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Author Message
25 new of 1177 responses total.
aruba
response 359 of 1177: Mark Unseen   Jan 6 15:56 UTC 2000

mammoth
kentn
response 360 of 1177: Mark Unseen   Jan 7 01:14 UTC 2000

  mammoth  1  (aruba)
albaugh
response 361 of 1177: Mark Unseen   Jan 7 18:37 UTC 2000

largely
aruba
response 362 of 1177: Mark Unseen   Jan 9 05:51 UTC 2000

avocado
prp
response 363 of 1177: Mark Unseen   Jan 10 20:26 UTC 2000

purpose
albaugh
response 364 of 1177: Mark Unseen   Jan 10 23:32 UTC 2000

liberal
aruba
response 365 of 1177: Mark Unseen   Jan 12 03:40 UTC 2000

Well, if Kevin can go twice, then so can I...

syllabi
kentn
response 366 of 1177: Mark Unseen   Jan 12 04:02 UTC 2000

  largely  0  (albaugh)
  avocado  0  (aruba)
  purpose  1  (prp)
  liberal  0  (albaugh)
  syllabi  0  (aruba)
prp
response 367 of 1177: Mark Unseen   Jan 12 21:10 UTC 2000

humbles
albaugh
response 368 of 1177: Mark Unseen   Jan 12 23:46 UTC 2000

Oops, sorry!  My offering:

apology
kentn
response 369 of 1177: Mark Unseen   Jan 13 02:49 UTC 2000

  humbles  4  (prp)
  apology  0  (albaugh)
aruba
response 370 of 1177: Mark Unseen   Jan 14 01:26 UTC 2000

kickoff
prp
response 371 of 1177: Mark Unseen   Jan 14 20:54 UTC 2000

hundred
kentn
response 372 of 1177: Mark Unseen   Jan 15 02:13 UTC 2000

  kickoff  0  (aruba)
  hundred  2  (prp)
aruba
response 373 of 1177: Mark Unseen   Jan 15 03:28 UTC 2000

shiatzu
kentn
response 374 of 1177: Mark Unseen   Jan 15 23:20 UTC 2000

  shiatzu  1  (aruba)
albaugh
response 375 of 1177: Mark Unseen   Jan 17 20:06 UTC 2000

mumbles
aruba
response 376 of 1177: Mark Unseen   Jan 18 14:42 UTC 2000

kibbitz
prp
response 377 of 1177: Mark Unseen   Jan 19 19:30 UTC 2000

humbird
kentn
response 378 of 1177: Mark Unseen   Jan 21 00:55 UTC 2000

  mumbles  3  (albaugh)
  kibbitz  2  (aruba)
  humbird  4  (prp)
 
While I couldn't find an OED dictionary defn of 'humbird' it does seem
to be used fairly frequently as a shortening of 'hummingbird' and
appears in some URLs...so I guess that's good enough for our standards
(I hope).
aruba
response 379 of 1177: Mark Unseen   Jan 21 17:38 UTC 2000

OK, the word is "humbuzz" - proof in the next response.
aruba
response 380 of 1177: Mark Unseen   Jan 21 18:11 UTC 2000

  peewees  0  (aruba)
  apology  0  (albaugh)
  humbugs  5  (aruba)
  mammoth  1  (aruba)
  kibbitz  2  (aruba)
  shiatzu  1  (aruba)

Let "-" represent a letter that is definitely wrong,
    s(w) be the score of a word or partial word w, and
    n(w) be the number of letters in w which are not -'s.

Then clearly
    0 <= s(w) <= n(w).

0 = s(peewees) >= s(------s), so s(------s) = 0.
0 = s(apology) >= s(-----g-), so s(-----g-) = 0.
5 = s(humbugs) = s(humbu--) + s(-----g-) + s(------s)
               = s(humbu--) + 0 + 0
               = s(humbu--)
               = s(h------) + s(-u-----) + s(--m----) + s(---b---) + s(----u--)
==> 1 = s(h------) = s(-u-----) = s(--m----) = s(---b---) = s(----u--),
since all five terms are between 0 and 1 and they add up to 5.

1 = s(mammoth) = s(ma-mo-h) + s(--m----) + s(-----t-)
               = s(ma-mo-h) + 1 + s(-----t-)
==> s(ma-mo-h) + s(-----t-) = 0
==> s(-----t-) = 0.

2 = s(kibbitz) = s(kib-i--) + s(---b---) + s(-----t-) + s(------z)
               = 0 + 1 + 0 + s(------z)
==> s(------z) = 1.

1 = s(shiatzu) = s(shiat--) + s(-----z-) + s(------u)
               = 0 + s(-----z-) + 0
==> s(-----z-) = 1.

So therefore
s(humbuzz) = s(humbu--) + s(-----z-) + s(------z)
           = 5 + 1 + 1 = 7.

Of course, I don't have any evidence that humbuzz is really a word.
albaugh
response 381 of 1177: Mark Unseen   Jan 21 18:24 UTC 2000

a buzz you get from drinking a hummer?  ;-)
prp
response 382 of 1177: Mark Unseen   Jan 21 19:31 UTC 2000

HUMBIRD: n. Now U.S. M17. The hummingbird.
from The New Shorter Oxford (1973,1993)

It is not in the Worldbook Dictionary.

HUMBUZZ is not in either.
orinoco
response 383 of 1177: Mark Unseen   Jan 21 19:41 UTC 2000

Not to further complicate matters, but isn't it "shiatsu"?
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