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| Author |
Message |
| 25 new of 1103 responses total. |
aruba
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response 475 of 1103:
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May 3 15:48 UTC 2002 |
stadia
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brighn
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response 476 of 1103:
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May 3 19:08 UTC 2002 |
durnit, I think I have this one, too... fine time for me to go on a vacation
(by my calculations, there are four possible frames, each missing two letters,
and onelook only gives one word that fits all the frames)
cyanid (should score 4) -- alternate spelling of cyanide
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brighn
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response 477 of 1103:
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May 3 19:09 UTC 2002 |
(I should have said: I can only find one word that fits any of the frames,
and isn't otherwise ruled out.)
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kentn
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response 478 of 1103:
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May 3 23:21 UTC 2002 |
stadia 2 (aruba)
cyanid 4 (brighn)
I'm going to be out of town on business for a few days...so if
someone doesn't get it tonight, be patient.
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aruba
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response 479 of 1103:
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May 4 13:53 UTC 2002 |
stupid
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brighn
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response 480 of 1103:
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May 5 04:24 UTC 2002 |
evince (worth 3?)
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kentn
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response 481 of 1103:
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May 9 07:15 UTC 2002 |
stupid 2 (aruba)
evince 3 (brighn)
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aruba
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response 482 of 1103:
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May 9 13:52 UTC 2002 |
Right, so the word is "evanid". Proof to follow.
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aruba
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response 483 of 1103:
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May 9 14:14 UTC 2002 |
ethnic 3 (aruba)
static 2 (aruba)
cyanid 4 (brighn)
stupid 2 (aruba)
evince 3 (brighn)
4 = score(cyanid) = score(cya-id) + score(---n--)
3 = score(evince) = score(evi-ce) + score(---n--)
==> 7 = score(cya-id) + score(evi-ce) + 2*score(---n--).
Now score(cya-id) + score(evi-ce) <= 5, since they have no letters in common,
and of course 2*score(---n--) <= 2. If either were less than their maximum,
their sum would be less than 7, a contradiction. So
score(---n--) = 1 and
score(cya-id) + score(evi-ce) = 5.
That means all the correct letters appear in cyanid or evince, and the
correct word matches [ce][vy][ai]n[ci][de]. So we can cross out a lot of
letters we know are wrong:
score(ethnic) = 3 ==> score(e--ni-) = 3 \
score(static) = 3 ==> score(--a-i-) = 2 | ==> score(e-anid) = 5.
score(stupid) = 2 ==> score(----id) = 2 /
Then 3 = score(evince) = score(e-ince) + score(-v----) = 2 + score(-v----)
==> score(-v----) = 1
==> score(evanid) = 6.
I have no idea what it means.
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brighn
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response 484 of 1103:
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May 9 14:26 UTC 2002 |
Liable to vanish or disappear; faint; weak; evanescent; as, evanid color.
(www.dictionary.com)
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brighn
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response 485 of 1103:
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May 9 15:29 UTC 2002 |
I arrived at evanid after erotic(2). Here's how:
Removing all the 0 answers and their letter overlaps, we have these clues:
etcher 1
etches 1
ethnic 3
_tatic 2
e_otic 2
This leads us to some other conclusions:
1a. ethnic 3 & e_otic 2 -> if _thn__ 2 then __ot__ 1 -> (_tht__ 3 || _ton__
3) & _tatic 2
1aa. _ton__ 3 & _tatic 2 -> _t__ic 2 -> ____ic 1 but ____ic 0 if _ton__ 3
since ethnic 3
ergo not _ton__ 3
1ab. _tht__ 3 & _tatic 2 -> _tht__ 3 (all clues satisfied)
1b. If _thn__ 3 then e___ic 0 & __ot__ 0 but e_otic 2 ergo not _thn__ 3
1c. If _thn__ 0 then e___ic 3 but e_otic 2 ergo not _thn__ 0
1d. If _thn__ 1 then e___ic 2 & __ot__ 0
Assume 1d and proceed.
