如何计算多瓣BSDF的PDF

Question

在PBRT中，BSDF表示为一个BxDF瓣数组。每个瓣都设置一个位标志，表示它是哪种类型的瓣。(镜面、漫射、反射、透射等)我正在研究PBRT是如何为输入方向采样BSDF的，wi。

按照标准的Monte抽样，我期望代码能够随机地采样一个适当的瓣，然后设置pdf = pdf_{picked}/n，其中n是适当类型的叶数。然而，这不是PBRT所做的。我的问题是为什么？为了进一步发展，这是PBRT函数：

Spectrum BSDF::Sample_f(const Vector3f &woWorld, Vector3f *wiWorld,
                        const Point2f &u, Float *pdf, BxDFType type,
                        BxDFType *sampledType) const {
    // Choose which _BxDF_ to sample
    int matchingComps = NumComponents(type);
    if (matchingComps == 0) {
        *pdf = 0;
        if (sampledType)
            *sampledType = BxDFType(0);
        return Spectrum(0);
    }
    int comp = std::min((int)std::floor(u[0] * matchingComps), matchingComps - 1);

    // Get _BxDF_ pointer for chosen component
    BxDF *bxdf = nullptr;
    int count = comp;
    for (int i = 0; i < nBxDFs; ++i)
        if (bxdfs[i]->MatchesFlags(type) && count-- == 0) {
            bxdf = bxdfs[i];
            break;
        }

    // Remap _BxDF_ sample _u_ to [0,1)^2#qcStackCode#
    Point2f uRemapped(std::min(u[0] * matchingComps - comp, OneMinusEpsilon),
                      u[1]);

    // Sample chosen _BxDF_
    Vector3f wi, wo = WorldToLocal(woWorld);
    if (wo.z == 0)
        return 0.;
    *pdf = 0;
    if (sampledType) *sampledType = bxdf->type;
    Spectrum f = bxdf->Sample_f(wo, &wi, uRemapped, pdf, sampledType);

    if (*pdf == 0) {
        if (sampledType) *sampledType = BxDFType(0);
        return 0;
    }
    *wiWorld = LocalToWorld(wi);

    // Compute overall PDF with all matching _BxDF_s
    if (!(bxdf->type & BSDF_SPECULAR) && matchingComps > 1)
        for (int i = 0; i < nBxDFs; ++i)
            if (bxdfs[i] != bxdf && bxdfs[i]->MatchesFlags(type))
                *pdf += bxdfs[i]->Pdf(wo, wi);
    if (matchingComps > 1)
        *pdf /= matchingComps;

    // Compute value of BSDF for sampled direction
    if (!(bxdf->type & BSDF_SPECULAR)) {
        bool reflect = Dot(*wiWorld, ng) * Dot(woWorld, ng) > 0;
        f = 0.;
        for (int i = 0; i < nBxDFs; ++i)
            if (bxdfs[i]->MatchesFlags(type) &&
                ((reflect && (bxdfs[i]->type & BSDF_REFLECTION)) ||
                 (!reflect && (bxdfs[i]->type & BSDF_TRANSMISSION))))
                f += bxdfs[i]->f(wo, wi);
    }

    return f;
}

在伪代码中，它看起来正在执行以下操作：

// Get the number of lobes that match the ones we asked for
numMatchingLobes = NumComponents(type);

// Early out check
if (numMatchingLobes == 0):
    return

// Randomly pick one of the matching lobes
pickedBxDF = Rand()

// Transform wo from world space to local space
// Then sample the chosen BxDF to get f, pdf, and wi
f, pdf, wi = pickedBxDF.sample_f(woWorld.ToLocal())

// Early out check
if (pdf == 0):
    return

// Transform wi to world space
wiWorld = wi.ToWorld()

// Add all of the pdfs of the matching lobes if the picked BxDF wasn't specular
if pickedBxDF.type not Specular && numMatchingLobes > 1:
    for BxDF in matchingLobes:
        pdf += BxDF.pdf(wi, wo)

// Divide the resulting pdf by the number of matching lobes
pdf /= numMatchingLobes;


// Calculate the combined f if the picked BxDF isn't specular
if pickedBxDF.type not Specular:
    // Reset f to zero
    f = 0

    // Add each BxDF's f, iff it matches reflection / transmission
    reflect = wo and wi on same hemisphere as geoNormal
    for BxDF in matchingLobes:
        // Add the 
        if reflect and BxDF.type == Reflection:
            f += BxDF.f(wo, wi)
        elif !reflect and BxDF.type == Transmission:
            f += BxDF.f(wo, wi)

我不理解的部分是，为什么它会在已经计算出来的pdf的基础上添加所有匹配的叶的pdf。这似乎会使摘取的叶的pdf数加倍。

例如，如果有两个匹配的叶，称为A和B。

我们用A计算f，pdf和wi，然后用一个for循环，我们把A和B的pdf相加，除以2。

\frac{A + A + B}{2}

这似乎不对。PBRT代码错误吗？它是否应该重置pdf，类似于下面的f计算代码？

Answer 1

求和不包括被挑选来取样的BxDF。再看看这一行：

if (bxdfs[i] != bxdf && bxdfs[i]->MatchesFlags(type))

在这里，bxdf是前面取样的那个，所以当它迭代到那个样本时，它会跳过它。