Part 5 · Chapter 18

One-Dimensional Search Methods

Nonlinear programming almost always comes down, in its innermost loop, to a single humble question: given a point and a direction to move, how far should we step? Answering it means minimising a function of one variable — a line search. This chapter builds the tools for that task from the ground up: how to trap a minimum inside an interval, how to shrink that interval as fast as possible with as few function evaluations as possible using Fibonacci and golden-section elimination, how to accelerate with interpolation when the function is smooth, and how to settle for a "good enough" step with the inexact conditions that real solvers actually use.

Optimization Techniques Prof. Mithun Mondal Reading time ≈ 45 min
i What you'll learn
  • The line-search problem — minimising \(f(\alpha)\) along a direction as the core step of nonlinear programming.
  • Unimodality and how to bracket a minimum inside an interval.
  • Region-elimination methods that discard part of the interval at each step.
  • Fibonacci search and its optimal reduction for a fixed number of evaluations.
  • The golden-section method, the golden ratio, and its single-evaluation-per-step economy.
  • Interpolation (quadratic and Newton) and inexact line search via the Armijo and Wolfe conditions.
Section 18-1

The Line-Search Problem

A one-dimensional search — or line search — is the problem of minimising a function of a single variable, \(\min_{x} f(x)\). It matters far beyond one dimension: almost every method for minimising a multivariable function \(F(\mathbf{x})\) works by repeatedly choosing a search direction \(\mathbf{d}_k\) at the current point \(\mathbf{x}_k\) and then asking how far to move along it. That step-length question is itself a one-variable minimisation.

The step-length subproblem
\[ \min_{\alpha \ge 0}\ \phi(\alpha), \qquad \phi(\alpha) = F\big(\mathbf{x}_k + \alpha\,\mathbf{d}_k\big) \]

Restricting \(F\) to the line through \(\mathbf{x}_k\) in direction \(\mathbf{d}_k\) collapses a hard multivariable problem into a scalar one in \(\alpha\). Because this subproblem is solved thousands of times inside a larger algorithm, it must be cheap — the whole subject of this chapter is how to find (or approximate) the minimising \(\alpha\) using as few evaluations of \(f\) as possible.

Section 18-2

Unimodality & Bracketing

The elimination methods rest on one assumption: the function is unimodal on the interval of interest. A function \(f\) is unimodal on \([a,b]\) if it has a single minimum there and is strictly decreasing before it and strictly increasing after it. Unimodality is what lets us throw away part of the interval: comparing two interior evaluations tells us which side the minimum cannot be on.

Before shrinking, we must first bracket the minimum — find an interval \([a,b]\) that is guaranteed to contain it. A simple bracketing scheme steps outward from a starting point with a growing step until the function stops decreasing and starts to rise; the last three points then straddle the minimum.

a x₁ x₂ b f(x₁) lower f(x₂) higher discard (x₂, b]
Unimodality in action: since \(f(x_1) < f(x_2)\), the minimum cannot lie in \((x_2, b]\), so that piece is eliminated
Why unimodality is the whole game. With two interior evaluations on a unimodal interval, the larger value always sits on the "wrong" side of the minimum. Discard everything beyond it and the remaining sub-interval is still guaranteed to contain the minimum. Repeat, and the bracket shrinks geometrically. Every method in this chapter is a different, cleverer way of choosing where to place those interior points so that each evaluation buys the most reduction.
Section 18-3

Region-Elimination Methods

All region-elimination (or interval-reduction) methods follow the same loop: place two points inside the current bracket \([a,b]\), evaluate \(f\) there, compare, and keep the sub-interval on the side of the smaller value. They differ only in where the points go and how many fresh evaluations each step costs.

MethodPlacement of interior pointsNew evaluations / step
Interval halvingSymmetric about the midpointTwo
Fibonacci searchSpacing set by Fibonacci ratios; needs \(n\) fixed in advanceOne (after the first step)
Golden sectionSymmetric at the golden ratio \(0.618\) of the intervalOne (after the first step)

The prize is reusing points. Halving spends two evaluations per step; the smarter methods place their points so that one of the two survives into the next, smaller interval and can be reused, costing only one new evaluation each iteration. That single idea is what makes Fibonacci and golden-section search so efficient.

