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Is this an incorrect way of back-propagating error with matrices?



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
2019 Moderator Election Q&A - Questionnaire
2019 Community Moderator Election ResultsBack-propagation through max pooling layersDeriving backpropagation equations “natively” in tensor formGradients for bias terms in backpropagationCustom c++ LSTM slows down at 0.36 cost is usual?A good reference for the back propagation algorithm?GradientChecking, can I blame float precision?Synthetic Gradients - what's the practical benefit?Questions about Neural Network training (back propagation) in the book PRML (Pattern Recognition and Machine Learning)Purpose of backpropagation in neural networksHow can I perform backpropagation directly in matrix form?










0












$begingroup$


I was watching a public available video from Stanford (https://youtu.be/d14TUNcbn1k?t=2720) on the mathematics behind back propagation. They proposed a graph:



Graph 1



that was then used as an example of back propagation using matrices. (The red text is the back propagated gradient, the green is the forward pass vales). The final gradients found for the initial matrices (of $[[0.1,0.5],[-0.3,0.8]]$ and $[[0.2],[0.4]]$) is below:
enter image description here



I agree with the values for $x$, however, I don't quite understand how they achieved the values of $W$. The equation (see circled in red) is the equation they used for calculating the gradients for $W$:enter image description here



With this equation (where $q$ = [[0.22],[0.26]], I would think that for $W_1,2$ (which has a value of 0.5, and $i$ = 1 and $j$ = 2) would be equal to $2*q_1*x_2$, which in this case is $2*0.22*0.4 = 0.176$, which is not what they got.



Intuitively, I thought the values of $W$ would be exactly what they calculated, but with $0.104$ and $0.176$ swapped. The way I calculated it was taking the top value of $q$ as the gradient on $0.22$ (which it is), and therefor, as $0.22 = 0.1*0.2 + 0.5*0.4$, taking $fracpartial qpartial W_1,1=0.2$, and then multiplying this by $q_1$, I got $0.2*0.44=0.088$, which aligns with their calculations.



However, applying the same logic to $W_1,2$, I get $fracpartial qpartial W_1,2=0.4$, and multiplying this by $q_1$, $0.4*0.44=0.176$. This conflicts with their value of $0.104$. If this logic is continued, the matrix of gradients matches theirs exactly, aside from the $0.104$ and $0.176$ being swapped.



For clarity, I also drew out a graph and deconstructed the matrix multiplication into simplified multiplication of two different regular equations. See below for the graph (where the numbers above nodes/inputs represent their gradient):





(Sorry for bad hand writing and quality, there's a reason I submit all work using latex)



I suspect I am using incorrect notation for $W$, as in, $W_1,2$ doesn't actually represent $0.5$, but actually $-0.3$, but then that doesn't align with their provided example of how the matrix $[[0.22],[0.26]]$ was constructed (the equation describing $q= W cdot x = ...$), and also doesn't explain how I got the values for the graph I hand drew, as that doesn't rely on notation.



If you read all this and have any idea what I'm doing wrong I would very much appreciate your effort. Thankyou!










share|improve this question







New contributor




Recessive is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
















    0












    $begingroup$


    I was watching a public available video from Stanford (https://youtu.be/d14TUNcbn1k?t=2720) on the mathematics behind back propagation. They proposed a graph:



    Graph 1



    that was then used as an example of back propagation using matrices. (The red text is the back propagated gradient, the green is the forward pass vales). The final gradients found for the initial matrices (of $[[0.1,0.5],[-0.3,0.8]]$ and $[[0.2],[0.4]]$) is below:
    enter image description here



    I agree with the values for $x$, however, I don't quite understand how they achieved the values of $W$. The equation (see circled in red) is the equation they used for calculating the gradients for $W$:enter image description here



    With this equation (where $q$ = [[0.22],[0.26]], I would think that for $W_1,2$ (which has a value of 0.5, and $i$ = 1 and $j$ = 2) would be equal to $2*q_1*x_2$, which in this case is $2*0.22*0.4 = 0.176$, which is not what they got.



    Intuitively, I thought the values of $W$ would be exactly what they calculated, but with $0.104$ and $0.176$ swapped. The way I calculated it was taking the top value of $q$ as the gradient on $0.22$ (which it is), and therefor, as $0.22 = 0.1*0.2 + 0.5*0.4$, taking $fracpartial qpartial W_1,1=0.2$, and then multiplying this by $q_1$, I got $0.2*0.44=0.088$, which aligns with their calculations.



    However, applying the same logic to $W_1,2$, I get $fracpartial qpartial W_1,2=0.4$, and multiplying this by $q_1$, $0.4*0.44=0.176$. This conflicts with their value of $0.104$. If this logic is continued, the matrix of gradients matches theirs exactly, aside from the $0.104$ and $0.176$ being swapped.



    For clarity, I also drew out a graph and deconstructed the matrix multiplication into simplified multiplication of two different regular equations. See below for the graph (where the numbers above nodes/inputs represent their gradient):





    (Sorry for bad hand writing and quality, there's a reason I submit all work using latex)



    I suspect I am using incorrect notation for $W$, as in, $W_1,2$ doesn't actually represent $0.5$, but actually $-0.3$, but then that doesn't align with their provided example of how the matrix $[[0.22],[0.26]]$ was constructed (the equation describing $q= W cdot x = ...$), and also doesn't explain how I got the values for the graph I hand drew, as that doesn't rely on notation.



