The Back Propagation Learning Algorithm
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The document describes the back propagation learning algorithm, which is an error correcting algorithm that solves the credit assignment problem for neural networks with hidden units. It introduces supervised learning and using a training set to adapt network weights to produce desired outputs. The algorithm is then derived, showing how to calculate the gradient of the error function with respect to weights using back propagation to determine how to update weights between hidden and output units as well as between input and hidden units.
The Back Propagation Learning Algorithm
- 1. The Back Propagation Learning Algorithm For networks with hidden units. Error Correcting algorithm. Solves the credit (blame) assignment problem. 1
- 2. What is supervised learning? Can we teach a network to learn to associate a pattern of inputs with corresponding outputs? i.e. given initial set of weights, how can they be adapted to produce the desired output? Use a training set: y a f? d payment b e? c w p workload person workload pay P(happy) a 0.1 0.9 0.95 b 0.3 0.7 0.8 c 0.07 0.2 0.2 d 0.9 0.9 0.3 e 0.7 0.5 ?? f 0.4 0.8 ?? After training, how does network generalise to patterns unseen during learning? 2
- 3. Learning by Error Correction In the perceptron there was a binary valued output Ý and a target Ø. x1 x2 xN w1 w2 wN output y target t y Æ 1 Ý step ÛÜ ¼ 0 Σwi xi i Define this error measure: ½ ´Ø Ý µ¾ ¾ It counts the number of incorrect outputs. We want to design a weight changing procedure that minimises . 3
- 4. Learning by Error Correction How do we change the weights Û¼ Û½ ÛÆ so that error decreases? E Suppose error slope slope varies with weight -ve +ve Û like this. wi If we could measure the slope Û then changing weights by the negative of the slope will minimise . slope +ve ¡Û -ve move towards minimum of slope -ve ¡Û +ve 4
- 5. More Perceptron Problems For the perceptron, can’t be differentiated with respect to weights Û¼ Û½ ÛÆ because involves output Ý which is not differentiable. ½ ´Ø Ý µ¾ Ý step Æ ÛÜ ¾ ¼ Threshold Unit: y ´ ÈÆ Û Ü 1 ½ if ¼ Ý ¼ if ÈÆ ¼ Û Ü ¼ ¼ 0 Σwi xi i Sigmoid Unit: y ½ 1 Ý ÈÆ ¡ ½ · ÜÔ ÛÜ 0 Σwi xi i 5
- 6. Gradient Descent E The error is now slope slope a differentiable -ve +ve function. wi Change weights using negative slope ¡Û Û Û +ve ¡Û -ve move towards minimum of Û -ve ¡Û +ve This approach is called Gradient Descent 6
