The coefficient matrix of AX + XB = C against the answer, to n = 100
At its defaults it draws the coefficient matrix of ax + xb = c against the answer, to n = 100. At n = 100 the unknown X has 10000 entries and the coefficient matrix of the linear map has 10⁸ — 0.80 gigabytes of doubles. Eliminating it costs 6.67·10¹¹ operations against the 6·10⁷ Bartels and Stewart's algorithm needs, a ratio of 11111. The Kronecker form is what the equation means and it is not a method.
kronecker-size is one function in lib/figures/sylvester.js —
matrix equations — the n²×n² coefficient matrix nobody forms, and sep. 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.
At n = 100 the unknown X has 10000 entries and the coefficient matrix of the linear map has 10⁸ — 0.80 gigabytes of doubles. Eliminating it costs 6.67·10¹¹ operations against the 6·10⁷ Bartels and Stewart's algorithm needs, a ratio of 11111. The Kronecker form is what the equation means and it is not a method.
nMax: 100
The arguments are the ones The coarse problem is a different problem passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
At n = 100 the unknown X has 10000 entries and the coefficient matrix of the linear map has 10⁸ — 0.80 gigabytes of doubles. Eliminating it costs 6.67·10¹¹ operations against the 6·10⁷ Bartels and Stewart's algorithm needs, a ratio of 11111. The Kronecker form is what the equation means and it is not a method.
nMax: 20
The arguments are the ones The elimination the matrix does not need passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
At n = 20 the unknown X has 400 entries and the coefficient matrix of the linear map has 1.6·10⁵ — 0.00 gigabytes of doubles. Eliminating it costs 4.27·10⁷ operations against the 4.8·10⁵ Bartels and Stewart's algorithm needs, a ratio of 89. The Kronecker form is what the equation means and it is not a method.
nMax: 50
The arguments are the ones The elimination the matrix does not need passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
At n = 50 the unknown X has 2500 entries and the coefficient matrix of the linear map has 6.25·10⁶ — 0.05 gigabytes of doubles. Eliminating it costs 1.04·10¹⁰ operations against the 7.5·10⁶ Bartels and Stewart's algorithm needs, a ratio of 1389. The Kronecker form is what the equation means and it is not a method.
nMax: 200
The arguments are the ones The elimination the matrix does not need passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
At n = 200 the unknown X has 40000 entries and the coefficient matrix of the linear map has 1.6·10⁹ — 12.80 gigabytes of doubles. Eliminating it costs 4.27·10¹³ operations against the 4.8·10⁸ Bartels and Stewart's algorithm needs, a ratio of 88889. The Kronecker form is what the equation means and it is not a method.
nMax: 400
The arguments are the ones The elimination the matrix does not need passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
At n = 400 the unknown X has 160000 entries and the coefficient matrix of the linear map has 2.56·10¹⁰ — 204.80 gigabytes of doubles. Eliminating it costs 2.73·10¹⁵ operations against the 3.84·10⁹ Bartels and Stewart's algorithm needs, a ratio of 711111. The Kronecker form is what the equation means and it is not a method.
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.
3 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 size the cost model is meant to cover
and the arithmetic ratio is well past ten at the largest size
the coefficient matrix has n⁴ entries
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.
The coarse problem is a different problem
In one dimension the Galerkin coarse operator is the coarse discretisation, entry for entry — this site asserted it. In two dimensions a five-point operator produces a nine-point coarse one, so the recursion solves a different discretisation at every level below the first, and converges at 0.20 a cycle regardless.
Structure, and the solver that cannot see itThe elimination the matrix does not need
The Kronecker form of AX + XB = C is dismissed with a hundred million entries and (2/3)n⁶ operations. Both price an elimination, and after the reduction both routes take, the matrix has exactly n³ nonzeros, none of them above the block diagonal, and nothing left to eliminate.