Fibonacci coding.

In [2]:
```@memoize
def fib(n):
if n < 2:
return n
return fib(n - 1) + fib(n - 2)```
In [3]:
`fib(42)`
Out [3]:
`267914296`
In [4]:
```@memoize
def largest_fib(target):
n = 2
while fib(n + 1) <= target:
n += 1
return n, fib(n)```
In [5]:
`largest_fib(255)`
Out [5]:
`(13, 233)`
In [6]:
```@memoize
def encode_num(N):
digits, f = largest_fib(N)
encoded_bits = [0] * digits
while N:
n, f = largest_fib(N)
N = N - f
encoded_bits[digits-n-2] = 1
encoded_bits[-1] = 1
return encoded_bits```

# Appendix A: Comparisons

Let’s compare Fibonacci coding with the baseline and some common solutions. We can plot how many bits are required as the number we encode grows.

In [7]:
```def compare(low, high):
plt.plot([len(encode_num(x)) for x in range(low, high)], label="Fibonacci")
plt.plot([len(varint(x)) * 8 for x in range(low, high)], label="varint")
plt.plot([math.ceil(math.log(x, 2)) for x in range(low, high)], label="Direct binary")

plt.legend()```
In [8]:
`compare(1, 256)`
Out:
In [9]:
`compare(1, 8196)`
Out: