Answer fractional powers of exact numbers past a double's range (closes #16)
check / check (push) Successful in 1m21s

A fractional power turned its base into a double first and refused a
base outside the normal range of a double, so (2^1200)^0.5, (2^1024)^0.5,
(2^-1200)^0.5 and 1e400^0.5 were refused although each answer is an
ordinary double. Such a base is now brought into that range by square
roots taken from its exact value in a big.Float, at most three under the
4096-bit limit, with the exponent doubled for each, and only a result
that is not a normal double is refused. The tests give these powers and
the edges of the range their values, and the bounded-work test covers a
base that needs three roots. The README's refusal sentence and example
are updated.

Model: opus-5-5
This commit is contained in:
2026-09-29 07:37:57 +00:00
parent 4b871c2b12
commit c4094fff7d
4 changed files with 55 additions and 14 deletions
+27 -6
View File
@@ -130,6 +130,21 @@ func TestEvaluatePowers(t *testing.T) {
// A power computed in float64 carries its rounding into the exact
// arithmetic after it, past the range of a double as within it.
"2^0.5 * 1e400": "1.4142135623730951e+400",
// A fractional power of an exact number outside the range of a
// double, taken from its exact value, up to the edges of that
// range.
"(2^1200)^0.5": "4.149515568880993e+180",
"(2^1024)^0.5": "1.3407807929942597e+154",
"(2^-1200)^0.5": "2.409919865102884e-181",
"1e400^0.5": "1e+200",
"1e-400^0.5": "1e-200",
"1e-310^0.5": "1e-155",
"(2^1200)^-0.5": "2.409919865102884e-181",
"(2^1200)^0.5 / 2^600": "1",
"1e400^-0.001": "0.39810717055349726",
"1e-400^0.001": "0.39810717055349726",
"(2^2047)^0.5": "1.2711610061536464e+308",
"(2^-2044)^0.5": "2.2250738585072014e-308",
})
}
@@ -252,15 +267,17 @@ func TestEvaluateOutOfRange(t *testing.T) {
"2^5000": calc.ErrOutOfRange,
"(-2)^5001": calc.ErrOutOfRange,
"0.5^-5000": calc.ErrOutOfRange,
// Powers computed in float64 whose base or result is not a
// normal double, and so has lost digits, or all of them.
// Powers computed in float64 whose result is not a normal double,
// and so has lost digits, or all of them, whatever the size of
// the base.
"2^-1073.5 * 2^1073": calc.ErrOutOfRange,
"1e400^-0.001": calc.ErrOutOfRange,
"1e-400^0.001": calc.ErrOutOfRange,
"1e-310^0.5": calc.ErrOutOfRange,
"2^1500.5": calc.ErrOutOfRange,
"(2^1200)^0.9": calc.ErrOutOfRange,
"1e-400^0.9": calc.ErrOutOfRange,
"(2^2048)^0.5": calc.ErrOutOfRange,
"(2^-2046)^0.5": calc.ErrOutOfRange,
"(0.5^1100)^4 / (0.5^1100)^4": calc.ErrOutOfRange,
"(1/3)^1e400": calc.ErrOutOfRange,
"(2^1200)^0.5": calc.ErrOutOfRange,
// go/constant holds numbers of this size rounded. A sum of them
// can lose the answer (this one would be 0), and so can a
// remainder or the sign of -1 to such a power.
@@ -343,6 +360,10 @@ func TestEvaluateBoundsWork(t *testing.T) {
{in: "3^2583", want: "2.5363018640659988e+1232"},
{in: "2^-4094", want: "3.8299909843808741e-1233"},
{in: "-1/3^2583", want: "-3.9427483540814775e-1233"},
// Fractional powers of numbers just below the limit, whose bases
// take the most square roots to bring into the range of a double.
{in: "(2^-4094)^0.125", want: "8.869511863657883e-155"},
{in: "(1/3^2583)^0.5", err: calc.ErrOutOfRange},
}
for _, c := range cases {