Keeping the variable 's value in recursive function, python 3.3 - function

Question

I managed to do it, some other way.
but I have a question, I had this code before
def jumphunt(start, mylist, count = 0):
if count < len(mylist):
place = mylist[start]
print(place)
if place == 0:
return True
elif start >= len(mylist) or start < 0:
return False
move_left = (start - place)
move_right = (start + place)
return jumphunt(move_right, mylist, count+1) or jumphunt(move_left, mylist, count+1)
else:
return False
but for some reason it's not trying both ways
to get to the last item on the list.
for example: [1,2,2,3,4,5,3,2,1,7,0] and ,start=mylist[0]
it supposed to jump like this (from 1-2-4-1-left to 2-left to 5-right to 0)
but it keeps trying to go right and then index is out of range etc.
I thought that if u use return of or this or that, it will try both until it reaches True, why won't it work here?
Thanks!

Answer 1

Include the value you want to keep as a default parameter for the method, like this:
def my_func(int, list, i=0):
a = (i + int)
if int == 0:
return True
elif a > len(list):
i -= int
else:
i += int
int = list[i]
my_func(int, list, i)
Bear in mind that it may not even always be possible to arrive at the end of the list doing the jumping pattern you describe, and even if it is possible, this method may choose the wrong branch.
A better algorithm would look like this:
def branching_search(list, start):
marks = [0]*len(list)
pos = start
while list[pos]!=0:
marks[pos]++
if marks[pos] % 2 == 0 and pos + list[pos] < len(list):
pos += list[pos]
elif marks[pos] % 2 == 1 and pos - list[pos] >= 0:
pos -= list[pos]
else:
return False
if all(item == 0 or item > 1 for item in list)
return False
return True
This way, if it comes to an item that it has already visited, it will decide to go the opposite direction that it went last time. Also, if it comes to an item that it can't leave without going out-of-bounds, or if there is not way to get to the end, it will give up and return.
EDIT: I realized there are a number of flaws in this algorithm! Although it is better than the first approach, it is not guaranteed to work, although the reasons are somewhat complicated.
Just imagine this array (the unimportant elements are left blank):
1, 2, , 5, , , , , 5, 0
The first two elements would get only one mark (thus the loop checking condition would not work), but it would still get stuck looping between the two fives.
Here is a method that will always work:
def flood_search(list):
marks = [[]]*len(list)
marks[0] = [0]
still_moving = True
while still_moving:
still_moving = False
for pos in range(0,len(list)):
if marks[pos]:
if pos + list[pos] < len(list) and not marks[pos + list[pos]]:
marks[pos + list[pos]] = marks[pos] + [list[pos]];
pos += list[pos]
still_moving = True
if pos - list[pos] >= 0 and not marks[pos - list[pos]]:
marks[pos - list[pos]] = marks[pos] + [-list[pos]];
pos -= list[pos]
still_moving = True
return marks[-1]
This works by taking every possible branch at the same time.
You can also use the method to get the actual route taken to get to the end. It can still be used as a condition, since it returns an empty list if no path is found (a falsy value), or a list containing the path if a path is found (a truthy value).
However, you can always just use list[-1] to get the last item.