What each way of eliminating a constraint inherits, over five decades of κ(A)
At its defaults it draws what each way of eliminating a constraint inherits, over five decades of κ(a). The same system solved twice, at 8 unknowns and 3 constraints with κ(H) = 100. The range-space method forms S = AH⁻¹Aᵀ and inherits κ(S), which rises from 12.99 to 3.981·10¹⁰ — the square of κ(A), for the reason the normal equations square it. The null-space method solves with the reduced Hessian ZᵀHZ, whose condition number is 21.13 at the start of the sweep and 21.13 at the end: it does not contain κ(A) at all. The two forward errors, measured against a solution computed in BigInt rationals, follow their own condition numbers: 5.314·10⁻⁶ against 5.788·10⁻¹² at the far end. Both methods are correct and one of them is usable.
two-eliminations is one function in lib/figures/kkt.js —
the matrix a constraint makes — an inertia known before assembly, and a failure with an address. 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.
The same system solved twice, at 8 unknowns and 3 constraints with κ(H) = 100. The range-space method forms S = AH⁻¹Aᵀ and inherits κ(S), which rises from 12.99 to 3.981·10¹⁰ — the square of κ(A), for the reason the normal equations square it. The null-space method solves with the reduced Hessian ZᵀHZ, whose condition number is 21.13 at the start of the sweep and 21.13 at the end: it does not contain κ(A) at all. The two forward errors, measured against a solution computed in BigInt rationals, follow their own condition numbers: 5.314·10⁻⁶ against 5.788·10⁻¹² at the far end. Both methods are correct and one of them is usable.
kappaH: 100
The arguments are the ones Two ways to remove a constraint passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
The same system solved twice, at 8 unknowns and 3 constraints with κ(H) = 100. The range-space method forms S = AH⁻¹Aᵀ and inherits κ(S), which rises from 12.99 to 3.981·10¹⁰ — the square of κ(A), for the reason the normal equations square it. The null-space method solves with the reduced Hessian ZᵀHZ, whose condition number is 21.13 at the start of the sweep and 21.13 at the end: it does not contain κ(A) at all. The two forward errors, measured against a solution computed in BigInt rationals, follow their own condition numbers: 5.314·10⁻⁶ against 5.788·10⁻¹² at the far end. Both methods are correct and one of them is usable.
kappaH: 1
The arguments are the ones Two ways to remove a constraint passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
The same system solved twice, at 8 unknowns and 3 constraints with κ(H) = 1. The range-space method forms S = AH⁻¹Aᵀ and inherits κ(S), which rises from 1 to 10·10⁹ — the square of κ(A), for the reason the normal equations square it. The null-space method solves with the reduced Hessian ZᵀHZ, whose condition number is 1 at the start of the sweep and 1 at the end: it does not contain κ(A) at all. The two forward errors, measured against a solution computed in BigInt rationals, follow their own condition numbers: 7.341·10⁻⁷ against 2.025·10⁻¹² at the far end. Both methods are correct and one of them is usable.
kappaH: 10
The arguments are the ones Two ways to remove a constraint passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
The same system solved twice, at 8 unknowns and 3 constraints with κ(H) = 10. The range-space method forms S = AH⁻¹Aᵀ and inherits κ(S), which rises from 3.169 to 1.585·10¹⁰ — the square of κ(A), for the reason the normal equations square it. The null-space method solves with the reduced Hessian ZᵀHZ, whose condition number is 4.149 at the start of the sweep and 4.149 at the end: it does not contain κ(A) at all. The two forward errors, measured against a solution computed in BigInt rationals, follow their own condition numbers: 2.278·10⁻⁷ against 3.002·10⁻¹² at the far end. Both methods are correct and one of them is usable.
kappaH: 10000
The arguments are the ones Two ways to remove a constraint passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
The same system solved twice, at 8 unknowns and 3 constraints with κ(H) = 10⁴. The range-space method forms S = AH⁻¹Aᵀ and inherits κ(S), which rises from 199.3 to 2.728·10¹¹ — the square of κ(A), for the reason the normal equations square it. The null-space method solves with the reduced Hessian ZᵀHZ, whose condition number is 514.7 at the start of the sweep and 514.7 at the end: it does not contain κ(A) at all. The two forward errors, measured against a solution computed in BigInt rationals, follow their own condition numbers: 7.183·10⁻⁶ against 5.856·10⁻¹² at the far end. Both methods are correct and one of them is usable.
kappaH: 1000000
The arguments are the ones Two ways to remove a constraint passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
The same system solved twice, at 8 unknowns and 3 constraints with κ(H) = 10⁶. The range-space method forms S = AH⁻¹Aᵀ and inherits κ(S), which rises from 1700 to 1.118·10¹² — the square of κ(A), for the reason the normal equations square it. The null-space method solves with the reduced Hessian ZᵀHZ, whose condition number is 7646 at the start of the sweep and 7646 at the end: it does not contain κ(A) at all. The two forward errors, measured against a solution computed in BigInt rationals, follow their own condition numbers: 2.467·10⁻⁶ against 4.419·10⁻¹² at the far end. Both methods are correct and one of them is usable.
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.
9 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.
a constraint no larger than the problem
a finite double, since an infinity is not a rational
a size the exact rational solve can afford
and κ(ZᵀHZ) does not move at all
fewer constraints than unknowns
fewer constraints than unknowns, so something is left to minimise
LU is for square matrices
matmul shapes agree
κ(S) grows like κ(A)² across the sweep
Against the rule
It draws a decomposition and prints its residual. It calls
rangeSpaceSolve, nullSpaceSolve, orthonormalNullSpace, exactKkt,
and every figure above carries the badge — which residualcheck verifies by looking
for it in the emitted SVG rather than by finding the call that builds one. A badge that is
constructed and then left out of the body is the failure that check exists for.
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.