Simplification of a circuit
I'll show you the general mathematical approach and you'll proceed. 

For the purpose of the tutorial lets take the upper part of your circuit. I rearranged it a bit and you need to check that there is no mistake there: 







simulate this circuit &ndash; Schematic created using CircuitLab

Lets define the current through each resistor as flowing downwards. In this case, the following matrix is the complete representation of the circuit (and if it isn't - find additional constraints and it will be):

$$\begin{bmatrix}
 1&amp;-1  &amp;-1  &amp;0  &amp;0  &amp;0  &amp;0  &amp;0  &amp;0 \\ 
 0&amp;1  &amp;0  &amp;-1  &amp;-1  &amp;0  &amp;0  &amp;0  &amp;0 \\ 
 0&amp;0  &amp;0  &amp;1  &amp;0  &amp;-1  &amp;0  &amp;-1  &amp;0 \\ 
 0&amp;0  &amp;0  &amp;0  &amp;1  &amp;0  &amp;1  &amp;0  &amp;-1 \\ 
 0&amp;0  &amp;1  &amp;0  &amp;0  &amp;1  &amp;-1  &amp;0  &amp;0 \\ 
 &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp; \\ 
 0&amp;R_2  &amp;-R_3  &amp;R_4  &amp;0  &amp;R_6  &amp;0  &amp;0  &amp;0 \\ 
 0&amp;-R_2  &amp;R_3  &amp;0  &amp;-R_5  &amp;0  &amp;R_7  &amp;0  &amp;0 \\
\end{bmatrix}* \begin{bmatrix}
I_1\\ 
I_2\\ 
I_3\\ 
I_4\\ 
I_5\\ 
I_6\\ 
I_7\\ 
I_8\\ 
I_9\\
\end{bmatrix} = 0$$

Now just use any program which is capable of solving the systems of linear equations and you can get all the currents in terms of any pair of currents (lets say \$I_1\$ and \$I_2\$ for the sake of this tutorial). You can even solve this system by hand (are you crazy???).

Afterwards, you'll have to do few more calculations because you did not provide any boundary conditions. The additional equations are:

$$V_{in}-V_{out1}=I_1R_1+I_2R_2+I_4R_4+I_8R_8$$

$$and$$

$$V_{in}-V_{out2}=I_1R_1+I_3R_3+I_7R_7+I_9R_9$$

EXPLANATION:

The first five equations in the matrix are KCLs for inner nodes (sum of currents in each node is zero), and the last two equations in the matrix are KVLs for inner loops which introduce voltage constraints (sum of voltages on each closed loop is zero).

Once all the currents are known in terms of \$I_1\$ and \$I_2\$, and the boundary conditions are given, the last two equations (not in matrix) may be solved to obtain \$I_1\$ and \$I_2\$.

Once you know all the currents (and consequently all the voltages) in the network - it is solved. 

IN CASE YOU STILL WANT SIMPLIFIED SCHEMATICS:

You won't be able to separate the branches leading to out1 and out2. The reason is that they share a common nodes, and the change in boundary conditions on one node will alter the currents in the branch of the other too. However, you might find some parts of your network which can be simplified. For example, in your original schematic, it seems like you might be able to represent \$I_{A}\$ and \$I_{B}\$ in terms of \$I_8\$ and \$I_9\$ in a quite simple manner, and reduce the bottom ladder of the resistors to just a few resistors. However, you still need to solve the above equations as prerequisite.

NOTE:

If either A or B can be defined a constant, it can simplify the issue a lot.