BFC5936编程代写、代做c/c++，Java程序、Python语言代做

日期：2020-11-01 11:23

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BFC5936 – Final assessment task

Pathfinding algorithm

Background:

Finding solutions to mazes is usually done as a leisurely activity but finding efficient

solutions to extremely intricate and complicated mazes has been a field that has seem many

uses in the real world. Google maps and Uber are among the many applications that we

interact with on a daily basis using such algorithms. Some famous pathfinding algorithms

include Dijkstra’s algorithm and A*. Whilst knowledge of these algorithms may help you to

understand pathfinding in more detail, you do not need to implement any of these

algorithms to solve this task.

Instructions:

There are three mazes for you to solve:

? easy

? medium

? hard

Each maze is a grid of cells that has been coloured either black, yellow, blue or red. Black

cells are the boundaries or walls of the maze, yellow cells are the path you can follow, the

blue cell is the starting point of the maze and the red cell/s is the finishing point.

Your algorithm will need start at the blue cell (top left hand corner) and move through each

yellow cell (test if the cell colour is yellow) until the red cell/s are found. Each cell you visit,

you must leave behind a sign that you have visited the cell by printing the number of steps it

took you to reach that cell from the starting cell (unless the path ends in a dead end, see

hard maze description). For example, if it is the fifth cell you have visited, you should print

“5” in that cell.

A solved maze is one where your path to the final red cell from the blue cell is laid out by a

series of numbers that denote your path without any jumps, missed yellow cells or any

deviations from the path (printing into black cells). An example of a solved maze is attached

as a screenshot.

A function has been set up and commented for you to complete that should lead to a

solution for the easy maze. Once your function can solve the easy maze, think about how

you can tweak it to accommodate the extra complexities of the medium and hard mazes. An

algorithm that solves the hard maze should be able to solve the medium and easy ones too.

Each maze is equally weighted meaning each is worth 1/3rd of the spreadsheet mark.

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Easy maze:

The yellow path will only go down or to the right. There are no dead ends and only one red

cell.

Medium maze:

The yellow path can go left, down, up, or right. There are no dead ends but multiple red

cells. You must print your path to all the red cells as if each path is considered from the

starting blue cell. This means that any cells that have multiple paths exiting it will have to

have numbers counting equally in each path. A screenshot of a crossroad is attached for

clarity.

Hard maze:

The yellow path can go left, down, up, or right. There are dead ends and multiple red cells.

You must print your path to all the red cells as if each path is considered from the starting

blue cell (just like the medium maze). Additionally, your finished maze should not contain

any numbers in a path that ends in a dead end (no red cell at the end). A screenshot is

attached for clarity. Lastly, this maze is not visible but can be referenced by it’s named

“Hard”.

Tips:

? START WITH A CODE PLAN

o What is the problem?

o How would you solve a maze like this by hand?

o Can you recreate your manual process in code?

? Testing the colour of a cell’s background can be done using the following snippet of

code:

o If range(“a1”).interior.color = RGB(255, 255, 0) then

o RGB stands for Red Green Blue and it is a standard way of reproducing

colours

o The colours relevant to this assessment are:

? Blue – RGB(0, 0, 255)

? Yellow – RGB(255, 255, 0)

? Red – RGB(255, 0, 0)

? Black – RGB(0, 0, 0)

? Each maze range can be referenced by its name

o The easy maze can be referenced by range(“Easy”)

o The medium maze can be referenced by range(“Medium”)

o The hard maze can be referenced by range(“Hard”)

If there are any questions,

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