Conjugate gradients — where it appears
Named by 43 essays across 9 fields — each of them below, with the objects they name alongside it.
A parameter that counts steps
The regularisation field's knob is a positive real number chosen by one of three rules. The iterative field's is an integer nobody called a knob — where to stop. On the same problem the best step is 20 and the best λ is 0.025, and they reach 0.1426 and 0.1406.
The rate the condition number predicts
Conjugate gradients converge at a rate governed by the square root of the condition number. That is a bound rather than an estimate, it is provable, and it is loose enough that provisioning iterations from it wastes nine out of ten.
The accuracy that is thrown away
A Newton step is the exact answer to a linearised problem, and the linearisation is wrong at second order. So there is a floor under how close the step can land, the floor is the square of where it started, and eleven decades of inner tolerance below it buy the same four digits at four times the price.
An orthogonalisation nobody calls one
Conjugate gradients are derived as a minimisation and behave as an orthogonalisation, which is why the finite-termination property in every textbook is not a property the method has in floating point.
A tolerance that reads its own residual
The cheapest constant forcing term costs 980 inner iterations and arrives with a hundred times the forward error of the dearest, which costs 9,358. A rule that sets each step's tolerance from the ratio of the last two residuals costs 1,009 and arrives with neither problem — and it is not a constant, so it does not appear on the curve the constants are compared on.
The part of a solver that may be rounded
A preconditioner computed and applied with a three-bit significand still returns thirteen correct digits — it costs seventeen extra iterations and nothing else. Round the working arithmetic instead and the step count barely moves while the answer loses exactly the digits the format dropped.
Changing the condition number on purpose
Preconditioning is usually introduced as a trick that makes an iteration converge faster. It is not a trick. It is solving a different system with the same solution and a condition number chosen rather than inherited, and the new condition number is computable.
A preconditioner that changes sign
Strang's circulant preconditioner takes Toeplitz conjugate gradients from 179 steps to 10 at n = 256. At n = 64 on the same family it takes 66 steps to 109 — worse than doing nothing. Between those rows the preconditioner's smallest eigenvalue crosses zero, and nothing in the published account of the method mentions that it can be negative.
The circulant that cannot be indefinite
The previous essay found a preconditioner taking 117 steps against an unpreconditioned 59, because its smallest eigenvalue was −0.173. Average the two diagonals instead of choosing between them and the count is 7, 8, 9, 10, 10 across a factor of sixteen in size.
Two dimensions, and the cluster that thins
The same kernel, the same averaging, the same transform — applied along two axes instead of one. In one dimension the preconditioned step count is 7, 10, 10, 10; on square grids with the same unknown counts it is 10, 18, 20, 21, and the share of the spectrum near one falls from 56% to 17%.
A rate that is known in advance
On the model problem, Jacobi contracts by cos(π/(n+1)) per step, Gauss–Seidel by its square, and optimally relaxed SOR by a number given in closed form. Three rates, all known before anything runs, and all measurable against what runs.
Where the drift lands
The standing rule for when a preconditioner has gone stale is to rebuild it once the matrix has changed by more than some fraction of itself. Two drifts of exactly the same relative size cost 19 iterations and 5 on the same matrix, and the quantity that separates them is not in the rule at all — the perturbation is divided by the eigenvalue it lands on.
What a rebuild is worth
One sequence, one drift, one preconditioner — and six different right answers, because the cheapest rebuild period depends on what a rebuild cost to build and on nothing else. The optimum walks from every member to every twelfth as the setup gets dearer, and the free rule that reads the iteration count beats it in the middle of that range and loses at both ends.
An expiry date the noise does not move
The polynomial description of conjugate gradients leaves the level of rounding at step 17 or 18 on this operator, at every noise level from 10% to 0.1%. The step worth stopping at moves from 3 to 44 across the same range. They coincide at about 1% noise, which is where the coincidence was first read, and it is a fact about the noise rather than about the method.
A preconditioner that arrives past the answer
On a system that is solved to convergence a preconditioner changes how fast the answer arrives and not what it is. On a problem regularised by stopping it changes where every step lands. Conjugate gradients preconditioned by AᵀA + αI reaches its best answer in one step at α = 10⁻³, and at α = 10⁻⁶ its best answer is its first step, with an error of 1.35 against the unpreconditioned run's 0.1426 — while the count of eigenvalues it has clustered at one rises from 22 to 32.
The accuracy worth paying for
Used as a preconditioner, a hierarchical representation gets better at every accuracy — the iteration count falls monotonically all the way to the tightest tolerance. The total work does not. Its minimum sits at a rank-one preconditioner on an easy problem and six decades further along on a hard one.
The penalty for keeping it is a ratio
A kept incomplete Cholesky costs 40 iterations against a rebuilt one's 10 on 64 unknowns, and 55 against 17 on 256. Across six grids the difference between the two rises by 27 per cent and the ratio between them falls by 19. Neither quantity is free of the problem's size, and the one a policy is paid in is the one that transfers worse.
