## Minimum Split

### December 4, 2015

Before you look at the suggested solution, check what your solution does when given the list (5 -5).

Our algorithm considers each split from left to right, beginning with the split that has no items in the left split and the entire list in the right split. At each step, we move one item from right to left, recompute totals, and if the current difference is less than the current minimum, update the minimum. The algorithm stops when the list is exhausted or the difference drops to zero.

```(define (min-split xs)
(let loop ((i 0) (lo 0) (los (list)) (hi (sum xs)) (his xs)
(best-point 0) (best-value (sum xs)))
(cond ((or (null? his) (= lo hi)) (split best-point xs))
((< (abs (- lo hi)) best-value)
(loop (+ i 1) (+ lo (car his)) (cons (car his) los)
(- hi (car his)) (cdr his) i (abs (- lo hi))))
(else (loop (+ i 1) (+ lo (car his)) (cons (car his) los)
(- hi (car his)) (cdr his) best-point best-value)))))```

Here are some examples:

```> (min-split '(2 7 3 1 4))
(2 7)
(3 1 4)
> (min-split '(1 2 4 7 3))
(1 2 4)
(7 3)
> (min-split '(5 -5))
()
(5 -5)```

We used `sum` and `split` from the Standard Prelude. You can run the program at http://ideone.com/2GOsm9.

Pages: 1 2

### 11 Responses to “Minimum Split”

1. Rutger said

In Python

```import sys

# returns tuple (minimum_split, index to split on)
# not so efficient w.r.t. memory/copying
def minimum_split_lazy_coding(l):
return min((abs(sum(l[:i]) - sum(l[i:])), i) for i in range(0, len(l)))

# returns tuple (minimum_split, index to split on)
# traverses list only twice.
def minimum_split_efficient(l):
sum_left = 0
sum_right = sum(l)
minimum = (abs(sum_right), 0)
for i in range(0, len(l)):
sum_left += l[i]
sum_right -= l[i]
difference = abs(sum_right - sum_left)
if difference < minimum:
minimum = (difference, i+1)
return minimum

examples = []
examples.append()
examples.append([3,7,4,2,1])
examples.append([-1, -1])
examples.append([-1, 1])
examples.append([1,2,3,3,2,1])

for e in examples:
print minimum_split_lazy_coding(e)
print minimum_split_efficient(e)
```
2. John Cowan said

It’s not very clear whether this means the minimum difference, or the minimum absolute value of the difference. The wording of page 2 suggests the latter, but page 1 doesn’t say so.

3. Francesco said

```import Data.List

f x = minimumBy (\(_,a) (_,b) -> compare (abs a) (abs b)) a
where a = zipWith (\a b -> ((a,b), sum a - sum b)) (inits x) (tails x)```
4. yaphats said

Python:

def bestSplit(lst):
diff = 100000000
idx = 0
for count in range(0, len(lst)):
if diff > abs(sum(lst[:count], 0) – sum(lst[count:], count)):
diff = abs(sum(lst[:count], 0) – sum(lst[count:], count))
idx = count
return idx

5. mcmillhj said
```fun split xs = let
val sum = List.foldr (op +) 0 xs
fun partial_sum [] total = []
| partial_sum (x::xs) total = let
val partial = total + x
in
partial :: partial_sum xs partial
end

fun min []      i res = ( res, i )
| min (x::xs) i res = let
val new_min = Int.abs(x - (sum - x))
in
if new_min < res then min xs (i+1) new_min
else min xs i res
end
in
min (partial_sum xs 0) 0 (Int.abs sum)
end
[hunter@apollo: 12]\$ poly --use minsplit.sml
Poly/ML 5.2 Release
> use "minsplit.sml";
val split = fn : Int.int LIST.list -> Int.int * int
val it = () : unit
> split [3,7,4,2,1];
val it = (3, 2) : Int.int * int
> split [~1,1];
val it = (0, 0) : Int.int * int
> split [1,2,3,4,5];
val it = (3, 3) : Int.int * int
```
6. sc120 said

Solution in Python

7. sc120 said

Third attempt in posting the code :p

```def list_sep(number_list):
for i in xrange(1,len(number_list)):
temp_diff=abs(sum(number_list[:i])-sum(number_list[i:]))
if i== 1:
diff=temp_diff
position=i
if temp_diff<diff:
diff= temp_diff
position=i
return diff, position

number_list=[3,7,4,2,1]

print('Difference: %s' %list_sep(number_list))
print('Split position: %s' %list_sep(number_list))
```
8. Paul said

In Python.

```from itertools import accumulate, chain
def ssum(L):
S = sum(L)
return min((abs(S-2*a), i) for i, a in enumerate(chain(, accumulate(L))))
```
9. Eric Hansen said

C++ attempt

#include “math.h”
#include
std::vector numberList;
numberList.push_back(3);
numberList.push_back(7);
numberList.push_back(4);
numberList.push_back(2);
numberList.push_back(1);

std::vector differences;
for(int split = 0; split < numberList.size(); ++split)
{
int leftSum = 0;
for(int left = 0; left < split; ++left) {
leftSum += numberList[left];
}
int rightSum = 0;
for(int right = split; right < numberList.size(); ++right) {
rightSum += numberList[right];
}
int difference = abs(rightSum – leftSum);
differences.push_back(difference);
}

int lowestDiffIndex = 0;
int lowestDiffValue = differences;
for(int i = 0; i < differences.size(); ++i)
{
lowestDiffIndex = (differences[i] < lowestDiffValue) ? i : lowestDiffIndex;
lowestDiffValue = (differences[i] < lowestDiffValue) ? differences[i] : lowestDiffValue;
}
// Split is at element lowestDiffIndex

10. r. clayton said

A solution in Racket.