    If you read all this and have any idea what I'm doing wrong I would very much appreciate your effort. Thankyou!










    share|improve this question







    New contributor




    Recessive is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$














      0












      0








      0





      $begingroup$


      I was watching a public available video from Stanford (https://youtu.be/d14TUNcbn1k?t=2720) on the mathematics behind back propagation. They proposed a graph:



      Graph 1



      that was then used as an example of back propagation using matrices. (The red text is the back propagated gradient, the green is the forward pass vales). The final gradients found for the initial matrices (of $[[0.1,0.5],[-0.3,0.8]]$ and $[[0.2],[0.4]]$) is below:
      enter image description here



      I agree with the values for $x$, however, I don't quite understand how they achieved the values of $W$. The equation (see circled in red) is the equation they used for calculating the gradients for $W$:enter image description here



      With this equation (where $q$ = [[0.22],[0.26]], I would think that for $W_1,2$ (which has a value of 0.5, and $i$ = 1 and $j$ = 2) would be equal to $2*q_1*x_2$, which in this case is $2*0.22*0.4 = 0.176$, which is not what they got.



      Intuitively, I thought the values of $W$ would be exactly what they calculated, but with $0.104$ and $0.176$ swapped. The way I calculated it was taking the top value of $q$ as the gradient on $0.22$ (which it is), and therefor, as $0.22 = 0.1*0.2 + 0.5*0.4$, taking $fracpartial qpartial W_1,1=0.2$, and then multiplying this by $q_1$, I got $0.2*0.44=0.088$, which aligns with their calculations.



      However, applying the same logic to $W_1,2$, I get $fracpartial qpartial W_1,2=0.4$, and multiplying this by $q_1$, $0.4*0.44=0.176$. This conflicts with their value of $0.104$. If this logic is continued, the matrix of gradients matches theirs exactly, aside from the $0.104$ and $0.176$ being swapped.



      For clarity, I also drew out a graph and deconstructed the matrix multiplication into simplified multiplication of two different regular equations. See below for the graph (where the numbers above nodes/inputs represent their gradient):





      (Sorry for bad hand writing and quality, there's a reason I submit all work using latex)



      I suspect I am using incorrect notation for $W$, as in, $W_1,2$ doesn't actually represent $0.5$, but actually $-0.3$, but then that doesn't align with their provided example of how the matrix $[[0.22],[0.26]]$ was constructed (the equation describing $q= W cdot x = ...$), and also doesn't explain how I got the values for the graph I hand drew, as that doesn't rely on notation.



      If you read all this and have any idea what I'm doing wrong I would very much appreciate your effort. Thankyou!










      share|improve this question







      New contributor




      Recessive is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      I was watching a public available video from Stanford (https://youtu.be/d14TUNcbn1k?t=2720) on the mathematics behind back propagation. They proposed a graph:



      Graph 1



      that was then used as an example of back propagation using matrices. (The red text is the back propagated gradient, the green is the forward pass vales). The final gradients found for the initial matrices (of $[[0.1,0.5],[-0.3,0.8]]$ and $[[0.2],[0.4]]$) is below:
      enter image description here



      I agree with the values for $x$, however, I don't quite understand how they achieved the values of $W$. The equation (see circled in red) is the equation they used for calculating the gradients for $W$:enter image description here



      With this equation (where $q$ = [[0.22],[0.26]], I would think that for $W_1,2$ (which has a value of 0.5, and $i$ = 1 and $j$ = 2) would be equal to $2*q_1*x_2$, which in this case is $2*0.22*0.4 = 0.176$, which is not what they got.



      Intuitively, I thought the values of $W$ would be exactly what they calculated, but with $0.104$ and $0.176$ swapped. The way I calculated it was taking the top value of $q$ as the gradient on $0.22$ (which it is), and therefor, as $0.22 = 0.1*0.2 + 0.5*0.4$, taking $fracpartial qpartial W_1,1=0.2$, and then multiplying this by $q_1$, I got $0.2*0.44=0.088$, which aligns with their calculations.



      However, applying the same logic to $W_1,2$, I get $fracpartial qpartial W_1,2=0.4$, and multiplying this by $q_1$, $0.4*0.44=0.176$. This conflicts with their value of $0.104$. If this logic is continued, the matrix of gradients matches theirs exactly, aside from the $0.104$ and $0.176$ being swapped.



      For clarity, I also drew out a graph and deconstructed the matrix multiplication into simplified multiplication of two different regular equations. See below for the graph (where the numbers above nodes/inputs represent their gradient):





      (Sorry for bad hand writing and quality, there's a reason I submit all work using latex)



      I suspect I am using incorrect notation for $W$, as in, $W_1,2$ doesn't actually represent $0.5$, but actually $-0.3$, but then that doesn't align with their provided example of how the matrix $[[0.22],[0.26]]$ was constructed (the equation describing $q= W cdot x = ...$), and also doesn't explain how I got the values for the graph I hand drew, as that doesn't rely on notation.



      If you read all this and have any idea what I'm doing wrong I would very much appreciate your effort. Thankyou!







      backpropagation matrix






      share|improve this question







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      Recessive is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











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      Recessive is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









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      Check out our Code of Conduct.









      asked 28 mins ago









      RecessiveRecessive

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      Recessive is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






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