Section 18-4

Fibonacci Search

Fibonacci search is the provably optimal region-elimination method for a fixed budget of \(n\) function evaluations: no other method guarantees a smaller final interval for the same number of evaluations. It uses the Fibonacci numbers \(F_0 = F_1 = 1,\ F_k = F_{k-1} + F_{k-2}\), giving \(1,1,2,3,5,8,13,21,34,\dots\)

Interval reduction after \(n\) evaluations
\[ \frac{L_{\text{final}}}{L_{0}} = \frac{1}{F_{n}} \]

At each step the two interior points are placed symmetrically using ratios of consecutive Fibonacci numbers, so one point is always reused in the next interval. The catch is that \(n\) must be chosen in advance to set the spacing — you decide how many evaluations you can afford, then the method squeezes the interval by the factor \(1/F_n\). To guarantee a final interval no longer than a tolerance \(\varepsilon\) starting from length \(L_0\), pick the smallest \(n\) with \(F_n \ge L_0/\varepsilon\).

Section 18-5

The Golden-Section Method

Fibonacci search is optimal but awkward: it needs \(n\) fixed ahead of time and a changing ratio at every step. The golden-section method sacrifices a hair of optimality for a constant, self-similar ratio that needs no advance planning. It comes from letting \(n \to \infty\) in the Fibonacci ratios, which converge to the golden ratio:

The golden ratio
\[ \tau = \frac{\sqrt{5} - 1}{2} \approx 0.618, \qquad 1 - \tau \approx 0.382 \]

On the current interval \([a,b]\) of length \(L = b - a\), place two symmetric points:

Golden-section points
\[ x_1 = a + (1-\tau)\,L, \qquad x_2 = a + \tau\,L \]
a x₁ x₂ b f(x₁) < f(x₂) → keep [a, x₂] reused as new x₂ only 1 new eval
Golden-section step: keep the sub-interval on the smaller side; the surviving interior point is reused, so each iteration costs a single new evaluation

Compare \(f(x_1)\) and \(f(x_2)\). If \(f(x_1) < f(x_2)\), the minimum lies in \([a, x_2]\); otherwise it lies in \([x_1, b]\). The golden ratio is chosen precisely so the surviving interior point is already positioned correctly for the next interval — only one new evaluation is needed per iteration, and the interval shrinks by a constant factor.

🔑
Constant reduction factor
\[ L_{k+1} = \tau\,L_k \approx 0.618\,L_k, \qquad L_k = \tau^{\,k}\,L_0 \]

Each iteration keeps \(61.8\%\) of the interval. To reach a tolerance \(\varepsilon\), the number of iterations is \(k \ge \dfrac{\ln(\varepsilon/L_0)}{\ln \tau}\). Golden-section needs no advance choice of \(n\) — you simply iterate until the bracket is small enough.

Section 18-6

Interpolation Methods

Elimination methods use only comparisons of function values and make no assumption of smoothness. When \(f\) is smooth, interpolation methods exploit its shape to converge much faster. The idea: fit a simple model — a parabola or a cubic — through a few sampled points, jump to the model's minimum, and repeat.

Quadratic interpolation fits a parabola through three points \((x_1,f_1), (x_2,f_2), (x_3,f_3)\) and takes its vertex as the next estimate of the minimum:

Three-point quadratic minimiser
\[ x^\star = x_2 - \frac{1}{2}\cdot\frac{(x_2-x_1)^2\,[\,f_2-f_3\,] - (x_2-x_3)^2\,[\,f_2-f_1\,]}{(x_2-x_1)\,[\,f_2-f_3\,] - (x_2-x_3)\,[\,f_2-f_1\,]} \]
x₁ x₂ x₃ x★ fitted parabola
Quadratic interpolation: fit a parabola through three points and take its vertex \(x^\star\) as the next estimate of the minimum

When derivatives are available, Newton's method for line search converges even faster by using curvature directly. Applying Newton's root-finding to \(f'(x) = 0\) gives:

Newton's method for line search
\[ x_{k+1} = x_k - \frac{f'(x_k)}{f''(x_k)} \]

Newton's method converges quadratically near a minimum and lands on the exact minimum of any quadratic in a single step — but it needs both derivatives and can diverge if \(f''\) is negative or the start is poor. Interpolation is powerful when \(f\) is smooth and well-behaved; elimination is the safe fallback when it is not.

