Algorithm 3 Decoding strategy at the destination node

1: Destination node $\mathit{D}$ receives a packet $\mathit{pkg}\mathit{3}$.

2: $\mathit{D}$ gets frame number $\mathit{N}$ from $\mathit{pkg}\mathit{3}$. 3: If $\mathit{N}\mathbf{=}\mathbf{=}\mathit{DFlag}$ 4:    $\mathit{D}$ discards $\mathit{pkg}\mathit{3}$. 5:  Else 6:    $\mathit{D}$ gets row vector ${\mathit{P}}_{\mathit{i}}$ and coded data slice ${\mathit{Y}}_{\mathit{i}}$ from $\mathit{pkg}\mathit{3}$. 7:    If $\mathit{N}\mathbf{=}\mathbf{=}\mathit{Flag}$

8:      $\mathit{r}\mathit{1}\mathbf{=}\mathit{rank}\mathit{\left(}\mathit{G}\mathit{\right)}$ 9:      ${\mathit{r}\mathit{2}\mathbf{=}\mathit{rank}\mathit{(}\mathit{P}}_{\mathit{i}}\cup \mathit{G}\mathit{)}$ 10:      If $\mathit{r}\mathit{2}\mathbf{>}\mathit{r}\mathit{1}$ do 11:        Put ${\mathit{P}}_{\mathit{i}}$ into $\mathit{G}$

12:        Put ${\mathit{Y}}_{\mathit{i}}$ into existing encoded data $\mathit{H}$.

13:        $\mathit{Count}\mathbf{+}\mathbf{+}$;

14:        If $\mathit{Count}\mathbf{=}\mathbf{=}\mathit{K}$

15:          $\mathit{D}$ calculates the inverse matrix ${\mathit{G}}^{-\mathit{1}}$ of $\mathit{G}$. 16:          $\mathit{D}$ uses ${\mathit{G}}^{-\mathit{1}}$ to decode $\mathit{H}$ into raw data $\mathit{P}$.

17:          $\mathit{D}$ sends $\mathit{P}$ to the play module. 18:          $\mathit{DFlag}\mathbf{=}\mathit{N}$ 19:        Endif 20:      Endif

21:    Else

22:      $\mathit{Flag}\mathbf{=}\mathit{N}$

23:        $\mathit{D}$ gets $\mathit{K}$ from $\mathit{pkg}\mathit{3}$. 24:        Clear matrix G and encoded data H

25:        $\mathit{Count}\mathbf{=}\mathit{0}$

26:        goto step 6

27: Endif