On 03/10/15 09:23, Anshu Kumar wrote:
Hi Alan,
I have given a wrong example of 16 . I am sorry for it. You are
correct it will take only 4 turns.
If i consider your solution for number 10
it will go like this 10-->10/2 =5 --> 5-1=4--> 4/2 =2-->2/2 =1 which
gives 4 as output but answer would be in 3 steps
10-->10-1=9-->9/3=3-->3/3=1
So we must consider every path /3, /2 and -1 and try to find out
shortest one
Ah, OK I see.
Here is my modified version which I think works as you want:
def findMinDepthPath(n):
if n <= 0: raise ValueError
elif n==1:
return 0
elif n==2 or n==3:
return 1
else:
d1 = findMinDepthPath(n-1)+1
d2 = d3 = (d1+1) # initialize to higher than d1
if n%3 == 0:
d3 = findMinDepthPath(n/3)+1
if n%2 == 0:
d2 = findMinDepthPath(n/2)+1
return min(d1,d2,d3)
n = int(raw_input('N? '))
print "Minimum depth = ", findMinDepthPath(n),'\n'
>I know list and sets can be used do we have any other data structure
in python
> which would be mutable and not a sequence?
You could use a dictionary but that is also a type of sequence...
So a class is the obvious non-sequential candidate.
class Store:
def __init__(self, value = 0):
self.value = value
You can then pass an instance of the class and modify its value.
--
Alan G
Author of the Learn to Program web site
http://www.alan-g.me.uk/
http://www.amazon.com/author/alan_gauld
Follow my photo-blog on Flickr at:
http://www.flickr.com/photos/alangauldphotos
_______________________________________________
Tutor maillist - Tutor@python.org
To unsubscribe or change subscription options:
https://mail.python.org/mailman/listinfo/tutor
