资 源 简 介
仿真1：首先把网络温度参数T固定在100，按工作规则共进行1000次状态更新,把这1000次状态转移中网络中的各个状态出现的次数Si（i＝1，2，…，16）记录下来 按下式计算各个状态出现的实际频率： Pi=Si/∑i=1,NSi=Si/M 同时按照Bo1tzmann分布计算网络各个状态出现概率的理论值： Q（Ei）=（1/Z）exp（-Ei/T） 仿真2：实施降温方案，重新计算 采用快速降温方案：T（t）= T0/（1+t） T从1000降到0.01，按工作规则更新网络状态 当T=0.01时结束降温，再让T保持在0.01进行1000次状态转移，比较两种概率-a simulation : First of all network parameters temperature T fixed at 100 and, according to the rules for a total of 1000 to update the state, this state of the 1000 network transfer of all states for the number of Si (i = 1, 2, ..., 16) all recorded determined by the formula state-of the actual frequency : Pi = Si/i = 1, NSi = Si/M in accordance with Bo1tzmann distributed computing network of states all probability the theoretical value : Q (Ei) = (1/Z) exp (- Ei/T) Simulation 2 : implementation of cooling, re-using rapid cooling programs : T (t) = T0/(1 t) T dropped to 0.01 from 1000 and, according to the rules updated network state when T = 0.01 at the end of cooling, let T at 0.01 for the 1000 state tran
