Thus, for any F $^{α}$, It implies that Thus which is a monotonic increasing function with α < 1. Lemma : if q ∈ (0 , 1 $_{3}$), for any large enough z we have $^{where}$− ∑ z x = 1 x z log x $_{z}$C x $_{z}$q $^{$^{x}$(1}$− q ) $^{z}$−x denotes the expectation of entropy of a node with the probability q to be chose and the total choosing time is z (as shown in Fig. 5). Proof : is a monotonic increasing function. For large enough z , normal distribution is a well approximation to binomial distribution then we have where σ 2 = zq (1 − q ) , µ = zq . Obviously, 3. Case 2: α > 1 lim $^{L}$→∞ q$_{F}$ $^{α}$, i = 0 . (11) lim $^{L}$→∞ ε$_{α}$ = log( c (2 − a ) 1 a + [ c (2 − a ) 1 a ]$^{2}$) . (12) lim $^{L}$→∞ E ( ε$_{α}$ ) = log( c (2 − a ) 1 a + [ c (2 − a ) 1 a ]$^{2}$) , (13) − q log q > − z ∑ x = 1 x z log x z C x $_{z}$q$^{x}$(1 − q ) $^{z}$−$^{x}$, (14) $^{According to Law of Large Numbers, lim$_{z}$}$→∞ − ∑ z x = 1 x z log x $_{z}$C x $_{z}$q$^{x}$(1 − q ) $^{z}$−x $^{=}$− q log q . Thus, we just need to prove g ( z ) $^{=}$− z ∑ x = 1 x z log x z C x $_{z}$q$^{x}$(1 − q ) $^{z}$−x (15) g ( z ) = ∫ z 1 1 σ √$_{2}$$_{π}$ e − ( $^{x}$− µ ) 2 2 σ 2 x z log x z dx , (16) g ′$_{z}$ ( z ) = √$_{2}$ 4 z $^{3}$q(1 − q ) √ π zq (1 − q ) ∫ z 1 [log x z ( q $^{2}$z2 − 3 q $^{2}$z + 3 qz − x $^{2}$) − 2 zq $^{2}$+ 2 zq ] e ( $^{x}$−zq ) 2 2 zq ( $^{q}$− 1) xdx (17) √$_{2}$ 4 z $^{3}$q(1 − q ) √ π zq (1 − q ) > 0 (18) 14