On Thu, Jul 10, 2008 at 3:43 PM, Elliott Hird
<[EMAIL PROTECTED]> wrote:
> (Watch as ais523 posts an INTERCAL version.)
If it's obfuscation you want, here's another Python version:
data = [32.0, 2.3992005224917388, 88.905740160097892, 14.1345137819239,
-42.619932779167272, -4.0249040677154548, 8.5713420535658571,
0.53058347581722731, -0.91336540292384771, -0.041376900359692048,
0.05858609461825362, 0.0020487457866678349, -0.0024376507408130767,
-6.7025280431881007e-005, 6.8863105521810549e-005,
1.4899994153630508e-006, -1.3572972875335265e-006,
-2.2906699892943481e-008, 1.8927622958353989e-008,
2.4515332527201483e-010, -1.8721661985584177e-010,
-1.815751028418898e-012, 1.3009748275012855e-012,
9.1008080235703277e-015, -6.1916890846465789e-015,
-2.9386749664827818e-017, 1.9166735804521151e-017,
5.5049605571884816e-020, -3.4684180396566863e-020,
-4.5369419960604571e-023, 2.7782760654811478e-023]
n = len(data) - 1
chars = [None] * (len(data) * 1000)
for i, x in enumerate(xrange(-n//2, n+1-n//2)):
chars[i::n+1] = [chr(int(round(sum(a * x ** i for i, a in
enumerate(data)))))] * 1000
print ''.join(chars),
-root