A step that is not a unit of work
Landweber's iteration reaches conjugate gradients' best answer on the same deconvolution — 0.1414 against 0.1426 — at step 1,778 instead of step 20, and at 0.1% noise at step 56,234 instead of 44. Each step costs the same two products. And within 10% of its best it runs from step 7 to step 6,310, where conjugate gradients runs from 4 to 26: the slow method is the one that forgives a late stop.
A speedup with a ceiling of its own
At ρ = 0.5 the averaged circulant takes 5 conjugate gradient steps at n = 512 against an unpreconditioned 30 — and that 30 is where the unpreconditioned count stops. It reads 29, 28, 30, 30 at n = 64 to 512 and then 29, 28, 26, 27, 25 at every doubling out to 16,384, because κ has reached 99.9% of Szegő's limit and the count has nothing left to grow with.
The method that cannot use a smooth answer
On the collection's own signal four regularisers reach the same floor to a few per cent. Score them instead against answers of increasing smoothness and one stops improving. Across six decades of noise Tikhonov's error falls with a fitted slope of 0.70 whether the answer is twice or four times as smooth, while truncation's rises to 0.94 — and at 10⁻⁶ noise Tikhonov's best is 38 times truncation's.
Where the augmentation puts the cost
Add γAᵀA to the objective block of a saddle-point system and its Schur complement tends to I/γ, so the cheapest possible approximation becomes the right one and the golden-ratio spectrum arrives — within 7.6·10⁻⁶ at γ = 10⁶. MINRES falls from 21 steps to 6. The inner solve with the augmented block rises from 14 conjugate gradient steps to 43, their product does not fall at all, and the answer loses seven and a half digits on the way.
A stopping test is a race
One matrix, one right-hand side, one tolerance, thirteen partition counts — and eleven different iteration counts between 674 and 690. Every run converged, every answer is right to the accuracy asked for, and what differs is the bill.
Four orders of conditioning, and four steps
On a 10×10 grid the two-dimensional kernel's condition number runs from 62 at ρ = 0.5 to 818,561 at ρ = 0.98. The preconditioned step count over the same range runs 18, 21, 21, 22, 21, 19, 18, and the count of eigenvalues the preconditioner actually brings within half a unit of one does not move at all — it is 9, 11, 13, 17 at every correlation the figure will draw.
A stopping rule that follows the run it is given
A preconditioner that reaches the answer four times sooner leaves four steps within 10% of its best instead of sixteen, and a rule that stops by the residual ought to miss so narrow a window more often. Over forty draws of the noise it misses it less: the discrepancy principle stops at 1.030 times the preconditioned run's best against 1.073 times the plain run's. And past the edge it stops within a factor of 1.7 of a run whose own best is 5.5 times Tikhonov's — faithful to a run that has already failed.
Spread resistances make the loops easy
Scaled to a unit diagonal, the loop equations on the least-resistance tree get easier as a network's resistances spread — from 120 to 5.44 over six decades — and stop depending on the grid's size, while the node equations of the same flow get harder, from 538 to 4.6·10⁴. The spread that ruins the range-space formulation rescues the null-space one, though the loops' density means the work saved is a factor of two, not the factor of nine the iteration counts suggest.
The tolerance that buys no agreement
Ask for four more orders of accuracy and you get them — the answers improve by a factor of 1.5 million. The ratio between the best and the worst run is 1.34, 1.48, 1.71 and 1.17 across the same sweep. The band falls and it does not close.
What a cheap preconditioner has to leave alone
A blur approximated by a matrix the cosine transform diagonalises agrees with the operator everywhere but its first and last seven rows. Made invertible by a shift, as the exact preconditioner was, it never reaches the unpreconditioned run's floor — at α = 10⁻³ its best iterate is 0.749 against 0.143. Made invertible by leaving every eigenvalue below τ alone, it reaches 0.141 in five steps instead of twenty, and the smallest τ that keeps the floor sits at a third to a half of the Tikhonov oracle's λ at three noise levels.
The rule that is wrong in the right direction
The preconditioner's cutoff is not a new parameter. It is the regularisation parameter this field already knows how to choose, halved — and the rule criticised for choosing λ a factor of two or three too large is the one whose cutoff keeps the floor on every draw, where the rule that chooses λ to within 3% has a worst draw two hundred times off it.
A guess worth two per cent
The previous Newton step looks like a free guess at the next one, and it is worth nothing. Started from it unscaled, the inner solve costs 4 to 61 per cent more than starting from zero, because the guess is 15 to 209 times too large. Scaled by the ratio of the two residual norms it is the right size and halves the starting residual — which buys a constant handful of inner iterations, not a share, because conjugate gradients costs the logarithm of its tolerance.
The parameter neither knob is
A preconditioned run has a cutoff and a step count, and neither is the regularisation parameter. The parameter is the effective dimension of the iterate: every cutoff that works puts its own best at 23.7 to 24.3 of it, where the unpreconditioned run's best sits at 23.2, and what the cutoff buys is the rate — 1.27 of it a step with no preconditioner and 3.53 with one. The edge is where a single stride is longer than the distance left.
