Multidimensional gradient descent in Tensorflow
What does Tensorflow really do when the Gradient descent optimizer is applied to a "loss" placeholder that is not a number (a tensor of size 1) but rather a vector (a 1-dimensional tensor of size 2, 3, 4, or more)?
Is it like doing the descent on the sum of the components?
python tensorflow gradient-descent
asked Nov 16 '18 at 8:10
AristodogAristodog
192
add a comment |
What does Tensorflow really do when the Gradient descent optimizer is applied to a "loss" placeholder that is not a number (a tensor of size 1) but rather a vector (a 1-dimensional tensor of size 2, 3, 4, or more)?
Is it like doing the descent on the sum of the components?
python tensorflow gradient-descent
asked Nov 16 '18 at 8:10
AristodogAristodog
192
add a comment |
What does Tensorflow really do when the Gradient descent optimizer is applied to a "loss" placeholder that is not a number (a tensor of size 1) but rather a vector (a 1-dimensional tensor of size 2, 3, 4, or more)?
Is it like doing the descent on the sum of the components?
python tensorflow gradient-descent
asked Nov 16 '18 at 8:10
AristodogAristodog
192
What does Tensorflow really do when the Gradient descent optimizer is applied to a "loss" placeholder that is not a number (a tensor of size 1) but rather a vector (a 1-dimensional tensor of size 2, 3, 4, or more)?
Is it like doing the descent on the sum of the components?
python tensorflow gradient-descent
python tensorflow gradient-descent
asked Nov 16 '18 at 8:10
AristodogAristodog
192
asked Nov 16 '18 at 8:10
AristodogAristodog
192
asked Nov 16 '18 at 8:10
AristodogAristodog
192
asked Nov 16 '18 at 8:10
AristodogAristodog
192
asked Nov 16 '18 at 8:10
AristodogAristodog
192
192
add a comment |
add a comment |
2 Answers
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oldest
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The answer to your second question is "no".
As for the second: just like in the one-dimensional case (e.g. y = f(x), x in R), where the direction the algorithm takes is defined by the derivative of the function with respect to its single variable, in the multidimensional case the 'overall' direction is defined by the derivative of the function with respect to each variable.
This means the size of the step you'll take in each direction will be determined by the value of the derivative of the variable corresponding to that direction.
Since there's no way to properly type math in StackOverflow, instead of messing around with it I'll suggest you take a look at this article.
answered Nov 16 '18 at 10:31
Lucas FariasLucas Farias
16610
Maybe my question was not clear. In y=f(x), I'm talking about the case where y in multidimensional.
– Aristodog
Nov 17 '18 at 11:25
add a comment |
Tensorflow first reduces your loss to a scalar and then optimizes that.
answered Nov 21 '18 at 16:23
Alexandre PassosAlexandre Passos
4,2211917
What does "reduce" a vector to a scalar mean?
– Aristodog
Nov 25 '18 at 14:16
Adding all its entries, as in tf.reduce_sum
– Alexandre Passos
Nov 26 '18 at 16:13
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The answer to your second question is "no".
As for the second: just like in the one-dimensional case (e.g. y = f(x), x in R), where the direction the algorithm takes is defined by the derivative of the function with respect to its single variable, in the multidimensional case the 'overall' direction is defined by the derivative of the function with respect to each variable.
This means the size of the step you'll take in each direction will be determined by the value of the derivative of the variable corresponding to that direction.
Since there's no way to properly type math in StackOverflow, instead of messing around with it I'll suggest you take a look at this article.
answered Nov 16 '18 at 10:31
Lucas FariasLucas Farias
16610
Maybe my question was not clear. In y=f(x), I'm talking about the case where y in multidimensional.
– Aristodog
Nov 17 '18 at 11:25
add a comment |
The answer to your second question is "no".
As for the second: just like in the one-dimensional case (e.g. y = f(x), x in R), where the direction the algorithm takes is defined by the derivative of the function with respect to its single variable, in the multidimensional case the 'overall' direction is defined by the derivative of the function with respect to each variable.
This means the size of the step you'll take in each direction will be determined by the value of the derivative of the variable corresponding to that direction.
Since there's no way to properly type math in StackOverflow, instead of messing around with it I'll suggest you take a look at this article.
answered Nov 16 '18 at 10:31
Lucas FariasLucas Farias
16610
Maybe my question was not clear. In y=f(x), I'm talking about the case where y in multidimensional.
