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计数的基本方法是确定边的数量，通过子图完全图边和点的关系来计算点数，最后得到答案

top n bounding box

首先处理出处理出top n的注意力权重和对应的bounding box

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def filter_most_important(self, n, boxes, attention):
    """ Only keep top-n object proposals, scored by attention weight """
    attention, idx = attention.topk(n, dim=1, sorted=False)
    idx = idx.unsqueeze(dim=1).expand(boxes.size(0), boxes.size(1), idx.size(1))
    boxes = boxes.gather(2, idx)
    return boxes, attention

boxes的shape为(n, 4, m)，attention的shape为(n, m)，其中n为batch size，m为一张图中bbox的数量

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torch.topk(input, k, dim=None, largest=True, sorted=True, out=None) → (Tensor, LongTensor)

k表示top k，dim为指定维度，如果未指定则默认最后一维，largest表示返回最大还是最小(True/False)，sorted表示返回的top k是否需要排序，out可选输出张量(Tensor, LongTensor)

返回值是一个元组(values,indices)

源码中指定的dim=1是bbox维度，即选取attention权值最大的k个bbox

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torch.unsqueeze(input, dim) → Tensor

unsqueeze用于维度扩充，dim为插入的维度下标

expand是将torch的各个维度扩展到更大的尺寸，在unsqueeze之后，idx变为(n,1,m)，然后扩展为(n,4,m)，和bbox的tensor尺寸相同

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torch.gather(input, dim, index, out=None, sparse_grad=False) → Tensor

gather函数在dim维度按照index对input进行索引把对应的数据提取出来的，这里是提取出top k的attention权值对应的bbox

Calculate A && D

然后计算注意矩阵和距离矩阵

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relevancy = self.outer_product(attention)
distance = 1 - self.iou(boxes, boxes)

看下矩阵外积(outer_produce)的实现

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def outer(self, x):
    size = tuple(x.size()) + (x.size()[-1],)
    a = x.unsqueeze(dim=-1).expand(*size)
    b = x.unsqueeze(dim=-2).expand(*size)
    return a, b

def outer_product(self, x):
    # Y_ij = x_i * x_j
    a, b = self.outer(x)
    return a * b

(x.size()[-1],)表示将x.size()的最后一维取出并将其表示为元组，元组相加是拼接而非按位相加，得到size为(n,m,m)，所以outer的作用就是将x变成两个(n,m,m)的tensor，将权值放在对应的倒数第一维和倒数第二维，outer_product将x转化后得到的a和b相乘得到外积

看一下Intersection over Union(IoU)的实现

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def iou(self, a, b):
    # this is just the usual way to IoU from bounding boxes
    inter = self.intersection(a, b)
    area_a = self.area(a).unsqueeze(2).expand_as(inter)
    area_b = self.area(b).unsqueeze(1).expand_as(inter)
    return inter / (area_a + area_b - inter + 1e-12)

先看一下 intersection

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def intersection(self, a, b):
    size = (a.size(0), 2, a.size(2), b.size(2))
    min_point = torch.max(
        a[:, :2, :].unsqueeze(dim=3).expand(*size),
        b[:, :2, :].unsqueeze(dim=2).expand(*size),
    )
    max_point = torch.min(
        a[:, 2:, :].unsqueeze(dim=3).expand(*size),
        b[:, 2:, :].unsqueeze(dim=2).expand(*size),
    )
    inter = (max_point - min_point).clamp(min=0)
    area = inter[:, 0, :, :] * inter[:, 1, :, :]
    return area

a[:, :2, :]表示将dim=1的前两个切片(左上角坐标)取出，对应的a[:, 2:, :]则为右下角坐标

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torch.max(input, other, out=None) → Tensor

torch.max可以将取两个tensor对应位置的最大值，返回是一个tensor，两个tensor不一样大时可以广播计算，不过这里源码直接将其拉伸成一样大的了

将左上角取最大值就得到了相交区域的上界，同理，右下角取最小值得到了相交区域的下界

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torch.clamp(input, min, max, out=None) → Tensor

clamp给tensor的元素限定上下界(min,max)，如果两个bbox没有交集，则可能计算出来的是负数，所以需要设定下界0，将两个inter对应的长宽相乘得到每两个bbox的相交面积area，得到面积交矩阵

然后看一下area

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def area(self, box):
    x = (box[:, 2, :] - box[:, 0, :]).clamp(min=0)
    y = (box[:, 3, :] - box[:, 1, :]).clamp(min=0)
    return x * y

显然计算的是bbox的面积，同时用了clamp做了极小限制防止出现负数

在iou中将得到的area通过unsqueeze在矩阵上横向扩展和纵向扩展，相加再减去面积交矩阵即可得到面积并矩阵，IoU = 面积交/面积并，计算即可

然后用得到的注意力矩阵A和距离矩阵D去计算$\bar{A}$

intra-object dedup

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score = self.f[0](relevancy) * self.f[1](distance)

将A和D通过两个不同的激活函数，然后点乘来消除intra-object

来看一下激活函数的实现

在__init__里有

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self.f = nn.ModuleList([PiecewiseLin(16) for _ in range(16)])

