Is there a way to obtain information reconciliation efficiency?

I tried something like

```python
summary.number_of_exposed_bits/(summary.output_length*h2(summary.input_error))
```


with 

```python
def h2(x: float) -> float:
    """Binary entropy function."""
    if x <= 0 or x >= 1:
        return 0
    return -x * np.log2(x) - (1 - x) * np.log2(1 - x)
```

but for cascade I get an efficiency bigger than 2 and for winnow smaller than 1.