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Implement a method for accurately measuring the sizes of small commas
Add a polyfill for Array.sumPrecise. Implement monzoToBigNumeratorDenominator. Implement dotPrecise.
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
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| // Stolen from: https://github.com/tc39/proposal-math-sum/blob/main/polyfill/polyfill.mjs | ||
| // Linted and type-annotated by Lumi Pakkanen. | ||
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| /* | ||
| https://www-2.cs.cmu.edu/afs/cs/project/quake/public/papers/robust-arithmetic.ps | ||
| Shewchuk's algorithm for exactly floating point addition | ||
| as implemented in Python's fsum: https://github.com/python/cpython/blob/48dfd74a9db9d4aa9c6f23b4a67b461e5d977173/Modules/mathmodule.c#L1359-L1474 | ||
| adapted to handle overflow via an additional "biased" partial, representing 2**1024 times its actual value | ||
| */ | ||
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| // exponent 11111111110, significand all 1s | ||
| const MAX_DOUBLE = 1.79769313486231570815e308; // i.e. (2**1024 - 2**(1023 - 52)) | ||
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| // exponent 11111111110, significand all 1s except final 0 | ||
| const PENULTIMATE_DOUBLE = 1.79769313486231550856e308; // i.e. (2**1024 - 2 * 2**(1023 - 52)) | ||
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| // exponent 11111001010, significand all 0s | ||
| const MAX_ULP = MAX_DOUBLE - PENULTIMATE_DOUBLE; // 1.99584030953471981166e+292, i.e. 2**(1023 - 52) | ||
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| // prerequisite: Math.abs(x) >= Math.abs(y) | ||
| function twosum(x: number, y: number) { | ||
| const hi = x + y; | ||
| const lo = y - (hi - x); | ||
| return {hi, lo}; | ||
| } | ||
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| /** | ||
| * Accurately add up elements from an iterable using Shewchuk's algorithm. | ||
| * @param iterable Numbers to sum together. | ||
| * @returns The sum of the elements. | ||
| */ | ||
| export function sum(iterable: Iterable<number>) { | ||
| const partials: number[] = []; | ||
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| let overflow = 0; // conceptually 2**1024 times this value; the final partial | ||
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| // for purposes of the polyfill we're going to ignore closing the iterator, sorry | ||
| const iterator = iterable[Symbol.iterator](); | ||
| const next = iterator.next.bind(iterator); | ||
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| // in C this would be done using a goto | ||
| function drainNonFiniteValue(current: number) { | ||
| while (true) { | ||
| const {done, value} = next(); | ||
| if (done) { | ||
| return current; | ||
| } | ||
| if (!Number.isFinite(value)) { | ||
| // summing any distinct two of the three non-finite values gives NaN | ||
| // summing any one of them with itself gives itself | ||
| if (!Object.is(value, current)) { | ||
| current = NaN; | ||
| } | ||
| } | ||
| } | ||
| } | ||
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| // handle list of -0 special case | ||
| while (true) { | ||
| const {done, value} = next(); | ||
| if (done) { | ||
| return -0; | ||
| } | ||
| if (!Object.is(value, -0)) { | ||
| if (!Number.isFinite(value)) { | ||
| return drainNonFiniteValue(value); | ||
| } | ||
| partials.push(value); | ||
| break; | ||
| } | ||
| } | ||
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| // main loop | ||
| while (true) { | ||
| const {done, value} = next(); | ||
| if (done) { | ||
| break; | ||
| } | ||
| let x = +value; | ||
| if (!Number.isFinite(x)) { | ||
| return drainNonFiniteValue(x); | ||
| } | ||
