帮我写一段用基础函数实现的可以成功运行的NTRU数字签名的MATLAB实验代码

好的，下面是基于基础函数实现的NTRU数字签名的MATLAB实验代码：

% NTRU数字签名基础函数实现代码
% 该代码实现了对明文的数字签名和验证操作

% 公钥和私钥生成
N = 191;
p = 3;
q = 64;
d = 47;
f = ntru_gen_poly(N, q-d, q, 1);  % 生成f(x)
g = ntru_gen_poly(N, d, q, 0);    % 生成g(x)
h = ntru_mod3(f, q, p);           % h(x) = 3*f(x)/(1-x^N) mod p

% 明文的数字签名
M = 'hello, world';
m = uint8(M);                     % 明文msg先转换成ASCII码
h_m = ntru_hash(m);               % 对msg进行哈希操作
rand_poly = ntru_rand_poly(N, q); % 随机生成多项式rand_poly
e = ntru_encrypt(rand_poly, h);   % 加密操作，得到e(x)
s = ntru_sign(h_m, rand_poly, f, g, q, p, N);  % 签名操作，得到s(x)

% 数字签名的验证
h_s = ntru_signverify(s, f, g, q, p, N);  % 对签名进行验证，得到h(s)
if (isequal(h_m, h_s))
    disp('数字签名验证成功!');
else
    disp('数字签名验证失败!');
end

% NTRU多项式生成函数
function f = ntru_gen_poly(N, df, q, small_coeff)
    rng('shuffle');
    f = zeros(1, N);
    num_small_coeffs = 0;
    while num_small_coeffs < df
        pos = randi([1, N], 1);
        if (f(pos) == 0)
            coeff = randi([-q, q], 1);
            if (coeff == 0 && small_coeff)
                coeff = 1;
            end
            f(pos) = coeff;
            if (abs(coeff) < q)
                num_small_coeffs = num_small_coeffs + 1;
            end
        end
    end
end

% NTRU多项式哈希函数
function h = ntru_hash(m)
    h = uint32(31);
    for i = 1:length(m)
        h = mod(h*uint32(131) + uint32(m(i)), 2^32-1);
    end
end

% NTRU多项式取模函数
function r = ntru_mod3(f, q, p)
    r = mod(f./3, p);
end

% NTRU多项式随机生成函数
function rand_poly = ntru_rand_poly(N, q)
    rand_coeffs = randi([-q, q], 1, N);
    rand_poly = poly2ntru(rand_coeffs, q, N);
end

% NTRU多项式加密函数
function e = ntru_encrypt(m, h)
    e = ntru_poly_mul(h, m);
end

% NTRU多项式签名函数
function s = ntru_sign(m, r, f, g, q, p, N)
    e = ntru_encrypt(r, f);
    a = ntru_rand_poly(N, q);
    b = ntru_encrypt(a, f);
    t = ntru_poly_add(m, ntru_poly_mul(b, g));
    d = ntru_decrypt(e, f, g, q, p, N);
    c = ntru_poly_sub(d, t);
    s = ntru_poly_add(c, ntru_poly_mul(a, g));
end

% NTRU多项式签名验证函数
function h_s = ntru_signverify(s, f, g, q, p, N)
    t = ntru_poly_mul(s, f);
    e = ntru_poly_sub(t, g);
    d = ntru_decrypt(e, f, g, q, p, N);
    h_s = ntru_hash(uint8(d));
end

% NTRU多项式相加函数
function c = ntru_poly_add(a, b)
    c = mod(a + b, 3);
end

% NTRU多项式相减函数
function c = ntru_poly_sub(a, b)
    c = mod(a - b, 3);
end

% NTRU多项式相乘函数
function c = ntru_poly_mul(a, b)
    c = mod(conv(a, b), 3);
end

% NTRU多项式解密函数
function m = ntru_decrypt(e, f, g, q, p, N)
    h = ntru_poly_mul(e, f);
    c = ntru_mod3(h, q, p);
    m = ntru_poly_add(c, g);
end

% NTRU多项式转换函数
function a_ntru = poly2ntru(a, q, N)
    a_ntru = zeros(1, N);
    for i = 1:N
        ai_mod_q = mod(a(i), q);
        if (ai_mod_q <= q/2)
            a_ntru(i) = max(-1, ai_mod_q - (q-1)/3);
        else
            a_ntru(i) = min(1, ai_mod_q - (2*q-1)/3);
        end
    end
end

请注意，此代码没有考虑保密性，因为密钥不能直接从代码中获取，因此本文在此不作讨论。