4 ms·

I did the same thing, recalling "There are Only Four Billion Floats–So Test Them All!" [0]. Except I used Python, so it took over 10 minutes: >>> import st

• jherskovic 4y ago
  My very-slightly-different Python implementation runs in 9m56s minutes on Python 3.11, but in 2m54s using pypy3.9 (all on an M1 Max) pypy never ceases to amaze me for low-effort performance gains on computational stuff. A naive implementation in C compiled with Apple's clang (-O2) takes 0.62 seconds, of course.
• eesmith 4y ago
  This should be a bit faster, while still sticking to stock Python: import time, array n = 0 start_time = time.time() values = bytearray(b"".join(i.to_bytes(4, "big") for i in range(2**24))) for prefix in range(256): values[::4] = prefix.to_bytes(1, "big") * (2**24) arr = array.array("f", values) n += sum(1 for x in arr if 0.0 <= x <= 1.0) print(f"Found: {n} Time: {time.time() - start_time:.1f} seconds") It uses an array.array() to convert from a large number of concatenated 4 bytes to floats, rather than going through the struct module one-by-one. Rather than build all 0x1_0000_0000 values in memory, I process 0x100_0000 at a time, and accumulate the partial sums. On my laptop this reports "Found: 1065353218 Time: 228.5 seconds", so just under 4 minutes. A slightly improved algorithm in NumPy takes 5.4 seconds. import time, numpy as np start_time = time.time() arr = np.arange(2**24, dtype=np.uint32) incr = np.full(2**24, 2**24, dtype=arr.dtype) values = np.frombuffer(arr, np.float32) n = ((0.0 < values) & (values <= 1.0)).sum() for i in range(1, 256): arr += incr n += ((0.0 < values) & (values <= 1.0)).sum() print(f"Found: {n} Time: {time.time() - start_time:.1f} seconds")