Use python jupyter lab, NO AI PLZZZZZZ:

Question $($check the pic$)$:

Declare intermediate variables, $q_1=x^3,q_2=y^4,q=q1+q2$ and so on$.$ Remember to use dfdx is needed when $x$ is used in

multiple paths of the computation$),$ similarly dfdy$.$

Code:

import sys

from packaging import version

import sklearn

import matplotlib.pyplot as plt

import numpy as np

import tensorflow as tf

from sklearn.preprocessing import add$\_$dummy$\_$feature

from sklearn.datasets import make$\_$blobs

from sklearn import metrics #Import scikit$-$learn metrics module for accuracy calculation

from sklearn.model$\_$selection import train$\_$test$\_$split

from sklearn.metrics import confusion$\_$matrix, ConfusionMatrixDisplay

from sklearn.datasets import make$\_$classification

print$("$Sklearn package",sys$.$version$\_$info$)$

print$("$Sklearn package",sklearn.$\_\_$version$\_\_)$

print$("$TensorFlow version:", tf$.\_\_$version$\_\_)$

assert sys$.$version$\_$info $>=(3,7)$

assert version.parse$($sklearn$.\_\_$version$\_\_)>=$ version.parse$("1\mathrm{.}0\mathrm{.}1")$

plt$.$rc$(\text{'}$font$\text{'},$ size$=12)$

plt$.$rc$(\text{'}$axes$\text{'},$ labelsize$=14,$ titlesize$=14)$

plt$.$rc$(\text{'}$legend$\text{'},$ fontsize$=14)$

plt$.$rc$(\text{'}$xtick$\text{'},$ labelsize$=10)$

plt$.$rc$(\text{'}$ytick$\text{'},$ labelsize$=10)$

##

x $=2$

y $=5$

#Forward pass

#Backward pass

#print$($f$)$

#print$($dfdx$,$dfdy$)$

#f $=12\mathrm{.}918367346938776$

#dfdx $=-3\mathrm{.}4460641399416914$

#dfdy $=6\mathrm{.}513119533527696$

##

#using computation graph from tensorflow

x $=$ tf$.$Variable$(2.)$

y $=$ tf$.$Variable$(5.)$

#print$($f$)$

#print$($grad$\_$x$,$grad$\_$y$)$

$-----$

Question $($check pic$)$:

Q$4(20$ points$)$

$6\mathrm{.}0\mathrm{.}1.$ Calculate the gradient of $f$ with respect to $x$ and $y$ using the backpropagation

algorithm

$f(x,y)={\displaystyle \underset{i=1}{\overset{i=m}{}}}(x_i**y_i^2)=(x_1**y_1^2+x_2**y_2^2+$cdots$+x_m**y_m^2)$

$q=y^2$

$z=x**q$

$f={\displaystyle \underset{i=1}{\overset{i=m}{}}}(z_i)$

Code:

# Using backpropagation, construct some intermediate

# variables for $q$=$y$^2$$ and z $=$ x$*$q$,$ then finally f $=$ np$.$sum$($z$)$

x $=$ np$.$array$([1,2,3,4])$

y $=$ np$.$array$([1\mathrm{.}5,2\mathrm{.}2,2\mathrm{.}4,4\mathrm{.}1])$

m $=$ x$.$shape$[0]$

print$($x$)$

print$($y$)$

#Forward Pass

#Backward Pass

#print$("$f$",$f$)$

#print$("$dfdx$",$dfdx$)$

#print$("$dfdy$",$dfdy$)$

#Sample output

#f $96\mathrm{.}44999999999999$

#dfdq $[1.1.1.1.]$

#dfdx $[2\mathrm{.}254\mathrm{.}845\mathrm{.}7616\mathrm{.}81]$

#dfdy $[3.8\mathrm{.}814\mathrm{.}432\mathrm{.}8]$

$-----$

Question $($check pic$)$:

$6\mathrm{.}0\mathrm{.}2.$ Calculate the gradient of $f$ with respect to $x$ and $y$ using the computation graph in

tensorflow

$\frac{df}{dx}=$

$\frac{df}{dy}=$

Code:

#using tensorflow

x $=$ tf$.$Variable$([1.,2.,3.,4.],$ trainable $=$ True, name$=\text{'}$x$\text{'})$

y $=$ tf$.$Variable$([1\mathrm{.}5,2\mathrm{.}2,2\mathrm{.}4,4\mathrm{.}1],$ trainable $=$ True, name$=\text{'}$y$\text{'})$

#tf$.$print$("$f$",$f$)$

#grad$\_$x$,$ grad$\_$y $=$ tp$.$gradient$($f$,[$x$,$y$])$

#tf$.$print$("$grad$\_$x$",$grad$\_$x$)$

#tf$.$print$("$grad$\_$y$",$grad$\_$y$)$

#Sample output

#f $96\mathrm{.}45$

#grad$\_$x $[2\mathrm{.}254\mathrm{.}845\mathrm{.}7616\mathrm{.}81]$

#grad$\_$y $[38\mathrm{.}814\mathrm{.}400000632\mathrm{.}8]$