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Feeling confident with your A Level Computer Science terminology?
Have a go at guessing the hidden keywords or expressions represented by the following dingbats…
Here is a list of all the keywords used in the above dingbats, in alphabetical order…
In this challenge, we will use a step by step approach to create a 2-player dice game with the following rules:
Step 1: Displaying a welcome banner
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Step 2: Retrieving the name of the first player
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Step 3: Throwing the first dice…
Here is the Python code to generate a random number from 1 to 10: import random number=random.randint(1,10) |
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Step 4: Throwing the second dice…
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Step 5: Carry on throwing the dice until a double six is rolled.
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Step 6: Counting the number of throws.
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Step 7: Player 2’s turn…
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Step 8: Deciding who wins the game!
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Feeling confident with your GCSE Computer Science terminology?
Have a go at guessing the hidden keywords or expressions represented by the following dingbats…
Here is a list of all the keywords used in the above dingbats, in alphabetical order…
Are you confident with your knowledge of key procedural programming concepts, including sequencing, selection, iteration and the use of subroutines?
You can find out more about the key terminology relevant to procedural programming using our Periodic Table of Procedural Programming Concepts.
Procedural Programming TerminologyOpen Crossword Puzzle in a New Window
The Morse Code was designed to quickly transfer messages using a series of “dots (.)” and “dashes (-)”. Morse code was named after Samuel Morse, one of the inventors of the telegraph.
The International Morse Code includes the 26 letters of the alphabet (there is no distinction between lowercase and uppercase characters), the 10 Arabic numerals (digits from 0 to 9) and a few punctuation and arithmetic signs such as +, -, = etc.
Each character consists of a series of 1 to 5 dots or dashes (also called “dits” and “dahs”). The code was designed taking into consideration the frequency of each character in the English language so that most frequent characters such as E and T consist of just 1 dot or dash (E = “.”, T = “–”) whereas less frequent characters may include 4 to 5 dots or dashes (e.g. Q = “– – . –” and J = “. – – –”)
Click on the above picture to open it in a new window.To find out the morse code for each character of the International Morse Code, we can use the following binary tree. Starting at the root of the tree, the succession of branches connecting the root node to the required character gives us the morse code for this character considering that:
For instance, the morse code for letter P is . – – . as explained on this diagram:
The following code is using a binary tree data structure to store the more code binary tree.
It then uses the tree to convert a message into morse code by locating each character in the message within the tree and working out the morse code equivalent for each character.
In the code below, our binary tree only includes the first four levels, needed to code all 26 letters of the alphabet.
Your task consists of completing the above program to take a user input in morse code, and use the binary tree to decode the message one character at a time.
A computer system consists of hardware and software. The hardware components are the physical components of the computer system whereas software refers to the programs that run on and control the computer hardware.
Some of the hardware components can be found inside the main unit/case of the computer, whereas some external peripherals can be externally plugged in. There are four main types of hardware components: input devices, output devices, processing components (including the CPU and GPU), and storage devices.
Are you confident with your knowledge of computer hardware? Can you identify the purpose and characteristics of the main hardware components of a computer system? Complete these crossword to test your knowledge!
Computer Hardware CrosswordOpen Crossword Puzzle in a New Window
Computer Hardware CrosswordOpen Crossword Puzzle in a New Window
The aim of this challenge was to write a recursive backtracking algorithm to solve a connect flow puzzle. The aim of the game is to connect all the pairs of dots of the same colours, positioned on a 2D grid, using continuous lines. The connection lines cannot cross over each other.
Here is a fully solved connect flow grid:
In order to try to reduce the number of steps (backtracks) needed by the algorithm to solve this puzzle we have applied the following heuristics:
Here is the full Python code for our backtracking algorithm. Our first option is not implementing the two heuristic strategies mentioned above, whereas our second option is based on these heuristic strategies.
This challenge is based on this Frog puzzle board game.
At the start of the game several green frogs and one red frog are positioned on the board/pond on different waterlilies based on a given configuration. (The game comes with a set of cards, each card describing one different configuration / puzzle to solve).
Here is an example of initial configuration:
The aim of this game is to remove all the green frogs from the board/pond so that only the red frog remains on the board. A frog is removed from the board when it falls off its waterlily. This happens when another frog hops above its head:
When a frog is on a waterlily, it can jump over adjacent waterlilies only when:
The aim of this programming challenge was to design and implement an algorithm to search for a solution to this puzzle using a trial-and-error approach. This can be done in Python using a backtracking algorithm which will identify the possible moves on the board, and try these moves until it either finds a solution (only one red frog remains on the board) or reaches a dead end (when several frogs are left on the board but can no longer jump around). If after moving frogs around a dead end, is met, the algorithm will backtrack to previous moves to try alternative possible moves.
The possible moves a frog can do depend on its position on the board as well as the position of other frogs. For instance, here are two different diagrams to show potential moves of a frog placed on the top-left waterlily (0) or on the central waterlily (6):
By investigating all the possible moves from each waterlily we can construct a weighted graph where:
This approach results in the following weighted graph:
In Python, a graph can be stored as a dictionary of lists. Each key of the dictionary represents a node, the value associated to each key is a list of edges. Each edge will contain a sub-list of two values: the node/waterlily being reached and the waterlily being jumped over (the weight of the edge).
