What if we send the total unchanged?

The last slide ended with hidden 1 holding a weighted total of .50.

If the unit sends .50 onward unchanged, its output weight of 25 makes hidden 1 contribute:

.50 x 25 = $12.50

We are isolating hidden 1 here. The complete price guess also receives contributions from the other three hidden units.

In that case, the middle unit is an unnecessary extra step. We can calculate the same $12.50 directly from the four tastes.

With the middle unit:

(1 x .20 + 0 x .10 + 1 x .30 + 0 x .20) x 25 = $12.50

Without the middle unit:

(1 x 5) + (0 x 2.5) + (1 x 7.5) + (0 x 5) = $12.50

Each direct weight combines one incoming weight with the outgoing 25. For example, the cabernet path becomes:

.20 x 25 = 5

Press fold weights. Watch the amount on the far right. Before the fold, hidden 1 adds $12.50. After the fold, the direct wires still add $12.50.

If every hidden unit only passed its total onward unchanged, we could fold away the entire middle row. We would be back to the old machine that could not separate the pairings.

The middle row needs to do something that cannot be replaced by one more set of direct weights. It needs a in the calculation.