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Note on backward precision over fp16: A float16 number has 10 bits of mantissa, 5 bits of exponent, and 1 bit for the sign. If the sign bit is positive, then with a mantissa $m$ and exponent $e$ represented in base 10, the number that the float16 format represents is $(1 + m / 1024) \exp2(e)$. ([source](https://en.wikipedia.org/wiki/Half-precision_floating-point_format)) Consider adding two numbers $a$ and $b$ which have arbitrary mantissas, and say their exponents are $e_a = 1$ (so $2 \le a \lt 4$) and $e_b=-3$ (so $0.175 \le b \lt 0.25$). Assume that the result has the same exponent as $a$. Since the exponents differ by 4, we'll effectively need to truncate the 4 rightmost bits of $b$'s mantissa, which would introduce a maximum error on the order of $(2^4 / 1024) \exp2(-3) \approx 0.002$. The error is nearly the same if $e_b = -2$ (so $0.25 \le b \lt 0.5$), where the 3 rightmost bits are truncated, giving a maximum error on the order of $(2^3 / 1024) \exp2(-2) \approx 0.002$. Same for $e_b=-1$. So if we're adding up nine different numbers that all have exponents -3, -2, or -1, and they sum to a number with exponent 1, then we would expect a maximum error of several times greater than 0.002. In my comments above, summing those particular nine numbers in different ways gave results that ranged between 3.1816 and 3.1758, a difference of $0.0058 \approx 2.9 * 0.002$. That's within the acceptable bounds, and we can safely just increase the error tolerance used in test_output_grad_match for the case of max_pool3d_backward with float16. Pull Request resolved: https://github.com/pytorch/pytorch/pull/157498 Approved by: https://github.com/malfet