border-approach

At its defaults it draws a rank-two sequence approaching a rank-three tensor: the residual falls like 1/n and the terms grow like n. A_n = n(e₁ + e₂/n)⊗³ − n·e₁⊗³ has rank two for every n and converges to a tensor of rank three. The falling curve is ‖A_n − A‖, which is √(3/n² + 1/n⁴) exactly and reaches 0.00169 at n = 1024; the rising one is the norm of the larger of its two rank-one terms, 1024. Their product runs 2.519, 1.917, 1.777 … 1.73205, descending onto √3 = 1.73205. So getting one digit closer costs a factor of ten in the size of the pieces, for ever, and the infimum of the distance is zero while no rank-two tensor attains it.

border-approach is one function in lib/figures/borderrank.js — border rank — a nearest point that is not there, and a rank that is a sign. 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.

A_n = n(e₁ + e₂/n)⊗³ − n·e₁⊗³ has rank two for every n and converges to a tensor of rank three. The falling curve is ‖A_n − A‖, which is √(3/n² + 1/n⁴) exactly and reaches 0.00169 at n = 1024; the rising one is the norm of the larger of its two rank-one terms, 1024. Their product runs 2.519, 1.917, 1.777 … 1.73205, descending onto √3 = 1.73205. So getting one digit closer costs a factor of ten in the size of the pieces, for ever, and the infimum of the distance is zero while no rank-two tensor attains it.

A_n = n(e₁ + e₂/n)⊗³ − n·e₁⊗³ has rank two for every n and converges to a tensor of rank three. The falling curve is ‖A_n − A‖, which is √(3/n² + 1/n⁴) exactly and reaches 0.00169 at n = 1024; the rising one is the norm of the larger of its two rank-one terms, 1024. Their product runs 2.519, 1.917, 1.777 … 1.73205, descending onto √3 = 1.73205. So getting one digit closer costs a factor of ten in the size of the pieces, for ever, and the infimum of the distance is zero while no rank-two tensor attains it.

A_n = n(e₁ + e₂/n)⊗³ − n·e₁⊗³ has rank two for every n and converges to a tensor of rank three. The falling curve is ‖A_n − A‖, which is √(3/n² + 1/n⁴) exactly and reaches 2.64·10⁻⁵ at n = 65536; the rising one is the norm of the larger of its two rank-one terms, 65540. Their product runs 2.519, 1.917, 1.777 … 1.73205, descending onto √3 = 1.73205. So getting one digit closer costs a factor of ten in the size of the pieces, for ever, and the infimum of the distance is zero while no rank-two tensor attains it.

A_n = n(e₁ + e₂/n)⊗³ − n·e₁⊗³ has rank two for every n and converges to a tensor of rank three. The falling curve is ‖A_n − A‖, which is √(3/n² + 1/n⁴) exactly and reaches 0.108 at n = 16; the rising one is the norm of the larger of its two rank-one terms, 16.09. Their product runs 2.519, 1.917, 1.777 … 1.74334, descending onto √3 = 1.73205. So getting one digit closer costs a factor of ten in the size of the pieces, for ever, and the infimum of the distance is zero while no rank-two tensor attains it.

A_n = n(e₁ + e₂/n)⊗³ − n·e₁⊗³ has rank two for every n and converges to a tensor of rank three. The falling curve is ‖A_n − A‖, which is √(3/n² + 1/n⁴) exactly and reaches 4.23·10⁻⁴ at n = 4096; the rising one is the norm of the larger of its two rank-one terms, 4096. Their product runs 2.519, 1.917, 1.777 … 1.73205, descending onto √3 = 1.73205. So getting one digit closer costs a factor of ten in the size of the pieces, for ever, and the infimum of the distance is zero while no rank-two tensor attains it.

A_n = n(e₁ + e₂/n)⊗³ − n·e₁⊗³ has rank two for every n and converges to a tensor of rank three. The falling curve is ‖A_n − A‖, which is √(3/n² + 1/n⁴) exactly and reaches 0.00677 at n = 256; the rising one is the norm of the larger of its two rank-one terms, 256. Their product runs 2.519, 1.917, 1.777 … 1.73209, descending onto √3 = 1.73205. So getting one digit closer costs a factor of ten in the size of the pieces, for ever, and the infimum of the distance is zero while no rank-two tensor attains it.

Every figure above asserted its own claims on the way to being drawn, and a claim that failed would have failed the build rather than drawn a wrong picture. Those assertions 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.

18 distinct claims across 6 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.

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 146 of 287 generators — 131 print a residual and 15 are exempt with a published reason; 141 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.