round3 扩展的时候，如何理解 to_coset_extended_lagrange 和 ZH_big 的对应关系？

[readygo67]

to_coset_extended_lagrange 扩展的方式如下：

def to_coset_extended_lagrange(self, offset):
        assert self.basis == Basis.LAGRANGE
        group_order = len(self.values)
        x_powers = self.ifft().values  # step1: 将多项式的点值形式通过ifft 转换成系数形式(f(x) = a_0+ a_1*x + a_2*x^2 + a_3*x^3 + ...+a_[order-1]*x^(order-1)
        x_powers = [(offset**i * x) for i, x in enumerate(x_powers)] + [Scalar(0)] * (
            group_order * 3
        ) # step2: 将多项式表示为f'(x) = a_0 + (a_1*offset) *x + (a_2*offsset^2)*x^2 + (a_3*offsset^3)*x^3 + ... +(a_[order-1] * offset^order-1)  *x^(order-1) + 0*x^(order) +....+ 0*x^(4*order-1)
        return Polynomial(x_powers, Basis.MONOMIAL).fft()  # 返回的结果用点值形式表示

ZH_big的扩展方式如下：

        ZH_big = Polynomial(
            [
                ((Scalar(r) * self.fft_cofactor) ** group_order - 1)
                for r in quarter_roots
            ],
            Basis.LAGRANGE,
        )
        # ZH_big:[10555632090856762420326882847893243614806512971877700564250452539433530666905, 12568874522361247563306689428466362806053021808013123783916208712770621909962, 11332610780982512801919522897364031473741851428538333779447751647142277828710, 9319368349478027658939716316790912282495342592402910559781995473805186585653, 10555632090856762420326882847893243614806512971877700564250452539433530666905, 12568874522361247563306689428466362806053021808013123783916208712770621909962, 11332610780982512801919522897364031473741851428538333779447751647142277828710, 9319368349478027658939716316790912282495342592402910559781995473805186585653, 10555632090856762420326882847893243614806512971877700564250452539433530666905, 12568874522361247563306689428466362806053021808013123783916208712770621909962, 11332610780982512801919522897364031473741851428538333779447751647142277828710, 9319368349478027658939716316790912282495342592402910559781995473805186585653, 10555632090856762420326882847893243614806512971877700564250452539433530666905, 12568874522361247563306689428466362806053021808013123783916208712770621909962, 11332610780982512801919522897364031473741851428538333779447751647142277828710, 9319368349478027658939716316790912282495342592402910559781995473805186585653, 10555632090856762420326882847893243614806512971877700564250452539433530666905, 12568874522361247563306689428466362806053021808013123783916208712770621909962, 11332610780982512801919522897364031473741851428538333779447751647142277828710, 9319368349478027658939716316790912282495342592402910559781995473805186585653, 10555632090856762420326882847893243614806512971877700564250452539433530666905, 12568874522361247563306689428466362806053021808013123783916208712770621909962, 11332610780982512801919522897364031473741851428538333779447751647142277828710, 9319368349478027658939716316790912282495342592402910559781995473805186585653, 10555632090856762420326882847893243614806512971877700564250452539433530666905, 12568874522361247563306689428466362806053021808013123783916208712770621909962, 11332610780982512801919522897364031473741851428538333779447751647142277828710, 9319368349478027658939716316790912282495342592402910559781995473805186585653, 10555632090856762420326882847893243614806512971877700564250452539433530666905, 12568874522361247563306689428466362806053021808013123783916208712770621909962, 11332610780982512801919522897364031473741851428538333779447751647142277828710, 9319368349478027658939716316790912282495342592402910559781995473805186585653]
        print(f"ZH_big:{ZH_big.values}")

如何理解他们之间的对应关系？

我的理解ZH(x)的点值表示应该是[0,0,0,0,0,0,0,0], 然后直接调用ff_expand()扩展就行。

        ZH1 = Polynomial([Scalar(0)] * (group_order), Basis.LAGRANGE)
        ZH1_big = self.fft_expand(ZH1)
        # ZH1_big:[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
        print(f"ZH1_big:{ZH1_big.values}")

[readygo67]

ZH(x)的表示式是X^N-1,

        # Compute Z_H = X^N - 1, also in evaluation form in the coset
        #  不考虑coset时，ZH的系数表示 = x^group_order -1,
        #  考虑coset时，  ZH的的系数表示是 = (offset *x)^group_order - 1 = (offset^group_order) * x^group_order - 1 ，
        #  有两种得到ZH_big 的方式：
        #  方式1：构造系数 = [-1,0,0,...,(offset^group_order),0,0... ]的多项式，对其做fft.
        #  方式2：直接将 x= (offset * w) 带入 x^group_order -1，获得ZH的点值表示
        #  这两种方式是等价的。

        #方式1：
        ZH_coeffs =[0] * (4 *group_order) # ZH_big 的阶数 = 4 * ground_order
        ZH_coeffs[0] = -1  # 常数项 = -1
        # [-1,0,0,...,(offset^group_order),0,0... ]
        ZH_coeffs[group_order] = self.fft_cofactor ** group_order #第group_order项系数= (offset^group_order)

        ZH_poly = Polynomial(list(map(Scalar, ZH_coeffs)), Basis.MONOMIAL)
        ZH_big = ZH_poly.fft()

        #方式2：
        ZH_big1 = Polynomial(
            [
                ((Scalar(r) * self.fft_cofactor) ** group_order - 1)
                for r in quarter_roots
            ],
            Basis.LAGRANGE,
        )

        assert ZH_big == ZH_big1 #验证ZH两种扩展方式是否等价。