One line that buys a quarter of the run
The adaptive forcing rule has a floor on it that no published statement of the rule carries: do not solve a step to an accuracy the outer loop will not use. Removing it costs 9 to 27 per cent of the whole inner run. Keeping it costs between 23 and 2,600 times the forward error — accuracy the residual test never asked for and both runs satisfy the test either way. The line is a trade between a residual and an error, and which of the two the caller meant decides whether it is a saving.
A count that marks the edge and not the pace
The number of directions a truncated preconditioner divides is counted for free when it is built, and it was proposed as a stand-in for the stride it buys. On three blurs it is not one — the runs leave the answer at strides of 5.71, 3.44 and 2.41. What the count does predict is the edge: on all three blurs, at two noise levels, a run stops landing on the answer's path within five per cent of the point where the count reaches the answer's own effective dimension. And the halved cutoff rule, measured on one blur, crosses that line on the narrowest.
The shift had an edge, and the approximation moved it
A fast-transform preconditioner made invertible by a shift was recorded as never reaching the unpreconditioned floor, and predicted to sit off the answer's path at every shift. At a large shift it sits on the path and reaches the floor to a tenth of a per cent. It has an edge like the truncated one — but on the exact operator that edge is where the shift's own effective dimension reaches the answer's, 1.02 to 1.05 of it on six problems, and on the fast approximation it arrives at 0.49 to 0.77. The difference is sixteen samples at the ends of the signal, where the approximation is wrong and a shift divides the error by α.
Three walks and one bound
A left-to-right sum, a chain of three thousand rotations and a conjugate gradient residual recurrence share no arithmetic and no vocabulary. Each has a standard bound that is linear in whatever it accumulates against. All three come out at a half — 0.486, 0.554 and 0.507 — and nothing is rescaled.
One arc, and what each filter pays to be on it
Conjugate gradients and Tikhonov stop at the same effective dimension, and that could have meant two curves crossing once or one curve. It is one curve over a stretch — on six problems, Tikhonov and truncation reach the iteration's error at the iteration's dimension to within 7.2 per cent from 0.7 of the answer to its top — and the two separate on either side. But the curve is shared by a trade, not by an identical answer: at the same dimension Tikhonov carries 22 to 42 per cent more noise than the iteration and up to five per cent less bias.
The reading that never moves
Across thirty runs — five grids from 36 to 196 unknowns, six working precisions from 53 significand bits down to 8 — the residual conjugate gradients stops on stays between 1.10·10⁻¹³ and 9.95·10⁻¹³. Over the same thirty runs the error of the answer spans a factor of 2.39·10¹¹, and the step count more than doubles. The one number the run publishes is the only one that responds to neither axis.
The division that cannot be done
Conjugate gradients divides by pᵀAp at every step, and on a matrix that is not positive definite that number can be zero or negative. The guard against it has been here from the first essay and described it as a failure. In the method that made conjugate gradients famous it is the single most valuable object the iteration can produce, and it costs six matrix–vector products.
The residual the method reports
Conjugate gradients prints a relative residual of 6.9·10⁻²¹. The unit roundoff is 1.1·10⁻¹⁶, so that is not a small residual and not a large one — it is not a residual. The vector the method is holding at that step has ‖b − Ax‖/‖b‖ = 5.1·10⁻¹⁰, and nothing in the run says so.
An iterate that must be made smaller
Applying a Kronecker-sum operator to a low-rank iterate multiplies its ranks by d and adding two of them adds their ranks, so a solver in a compressed format cannot keep what it produces. Every step is followed by a truncation — and whether that truncation is a floor on the residual depends on the right-hand side rather than on the truncation.
A run that is over at step five
A conjugate gradient whose every iterate is cut to a rank budget reaches the floor that budget allows at step 5, 36, 42 or 59, and then does nothing for the rest of the run. Four times the iterations move the floor by a factor of 1.8, and past the answer's own rank they move it the wrong way.
A proof that does not ask how large the matrix is
Proving a Hessian indefinite costs three matrix–vector products when the negative eigenvalue is 3 and nine to eleven when it is a thousandth, and that pair of numbers barely moves across a fourfold range in n. The factorisation that settles the same question costs a third of n³, which grows by a factor of sixty-four over the same range.
A walk needs a length
The gap between the two residuals grows as the square root of something, and a square root needs a length. Two quantities are candidates — how far the iterates travelled and how many steps were taken — and only a second sweep separates them. Across a fourfold change in size the iteration count goes from 39 to 96 and the gap goes from 5.04·10⁻¹⁵ to 5.33·10⁻¹⁵.
The certificate that arrives soonest is worth least
The more negative a Hessian's smallest eigenvalue, the sooner conjugate gradients meets a direction of negative curvature — and the less of the exact trust-region decrease that direction turns out to be worth. At λₘᵢₙ = −10 the step arrives after two products and gets 39.6 per cent; at −10⁻³ the same two products get 89.8, and the whole sweep costs eight.
Named alongside it
The objects these essays reach for when they reach for this one.
PreconditioningStopping criterionCondition numberKrylov subspaceIterative regularisationCirculant preconditionerFlop countTikhonov regularisationFilter factorsResidualSemi-convergenceClustered spectrum