its own unique packet. Therefore, although node \(\mathbf{i}\) and node \(\mathbf{j}\) may receive the same broadcasting packet from a common neighboring node, their packets will still be different. Therefore, none of the rows are linearly dependent. \(\Box\)  
In conclusion, the rows of the measurement matrix \(\mathbf{\Phi}\) are linearly independent; consequently, in CS-SSDG, \(\mathbf{X}\) can be reconstructed from \(\mathbf{Y}\) with a very high probability.