Compute a lower bound for the adjusted exponent of self.log10(). In other words, find r such that self.log10() >= 10**r. Assumes that self is finite and positive and that self != 1.
(self)
| 3202 | return ans |
| 3203 | |
| 3204 | def _log10_exp_bound(self): |
| 3205 | """Compute a lower bound for the adjusted exponent of self.log10(). |
| 3206 | In other words, find r such that self.log10() >= 10**r. |
| 3207 | Assumes that self is finite and positive and that self != 1. |
| 3208 | """ |
| 3209 | |
| 3210 | # For x >= 10 or x < 0.1 we only need a bound on the integer |
| 3211 | # part of log10(self), and this comes directly from the |
| 3212 | # exponent of x. For 0.1 <= x <= 10 we use the inequalities |
| 3213 | # 1-1/x <= log(x) <= x-1. If x > 1 we have |log10(x)| > |
| 3214 | # (1-1/x)/2.31 > 0. If x < 1 then |log10(x)| > (1-x)/2.31 > 0 |
| 3215 | |
| 3216 | adj = self._exp + len(self._int) - 1 |
| 3217 | if adj >= 1: |
| 3218 | # self >= 10 |
| 3219 | return len(str(adj))-1 |
| 3220 | if adj <= -2: |
| 3221 | # self < 0.1 |
| 3222 | return len(str(-1-adj))-1 |
| 3223 | op = _WorkRep(self) |
| 3224 | c, e = op.int, op.exp |
| 3225 | if adj == 0: |
| 3226 | # 1 < self < 10 |
| 3227 | num = str(c-10**-e) |
| 3228 | den = str(231*c) |
| 3229 | return len(num) - len(den) - (num < den) + 2 |
| 3230 | # adj == -1, 0.1 <= self < 1 |
| 3231 | num = str(10**-e-c) |
| 3232 | return len(num) + e - (num < "231") - 1 |
| 3233 | |
| 3234 | def log10(self, context=None): |
| 3235 | """Returns the base 10 logarithm of self.""" |