# Lesson 7: Behavior of `find_better_split` when `min_leaf` >1

In lesson 7, we’re introduced to the non-naive version of the find_better_split() function.

``````def std_agg(cnt, s1, s2): return math.sqrt((s2/cnt) - (s1/cnt)**2)

def find_better_split_foo(self, var_idx):
x,y = self.x.values[self.idxs,var_idx], self.y[self.idxs]
sort_idx = np.argsort(x)
sort_y,sort_x = y[sort_idx], x[sort_idx]
rhs_cnt,rhs_sum,rhs_sum2 = self.n, sort_y.sum(), (sort_y**2).sum()
lhs_cnt,lhs_sum,lhs_sum2 = 0,0.,0.

for i in range(0,self.n-self.min_leaf-1):
xi,yi = sort_x[i],sort_y[i]
lhs_cnt += 1; rhs_cnt -= 1
lhs_sum += yi; rhs_sum -= yi
lhs_sum2 += yi**2; rhs_sum2 -= yi**2
if i<self.min_leaf or xi==sort_x[i+1]:
continue

lhs_std = std_agg(lhs_cnt, lhs_sum, lhs_sum2)
rhs_std = std_agg(rhs_cnt, rhs_sum, rhs_sum2)
curr_score = lhs_std*lhs_cnt + rhs_std*rhs_cnt
if curr_score<self.score:
self.var_idx,self.score,self.split = var_idx,curr_score,xi
``````

As I understand it, if we had `self.n == 100`, when `min_leaf = 1`, in the for loop we will iterate through all the elements of the selected rows. However if `min_leaf = 2`, then we will iterate through the first 99 items, but we will also skip the first element (due to the if statement: `if<self.min_leaf or xi==sort_x[i+1]: continue`). To take it further, if we had `min_leaf = 10`, then we would iterate through the first 90 elements, but also skip the first 10 items.

I could totally have misread this, and if so please correct me. But right now it seems that the `min_leaf` parameter is forcing us to perform the split in the middle of the rows, but leaving the possibility open that the “true” best split element might be skipped out on. Is this the correct way to read this code? Thank you.

I might be late to your call for help here(!) Paul, but I’ve been pondering this question recently myself and in case it might help anyone would like to include a link to a short post with my explanation of what’s going on here: