# Exercise 11.4.5

Consider an open-address hash table with a load factor $\alpha$. Find the nonzero value $\alpha$ for which the expected number of probes in an unsuccessful search equals twice the expected number of problems in a successful search. Use the upper bounds given by Theorems 11.6 and 11.8 for these expected number of probes.

Using the theorems, we get the following equation:

$$ \frac{1}{1 - \alpha} = 2 \frac{1}{\alpha} \ln{\left(\frac{1}{1 - \alpha}\right)} $$

We can try to simplify it, but we get to a $x = 2 \ln x - 1$, or something similar, which I don't know how to solve in closed form.

Luckily, that's why SciPY exists.

### Python runner output

alpha = 0.7153318629590794

### Python code

import numpy as np from scipy.optimize import root def fn(a): one = np.array([1.0]) two = np.array([2.0]) return (two / a) * np.log(one / (one - a)) - one/(one - a) print("alpha =", root(fn, 0.6).x[0])