We do not need a fallback.

include/fast_float/decimal_to_binary.h

@@ -117,16 +117,11 @@ adjusted_mantissa compute_float(int64_t q, uint64_t w)  noexcept  {
   // 3. We might lose a bit due to the "upperbit" routine (result too small, requiring a shift)
 
   value128 product = compute_product_approximation<binary::mantissa_explicit_bits() + 3>(q, w);
-  if(product.low == 0xFFFFFFFFFFFFFFFF) { //  could guard it further
-    // In some very rare cases, this could happen, in which case we might need a more accurate
-    // computation that what we can provide cheaply. This is very, very unlikely.
-    //
-    const bool inside_safe_exponent = (q >= -27) && (q <= 55); // always good because 5**q <2**128 when q>=0, 
-    // and otherwise, for q<0, we have 5**-q<2**64 and the 128-bit reciprocal allows for exact computation.
-    if(!inside_safe_exponent) {
-      return compute_error_scaled<binary>(q, product.high, lz);
-    }
-  }
+  // The computed 'product' is always sufficient.
+  // Mathematical proof:
+  // Noble Mushtak and Daniel Lemire, Fast Number Parsing Without Fallback (to appear)
+  // See script/mushtak_lemire.py
+
   // The "compute_product_approximation" function can be slightly slower than a branchless approach:
   // value128 product = compute_product(q, w);
   // but in practice, we can win big with the compute_product_approximation if its additional branch

script/mushtak_lemire.py

@@ -0,0 +1,75 @@
+#
+# Reference : 
+# Noble Mushtak and Daniel Lemire, Fast Number Parsing Without Fallback (to appear)
+#
+
+all_tqs = []
+
+# Generates all possible values of T[q]
+# Appendix B of  Number parsing at a gigabyte per second. 
+# Software: Practice and Experience 2021;51(8):1700–1727.
+for q in range(-342, -27):
+    power5 = 5**-q
+    z = 0
+    while (1 << z) < power5:
+        z += 1
+    b = 2 * z + 2 * 64
+    c = 2**b // power5 + 1
+    while c >= (1 << 128):
+        c //= 2
+    all_tqs.append(c)
+for q in range(-27, 0):
+    power5 = 5**-q
+    z = 0
+    while (1 << z) < power5:
+        z += 1
+    b = z + 127
+    c = 2**b // power5 + 1
+    all_tqs.append(c)
+for q in range(0, 308 + 1):
+    power5 = 5**q
+    while power5 < (1 << 127):
+        power5 *= 2
+    while power5 >= (1 << 128):
+        power5 //= 2
+    all_tqs.append(power5)
+
+# Returns the continued fraction of numer/denom as a list [a0; a1, a2, ..., an]
+def continued_fraction(numer, denom):
+    # (look at page numbers in top-left, not PDF page numbers)
+    cf = []
+    while denom != 0:
+        quot, rem = divmod(numer, denom)
+        cf.append(quot)
+        numer, denom = denom, rem
+    return cf
+
+# Given a continued fraction [a0; a1, a2, ..., an], returns 
+# all the convergents of that continued fraction
+# as pairs of the form (numer, denom), where numer/denom is 
+# a convergent of the continued fraction in simple form.
+def convergents(cf):
+    p_n_minus_2 = 0
+    q_n_minus_2 = 1
+    p_n_minus_1 = 1
+    q_n_minus_1 = 0
+    convergents = []
+    for a_n in cf:
+        p_n = a_n * p_n_minus_1 + p_n_minus_2
+        q_n = a_n * q_n_minus_1 + q_n_minus_2
+        convergents.append((p_n, q_n))
+        p_n_minus_2, q_n_minus_2, p_n_minus_1, q_n_minus_1 = p_n_minus_1, q_n_minus_1, p_n, q_n
+    return convergents
+
+
+# Enumerate through all the convergents of T[q] / 2^137 with denominators < 2^64
+found_solution = False
+for j, tq in enumerate(all_tqs):
+    for _, w in convergents(continued_fraction(tq, 2**137)):
+        if w >= 2**64:
+            break
+            if (tq*w) % 2**137 > 2**137 - 2**64:
+                print(f"SOLUTION: q={j-342} T[q]={tq} w={w}")
+                found_solution = True
+if not found_solution:
+    print("No solutions!")