Pagerank概率收敛为零？

Question

我正在尝试从头开始编写pagerank算法，到目前为止我构建了随机矩阵@implementation NSObject (HACK) - (id)responderToSelector:(SEL)selector { if ([self respondsToSelector:selector] && [self.class instancesRespondToSelector:selector]) { return self; // the class and the instance actually will respond to the selector when called } if ([self respondsToSelector:selector]) { // invocations are forwarded id forward = [self forwardingTargetForSelector:selector]; if (forward && forward != self) { return [forward responderToSelector:selector]; } NSMethodSignature *ms = [self methodSignatureForSelector:selector]; if (ms) { MockInvocation *mockInvocation = [MockInvocation invocationWithMethodSignature:ms]; mockInvocation.selector = selector; [self forwardInvocation:mockInvocation]; return mockInvocation.target ?: mockInvocation.innerTarget; } } return nil; } @interface MockInvocation : NSInvocation @property (nonatomic, weak) id innerTarget; @end @implementation IIOMockInvocation - (void)invoke { } - (void)invokeWithTarget:(id)target { _innerTarget = target; } @end ，现在我试图找到限制概率。限制概率计算如下：

G

结果应该是另一个具有正元素且总和为1的向量。当我进行计算时，我的结果收敛为零。这是代码：

- choose a random vector pi, such that each element is positive and sums to 1
- compute pi * G
- compute pi * G^2
- compute pi * G^3
- repeat until convergence

算法应该收敛到一个结果，结果的总和应该是1.但是我得到import numpy as np # Define stochastic matrix (rows sum to 1), dimensions [6, 6] G = np.array([[ 8.29187396e-04, 4.98341625e-01, 8.29187396e-04, 4.98341625e-01, 8.29187396e-04, 8.29187396e-04], [ 3.32779623e-01, 5.53709856e-04, 3.32779623e-01, 5.53709856e-04, 3.32779623e-01, 5.53709856e-04], [ 1.66666667e-01, 1.66666667e-01, 1.66666667e-01, 1.66666667e-01, 1.66666667e-01, 1.66666667e-01], [ 4.98341625e-01, 4.98341625e-01, 8.29187396e-04, 8.29187396e-04, 8.29187396e-04, 8.29187396e-04], [ 1.66666667e-01, 1.66666667e-01, 1.66666667e-01, 1.66666667e-01, 1.66666667e-01, 1.66666667e-01], [ 1.65016502e-03, 1.65016502e-03, 1.65016502e-03, 1.65016502e-03, 9.91749175e-01, 1.65016502e-03]]) # Compute probability vector by repeated multiplication n_iter = 200 # Number of iterations pi = np.random.random(size=6) # Random starting vector pi /= sum(pi) # Normalise evolution = [np.dot(pi, G**i) for i in range(1, n_iter + 1)] # Stores each step 。此外，如果我在每次迭代中绘制结果，很明显结果将最终收敛为零：

我哪里错了？我确信转换矩阵是正确的，所以我认为问题出在最后的计算中。 任何帮助将不胜感激！