I recently found out about the technique called dynamic programming and I stumbled upon a problem which I can't figure out. You are given a list of arguments in the beginning and you need to do sums on as if you were cutting it. If the list has only one element, you don't sum it. If it has more, you sum the elements and cut it in every possible way. So if list has n elements, there are just n-1 ways to cut it. The picture will explain:
I first wanted to sum up all of the sumable parts and I expected the result 20( 11 + 9 ) ( even thought the correct answer is 9 ), but I thought it would be a good start. But my code returns number 37 and I have no idea why. What am I doing wrong?
summ = 0
def Opt( n ):
global summ
if len( n ) == 1:
return 0
else:
summ += sum( n )
for i in range( 1,len( n ) ):
summ += Opt( n[ :i ] ) + Opt( n[ i: ] )
return summ
print( Opt( [ 1,2,3 ] ) )
Thank you for your time and any answer!
I think this is what you want:
Example:
Dynamic programming is about dividing the "big problem" into small subproblems.
So, first of all, you should identify how the big problem is related to the subproblems. You do this by writing a recurrence relation. In this case:
You also need a starting point:
As you can see, once you have wrote the recurrence relation, transforming it into Python code is often straightforward.
It's important to understand that each subproblem is self-contained, and should not need external input. Your use of
global
variables was not only producing the wrong result, but was against the spirit of dynamic programming.Your use of trees for expressing
Opt()
is nice. What you forgot to do is writing the relationship between each node and its children. If you did, I'm almost sure that you would have found the correct solution yourself.We are not finished yet (thanks Stefan Pochmann for noticing). In order to build a true dynamic programming solution, you also need to avoid solving the same problem more than once. Currently, running
Opt([1,2,3,4])
results in callingOpt([1,2])
more than once. One way to prevent that is by using memoization:By the way, remember to handle the case where
n
is empty (i.e.len(n) == 0
).