代数法求骨架矩阵
此处输入要素的个数:
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骨架矩阵:指的是系统里存在环路,在进行缩点处理的后,得到的缩点矩阵再把其所有的向前边全部删除。得到的矩阵叫骨架矩阵
如果原始矩阵不存在回路,原始矩阵就是个DAG图,骨架矩阵符合一个简单的代数公式。
DAG图求骨架矩阵的代数公式:S=R-(R-I)2-I
显示的是一个随机 12 * 12 的方阵
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子 | 丑 | 寅 | 卯 | 辰 | 巳 | 午 | 未 | 申 | 酉 | 戌 | 亥 |
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丑 |
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寅 |
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1 |
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1 |
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卯 |
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辰 |
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1 |
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巳 |
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1 |
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午 |
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1 |
未 |
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申 |
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1 |
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酉 |
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1 |
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1 |
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戌 |
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1 |
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1 |
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亥 |
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1 |
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子 | 巳+午+亥 | 未 | 寅+酉 | 申 | 丑 | 卯 | 辰 | 戌 |
子 |
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巳+午+亥 |
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1 |
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未 |
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寅+酉 |
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1 |
1 |
1 |
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申 |
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1 |
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丑 |
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1 |
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1 |
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卯 |
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辰 |
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1 |
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戌 |
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1 |
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1 |
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第二步,对新矩阵求可达矩阵,可达矩阵如下
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子 | 巳+午+亥 | 未 | 寅+酉 | 申 | 丑 | 卯 | 辰 | 戌 |
子 |
1 |
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巳+午+亥 |
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1 |
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未 |
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1 |
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寅+酉 |
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1 |
1 |
1 |
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申 |
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1 |
1 |
1 |
1 |
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丑 |
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1 |
1 |
1 |
1 |
1 |
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卯 |
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1 |
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辰 |
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1 |
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1 |
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戌 |
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1 |
1 |
1 |
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1 |
1 |
子 |
子、 |
巳+午+亥 |
巳+午+亥、 |
未 |
未、 |
寅+酉 |
巳+午+亥、未、寅+酉、 |
申 |
巳+午+亥、未、寅+酉、申、 |
丑 |
巳+午+亥、未、寅+酉、申、丑、 |
卯 |
卯、 |
辰 |
未、辰、 |
戌 |
巳+午+亥、未、寅+酉、辰、戌、 |
第三步,单位矩阵
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子 | 巳+午+亥 | 未 | 寅+酉 | 申 | 丑 | 卯 | 辰 | 戌 |
子 |
1 |
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巳+午+亥 |
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1 |
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未 |
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1 |
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寅+酉 |
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1 |
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申 |
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1 |
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丑 |
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1 |
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卯 |
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1 |
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辰 |
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1 |
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戌 |
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1 |
子 |
子、 |
巳+午+亥 |
巳+午+亥、 |
未 |
未、 |
寅+酉 |
寅+酉、 |
申 |
申、 |
丑 |
丑、 |
卯 |
卯、 |
辰 |
辰、 |
戌 |
戌、 |
第四步,可达矩阵减去单位矩阵后的矩阵
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子 | 巳+午+亥 | 未 | 寅+酉 | 申 | 丑 | 卯 | 辰 | 戌 |
子 |
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巳+午+亥 |
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未 |
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寅+酉 |
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1 |
1 |
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申 |
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1 |
1 |
1 |
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丑 |
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1 |
1 |
1 |
1 |
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卯 |
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辰 |
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1 |
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戌 |
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1 |
1 |
1 |
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1 |
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寅+酉 |
巳+午+亥、未、 |
申 |
巳+午+亥、未、寅+酉、 |
丑 |
巳+午+亥、未、寅+酉、申、 |
辰 |
未、 |
戌 |
巳+午+亥、未、寅+酉、辰、 |
第五步,两个矩阵相乘
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子 | 巳+午+亥 | 未 | 寅+酉 | 申 | 丑 | 卯 | 辰 | 戌 |
子 |
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巳+午+亥 |
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未 |
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寅+酉 |
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申 |
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1 |
1 |
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丑 |
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1 |
1 |
1 |
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卯 |
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辰 |
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戌 |
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1 |
1 |
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申 |
巳+午+亥、未、 |
丑 |
巳+午+亥、未、寅+酉、 |
戌 |
巳+午+亥、未、 |
第六步,可达矩阵减去上面的矩阵
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子 | 巳+午+亥 | 未 | 寅+酉 | 申 | 丑 | 卯 | 辰 | 戌 |
子 |
1 |
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巳+午+亥 |
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1 |
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未 |
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1 |
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寅+酉 |
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1 |
1 |
1 |
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申 |
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1 |
1 |
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丑 |
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1 |
1 |
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卯 |
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1 |
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辰 |
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1 |
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1 |
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戌 |
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1 |
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1 |
1 |
子 |
子、 |
巳+午+亥 |
巳+午+亥、 |
未 |
未、 |
寅+酉 |
巳+午+亥、未、寅+酉、 |
申 |
寅+酉、申、 |
丑 |
申、丑、 |
卯 |
卯、 |
辰 |
未、辰、 |
戌 |
寅+酉、辰、戌、 |
第七步,再减去单位矩阵后就是骨架矩阵
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子 | 巳+午+亥 | 未 | 寅+酉 | 申 | 丑 | 卯 | 辰 | 戌 |
子 |
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巳+午+亥 |
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未 |
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寅+酉 |
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1 |
1 |
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申 |
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1 |
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丑 |
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1 |
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卯 |
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辰 |
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1 |
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戌 |
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1 |
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1 |
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寅+酉 |
巳+午+亥、未、 |
申 |
寅+酉、 |
丑 |
申、 |
辰 |
未、 |
戌 |
寅+酉、辰、 |
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