## Issue

My question is very simple.

I wrote this program for pure entertainment. It takes a numerical input and finds the length of every Collatz Sequence up to and including that number.

I want to make it faster algorithmically or mathematically (i.e. I know I could make it faster by running multiple versions parallel or by writing it in C++, but where’s the fun in that?).

Any and all help is welcome, thanks!

EDIT:

Code further optimized with the help of dankal444

```
from matplotlib import pyplot as plt
import numpy as np
import numba as nb
# Get Range to Check
top_range = int(input('Top Range: '))
@nb.njit('int64[:](int_)')
def collatz(top_range):
# Initialize mem
mem = np.zeros(top_range + 1, dtype = np.int64)
for start in range(2, top_range + 1):
# If mod4 == 1: (3x + 1)/4
if start % 4 == 1:
mem[start] = mem[(start + (start >> 1) + 1) // 2] + 3
# If 4mod == 3: 3(3x + 1) + 1 and continue
elif start % 4 == 3:
num = start + (start >> 1) + 1
num += (num >> 1) + 1
count = 4
while num >= start:
if num % 2:
num += (num >> 1) + 1
count += 2
else:
num //= 2
count += 1
mem[start] = mem[num] + count
# If 4mod == 2 or 0: x/2
else:
mem[start] = mem[(start // 2)] + 1
return mem
mem = collatz(top_range)
# Plot each starting number with the length of it's sequence
plt.scatter([*range(1, len(mem) + 1)], mem, color = 'black', s = 1)
plt.show()
```

## Solution

Applying numba on your code does help by much.

I removed tqdm since it does not help with performance.

```
import time
from matplotlib import pyplot as plt
from tqdm import tqdm
import numpy as np
import numba as nb
@nb.njit('int64[:](int_)')
def collatz2(top_range):
mem = np.zeros(top_range + 1, dtype=np.int64)
for start in range(2, top_range + 1):
# If mod(4) == 1: Value 2 or 3 Cached
if start % 4 == 1:
mem[start] = mem[(start + (start >> 1) + 1) // 2] + 3
# If mod(4) == 3: Use Algorithm
elif start % 4 == 3:
num = start
count = 0
while num >= start:
if num % 2:
num += (num >> 1) + 1
count += 2
else:
num //= 2
count += 1
mem[start] = mem[num] + count
# If mod(4) == 2 or 4: Value 1 Cached
else:
mem[start] = mem[(start // 2)] + 1
return mem
def collatz(top_range):
mem = [0] * (top_range + 1)
for start in range(2, top_range + 1):
# If mod(4) == 1: Value 2 or 3 Cached
if start % 4 == 1:
mem[start] = mem[(start + (start >> 1) + 1) // 2] + 3
# If mod(4) == 3: Use Algorithm
elif start % 4 == 3:
num = start
count = 0
while num >= start:
if num % 2:
num += (num >> 1) + 1
count += 2
else:
num //= 2
count += 1
mem[start] = mem[num] + count
# If mod(4) == 2 or 4: Value 1 Cached
else:
mem[start] = mem[(start // 2)] + 1
return mem
# profiling here
def main():
top_range = 1_000_000
mem = collatz(top_range)
mem2 = collatz2(top_range)
assert np.allclose(np.array(mem), mem2)
```

For top_range = 1_000 optimized function is ~100x faster. For top_range = 1_000_000, the optimized function is about 600x faster:

```
79 def main():
81 1 3.0 3.0 0.0 top_range = 1_000_000
83 1 24633045.0 24633045.0 98.7 mem = collatz(top_range)
85 1 39311.0 39311.0 0.2 mem2 = collatz2(top_range)
```

Answered By – dankal444

Answer Checked By – David Marino (BugsFixing Volunteer)