Section 18-7

Inexact Line Search

Inside a larger optimiser, solving the step-length subproblem exactly is wasteful — a rough but sufficient step is enough to keep the outer method converging. An inexact line search accepts any \(\alpha\) that makes "enough" progress, judged by two conditions on \(\phi(\alpha) = F(\mathbf{x}_k + \alpha\,\mathbf{d}_k)\).

Armijo (sufficient-decrease) condition
\[ F(\mathbf{x}_k + \alpha\,\mathbf{d}_k) \le F(\mathbf{x}_k) + c_1\,\alpha\,\nabla F(\mathbf{x}_k)^{\!\top}\mathbf{d}_k, \qquad 0 < c_1 < 1 \]

The Armijo condition demands that the actual decrease be at least a fraction \(c_1\) (often \(10^{-4}\)) of the decrease the slope predicts — it rules out steps that are too long. On its own it can be satisfied by tiny useless steps, so it is paired with the curvature condition, which rules out steps that are too short:

Wolfe curvature condition
\[ \nabla F(\mathbf{x}_k + \alpha\,\mathbf{d}_k)^{\!\top}\mathbf{d}_k \ge c_2\,\nabla F(\mathbf{x}_k)^{\!\top}\mathbf{d}_k, \qquad c_1 < c_2 < 1 \]

Together these are the Wolfe conditions. The simplest way to satisfy Armijo is backtracking: start from a full step \(\alpha = 1\) and repeatedly shrink \(\alpha \leftarrow \rho\,\alpha\) (say \(\rho = 0.5\)) until sufficient decrease holds. Backtracking with the Armijo test is the line search buried inside most practical gradient and quasi-Newton solvers.

Section 18-8

Worked Examples

1 Checking unimodality

Problem. Is \(f(x) = (x-2)^2\) unimodal on \([0,5]\)? Where is its minimum?

Solution. \(f'(x) = 2(x-2)\) is negative for \(x < 2\) and positive for \(x > 2\), so \(f\) strictly decreases then strictly increases with a single minimum at \(x = 2\). It is unimodal on \([0,5]\), so elimination methods apply.

2 One golden-section step

Problem. Minimise \(f(x) = (x-2)^2\) on \([0,5]\). Take one golden-section step and give the new interval.

Solution. With \(L = 5\): \(x_1 = 0 + 0.382(5) = 1.910\) and \(x_2 = 0 + 0.618(5) = 3.090\). Then \(f(x_1) = (-0.090)^2 = 0.008\) and \(f(x_2) = (1.090)^2 = 1.188\). Since \(f(x_1) < f(x_2)\), keep \([0,\,3.090]\). The new length \(3.090 = 0.618 \times 5\) — a \(61.8\%\) survivor, as expected.

3 How many golden-section iterations?

Problem. Starting from \(L_0 = 5\), how many golden-section iterations reduce the interval to length \(\le 0.2\)?

Solution. Need \(5\,(0.618)^k \le 0.2\), i.e. \((0.618)^k \le 0.04\). Taking logs, \(k \ge \dfrac{\ln 0.04}{\ln 0.618} = \dfrac{-3.219}{-0.481} \approx 6.69\). So \(k = 7\) iterations suffice.

4 Fibonacci budget

Problem. Using Fibonacci search on \(L_0 = 5\), how many evaluations \(n\) guarantee a final interval \(\le 0.2\)?

Solution. Require \(F_n \ge L_0/\varepsilon = 5/0.2 = 25\). From \(1,1,2,3,5,8,13,21,34\), the first Fibonacci number \(\ge 25\) is \(F_8 = 34\). So \(n = 8\) evaluations suffice — slightly fewer than golden-section for the same guarantee, which is why Fibonacci is the optimal elimination method.

5 Quadratic interpolation

Problem. Fit a parabola to \(f(x) = (x-2)^2\) at \(x_1 = 1,\ x_2 = 2.5,\ x_3 = 4\) and find its minimiser.