Here is how the above graph is implemented in Python:
#The graph representing all the potential jumps a frog can do
pond = {0:[[2,1],[6,3],[10,5]],
1:[[5,3],[11,6],[7,4]],
2:[[0,1],[6,4],[12,7]],
3:[[9,6]],
4:[[8,6]],
5:[[1,3],[7,6],[11,8]],
6:[[0,3],[2,4],[12,9],[10,8]],
7:[[1,4],[5,6],[11,9]],
8:[[4,6]],
9:[[3,6]],
10:[[0,5],[6,8],[12,11]],
11:[[5,8],[1,6],[7,9]],
12:[[10,11],[6,9],[2,7]]
}
You can investigate here the full Python code used to implement our backtracking algorithm using a recursve function called jumpAround().
In computer science, sorting algorithms are used to sort the elements of a data sets in numerical or alphabetical order. In this blog post will investigate the Python code used to implement 4 of the most frequently used sorting algorithms:
We will implement the first two algorithm using nested loops. (Iterative algorithms) whereas the Merge sort and Quick sort algorithms will be based on using recursive functions to implement the divide-and-conquer approach these algorithms are based on.
def insertionSort(list):
#Iterate through each value of the list
for i in range(1,len(list)):
pointer = i
#Slide value on the left to insert it in postion
while list[pointer]<list[pointer-1] and pointer>0:
#Swap values with the previous value...
tmp=list[pointer]
list[pointer]=list[pointer-1]
list[pointer-1] = tmp
pointer = pointer - 1
#Print list at the end of each iteration
print(list)
def bubbleSort(list):
counter = len(list)-1
sorted=False
#The Bubble sort stops iterating if it did not perform any "swap" in the previous iteration.
while sorted==False:
sorted=True
for pointer in range(0,counter):
if list[pointer]>list[pointer+1]:
#Swap values with the next value...
tmp = list[pointer]
list[pointer] = list[pointer+1]
list[pointer+1] = tmp
sorted=False
#Print list at the end of each iteration
print(list)
counter-=1
#Merge Sort: A divide-and-conquer algorithm implemented using a recursive function!
def mergeSort(list):
midPointer = len(list) // 2
leftList = list[0:midPointer]
rightList = list[midPointer:len(list)]
if len(leftList)>2:
leftList = mergeSort(leftList)
else:
if len(leftList)==2:
if leftList[0]>leftList[1]:
tmp = leftList[0]
leftList[0] = leftList[1]
leftList[1] = tmp
if len(rightList)>2:
rightList = mergeSort(rightList)
else:
if len(rightList)==2:
if rightList[0]>rightList[1]:
tmp = rightList[0]
rightList[0] = rightList[1]
rightList[1] = tmp
#Merge left and right...'
leftPointer = 0
rightPointer = 0
sortedList=[]
while leftPointer<len(leftList) and rightPointer<len(rightList):
if leftList[leftPointer] < rightList[rightPointer]:
sortedList.append(leftList[leftPointer])
leftPointer += 1
else:
sortedList.append(rightList[rightPointer])
rightPointer += 1
while leftPointer<len(leftList):
sortedList.append(leftList[leftPointer])
leftPointer+=1
while rightPointer<len(rightList):
sortedList.append(rightList[rightPointer])
rightPointer+=1
return sortedList
#Quick Sort: Another divide-and-conquer algorithm implemented using a recursive function!
def quickSort(list, leftPointer, rightPointer):
# Use the mid value as the pivot
pivotPointer = leftPointer + ((rightPointer-leftPointer) // 2)
# Alternative approach: use the first value as the pivot
#pivotPointer = leftPointer
# Alternative approach: use the last value as the pivot
#pivotPointer = rightPointer
print("Pivot: " + str(list[pivotPointer]))
left_Pointer = leftPointer
right_Pointer = rightPointer
while leftPointer<rightPointer:
#Find the next value to swap (left side of the pivot)
while list[leftPointer]<=list[pivotPointer] and leftPointer<pivotPointer:
leftPointer = leftPointer+1
#Find the next value to swap (right side of the pivot)
while list[rightPointer]>=list[pivotPointer] and rightPointer>pivotPointer:
rightPointer = rightPointer-1
if leftPointer<rightPointer:
#Swap both values
tmp = list[leftPointer]
list[leftPointer] = list[rightPointer]
list[rightPointer] = tmp
#Has the pivot moved?
if leftPointer == pivotPointer:
pivotPointer = rightPointer
elif rightPointer == pivotPointer:
pivotPointer = leftPointer
print(list)
# Let's implement the double recursion to quick sort both sub lists on the left and the right of the pivot Pointer
if pivotPointer-1-left_Pointer>=1: #Stop the recursion if there is only on value left in the sublist to sort
quickSort(list,left_Pointer,pivotPointer-1)
if right_Pointer - (pivotPointer+1)>=1: #Stop the recursion if there is only on value left in the sublist to sort
quickSort(list,pivotPointer+1,right_Pointer)
See below the full code for all four sorting algorithms.
The quick sort algorithm is a divide-and-conquer algorithm that can be implemented using a recursive function. The following graphic explains the different steps of a quick sort algorithm.
Note, that the quick sort algorithm starts by identifying a pivot. Depending on the implementation, this pivot can be one of the following three values:
The example below uses the mid-value of the list (value in the middle position):
Here is our Python implementation of a quick sort algorithm using a recursive function. This implementation uses the mid-value as the pivot.
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