In two previous posts I discussed the simple and composite kinematics of three-body problems. The composite Sudakov decomposition is done in two two-body stages. First, you write
$$ p_{1} = n_{1} + \frac{m_{1}^{2}}{S_{12}} n_{2}, \qquad p_{2} = n_{2} + \frac{m_{2}^{2}}{S_{12}} n_{1}.$$
This is just a two-body Sudakov decomposition of two of the three momenta. Next, you perform another two-body decomposition with a 2-composite vector and the remaining simple vector:
$$ p_{12} = n_{12} + \frac{s_{12}}{S_{(12)3}} n_{3}, \qquad p_{3} = n_{3} + \frac{m_{3}^{2}}{S_{(12)3}} n_{12}.$$
Here \(n_{1}\), \(n_{2}\), \(n_{12}\), and \(n_{3}\) are null vectors, and I have introduced two 2-Sudakov invariants:
$$ S_{12} \equiv \left\Vert n_{1} + n_{2} \right\Vert^{2} = 2 \left( n_{1} \cdot n_{2} \right), \qquad S_{(12)3} \equiv \left\Vert n_{12} + n_{3} \right\Vert^{2} = 2 \left( n_{12} \cdot n_{3} \right).$$
From the familiar two-body Sudakov decomposition you have
$$ S_{12} = \frac{1}{2} \left( s_{12} - m_{1}^{2} - m_{2}^{2} + \sqrt{ \left( s_{12} - m_{1}^{2} - m_{2}^{2} \right)^{2} - 4 m_{1}^{2} m_{2}^{2} } \right); $$
$$ S_{(12)3} = \frac{1}{2} \left( s_{123} - s_{12} - m_{3}^{2} + \sqrt{ \left( s_{123} - s_{12} - m_{3}^{2} \right)^{2} - 4 s_{12} m_{3}^{2} } \right). $$
But you can further introduce two other 2-Sudakov invariants:
$$ S_{23} \equiv \left\Vert n_{2} + n_{3} \right\Vert^{2} = 2 \left( n_{2} \cdot n_{3} \right), \qquad S_{31} \equiv \left\Vert n_{3} + n_{1} \right\Vert^{2} = 2 \left( n_{3} \cdot n_{1} \right).$$
The goal here is to write these 2-Sudakov invariants in terms of the Mandelstam invariants and the masses. Using
$$ p_{12} \equiv p_{1} + p_{2} = \left(1 + \frac{m_{2}^{2}}{S_{12}} \right) n_{1} + \left(1 + \frac{m_{1}^{2}}{S_{12}} \right) n_{2} \quad \Longrightarrow \quad n_{12} = \left(1 + \frac{m_{2}^{2}}{S_{12}} \right) n_{1} + \left(1 + \frac{m_{1}^{2}}{S_{12}} \right) n_{2} - \frac{s_{12}}{S_{(12)3}} n_{3}, $$
you find that
$$ p_{3} = \frac{m_{3}^{2}}{S_{(12)3}} \left(1 + \frac{m_{2}^{2}}{S_{12}} \right) n_{1} + \frac{m_{3}^{2}}{S_{(12)3}} \left(1 + \frac{m_{1}^{2}}{S_{12}} \right) n_{2} + \left(1 - \frac{s_{12} m_{3}^{2}}{S_{(12)3}^{2}} \right) n_{3}. $$
At this stage you introduce Regge-Sudakov invariants:
$$ R_{12} \equiv \frac{S_{12}}{m_{1} m_{2}}, \qquad R_{23} \equiv \frac{S_{23}}{m_{2} m_{3}}, \qquad R_{31} \equiv \frac{S_{31}}{m_{3} m_{1}}, \qquad R_{(12)3} \equiv \frac{S_{(12)3}}{\sqrt{s_{12}} m_{3}}. $$
Using the identities
$$ r_{23} = \frac{p_{2} \cdot p_{3}}{m_{2} m_{3}}, \qquad r_{31} = \frac{p_{3} \cdot p_{1}}{m_{3} m_{1}}; $$
you find the relations
$$ 2r_{23} = \frac{m_{2}}{ \sqrt{s_{12}} R_{(12)3}} \left( 2 + \frac{m_{1} R_{12}}{m_{2}} + \frac{m_{1}}{m_{2} R_{12}} \right) + \left( 1 - \frac{1}{R_{(12)3}^{2}}\right) \left( R_{23} + \frac{R_{31}}{R_{12}} \right), $$
$$ 2r_{31} = \frac{m_{1}}{ \sqrt{s_{12}} R_{(12)3}} \left( 2 + \frac{m_{2} R_{12}}{m_{1}} + \frac{m_{2}}{m_{1} R_{12}} \right) + \left( 1 - \frac{1}{R_{(12)3}^{2}}\right) \left( R_{31} + \frac{R_{23}}{R_{12}} \right). $$
Note that \(R_{23}\) and \(R_{31}\) appear only in a linear way. You can easily solve for \(R_{23}\) and \(R_{31}\), but the important point is that this "recursive" decomposition is very different from the contiguous decomposition presented earlier.
