xref: /aosp_15_r20/external/XNNPACK/src/math/expm1minus-f16-neonfp16arith-rr1-p3.c (revision 4bdc94577ba0e567308109d787f7fec7b531ce36)
1 // Copyright 2022 Google LLC
2 //
3 // This source code is licensed under the BSD-style license found in the
4 // LICENSE file in the root directory of this source tree.
5 
6 #include <assert.h>
7 #include <stddef.h>
8 
9 #include <arm_neon.h>
10 
11 #include <xnnpack/math-stubs.h>
12 
13 
xnn_math_f16_expm1minus__neonfp16arith_rr1_p3(size_t n,const void * input,void * output)14 void xnn_math_f16_expm1minus__neonfp16arith_rr1_p3(
15     size_t n,
16     const void* input,
17     void* output)
18 {
19   assert(n % (8 * sizeof(__fp16)) == 0);
20 
21   // The largest x for which expm1f(x) is saturated at -1.0f.
22   const float16x8_t vsat_cutoff = vmovq_n_f16(-0x1.0A4p+3f);
23   // Large number such that ulp(magic bias) == 1 and magic bias === 15 mod 2**9.
24   const float16x8_t vmagic_bias = vmovq_n_f16(0x1.83Cp+10f);
25   const float16x8_t vlog2e = vmovq_n_f16(0x1.714p+0f);
26   const float16x8_t vminus_ln2 = vmovq_n_f16(-0x1.630p-1f);
27   // Coefficient of polynomial approximation
28   //   exp(t) - 1 ~ t * (1 + t * (c2 + t * c3))
29   // on [-log(2)/2, log(2)/2]
30   const float16x8_t vc3 = vmovq_n_f16(0x1.56Cp-3f);
31   const float16x8_t vc2 = vmovq_n_f16(0x1.020p-1f);
32   const float16x8_t vone = vmovq_n_f16(1.0f);
33 
34   const __fp16* i = (const __fp16*) input;
35   __fp16* o = (__fp16*) output;
36   for (; n != 0; n -= 8 * sizeof(__fp16)) {
37     float16x8_t vx = vld1q_f16(i); i += 8;
38 
39     // The function saturates at -1 for large negative inputs: expm1h(x) == -1.0h for x <= sat_cutoff ~= -8.3203125.
40     // To guarantee this behaviour, we clip input at sat_cutoff, and leverage the fact that for our implementation
41     // expm1m(sat_cutoff) == -1.0f. NaN inputs are passed unchanged.
42     vx = vmaxq_f16(vx, vsat_cutoff);
43 
44     // Compute reduced argument n := round(x / log(2)).
45     // We do it by adding a large number (magic bias), which cause rounding of the result to integer, then subtracing
46     // the large number back. The addition is combined with multiplication by log2e into a single FMA instruction. The
47     // trick with adding large number is valid only within certain bounds (|x / log(2)| <= 2**9, i.e.
48     // |x| <= 0x1.630p+8 = 355.0), but that is acceptable, because inputs x are restricted to [-8.3203125, 0].
49     // Note that addition-subtraction of the large number doesn't cause overflow for inputs in this range.
50     float16x8_t vn = vfmaq_f16(vmagic_bias, vx, vlog2e);
51 
52     // Create a floating-point number s (scale) such that s == 2**n for valid inputs, i.e.
53     // -8.3203125 <= x <= 0.0, and -12 <= n <= 0 accordingly.
54     // For NaN inputs, s would have zero mantissa and can have arbitrary sign and exponent, depending on the input
55     // NaN payload. In these cases, n and t are NaNs with the same payload as input while s is non-NaN, and thus
56     // input payload would be propagated in all computations.
57     const float16x8_t vs = vreinterpretq_f16_s16(vshlq_n_s16(vreinterpretq_s16_f16(vn), 10));
58 
59     // Subtract the large number back to get final n := round(x / log(2)).
60     vn = vsubq_f16(vn, vmagic_bias);
61 
62     // Compute reduced argument t := x - n * log(2).
63     float16x8_t vt = vfmaq_f16(vx, vn, vminus_ln2);
64 
65     // Compute degree-3 polynomial approximation for exp(t) - 1 on [-log(2)/2, log(2)/2].
66     //   P(t) = t * (1 + t * (c2 + t * c3))
67     //        = t + t * (t * (c2 + t * c3)) = t + t * p
68     float16x8_t vp = vfmaq_f16(vc2, vc3, vt);
69     vp = vmulq_f16(vp, vt);
70 
71     // Reconstruct the exp(x) - 1 value:
72     //   exp(x) - 1 = s * (1 + t * (1 + t * (c2 + t * c3))) - 1
73     //              = (s - 1) + s * (t + t * p)
74     //              = ((t * s) + (t * s) * p) + (s - 1)
75     vt = vmulq_f16(vt, vs);
76     const float16x8_t vsm1 = vsubq_f16(vs, vone);
77     vp = vfmaq_f16(vt, vp, vt);
78     const float16x8_t vf = vaddq_f16(vp, vsm1);
79 
80     vst1q_f16(o, vf); o += 8;
81   }
82 }
83