Solution. Values: \(f_1 = 1,\ f_2 = 0.25,\ f_3 = 4\). With \(x_2-x_1 = 1.5,\ x_2-x_3 = -1.5,\ f_2-f_3 = -3.75,\ f_2-f_1 = -0.75\): numerator \(=(1.5)^2(-3.75)-(-1.5)^2(-0.75) = -8.4375+1.6875 = -6.75\); denominator \(=(1.5)(-3.75)-(-1.5)(-0.75) = -5.625-1.125 = -6.75\). So \(x^\star = 2.5 - 0.5(-6.75/-6.75) = 2.5 - 0.5 = 2.0\) — the exact minimum, since a parabola fits a quadratic perfectly.

6 Newton's method in one step

Problem. Apply one Newton step to \(f(x) = (x-2)^2\) from \(x_0 = 5\).

Solution. \(f'(x) = 2(x-2)\) and \(f''(x) = 2\). Then \(x_1 = 5 - \dfrac{2(5-2)}{2} = 5 - 3 = 2\). Newton lands on the exact minimiser in a single step because the function is quadratic — its second derivative captures the curvature perfectly.

Review

Chapter Summary

The role

Line search finds the step length \(\alpha\) along a direction — the inner loop of nonlinear programming.

Unimodality

Bracket the minimum, then compare two interior points and discard the side that cannot contain it.

Elimination

Fibonacci is optimal for a fixed budget; golden section keeps \(0.618\) each step with one new evaluation.

Interpolation

Quadratic fitting and Newton's step \(x_{k+1}=x_k - f'/f''\) converge fast on smooth functions.

Inexact search

Armijo (not too long) and Wolfe curvature (not too short) accept a "good enough" step via backtracking.

Choosing

Elimination is robust and derivative-free; interpolation is faster when \(f\) is smooth and well-behaved.

Practice

Problems

For each, state the interval, apply the method step by step, and track how the bracket shrinks. Difficulty rises down the list.

  1. Define a unimodal function and explain why unimodality is required for region-elimination methods.
  2. Show that \(f(x) = x^2 - 4x + 7\) is unimodal on \([0,4]\) and find its minimum analytically.
  3. Describe a bracketing procedure that finds an interval containing the minimum, starting from a single point.
  4. State the golden-ratio value \(\tau\) and derive the two golden-section point formulas for an interval \([a,b]\).
  5. Minimise \(f(x) = (x-3)^2\) on \([0,6]\): carry out two golden-section iterations and report the interval after each.
  6. For golden-section search, how many iterations reduce an interval of length \(4\) to \(\le 0.1\)? Show the calculation.
  7. Using Fibonacci search, find the smallest number of evaluations to reduce a length-\(4\) interval to \(\le 0.1\).
  8. Explain why Fibonacci search needs \(n\) fixed in advance while golden-section does not, and state the trade-off.
  9. Fit a parabola through three points of \(f(x) = x^2 - 6x + 10\) and confirm the interpolated minimiser is \(x = 3\).
  10. Apply one Newton line-search step to \(f(x) = x^4 - 3x\) from \(x_0 = 1\), and comment on whether it moved toward the minimum.
  11. Write the Armijo and Wolfe conditions for \(\phi(\alpha) = F(\mathbf{x}_k + \alpha\,\mathbf{d}_k)\) and explain what each one prevents.
  12. Given \(F(\mathbf{x}_k) = 10\), \(\nabla F^{\top}\mathbf{d}_k = -4\), and \(c_1 = 10^{-4}\), find the largest \(F(\mathbf{x}_k + \alpha\mathbf{d}_k)\) that passes the Armijo test at \(\alpha = 0.5\).
Tip: keep the two families straight by what they assume. Elimination methods (Fibonacci, golden section) ask only "which point is lower?" — they need nothing but unimodality, never a derivative, and they cannot be fooled by a badly-scaled or noisy function, which is why they are the safe default. Interpolation methods (quadratic, Newton) assume the function is smooth and shaped roughly like the model they fit; when that assumption holds they are dramatically faster, but a poor start or negative curvature can send them the wrong way. In practice the two are combined: bracket robustly with golden-section-style elimination, then switch to interpolation to polish the minimum once the bracket is small and the function looks locally quadratic.