- 7. Derivation of Back Propagation x1 v1 y1 x2 v2 y2 xk vj yi uj k wi j xN vN yN inputs hidden outputs xk vj yi È ¡ output Ý sig Û Ú È ¡ hidden Ú sig Ù Ü error ½È È Ø Ý ¡¾ ¾ We need to find the derivatives of with respect to weights Û and Ù . 7
- 8. Preliminaries xk ujk vj wij yi On a single pattern (drop ) ½ ¡¾ ¾ Ø Ý and ½ Ý ÈÆ ¡ ½ · ÜÔ Û Ú Note that: Ý ¡ Ú Ý ½ Ý Û Ý ¡ Û Ý ½ Ý Ú since if Ý ½ ½ · ÜÔ´ ܵ Ý then Ý ´½ Ý µ Ü 8
- 9. Between Hidden and Output Û xk ujk vj wij yi For weights between hidden units and output units. ½ ¡¾ ¾ Ø Ý Ý Û Ý Û ¡ Ý Ý Ø Ý Û Ý ´½ ÝµÚ ¡ Û Ý Ø ßÞ ´½ Ý µ Ú Ý call this Æ 9
- 10. Between Input and Hidden Ù xk ujk vj wij yi For weights between input units and hidden units. ½ ¡¾ ¾ Ø Ý Ý Ú Ù Ý Ú Ù ¡ Ý Ý Ø Ý Ú Ý ´½ ÝµÛ Ú Ù Ú ´½ Ú µ Ü ¡ Ù Ý Ø Ý ´½ Ý µ Û Ú ´½ Ú µ Ü Ù ÆÛ Ú ´½ Ú µ Ü 10
- 11. Between Hidden and Output ¡Û xk ujk vj wij yi Modifying weights between hidden units and output units using gradient descent. ¡Û Û ¡ Ý ßÞ Ø Ý ´½ ßÞ Ý µ Ú close to ¼ ½ small for Ý Learning constant “input” error ßÞ Æ 11
- 12. Between Input and Hidden ¡Ù xk ujk vj wij yi Modifying weights between input units and hidden units using gradient descent. ¡Ù Ù Æ Û Ú ´½ Ú µÜ back propagation of error The same procedure is applicable to a net with many hidden layers. 12
- 13. An Example x1 u x2 =0 2.0 21 .8 = u 11 =2.0 u 12 u 22 =0.8 ܽ ܾ target Ø u 10 = -1.0 u 20 = -1.0 0 0 0 v1 v2 1 1 0 1 1 1 0 1 w1 =2.0 w2 = -1.0 1 1 0 1 y w0 = -1.0 ¡ hidden Ú½ sig Ù½½Ü½ · Ù½¾Ü¾ · Ù½¼ 0.9526 ¡ Ú¾ sig Ù¾½Ü½ · Ù¾¾Ü¾ · Ù¾¼ 0.6457 ¡ output Ý sig Û½Ú½ · Û¾Ú¾ · Û¼ 0.5645 error ½ Ø Ý ¡¾ ¾ 0.1593 13
- 14. An Example: updating the weights Learning constant ½¼ output Æ ´Ý ص Ý´½ ݵ 0.1388 ¡Û¼ ƽ ¼ -0.1388 ¡Û½ ÆÚ½ -0.1322 ¡Û¾ ÆÚ¾ -0.0896 hidden (to Ú½) hidden (to Ú¾) ¡Ù½¼ ÆÛ½ Ú½´½ Ú½µ½ ¼ ¡Ù¾¼ ÆÛ¾ Ú¾´½ Ú¾µ½ ¼ -0.0125 0.0318 ¡Ù½½ ÆÛ½ Ú½´½ Ú½µÜ½ ¡Ù¾½ ÆÛ¾ Ú¾´½ Ú¾µÜ½ -0.0125 0.0318 ¡Ù½¾ ÆÛ½ Ú½´½ Ú½µÜ¾ ¡Ù¾¾ ÆÛ¾ Ú¾´½ Ú¾µÜ¾ -0.0125 0.0318 14
- 15. An Example: a New Error x1 u x2 8 =0 1.9 21 .83 = u 11 =1.98 u 12 u 22 =0.83 ܽ ܾ target Ø u 10 = -1.01 u 20 = -0.96 0 0 0 v1 v2 1 1 0 1 1 1 0 1 w1 =1.86 w2 = -1.08 1 1 0 1 y w0 = -1.13 ¡ hidden Ú½ sig Ù½½Ü½ · Ù½¾Ü¾ · Ù½¼ 0.9509 ¡ Ú¾ sig Ù¾½Ü½ · Ù¾¾Ü¾ · Ù¾¼ 0.6672 ¡ output Ý sig Û½Ú½ · Û¾Ú¾ · Û¼ 0.4776 error ½ Ø Ý ¡¾ ¾ 0.1140 The error has reduced for this pattern. 15
- 16. Summary Credit-assignment problem solved for hidden units: Input Output ƽ Û½ Û¾ Æ Æ¾ Û¿ Æ ¼ ´ µÈ Û Æ Æ¿ Errors total input to unit ; 1st derivative of acti- ¼ vation function (sigmoid) Outstanding issues: 1. Number of layers; number and type of units in layer 2. Learning rates 3. Local or distributed representations 16