Numerical Linear Algebra
where the algebra stops being the arithmetic
Generator

Recovering 355/113 from its residue: nothing below 2^18, everything above

One function in the integer library, called 4 times across 1 essay. Below: what it draws at its defaults, what it draws at every value an essay asks for, the 6 claims it put to the test while drawing them, and where it stands against the rule this site is named for.

At its defaults it draws recovering 355/113 from its residue: nothing below 2^18, everything above. The fraction 355/113 is reduced modulo a prime just above 2^e and then reconstructed from that one residue by the extended Euclidean algorithm. Below e = 18 the reconstruction returns a different fraction or refuses; at and above it, it returns 355/113 exactly, at every e tried. The crossover is not a property of the algorithm: a residue determines a rational uniquely once the modulus exceeds twice the square of the LARGER of the two parts, which here is 2^17.94, and below that there are two fractions with the same residue and both parts small enough. The measured first success is 2^18. There is no band in which the answer is is nearly right — the value returned below the cliff is a perfectly good fraction that is not this one.

reconstruct-cliff is one function in lib/figures/integer.js — exact arithmetic — no residual to print, and a cost measured in the length of the numbers. Everything below came out of it during this build, at arguments taken from the essays rather than invented for this page. A figure here is the figure a reader meets in an essay, and if the generator changes, this page changes with it.

At its defaults

Drawn even though every essay passes arguments — which on this site is every essay, at 100% of placements since the standard pass. A default nothing exercises is a trap for the next essay to call this with none, and this is the page where a default that has drifted from the figures around it becomes visible.

Recovering 355/113 from its residue: nothing below 2^18, everything aboveThe fraction 355/113 is reduced modulo a prime just above 2^e and then reconstructed from that one residue by the extended Euclidean algorithm. Below e = 18 the reconstruction returns a different fraction or refuses; at and above it, it returns 355/113 exactly, at every e tried. The crossover is not a property of the algorithm: a residue determines a rational uniquely once the modulus exceeds twice the square of the LARGER of the two parts, which here is 2^17.94, and below that there are two fractions with the same residue and both parts small enough. The measured first success is 2^18. There is no band in which the answer is is nearly right — the value returned below the cliff is a perfectly good fraction that is not this one.2^18modulus, as a power of twoa lattice with one short vectornumerator355denominator1132·max(n, d)², bits18first recovered at18moduli tried42below the bound there are two answersand the algorithm cannot prefer one

The fraction 355/113 is reduced modulo a prime just above 2^e and then reconstructed from that one residue by the extended Euclidean algorithm. Below e = 18 the reconstruction returns a different fraction or refuses; at and above it, it returns 355/113 exactly, at every e tried. The crossover is not a property of the algorithm: a residue determines a rational uniquely once the modulus exceeds twice the square of the LARGER of the two parts, which here is 2^17.94, and below that there are two fractions with the same residue and both parts small enough. The measured first success is 2^18. There is no band in which the answer is is nearly right — the value returned below the cliff is a perfectly good fraction that is not this one.

num: 355, den: 113

The arguments are the ones A fraction recovered from one remainder passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

Recovering 355/113 from its residue: nothing below 2^18, everything aboveThe fraction 355/113 is reduced modulo a prime just above 2^e and then reconstructed from that one residue by the extended Euclidean algorithm. Below e = 18 the reconstruction returns a different fraction or refuses; at and above it, it returns 355/113 exactly, at every e tried. The crossover is not a property of the algorithm: a residue determines a rational uniquely once the modulus exceeds twice the square of the LARGER of the two parts, which here is 2^17.94, and below that there are two fractions with the same residue and both parts small enough. The measured first success is 2^18. There is no band in which the answer is is nearly right — the value returned below the cliff is a perfectly good fraction that is not this one.2^18modulus, as a power of twoa lattice with one short vectornumerator355denominator1132·max(n, d)², bits18first recovered at18moduli tried42below the bound there are two answersand the algorithm cannot prefer one

The fraction 355/113 is reduced modulo a prime just above 2^e and then reconstructed from that one residue by the extended Euclidean algorithm. Below e = 18 the reconstruction returns a different fraction or refuses; at and above it, it returns 355/113 exactly, at every e tried. The crossover is not a property of the algorithm: a residue determines a rational uniquely once the modulus exceeds twice the square of the LARGER of the two parts, which here is 2^17.94, and below that there are two fractions with the same residue and both parts small enough. The measured first success is 2^18. There is no band in which the answer is is nearly right — the value returned below the cliff is a perfectly good fraction that is not this one.

num: 355, den: 7

The arguments are the ones A fraction recovered from one remainder passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

