Journal of Systems Engineering and Electronics ›› 2018, Vol. 29 ›› Issue (6): 1209-1227.doi: 10.21629/JSEE.2018.06.09
• Systems Engineering • Previous Articles Next Articles
Chuchao HE(), Xiaoguang GAO*(), Zhigao GUO()
Received:
2017-09-11
Online:
2018-12-25
Published:
2018-12-26
Contact:
Xiaoguang GAO
E-mail:xomrssh@163.com;cxg2012@nwpu.edu.cn;guozhigao2004@163.com
About author:
HE Chuchao was born in 1992. He is a Ph.D. candidate in the Department of System Engineering, Northwestern Polytechnical University. His research interests focus on Bayesian network learning, especially on structure learning. E-mail: Supported by:
Chuchao HE, Xiaoguang GAO, Zhigao GUO. Structure learning on Bayesian networks by finding the optimal ordering with and without priors[J]. Journal of Systems Engineering and Electronics, 2018, 29(6): 1209-1227.
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Table 1
"
Input: Data, node sizes, a random order |
Output: the final order |
current ord——the current order who holds the best score |
best score——the score of the current order |
best node sc——the family score of each node in the current order |
K2——K2 algorithm |
sc——the score of the order obtained by transformation |
node sc——the family score of each node in the obtained order |
t——the subscript variable |
m——the intermediate variable |
1. the final order $\leftarrow $ a random order; |
2. current_ord $\leftarrow $ the final order; |
3. [best_score, best_node_$sc$] $\leftarrow $ K2(current_ord); |
4.$t\leftarrow 0$; |
5.Repeat |
6. For $i = t + 2 : n$ |
7. Update the current_ord by swapping two nodes whose subscript are $i$ and $t+1$ respectively; |
8. [$sc$, node_$sc$] $\leftarrow $ K2(current_ord); |
9. if $sc$ $ > $ best_score |
10. best_score $\leftarrow $ $sc$; |
11. best_node_$sc$ $\leftarrow $ node_$sc$; |
12. the final order $\leftarrow $ current_ord; |
13. $t\leftarrow 0$; |
14. else |
15. current_ord $\leftarrow $ the final order; |
16. end if |
17. End for |
18. $t \leftarrow t + 1$; |
19.Until $t = = n-2$ |
20.Return the final order |
Table 2
"
Input: $X = \{X_1, X_2, \ldots, X_n \}$, Data, order |
Output: $PA$ |
1.For $j = 1$ to $n$ |
2. $PA_j \leftarrow \varphi; $ |
3. $V_{{{{\rm{old}}}}} \leftarrow {{{\rm{CH}}}}(\langle X_j, PA_j \rangle\vert D)$; \quad // CH is the family score of node $X_j $ |
4. While (true) |
5. $i\leftarrow \mathop {\rm argmax}\limits_{1{\leqslant} i{\leqslant} j, X_i \notin PA_j } {{{\rm{CH}}}}(\langle X_j, PA_j \cup \{X_i \}\rangle\vert D)$ // $X_i $ precedes $X_j $ in the given order |
6. $V_{{{{\rm{new}}}}} \leftarrow {{{\rm{CH}}}}(\langle X_j, PA_j \cup \{X_i \}\rangle \vert D)$ |
7. if $(V_{{{{\rm{new}}}}} > V_{{{{\rm{old}}}}} $ and $PA_j < j-1)$ |
8. $V_{{{{\rm{old}}}}} \leftarrow V_{{{{\rm{new}}}}}; $ |
9. $PA_j \leftarrow PA_j \cup \{X_i \}; $ |
10. else |
11. break; |
12. end if |
13. End while |
14.End for |
15.Return $PA$ |
"
Input: Data, node sizes, a random order, priors, confidence factor $p$, prior function SU$'$ |
Output: the final order |
$pc_accept$——the probability to accept the transformation when consistent with the prior |
$pc_reject$——the probability to reject the transformation when inconsistent with the prior |
current_ord——the current order who holds the best score |
best_score——the score of the current order |
$t$——the subscript variable $sc$——the score of the order obtained by transformation |
1. $pc_accept$ $\leftarrow $ interpolate SU$'$ at value $p$; $pc_reject$ $\leftarrow $ interpolate SU$'$ at value $(1-p)$; the final order $\leftarrow $ a random order; current_ord $\leftarrow $ the final order; best_score $\leftarrow $ K2(current_ord); $t \leftarrow 0$; |
2.Repeat |
3. For $i = t + 2: n$ |
4. $trans_prob$ $\leftarrow $ a random value between 0 and 1; |
5. node1 $\leftarrow $ the $i$th node in the current_ord; |
6. node2 $\leftarrow $ the $(t+1)$th node in the current_ord; |
7. if there's a directed path from node1 to node2 |
8. if $trans_prob > pc_accept$ //accept the transformation in a large probability |
