data 程序代做、代写 Python 编程

日期：2024-03-16 10:50

AIML 2023-2024 Coursework

March 12, 2024

x1 w1

w2 z1 v1

x2 w1

w2 z3 v3

w1 z4 v4 w2

     v2 2

w z2

x3 w1 y

     x4

x5

 Figure 1: Convolutional neural network for coursework assignment.

Problem The goal of this take-home assignment is to implement, in Python, a simple two-layer convolutional neural network (CNN) with five inputs x1, . . . , x5, four hidden nodes z1, . . . , z4 and one output y with ReLU activations, according to the diagram shown in Figure 1. The hidden layer and output of the CNN is to be computed along with the gradient of the hidden layer and output with respect to parameter w1. The values oftheparameterswillbew1 =1.2,w2 =?0.2,v1 =?0.3,v2 =0.6,v3 =1.3andv4 =?1.5.

Instructions The CNN implementation is to be computed using a single Python function in single Python file. The interface to the function should be in the precise format,

y, z = convnet(x) (1)

where x = [x1, x2, x3, x4, x5] is a list of five numerical inputs (for example, a set of real numbers x=[0.3,?1.5,0.7,2.1,0.1]), and it should return the value of y as a number of the type dual and, z=[z1,z2,z3,z4] as a list of four numbers of type dual defined in the course code module ad.py. Therefore, when testing, you should expect to import this module. The implementation should use the specific values of the weight parameters given above.

Submission TopreparethePythoncodefileforsubmission,itmustbenamedintheformatinitials_studentid.py, for instance if your initials are ’AJD’ and your ID is 5716631 then your file should be named ajd_5716631.py. Submit the file through the Assignments page on Canvas. The deadline for submissions is 12pm UK time, 21st March 2024.

Marking The function will be marked automatically by calling it inside Python, and checking the results against a model solution. A fully correct solution will receive 20 marks. A solution which has a partially correct

1

Inputs

Output

Hidden

interface, or produces only partially correct output values for y and/or z, will lose marks accordingly. A solution which does not have the correct interface may score 0 marks, therefore it is critical that the interface (1) is correctly implemented.

2