Recovering 355/7 from its residue: nothing below 2^18, everything aboveThe fraction 355/7 is reduced modulo a prime just above 2^e and then reconstructed from that one residue by the extended Euclidean algorithm. Below e = 18 the reconstruction returns a different fraction or refuses; at and above it, it returns 355/7 exactly, at every e tried. The crossover is not a property of the algorithm: a residue determines a rational uniquely once the modulus exceeds twice the square of the LARGER of the two parts, which here is 2^17.94, and below that there are two fractions with the same residue and both parts small enough. The measured first success is 2^18. There is no band in which the answer is is nearly right — the value returned below the cliff is a perfectly good fraction that is not this one.2^18modulus, as a power of twoa lattice with one short vectornumerator355denominator72·max(n, d)², bits18first recovered at18moduli tried42below the bound there are two answersand the algorithm cannot prefer one

The fraction 355/7 is reduced modulo a prime just above 2^e and then reconstructed from that one residue by the extended Euclidean algorithm. Below e = 18 the reconstruction returns a different fraction or refuses; at and above it, it returns 355/7 exactly, at every e tried. The crossover is not a property of the algorithm: a residue determines a rational uniquely once the modulus exceeds twice the square of the LARGER of the two parts, which here is 2^17.94, and below that there are two fractions with the same residue and both parts small enough. The measured first success is 2^18. There is no band in which the answer is is nearly right — the value returned below the cliff is a perfectly good fraction that is not this one.

num: 355, den: 99991

The arguments are the ones A fraction recovered from one remainder passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

Recovering 355/99991 from its residue: nothing below 2^35, everything aboveThe fraction 355/99991 is reduced modulo a prime just above 2^e and then reconstructed from that one residue by the extended Euclidean algorithm. Below e = 35 the reconstruction returns a different fraction or refuses; at and above it, it returns 355/99991 exactly, at every e tried. The crossover is not a property of the algorithm: a residue determines a rational uniquely once the modulus exceeds twice the square of the LARGER of the two parts, which here is 2^34.22, and below that there are two fractions with the same residue and both parts small enough. The measured first success is 2^35. There is no band in which the answer is is nearly right — the value returned below the cliff is a perfectly good fraction that is not this one.2^35modulus, as a power of twoa lattice with one short vectornumerator355denominator10·10⁴2·max(n, d)², bits34first recovered at35moduli tried42below the bound there are two answersand the algorithm cannot prefer one

The fraction 355/99991 is reduced modulo a prime just above 2^e and then reconstructed from that one residue by the extended Euclidean algorithm. Below e = 35 the reconstruction returns a different fraction or refuses; at and above it, it returns 355/99991 exactly, at every e tried. The crossover is not a property of the algorithm: a residue determines a rational uniquely once the modulus exceeds twice the square of the LARGER of the two parts, which here is 2^34.22, and below that there are two fractions with the same residue and both parts small enough. The measured first success is 2^35. There is no band in which the answer is is nearly right — the value returned below the cliff is a perfectly good fraction that is not this one.

What it checked while drawing

Every figure above checked its own claims on the way to being drawn, and a claim that failed would have stopped the picture rather than shipped a wrong one. Those checks used to leave no trace at all: a passing one returned true and the only evidence the figure had checked anything was that nothing crashed. The list below is what they actually said, collected by running this generator with an observer installed — not a description of what it is believed to check.

6 distinct claims across 4 sets of arguments, grouped below by shape — because most of them are one sentence with a different number in it, and how many separate times that sentence was put to the test is the informative part.

a denominator to recover

a numerator to recover

and the crossover is where the lattice bound puts it, within two bits

the assertion refuses a counterexample

the fraction is in lowest terms once its common factor is removed

the fraction is recovered once the modulus is large enough

Against the rule

The rule does not apply to it. It factorises nothing, so there is no residual it could be withholding. That is worth stating rather than leaving blank: a site that reported the rule as satisfied by every generator would be counting mostly generators the rule never reached.

Across the library: the rule bites on 217 of 397 generators — 199 print a residual and 18 are exempt with a published reason; 180 factorise nothing. Read from lib/residual-rule.js, which is the same body the gate enforces from, and the gate's last check fails the build if this page and it disagree about any generator.

Where it is called

Changing this generator changes every figure on this list. That is what makes the list worth publishing rather than keeping in a check script.

Exact arithmetic, and what it costs instead

A fraction recovered from one remainder

A solution over the rationals can be computed modulo a prime power and then recovered — the residue determines the fraction uniquely, but only once the modulus is twice the square of the fraction's longer part. Below that there is no partial credit: the algorithm returns a different fraction with the same residue, and it is a perfectly good one.

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