《算法导论》第四章-第6节_练习（参考答案）

算法导论（第三版）参考答案：练习4.6-1，练习4.6-2，练习4.6-3

Exercise 4.6-1

⋆ Give a simple and exact expression for nj in equation (4.27) for the case in which b is a positive integer instead of an arbitrary real number.

nj=⌈n/bj⌉ 。( 一个很好的证明 )

Exercise 4.6-2

⋆ Show that if f(n)=Θ(nlogbalgkn), where k≥0, then the master recurrence has solution T(n)=Θ(nlogbalgk+1n). For simplicity, confine your analysis to exact powers of b.

g(n)=∑j=0logbn−1ajf(n/bj)

将 f(n/bj)=Θ((n/bj)logbalgk(n/bj)) 代入 g(n)

g(n)=Θ(∑j=0logbn−1aj(nbj)logbalgk(nbj))∑j=0logbn−1aj(nbj)logbalgk(nbj)=nlogba∑j=0logbn−1(ablogba)jlgknbj=nlogba∑j=0logbn−1lgknbj∑j=0logbn−1lgknbj=∑j=0logbn−1(lgkn−o(lgkn))=logbnlgkn+logbn⋅o(lgkn)=Θ(logbnlgkn)=Θ(lgk+1n)⇒g(n)=Θ(nlogbalgk+1n)

所以

T(n)=Θ(nlogba)+g(n)=Θ(nlogba)+Θ(nlogbalgk+1n)=Θ(nlogbalgk+1n)

Exercise 4.6-3

⋆ Show that case 3 of the master method is overstated, in the sense that the regularity condition af(n/b)≤cf(n) for some constant c<1 implies that there exists a constant ϵ>0 such that f(n)=Ω(nlogba+ϵ).

令 α=a/c

af(n/b)≤cf(n)⇒αf(n/b)≤f(n)⇒αf(n)≤f(nb)⇒αif(n)≤f(nbi)⇒αif(1)≤f(bi)

令 n=bi,i=logbn，所以

f(n)≥αlogbnf(1)=Ω(nlogbα)α>a⇒α=a+d(c<1,d>0)⇒f(n)=nlogba+logbd=nlogba+ϵ(ϵ=logbd)

解答参考。