Level 8: 8-bit Adder

Try building the circuit yourself first. The solution below is a way to check yourself if you're stuck or have already solved it.

The in-game hint on this level inserts a scaffold: the circuit without two wires — the A bus input to the first FullAdder and the carry between the first and second adders. Connect them yourself: the scaffold deducts half the penalty and shows where the level principle lives.

Notice the carry chain: bit 7 (the most significant) is computed last — the carry must "run" through all 8 adders. That is propagation delay: the longer the chain, the later the result settles and the lower the maximum clock frequency. See the Propagation Delay article.

What you should get: buses A and B add bit by bit, and the carry passes from the low digit to the high one.

1. Place two Splitters and feed buses A and B into them. Each outputs 8 bits: Bit0 is the low bit, Bit7 the high one.
2. Place 8 FullAdders vertically, one under another. Feed each its bits: FullAdder#1 — A.Bit0 and B.Bit0, FullAdder#2 — A.Bit1 and B.Bit1, and so on up to FullAdder#8.
3. Connect the carry chain: the first adder's CarryOut goes into the second's CarryIn, and so on. Leave the first CarryIn unconnected (it defaults to 0).
4. Feed the Sum outputs of all 8 adders to the Maker inputs (Sum#1→Bit0 … Sum#8→Bit7). The last adder's CarryOut is Carry.

Boundary vectors deserve special attention: 127+1 gives Sum=128 and Carry=0 (signed-range overflow: the top bit of the result is the sign bit), 200+100 gives Sum=44 and Carry=1 (an unsigned carry past 255), and 255+1 gives Sum=0 and Carry=1. The same adder — two different meanings for its outputs.