9. Update the {current_ord} by swapping node1 and node2; |
10. $sc$ $\leftarrow $ K2(current_ord); |
11. if $sc$ $ > $ {best_score} |
12. best_score $\leftarrow $ $sc$; |
13. the final order $\leftarrow $ {current_ord}; |
14. $t\leftarrow 0$; |
15. else |
16. end if |
17. $t\leftarrow 0$; |
18. end if |
19. else//if there's no direct path from node1 to node2 |
20. if $trans_prob > pc_reject$ //accept the transformation in a small probability |
21. Update the {current_ord} by swapping node1 and node2; |
22. $sc$ $\leftarrow $ K2({current_ord}); |
23. if $sc$ $ > $ best_score |
24. best_score $\leftarrow $ $sc$; |
25. the final order $\leftarrow $ {current_ord}; |
26. $t\leftarrow 0$; |
27. else |
28. current_ord $\leftarrow $ the final order; |
29. end if |
30. end if |
31. end if |
32. End for |
33. $t \leftarrow t + 1$; |
34. Until $t = = n - 2$ |
35.Return the final order |
Table 5
"
BN | Node | Edge | Sample size | Original network | HC_bo+C | HC | HC_bo |
Cancer | 5 | 4 | 500 | –1 106.0 | –1 113.0 | –1 113.0 | –1 113.0 |
2 000 | –4 212.0 | –4 218.0 | –4 219.0 | –4 218.0 | |||
5 000 | –10 485.0 | –10 488.0 | –10 499.0 | –10 488.0 | |||
10 000 | –21 138.0 | –21 143.0 | –21 149.0 | –21 143.0 | |||
Asia | 5 | 4 | 500 | –1 157.0 | –1 157.0 | –1 161.0 | –1 160.0 |
2 000 | –4 527.0 | –4 528.0 | –4 549.0 | –4 531.0 | |||
5 000 | –11 148.0 | –11 155.0 | –11 212.0 | –11 160.0 | |||
10 000 | –22 394.0 | –22 394.0 | –22 436.0 | –22 394.0 | |||
Sachs | 11 | 17 | 500 | –4 043.0 | –4 050.0 | –4 182.0 | –4 162.0 |
2 000 | –14 853.0 | –14 853.0 | –15 088.0 | –14 985.0 | |||
5 000 | –36 353.0 | –36 353.0 | –36 583.0 | –36 353.0 | |||
10 000 | –72 153.0 | –72 153.0 | –72 837.0 | –72 153.0 | |||
Child | 20 | 25 | 500 | –6 830.0 | –6 942.0 | –7 350.0 | –7 030.0 |
2 000 | –25 270.0 | –25 263.0 | –27 260.0 | –25 480.0 | |||
5 000 | –62 420.0 | –62 655.0 | –67 440.0 | –63 400.0 | |||
10 000 | –123 860.0 | –124 281.0 | –129 760.0 | –124 640.0 | |||
Insurance | 27 | 52 | 500 | –8 490.0 | –8 605.0 | –9 820.0 | –8 820.0 |
2 000 | –30 380.0 | –30 931.0 | –32 330.0 | –31 290.0 | |||
5 000 | –73 310.0 | –73 522.0 | –79 510.0 | –74 100.0 | |||
10 000 | –143 680.0 | –144 077.0 | –158 250.0 | –144 270.0 | |||
Alarm | 37 | 46 | 500 | –6 600.0 | –6 650.0 | –7 100.0 | –7 000.0 |
2 000 | –23 350.0 | –23 465.0 | –24 400.0 | –23 610.0 | |||
5 000 | –56 850.0 | –57 032.0 | –60 530.0 | –57 470.0 | |||
10 000 | –111 450.0 | –111 573.0 | –114 470.0 | –112 050.0 | |||
Hailfinder | 56 | 66 | 500 | –28 150.0 | –28 571.0 | –31 880.0 | –28 730.0 |
2 000 | –103 490.0 | –103 698.0 | –108 450.0 | –106 940.0 | |||
5 000 | –225 310.0 | –252 862.0 | –256 970.0 | –252 950.0 | |||
10 000 | –499 850.0 | –500 515.0 | –507 920.0 | –502 150.0 |
Table 6
"
Data set | Node | Sample size | HC_bo+ | HC_bo | HC |
monks_T | 7 | 432 | –2 876.5 | –2 876.5 | –2 876.5 |
car | 7 | 1 728 | –13 646 | –13 684 | –13 696 |
diabetes | 9 | 768 | –12 248 | –12 271 | –12 276 |
abalone | 9 | 4 177 | –65 845 | –66 088 | –66 563 |
nursery | 9 | 12 960 | –125 990 | –126 130 | –126 190 |
contrasep | 10 | 1 473 | –15 347 | –15 387 | –15 388 |
wine | 14 | 178 | –2 880.4 | –2 932.3 | –2 897.7 |
heart | 14 | 270 | –4 294.1 | –4 311.5 | –4 299.0 |
australian | 15 | 690 | –11 098 | –11 114 | –11 116 |
zoo | 17 | 101 | –645.05 | –681.54 | –674.16 |
pen_A | 17 | 7 494 | –192 350 | –200 580 | –210 730 |
letter_A | 17 | 15 000 | –452 530 | –452 610 | –471 520 |
segment | 20 | 2 310 | –58 599 | –60 537 | –69 570 |
german | 25 | 1 000 | –21 154 | –21 269 | –21 189 |
Table 7
"
Model | Node | Sample size | HC_bo+C | HC_bo | OMRMRG |
kdd_test | 64 | 34 955 | –75 053.0 | –75 271.0 | –75 439.0 |
plants_ts | 69 | 17 412 | –231 600.0 | –232 270.0 | –232 790.0 |
Hepar2 | 70 | 10 000 | –329 860.0 | –330 820.0 | –331 560.0 |
Win95pts | 76 | 10 000 | –104 200.0 | –104 500.0 | –104 730.0 |
baudio_ts | 100 | 15 000 | –618 110.0 | –619 910.0 | –621 300.0 |
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