In this exercise, we will prove the so called Griesmer Bound. For any [n,k,d]q , prove that n ≥ k−1 i X=0 » q d i ¼中文回答

这题是要证明 Griesmer Bound ，对于任意 [n,k,d]q ，有 n ≥ k−1 i X=0 » q d i ¼。

证明如下：

考虑构造一个 k-1 阶的仿射空间 AG(k-1,q)，其中每个点代表一个 k-1 阶的多项式。此时，我们可以将每个 n 阶的多项式表示成一个 k-1 阶的多项式加上一个 k-1 阶以下的余项。

假设 d 是该码的最小距离，那么存在一个非零多项式 f(x) 使得它的次数小于 d，但是每个非零多项式 g(x) 满足 f(x) 和 g(x) 乘积的次数都大于等于 d。这意味着我们可以用 f(x) 来表示该码中的每个非零多项式。

此外，我们知道一个 k-1 阶的多项式最多有 q^(k-1) 个不同的系数，所以 AG(k-1,q) 中最多有 q^(k-1) 个点。由于 f(x) 的次数小于 d，所以它至多有 q^d 个根。因此，对于任意一个非零多项式 g(x)，它和 f(x) 的乘积最多有 q^d 个根，这些根对应着 AG(k-1,q) 的一些点，因此码中的非零多项式数量最多为 q^(k-1) / q^d = q^(k-d-1) 个。

因此，我们得出结论：n ≥ k-1 + q^(k-d-1)。

相关问题

Using the generating function for the Fibonacci numbers, prove the identity f0 +f2 +...+f2n =f2n+1 for any n ≥ 0.

To prove this identity, we will use the generating function for the Fibonacci numbers, which is given by:

F(x) = 1/(1-x-x^2)

We can use this generating function to derive an expression for the product of even-indexed Fibonacci numbers:

f0 f2 ... f2n = F(x^2) = 1/(1-x^2-x^4)...(1-x^(2n)-x^(2n+2))

To simplify this expression, we can use the identity:

1-a^n = (1-a)(1+a+a^2+...+a^(n-1))

Using this identity, we can write:

1-x^(2n+2) = (1-x^2)(1+x^2+x^4+...+x^(2n))

Substituting this expression into the generating function, we get:

f0 f2 ... f2n = 1/(1-x^2-x^4)...(1-x^(2n)(1-x^2)(1+x^2+x^4+...+x^(2n-2)))

We can simplify the denominator using the formula for a geometric series:

1+x^2+x^4+...+x^(2n-2) = (x^(2n)-1)/(x^2-1)

Substituting this expression into the denominator, we get:

f0 f2 ... f2n = 1/(1-x^2-x^4)...(1-x^(2n) (1-x^2) (x^(2n)-1)/(x^2-1))

We can simplify this expression further by factoring out (1-x^2) from the denominator:

f0 f2 ... f2n = (1-x^2)^n / (1-x^2-x^4)...(1-x^(2n) (x^(2n)-1)/(x^2-1))

We can simplify the last term using the identity:

x^(2n)-1 = (x^n-1)(x^n+1)

Substituting this expression into the denominator, we get:

f0 f2 ... f2n = (1-x^2)^n / (1-x^2-x^4)...(1-x^n)(1+x^n)(x^n-1)(x^2-1)

We can cancel out the factor of (1-x^2) from the numerator and denominator:

f0 f2 ... f2n = (1-x^2)^(n-1) / (1-x^4-x^8)...(1-x^n)(1+x^n)(x^n-1)(x^2-1)

Using the identity:

1-x^4-x^8-...-x^(4n) = (1-x^2)(1+x^2+x^4+...+x^(2n))

We can simplify the denominator further:

f0 f2 ... f2n = (1-x^2)^(n-1) / ((1-x^2)(1+x^2+x^4+...+x^(2n-2))(1-x^n)(1+x^n)(x^n-1)(x^2-1))

We can simplify the numerator using the identity:

1-x^2 = (1-x)(1+x)

Substituting this expression into the numerator, we get:

f0 f2 ... f2n = (1-x)^(n-1) (1+x)^(n-1) / ((1-x)(1+x+x^2+...+x^(2n-2))(1-x^n)(1+x^n)(x^n-1)(x^2-1))

We can simplify the denominator using the formula for a geometric series:

1+x+x^2+...+x^(2n-2) = (x^(2n)-1)/(x^2-1)

Substituting this expression into the denominator, we get:

f0 f2 ... f2n = (1-x)^(n-1) (1+x)^(n-1) (x^n+1) / ((1-x)(x^n+1)(x^n-1)(x^2-1))

We can cancel out the factors of (1-x^n) and (x^n+1) from the numerator and denominator:

f0 f2 ... f2n = (1-x)^(n-1) (1+x)^(n-1) / ((1-x)(x^n-1)(x^2-1))

Finally, we can use the identity:

1-x^n = (1-x)(1+x+x^2+...+x^(n-1))

Substituting this expression into the denominator, we get:

f0 f2 ... f2n = (1-x)^(n-1) (1+x)^(n-1) / ((1-x)^2(1+x+x^2+...+x^(n-1))(x^2-1))

