$(10$ points$)$ Finding cycles Press shift enter to test your code. Ensure that your code has been saved first by pressing shift$+$enter on the previous cell.

from IPython.core.display import display, HTML

def part$\_$of$\_$a$\_$cycle$\_$test$()$:test$\_$cases $=[([[1,3],[],[1,3],[2]],0,$ False$),([[],[0,2],[3],[1]],0,$ False$),([[1,2],[4,5],[3,4],[8,9],[7,8],[6,7],[],[],[],[]],0,$ False$)$for $($test$\_$graph, starting$\_$node, solution$)$ in test$\_$cases: if $($solution $!=$ output$)$: s$2=\text{'}\text{'}+$ str$($solution$)+\text{'}$ Your code output: $\text{'}+$ str$($output$)+""$ failed $=$ True display$($HTML$(\text{'}\text{'}))$ display$($HTML$(\text{'}\text{'}))$

part$\_$of$\_$a$\_$cycle$\_$test$()/$tmp$/$ipykernel$\_124/1471569436.$py:$3$: DeprecationWarning: Importing display from IPython.core.display is deprecated since IPython $7\mathrm{.}14,$ please import from IPython display

from IPython.core.display import display, HTML

Failed $-$ test case: Inputs: graph $=[1,3],[1,[1,3],[2]]$ node $2$

Expected Output: True Your code output: False

One or more tests failed.

Write a function that returns whether a node is part of a cycle.

HINT: Modify you DFS to return early when it finds a cycledef part$\_$of$\_$a$\_$cycle$($a$,$ u$)$: color$[$u$]=$ 'GRAY' if color$[$v$]==$ 'WHITE': return True color$[$u$]=$ 'BLACK'n $=$ len$($a$)$return dfs$($a$,$ u$,$ color, $-1)$

def detect$\_$cycle$($a$)$:color $=[\text{'}$WHITE$\text{'}]*$ n if color$[$u$]==$ 'WHITE': return True

Testing below