he code provided to me when I asked this question is my own code from a preious question I asked on how to make it faster. My code is too slow and times out. I am looking for guidance on how to make it faster or a suggestion for a different approach. The question and my code are below. ************** Question Colour trio def colour_trio(colours): This problem was inspired by the fun little Mathologer video “Secret of Row 10” whose fractal animation once again remind us about how moving from two to three often opens the barn door for the chaos horse to emerge and wildly gallop away. To start, look at three values imaginatively named “red”, “yellow” and “blue”. These names serve as colourful (heh) mnemonics that could just as well have been “foo”, “bar” and “baz”, so no connection to actual physical colours is intended or implied. Next, de0ine a simple mixing rule between these colours with a rule that says that whenever any colour is mixed with itself, the result is that same colour, whereas mixing two different colours always gives the third. For example, mixing blue to blue gives that same blue, whereas mixing blue to yellow gives red. Given the 0irst row of colours as a string of lowercase letters to denote these colours, this function should construct the rows below the 0irst row one at the time according to the following discipline. Each row always contains one fewer element than the previous row above it. The i:th element of each row is calculated by mixing the colours of the previous row in positions i and i + 1. The singleton element of the bottom row is returned as the final answer. For example, starting from the 0irst row 'rybyr' leads to 'brrb', which leads to 'yry', which leads to 'bb', which leads to 'b' as the 0inal answer, Regis. When the Python virtual machine goes 'brrrrr', that in turn leads to 'yrrrr', 'brrr' , 'yrr', 'br' for the 0inal answer 'y' for “Yes, please!” colours Expected result 'y' 'y' 'bb' 'b' 'rybyry' 'r' 'brybbr' 'r' 'rbyryrrbyrbb' 'y' 'yrbbbbryyrybb' 'b' (Today's five-dollar power word to astonish your friends and co-workers is "quasigroup".) check_circle Expert Answer thumb_up thumb_down Step 1 According to the Information given:- We have to create and define Color trio here we mix the colors and find the singleton element in a row and rest follows the given instruction. Pyhton Code:- def colour_trio(colours): ##Recursive. #Base case if len(colours) == 1: return colours else: new_string = "" for letter in range(0,len(colours)-1): if colours[letter] == 'b': if colours [letter+1] == 'b': new_string+="b" elif colours [letter+1] == 'r': new_string+="y" elif colours [letter+1] == 'y': new_string+="r" elif colours[letter] == 'y': if colours [letter+1] == 'y': new_string+="y" elif colours [letter+1] == 'r': new_string+="b" elif colours [letter+1] == 'b': new_string+="r" elif colours[letter] == 'r': if colours [letter+1] == 'r': new_string+="r" elif colours [letter+1] == 'y': new_string+="b" elif colours [letter+1] == 'b': new_string+="y" return colour_trio(new_string) if __name__ == '__main__': colours="yrbbbbryyrybb" print(colour_trio(colours))
Colour trio
def colour_trio(colours):This problem was inspired by the fun little Mathologer video “Secret of Row 10” whose fractal animation once again remind us about how moving from two to three often opens the barn door for the chaos horse to emerge and wildly gallop away. To start, look at three values imaginatively named “red”, “yellow” and “blue”. These names serve as colourful (heh) mnemonics that could just as well have been “foo”, “bar” and “baz”, so no connection to actual physical colours is intended or implied. Next, de0ine a simple mixing rule between these colours with a rule that says that whenever any colour is mixed with itself, the result is that same colour, whereas mixing two different colours always gives the third. For example, mixing blue to blue gives that same blue, whereas mixing blue to yellow gives red.
Given the 0irst row of colours as a string of lowercase letters to denote these colours, this function should construct the rows below the 0irst row one at the time according to the following discipline. Each row always contains one fewer element than the previous row above it. The i:th element of each row is calculated by mixing the colours of the previous row in positions i and i + 1. The singleton element of the bottom row is returned as the final answer. For example, starting from the 0irst row 'rybyr' leads to 'brrb', which leads to 'yry', which leads to 'bb', which leads to 'b' as the 0inal answer, Regis. When the Python virtual machine goes 'brrrrr', that in turn leads to 'yrrrr', 'brrr' , 'yrr', 'br' for the 0inal answer 'y' for “Yes, please!”
|
colours |
Expected result |
|
'y' |
'y' |
|
'bb' |
'b' |
|
'rybyry' |
'r' |
|
'brybbr' |
'r' |
|
'rbyryrrbyrbb'
|
'y' |
|
'yrbbbbryyrybb'
|
'b' |
(Today's five-dollar power word to astonish your friends and co-workers is "quasigroup".)
Expert Answer
According to the Information given:-
We have to create and define Color trio here we mix the colors and find the singleton element in a row and rest follows the given instruction.
def colour_trio(colours):
##Recursive.
#Base case
if len(colours) == 1:
return colours
else:
new_string = ""
for letter in range(0,len(colours)-1):
if colours[letter] == 'b':
if colours [letter+1] == 'b':
new_string+="b"
elif colours [letter+1] == 'r':
new_string+="y"
elif colours [letter+1] == 'y':
new_string+="r"
elif colours[letter] == 'y':
if colours [letter+1] == 'y':
new_string+="y"
elif colours [letter+1] == 'r':
new_string+="b"
elif colours [letter+1] == 'b':
new_string+="r"
elif colours[letter] == 'r':
if colours [letter+1] == 'r':
new_string+="r"
elif colours [letter+1] == 'y':
new_string+="b"
elif colours [letter+1] == 'b':
new_string+="y"
return colour_trio(new_string)
if __name__ == '__main__':
colours="yrbbbbryyrybb"
print(colour_trio(colours))
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