Fix division by zero in StopWhenReturnsDiminish criterion

[stepankonrad]

Hi @dthuerck, this happens for example if the graph doesn't have any edges. Is this the right way to terminate? Can the cost go below 0? Best, Stepan

[dthuerck]

Hi @stepankonrad,

thanks for contributing! You're absolutely right, a 0-objective would cause a 0-division - shame on me :/ Good point about negative costs. Though I always assumed costs are nonnegative, in general I see no issues with actually using negative costs...

On the modelling side, the best possible solution is adding a bias pushing the global minimum to 0 or above. Otherwise, there's all sorts of stuff that could go wrong (think sequences 20 -> 10 -> 0 -> -10 -> -20 or 10 -> -10). After considering some cases, I would propose the following to stick as close as possible to the previous meaning of the termination criterion: let's use the following:

    /* avoid nondecreasing objective */
    if(newest_val >= oldest_val)
        return true;

    /* in order to support negative costs, handle the magnitudes only */
    const _s_t<COSTTYPE, SIMDWIDTH> max_val = std::max(
        oldest_val, newest_val);
    const _s_t<COSTTYPE, SIMDWIDTH> min_val = std::min(
        oldest_val, newest_val);

    /* avoid 0-division when objective stays constant */
    const _s_t<COSTTYPE, SIMDWIDTH> improv = 
        (max_val - min_val) / (static_cast<_s_t<COSTTYPE, SIMDWIDTH>(0.5) * 
        (std::abs(max_val) + std::abs(min_val)));

    return (improv < m_improv_threshold);

This would gracefully handle negative costs by biasing the expected loss the average. Does it solve you problem?