# A shortest k-sequence need not form a basis of the lattice

I'm reading Fundamental Problems of Algorithmic Algebra by Chee-Keng Yap and could not solve the following problem.

The shortest k-sequence (where k > 2 is the dimension of the lattice) need not form a basis of the lattice.

With following definition of a shortest k-sequence:

We define $$u ∈ Λ$$ to be a shortest vector in $$Λ$$ if it has the shortest length among the non-zero vectors of $$Λ$$. More generally, we call a sequence $$(u_1 , u_2 , . . . , u_k )$$ for $$k ≥ 1$$, of vectors a shortest k-sequence of $$Λ$$ if for each $$i = 1, . . . , k$$; $$u_i$$ is a shortest vector in the set $$Λ \setminus Λ(u_1 , u_2 , . . . , u_{i−1} )$$.

• just do exercise 1.7 on page 222 for $n=5$ Commented Apr 4, 2022 at 1:00
• Thank you, i hope my solution is correct.
– ASP
Commented Apr 4, 2022 at 14:09

Exercise 1.7: (Dubé) Consider $$e_1 , e_2 , . . . , e_{n−1} , h$$where $$e_i$$ is the elementary n-vector whose ith component equals 1 and all other components equal zero, and $$h = (\frac{1}{2} , \frac{1}{2}, . . . ,\frac{1}{2} )$$. Show that this set of vectors form a basis for the lattice $$Λ = \mathbb{Z} ∪ ( \frac{1}{2} + \mathbb{Z})$$. What is the shortest n-sequence of Λ? Show that for $$n ≥ 5$$, this shortest n-sequence is not a basis for $$Λ$$.
For $$n=5$$ the shortest 5-sequence is $$(e_1, e_2, e_3, e_4, e_5)$$, since $$\|h\| = \sqrt{0.5^5} \approx 1.118$$ is bigger than $$\|e_5\| = 1$$ and $$e_5 = -e_1 -e_2 -e_3 -e_4 + 2*h$$. The shortest sequence is not a base for the lattice since you cant generate $$h$$ with just integer coefficients.