Algorithm to find which number in a list sum up to a certain number

Question

I have a list of numbers. I also have a certain sum. The sum is made from a few numbers from my list (I may/may not know how many numbers it\'s made from). Is there a fast a

Answer 1

So, the logic is to reverse sort the numbers,and suppose the list of numbers is l and sum to be formed is s.

   for i in b:
            if(a(round(n-i,2),b[b.index(i)+1:])):
                r.append(i)    
                return True
        return False

then, we go through this loop and a number is selected from l in order and let say it is i . there are 2 possible cases either i is the part of sum or not. So, we assume that i is part of solution and then the problem reduces to l being l[l.index(i+1):] and s being s-i so, if our function is a(l,s) then we call a(l[l.index(i+1):] ,s-i). and if i is not a part of s then we have to form s from l[l.index(i+1):] list. So it is similar in both the cases , only change is if i is part of s, then s=s-i and otherwise s=s only.

now to reduce the problem such that in case numbers in l are greater than s we remove them to reduce the complexity until l is empty and in that case the numbers which are selected are not a part of our solution and we return false.

if(len(b)==0):
    return False    
while(b[0]>n):
    b.remove(b[0])
    if(len(b)==0):
        return False    

and in case l has only 1 element left then either it can be part of s then we return true or it is not then we return false and loop will go through other number.

if(b[0]==n):
    r.append(b[0])
    return True
if(len(b)==1):
    return False

note in the loop if have used b..but b is our list only.and i have rounded wherever it is possible, so that we should not get wrong answer due to floating point calculations in python.

r=[]
list_of_numbers=[61.12,13.11,100.12,12.32,200,60.00,145.34,14.22,100.21,14.77,214.35,200.32,65.43,0.49,132.13,143.21,156.34,11.32,12.34,15.67,17.89,21.23,14.21,12,122,134]
list_of_numbers=sorted(list_of_numbers)
list_of_numbers.reverse()
sum_to_be_formed=401.54
def a(n,b):
    global r
    if(len(b)==0):
        return False    
    while(b[0]>n):
        b.remove(b[0])
        if(len(b)==0):
            return False    
    if(b[0]==n):
        r.append(b[0])
        return True
    if(len(b)==1):
        return False
    for i in b:
        if(a(round(n-i,2),b[b.index(i)+1:])):
            r.append(i)    
            return True
    return False
if(a(sum_to_be_formed,list_of_numbers)):
    print(r)

this solution works fast.more fast than one explained above. However this works for positive numbers only. However also it works good if there is a solution only otherwise it takes to much time to get out of loops.

an example run is like this lets say

    l=[1,6,7,8,10]

and s=22 i.e. s=1+6+7+8
so it goes through like this 

1.) [10, 8, 7, 6, 1] 22
i.e. 10  is selected to be part of 22..so s=22-10=12 and l=l.remove(10)
2.) [8, 7, 6, 1] 12
i.e. 8  is selected to be part of 12..so s=12-8=4 and l=l.remove(8)
3.) [7, 6, 1] 4  
now 7,6 are removed and 1!=4 so it will return false for this execution where 8 is selected.
4.)[6, 1] 5
i.e. 7  is selected to be part of 12..so s=12-7=5 and l=l.remove(7)
now 6 are removed and 1!=5 so it will return false for this execution where 7 is selected.
5.)[1] 6
i.e. 6  is selected to be part of 12..so s=12-6=6 and l=l.remove(6)
now 1!=6 so it will return false for this execution where 6 is selected.
6.)[] 11
i.e. 1 is selected to be part of 12..so s=12-1=1 and l=l.remove(1)
now l is empty so all the cases for which 10 was a part of s are false and so 10 is not a part of s and we now start with 8 and same cases follow.
7.)[7, 6, 1] 14
8.)[6, 1] 7
9.)[1] 1

just to give a comparison which i ran on my computer which is not so good. using

l=[61.12,13.11,100.12,12.32,200,60.00,145.34,14.22,100.21,14.77,214.35,145.21,123.56,11.90,200.32,65.43,0.49,132.13,143.21,156.34,11.32,12.34,15.67,17.89,21.23,14.21,12,122,134]

and

s=2000

my loop ran 1018 times and 31 ms.

and previous code loop ran 3415587 times and took somewhere near 16 seconds.

however in case a solution does not exist my code ran more than few minutes so i stopped it and previous code ran near around 17 ms only and previous code works with negative numbers also.

so i thing some improvements can be done.