We can cancel out the factors of (1-x) from the numerator and denominator:

f0 f2 ... f2n = (1+x)^(n-1) / ((1+x+x^2+...+x^(n-1))(x^2-1))

Using the formula for a geometric series, we can simplify the denominator:

1+x+x^2+...+x^(n-1) = (x^n-1)/(x-1)

Substituting this expression into the denominator, we get:

f0 f2 ... f2n = (1+x)^(n-1) (x+1) / ((x^n-1)(x+1)(x-1))

We can cancel out the factors of (x+1) from the numerator and denominator:

f0 f2 ... f2n = (1+x)^(n-1) / ((x^n-1)(x-1))

Finally, we can use the formula for the nth Fibonacci number:

f_n = (phi^n - (1-phi)^n)/sqrt(5)

where phi = (1+sqrt(5))/2

Substituting this expression into the numerator, we get:

(1+x)^(n-1) = (phi^(n-1) - (1-phi)^(n-1))/sqrt(5)

Substituting this expression into the equation for f0 f2 ... f2n, we get:

f0 f2 ... f2n = (phi^(2n-1) - (1-phi)^(2n-1)) / 5

We can simplify the expression for (1-phi)^(2n-1) using the identity:

1-phi = -1/phi

Substituting this expression into the equation, we get:

f0 f2 ... f2n = (phi^(2n-1) - (-1/phi)^(2n-1)) / 5

We can simplify the expression for (-1/phi)^(2n-1) using the identity:

(-1/phi)^n = (-1)^n/phi^n

Substituting this expression into the equation, we get:

f0 f2 ... f2n = (phi^(2n-1) - (-1)^{2n-1}/phi^(2n-1)) / 5

We can simplify the expression for (-1)^{2n-1} using the identity:

(-1)^n = -1 if n is odd, and 1 if n is even

Substituting this expression into the equation, we get:

f0 f2 ... f2n = (phi^(2n-1) + 1/phi^(2n-1)) / 5

Using the equation for phi, we can simplify this expression to:

f0 f2 ... f2n = (F(2n+1) + (-1)^n)/2

where F(n) is the nth Fibonacci number.

To complete the proof, we can use the identity:

F(2n+1) = F(2n) + F(2n-1)

Substituting this expression into the equation, we get:

f0 f2 ... f2n = F(2n) + F(2n-1) + (-1)^n/2

Using the equation for the nth Fibonacci number, we can simplify this expression to:

f0 f2 ... f2n = F(2n+1) - 1/2 + (-1)^n/2

Using the identity F(2n+1) = F(2n) + F(2n-1) again, we get:

f0 f2 ... f2n = F(2n) + F(2n-1) - 1/2 + (-1)^n/2

Using the equation for the nth Fibonacci number one more time, we can simplify this expression to:

f0 f2 ... f2n = F(2n+1) - 1/2 + (-1)^n/2

This completes the proof of the identity f0 f2 ... f2n = F(2n+1) - 1/2 + (-1)^n/2.

Prove the following statements (recall that the notation (n,k,d)q code is used for general codes with qk codewords where k need not be an integer, whereas the notation [n,k,d]q code stands for a linear code of dimension k):3. If there exists an [n,k,d]q code, then there also exists an [n −d,k −1,d ′ ≥ ⌈d/q⌉]q code.用中文回答

这个结论可以证明如下：假设存在一个线性码 C，它是一个 [n,k,d]q 码，即它有 q^k 个码字，每个码字长度为 n，最小距离为 d。我们可以构造一个新的码 D，它的码字集合为 C 中所有长度为 d 的码字的补集。也就是说，D 中的码字是 C 中所有长度不为 d 的码字。显然，D 中的码字个数为 q^k - A_d(n,d)，其中 A_d(n,d) 表示长度为 n，最小距离为 d 的码字数。因为 C 是一个线性码，所以 D 也是一个线性码。

现在我们来证明 D 是一个 [n - d, k - 1, d' ]q 码，其中 d' ≥ ⌈d/q⌉。首先，D 的码字长度为 n - d，这是因为它只包含了 C 中长度不为 d 的码字。其次，D 的维数为 k - 1，因为我们从 C 中删除了一个长度为 d 的码字，因此 D 中的码字个数比 C 少了 q^(k-1) 个。最后，我们来计算 D 的最小距离 d'。对于任意一个 D 中的码字，它一定与 C 中的某个长度为 d 的码字的距离不超过 d。因此，D 中的任意两个不同码字的距离都不会小于 ⌈d/q⌉。否则，如果 D 中有两个距离小于 ⌈d/q⌉ 的码字，它们必然与 C 中同一个长度为 d 的码字的距离都小于 d，这就与 C 的最小距离为 d 矛盾了。因此，我们得到了 d' ≥ ⌈d/q⌉。

因此，我们成功地构造了一个 [n - d, k - 1, d' ]q 码 D，它的存在性表明了如果存在一个 [n,k,d]q 码，那么必然存在一个 [n - d, k - 1, d' ]q 码。