查看作者写的分段线性函数PiecewiseLin

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class PiecewiseLin(nn.Module):
    def __init__(self, n):
        super().__init__()
        self.n = n
        self.weight = nn.Parameter(torch.ones(n + 1))
        # the first weight here is always 0 with a 0 gradient
        self.weight.data[0] = 0

    def forward(self, x):
        # all weights are positive -> function is monotonically increasing
        w = self.weight.abs()
        # make weights sum to one -> f(1) = 1
        w = w / w.sum()
        w = w.view([self.n + 1] + [1] * x.dim())
        # keep cumulative sum for O(1) time complexity
        csum = w.cumsum(dim=0)
        csum = csum.expand((self.n + 1,) + tuple(x.size()))
        w = w.expand_as(csum)

        # figure out which part of the function the input lies on
        y = self.n * x.unsqueeze(0)
        idx = Variable(y.long().data)
        f = y.frac()

        # contribution of the linear parts left of the input
        x = csum.gather(0, idx.clamp(max=self.n))
        # contribution within the linear segment the input falls into
        x = x + f * w.gather(0, (idx + 1).clamp(max=self.n))
        return x.squeeze(0)

cumsum是累加

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torch.cumsum(input, dim, out=None, dtype=None) → Tensor

返回的tensor y满足$y_i=x_1+x_2+\dots+x_i$

通过w=w/w.sum()和self.weight.data[0] = 0保证$f_k(0)=0，f_k(1)=1$

inter-object dedup

然后消除inter-object，方法是等比例缩小边的权重

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dedup_score = self.f[3](relevancy) * self.f[4](distance)
dedup_per_entry, dedup_per_row = self.deduplicate(dedup_score, attention)
score = score / dedup_per_entry

dedup_score就是计算了式中的X，$X = f_4(A)\odot f_5(D)$

看一下deduplicate函数

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def deduplicate(self, dedup_score, att):
    # using outer-diffs
    att_diff = self.outer_diff(att)
    score_diff = self.outer_diff(dedup_score)
    sim = self.f[2](1 - score_diff).prod(dim=1) * self.f[2](1 - att_diff)
    # similarity for each row
    row_sims = sim.sum(dim=2)
    # similarity for each entry
    all_sims = self.outer_product(row_sims)
    return all_sims, row_sims

其中用了一个outer_diff函数

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def outer_diff(self, x):
    # like outer products, except taking the absolute difference instead
    # Y_ij = | x_i - x_j |
    a, b = self.outer(x)
    return (a - b).abs()

和outer_product的功能类似，用这个函数求了attention的两两权重差和dedup_score的两两差，按照式子计算了proposal两两之间的相似度

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torch.prod(input, dim, keepdim=False, dtype=None) → Tensor

prod是对指定维度dim求元素积，对应公式中的$\prod$

$row_sims$计算了proposal i和其它proposal相似的次数

对于$edge_{ij}$来说值需要除以${row_sims}_i$和${row_sims}_j$，所以这里处理了$row_sims$的$outer_product$，即$all_sims$

然后对score进行了缩放，即得到了$\bar{A}\odot ss^T$

aggregate the score

答案统计用的是$|V|=\sqrt{|E|}$，因此还需要将在intra-object dedup中消去的自环加回来

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correction = self.f[0](attention * attention) / dedup_per_row
score = score.sum(dim=2).sum(dim=1, keepdim=True) + correction.sum(dim=1, keepdim=True)
score = (score + 1e-20).sqrt()
one_hot = self.to_one_hot(score)

算自环直接attention点积然后除倍率，矩阵点积对于边权值是线性变化，所以乘s而不是乘$s\odot s$，原式中diag是将数值填入对角矩阵的对角线，因为代码中是直接计数所以不需要这步操作

直接求和得到了|E|，开方即可得到|V|，也就是正确的计数

并将答案转化为特殊的one-hot编码$o=[o_0,o_1,\dots,o_n]^T$

$o_i=max(0,1-|c-i|)$

若答案为整数，则one-hot编码结果为对应答案位置为1，其余位置为0，否则o对应两个one-hot编码的线性插值

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def to_one_hot(self, scores):
    """ Turn a bunch of non-negative scalar values into a one-hot encoding.
    E.g. with self.objects = 3, 0 -> [1 0 0 0], 2.75 -> [0 0 0.25 0.75].
    """
    # sanity check, I don't think this ever does anything (it certainly shouldn't)
    scores = scores.clamp(min=0, max=self.objects)
    # compute only on the support
    i = scores.long().data
    f = scores.frac()
    # target_l is the one-hot if the score is rounded down
    # target_r is the one-hot if the score is rounded up
    target_l = scores.data.new(i.size(0), self.objects + 1).fill_(0)
    target_r = scores.data.new(i.size(0), self.objects + 1).fill_(0)

    target_l.scatter_(dim=1, index=i.clamp(max=self.objects), value=1)
    target_r.scatter_(dim=1, index=(i + 1).clamp(max=self.objects), value=1)
    # interpolate between these with the fractional part of the score
    return (1 - f) * Variable(target_l) + f * Variable(target_r)

最后考虑尽可能用A和D矩阵中为1和0的数字来计算答案，计算两个置信系数，然后返回最终的计数特征

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one_hot = self.to_one_hot(score)
att_conf = (self.f[5](attention) - 0.5).abs()
dist_conf = (self.f[6](distance) - 0.5).abs()
conf = self.f[7](att_conf.mean(dim=1, keepdim=True) + dist_conf.mean(dim=2).mean(dim=1, keepdim=True))
return one_hot * conf