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One accepted Rosenbrock step gives different FR and PR+ coefficients

Example

For the Rosenbrock function f(x,y)=(1−x)2+100(y−x2)2, take x0=(−6/5,1) and d0=−g0. With c1=1/1000, c2=49/100, bounded search budgets E≥1, Z≥10, and outer budget K≥2, the first accepted step is t0=1/1024. Its FR and PR+ coefficients differ. The PR+ candidate ascends, so the safeguarded method restarts; for definiteness take σ=1/2 and C=2.

Facts & Assumptions

Given: The function, start and parameters in the Example, with exact arithmetic.

[F1]

Fletcher–Reeves and safeguarded Polak–Ribière nonlinear conjugate gradients: d0=−g0; FR uses β0FR=∥g1∥22/∥g0∥22, whereas the PR+ candidate uses β0PR+=max⁡{0,g1T(g1−g0)/∥g0∥22}; a failed descent or norm test resets it to −g1.

[F2]

Wolfe and strong Wolfe line-search conditions: strong Wolfe requires Armijo decrease and ∣ϕ′(t)∣≤c2∣ϕ′(0)∣.

[F3]

Finite-budget bracketing and zoom for a strong Wolfe step: the first expansion trial is 1; an Armijo failure calls zoom with endpoints (0,1), whose midpoint trials halve the positive endpoint while the same failure persists.

Verification

Given: The objective, initial point and budgets of the Example.

1.1givenF1algebra

Differentiation gives g0=(−1078/5,−88), d0=(1078/5,88), f(x0)=121/5, and ϕ′(0)=−G0 with G0=1,355,684/25=54,227.36. At a parameter t, write u=−6/5+(1078/5)t, v=1+88t, and r=v−u2. Then ϕ(t)=(1−u)2+100r2 and ϕ′(t)=(−400ur+2(u−1))(1078/5)+200r(88).

2.1givenF3step 1.1algebra

Exact substitution into step 1.1 gives integer lower bounds for ϕ(2−j), in order j=0,…,9: 210482437168, 12810991198, 758614013, 42388850, 2078788, 72456, 543, 211, 149, and 35. Each lower bound exceeds 121/5, whereas every positive-trial Armijo right side is below 121/5. Thus the initial trial fails Armijo and the first nine zoom trials fail it as well. By [F3], a=0 remains fixed and the tenth zoom trial is t0=2−10=1/1024.

3.1givenF2F3step 1.1step 2.1algebra

At this trial, x1=(−2533/2560,139/128) and r=700711/6553600. Exactly, f(x1)=43818224127785899345920<511100,g1=(160801353941943040,70071132768),h:=g1Td0=1064039515121104857600. The Armijo right side is 121/5−G0/(1000⋅1024)>24, so Armijo holds. Also 0<h<10148<26571<(49/100)G0=26571.4064, so strong curvature holds. The search therefore returns this tenth midpoint.

4.1givenF1step 3.1algebra

Put G1=∥g1∥22=3390155233212910921/1759218604441600. Because g0=−d0 and h>0, the coefficients are β0FR=G1/G0 and β0PR+=(G1+h)/G0>β0FR. The FR direction is −g1+(G1/G0)d0; the PR+ candidate is d1cand=−g1+((G1+h)/G0)d0. Exact values in step 3.1 give 1927<G1<1928, 10147<h<10148, and 54227<G0<54228. Hence g1Td1cand=−G1+(G1+h)hG0>−1928+(1927+10147)1014754228>331>0. It fails the descent safeguard for every σ>0. Thus the actual selected PR+ direction is d1=−g1 and its restart flag is true; the ascending candidate is not the selected direction.

5.1givenF1F2step 1.1algebra∎

For comparison, the originally displayed trial t^=1/1000 is also a mathematically valid strong Wolfe step, but is not an output of this dyadic search. It gives x^=(−2461/2500,136/125), r^=743479/6250000, f(x^)=5.35291158000896, g^=(42.8715665664,23.791328), and h^=g^Td0=11336.74661571584. Its Armijo bound is 24.14577264 and ∣h^∣<11400<26571.4064, verifying both inequalities directly. With G^=∥g^∥22, its distinct coefficients are G^/G0 and (G^+h^)/G0; these describe the hypothetical accepted update at t^, rather than the actual bounded-search trace above.

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