Implementing a (5,5;4) counter with FA and HA

Question

I am trying to implement a (5,5;4)-counter using only FA's and HA's as building blocks. I tried using both Wallace and Dadda tree structures, but I was never able to end up with a 4-bit number as a result. I keep on obtaining 5-bit outputs.

The counter that I would like to implement looks like this: Dot notation for a (5,5;4) counter

Here is my attempt at implementing it with FA's and HA's. A diagonal stripe across a connection between two dots is the result of a half adder, other connections are full adder results.

See My attempt at a (5,5;4) counter implementation.

I know it is possible to obtain only a 4-bit word at the end, because the addition of five numbers of $(11)_{two}$ results in $(1111)_{two}$.

Thanks in advance.

Answer 1

I managed to solve the problem by applying only 2 FA at the first step, then in the next iteration again 2 FA, then 1 FA, and finally another FA. I then successfully obtain a 4-bit wide number.