– Aristodog
Nov 17 '18 at 11:25
add a comment |
The answer to your second question is "no".
As for the second: just like in the one-dimensional case (e.g. y = f(x), x in R), where the direction the algorithm takes is defined by the derivative of the function with respect to its single variable, in the multidimensional case the 'overall' direction is defined by the derivative of the function with respect to each variable.
This means the size of the step you'll take in each direction will be determined by the value of the derivative of the variable corresponding to that direction.
Since there's no way to properly type math in StackOverflow, instead of messing around with it I'll suggest you take a look at this article.
answered Nov 16 '18 at 10:31
Lucas FariasLucas Farias
16610
The answer to your second question is "no".
As for the second: just like in the one-dimensional case (e.g. y = f(x), x in R), where the direction the algorithm takes is defined by the derivative of the function with respect to its single variable, in the multidimensional case the 'overall' direction is defined by the derivative of the function with respect to each variable.
This means the size of the step you'll take in each direction will be determined by the value of the derivative of the variable corresponding to that direction.
Since there's no way to properly type math in StackOverflow, instead of messing around with it I'll suggest you take a look at this article.
answered Nov 16 '18 at 10:31
Lucas FariasLucas Farias
16610
answered Nov 16 '18 at 10:31
Lucas FariasLucas Farias
16610
answered Nov 16 '18 at 10:31
Lucas FariasLucas Farias
16610
answered Nov 16 '18 at 10:31
Lucas FariasLucas Farias
16610
16610
Maybe my question was not clear. In y=f(x), I'm talking about the case where y in multidimensional.
– Aristodog
Nov 17 '18 at 11:25
add a comment |
Maybe my question was not clear. In y=f(x), I'm talking about the case where y in multidimensional.
– Aristodog
Nov 17 '18 at 11:25
Maybe my question was not clear. In y=f(x), I'm talking about the case where y in multidimensional.
– Aristodog
Nov 17 '18 at 11:25
Maybe my question was not clear. In y=f(x), I'm talking about the case where y in multidimensional.
– Aristodog
Nov 17 '18 at 11:25
add a comment |
Tensorflow first reduces your loss to a scalar and then optimizes that.
answered Nov 21 '18 at 16:23
Alexandre PassosAlexandre Passos
4,2211917
What does "reduce" a vector to a scalar mean?
– Aristodog
Nov 25 '18 at 14:16
Adding all its entries, as in tf.reduce_sum
– Alexandre Passos
Nov 26 '18 at 16:13
add a comment |
Tensorflow first reduces your loss to a scalar and then optimizes that.
answered Nov 21 '18 at 16:23
Alexandre PassosAlexandre Passos
4,2211917
What does "reduce" a vector to a scalar mean?
– Aristodog
Nov 25 '18 at 14:16
Adding all its entries, as in tf.reduce_sum
– Alexandre Passos
Nov 26 '18 at 16:13
add a comment |
Tensorflow first reduces your loss to a scalar and then optimizes that.
answered Nov 21 '18 at 16:23
Alexandre PassosAlexandre Passos
4,2211917
Tensorflow first reduces your loss to a scalar and then optimizes that.
answered Nov 21 '18 at 16:23
Alexandre PassosAlexandre Passos
4,2211917
answered Nov 21 '18 at 16:23
Alexandre PassosAlexandre Passos
4,2211917
answered Nov 21 '18 at 16:23
Alexandre PassosAlexandre Passos
4,2211917
answered Nov 21 '18 at 16:23
Alexandre PassosAlexandre Passos
4,2211917
4,2211917
What does "reduce" a vector to a scalar mean?
– Aristodog
Nov 25 '18 at 14:16
Adding all its entries, as in tf.reduce_sum
– Alexandre Passos
Nov 26 '18 at 16:13
add a comment |
What does "reduce" a vector to a scalar mean?
– Aristodog
Nov 25 '18 at 14:16
Adding all its entries, as in tf.reduce_sum
– Alexandre Passos
Nov 26 '18 at 16:13
What does "reduce" a vector to a scalar mean?
– Aristodog
Nov 25 '18 at 14:16
What does "reduce" a vector to a scalar mean?
– Aristodog
Nov 25 '18 at 14:16
Adding all its entries, as in tf.reduce_sum
– Alexandre Passos
Nov 26 '18 at 16:13
Adding all its entries, as in tf.reduce_sum
– Alexandre Passos
Nov 26 '18 at 16:13
add a comment |
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