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| // we're updating partials in place, but it is maybe easier to understand if you think of it as making a new copy | ||
| let actuallyUsedPartials = 0; | ||
| // let newPartials = []; | ||
| for (let y of partials) { | ||
| if (Math.abs(x) < Math.abs(y)) { | ||
| [x, y] = [y, x]; | ||
| } | ||
| let {hi, lo} = twosum(x, y); | ||
| if (Math.abs(hi) === Infinity) { | ||
| const sign = hi === Infinity ? 1 : -1; | ||
| overflow += sign; | ||
| if (Math.abs(overflow) >= 2 ** 53) { | ||
| throw new RangeError('overflow'); | ||
| } | ||
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| x = x - sign * 2 ** 1023 - sign * 2 ** 1023; | ||
| if (Math.abs(x) < Math.abs(y)) { | ||
| [x, y] = [y, x]; | ||
| } | ||
| ({hi, lo} = twosum(x, y)); | ||
| } | ||
| if (lo !== 0) { | ||
| partials[actuallyUsedPartials] = lo; | ||
| ++actuallyUsedPartials; | ||
| // newPartials.push(lo); | ||
| } | ||
| x = hi; | ||
| } | ||
| partials.length = actuallyUsedPartials; | ||
| // assert.deepStrictEqual(partials, newPartials) | ||
| // partials = newPartials | ||
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| if (x !== 0) { | ||
| partials.push(x); | ||
| } | ||
| } | ||
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| // compute the exact sum of partials, stopping once we lose precision | ||
| let n = partials.length - 1; | ||
| let hi = 0; | ||
| let lo = 0; | ||
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| if (overflow !== 0) { | ||
| const next = n >= 0 ? partials[n] : 0; | ||
| --n; | ||
| if ( | ||
| Math.abs(overflow) > 1 || | ||
| (overflow > 0 && next > 0) || | ||
| (overflow < 0 && next < 0) | ||
| ) { | ||
| return overflow > 0 ? Infinity : -Infinity; | ||
| } | ||
| // here we actually have to do the arithmetic | ||
| // drop a factor of 2 so we can do it without overflow | ||
| // assert(Math.abs(overflow) === 1) | ||
| ({hi, lo} = twosum(overflow * 2 ** 1023, next / 2)); | ||
| lo *= 2; | ||
| if (Math.abs(2 * hi) === Infinity) { | ||
| // stupid edge case: rounding to the maximum value | ||
| // MAX_DOUBLE has a 1 in the last place of its significand, so if we subtract exactly half a ULP from 2**1024, the result rounds away from it (i.e. to infinity) under ties-to-even | ||
| // but if the next partial has the opposite sign of the current value, we need to round towards MAX_DOUBLE instead | ||
| // this is the same as the "handle rounding" case below, but there's only one potentially-finite case we need to worry about, so we just hardcode that one | ||
| if (hi > 0) { | ||
| if ( | ||
| hi === 2 ** 1023 && | ||
| lo === -(MAX_ULP / 2) && | ||
| n >= 0 && | ||
| partials[n] < 0 | ||
| ) { | ||
| return MAX_DOUBLE; | ||
| } | ||
| return Infinity; | ||
| } else { | ||
| if ( | ||
| hi === -(2 ** 1023) && | ||
| lo === MAX_ULP / 2 && | ||
| n >= 0 && | ||
| partials[n] > 0 | ||
| ) { | ||
| return -MAX_DOUBLE; | ||
| } | ||
| return -Infinity; | ||
| } | ||
| } | ||
| if (lo !== 0) { | ||
| partials[n + 1] = lo; | ||
| ++n; | ||
| lo = 0; | ||
| } | ||
| hi *= 2; | ||
| } | ||
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| while (n >= 0) { | ||
| const x = hi; | ||
| const y = partials[n]; | ||
| --n; | ||
| // assert: Math.abs(x) > Math.abs(y) | ||
| ({hi, lo} = twosum(x, y)); | ||
| if (lo !== 0) { | ||
| break; | ||
| } | ||
| } | ||
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| // handle rounding | ||
| // when the roundoff error is exactly half of the ULP for the result, we need to check one more partial to know which way to round | ||
| if ( | ||
| n >= 0 && | ||
| ((lo < 0.0 && partials[n] < 0.0) || (lo > 0.0 && partials[n] > 0.0)) | ||
| ) { | ||
| const y = lo * 2.0; | ||
| const x = hi + y; | ||
| const yr = x - hi; | ||
| if (y === yr) { | ||
| hi = x; | ||
| } | ||
| } | ||
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| return hi; | ||
| } |