etcher 1 \ etche_ 1
etches 1 /
Update list for __ot__ 0 and _____r 0 and _____s 0:
etche_ 1
etche_ 1
ethnic 3
_ta_ic 2
e_otic 2
2. e___ic 2 -> e____c 2 || e___i_ 2 || ____ic 2
2a. ____ic 2 & _ta_ic 2 -> _ta___ 0 & etche_ 1 & e_____ 0 & ____ic 2 -> __ch__
1; _ta___0 & _thn__1 -> __hn__ 1; __ch__ 1 & __hn__ 1 -> __cn__ 2 || __hh__
2
ergo __cnic 4 || __hhic 4; all clues satisfied
2b. (e____c 2 || e___i_ 2) & _ta_ic 2 -> _ta___ 1; e_____ 1 & etche_ 1 ->
_t____ 0; ergo e____ 1 -> __a___ 1; ergo if ____ic 1 then e_a___ 2; _thn__
1 & e_a___ 2 & _t____ 0 -> e_an__ 3 -> (e_ani_ 4 || e_an_c 4); all clues
satisfied
So the clues to "erotic" gives these possible patterns:
_tht__ 3
__cnic 4
__hhic 4
e_ani_ 4
e_an_c 4
Onelook pattern-matching returns:
_tht__: No possibilities
__cnic: picnic
__hhic: No possibilities
e_ani_: evanid
e_an_c: No possibilities
Looking over the logic, I don't recall seeing __cnic in my initial analysis
(an oversight). In retrospect, had I seen it, I'd've assumed that Kent
wouldn't be using a word that common anyway. At any rate, cyanid 4 resolved
the issue, had I seen it (and I would have been truly embarrassed if, after
all that work to get to evanid, it had been picnic after all)..
|
rcurl
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response 486 of 1103:
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May 9 15:51 UTC 2002 |
I'm impressed. Can you reduce that to an algoithm so a computer could play
this game with itself?
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aruba
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response 487 of 1103:
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May 9 19:39 UTC 2002 |
It would be extremely simple to write a program to search through a
dictionary for possible remaining words - just compare each word with those
guessed so far, and see if they score the same.
A program without access to a dictionary could do somethings easily, but it
would be difficult to make it do as well as we humans can do.
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brighn
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response 488 of 1103:
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May 9 20:30 UTC 2002 |
#486> I'm sure it's possible. I may consider the challenge of constructing
the algorithm. Obviously, without access to a dictionary, the program could
only output possible frames and disallowed letters (or, more extensively, a
complete list of possible strings that match the remaining possibilities,
leaving the user to slog through the likes of qthtzws).
I did write an algorithm to solve Paint by Numbers puzzles (from Games
magazine), that seems to work pretty good. This feels like it might be
similar, though hardly identical.
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brighn
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response 489 of 1103:
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May 9 21:33 UTC 2002 |
Actually, when I sat down to reason out how this might be done with a computer
algorithm, I strong-armed it instead of trying to be logical, and I managed
to find some errors in my analysis above as a result. Here's the pseudocode
I came up with for returning all the possible frames, although it doesn't take
into account "ruled out" letters. The pseudocode doesn't exactly match the
analysis, and I used lots of dopey little shorthand, and I don't vouch for
the pseudocode because I haven't actually written the program, but...
anyway... things I do when I'm supposed to be working. ;}
Algorithm (pseudocode):
input(1..n) = strings
points(1..n) = value(input(1..n))
Dim frame(26, len(guess))
If points(1..n) = 0: remove from input(1..n), replacing common letters in
input(1..n) where points > 0 with _
For i = 1 to len(guess) {
Create possibles() as distinct $ in mid(guess, i, 1)
Add _ to possibles
For j = 1 to size(possibles) {
mid(frame(j,i), i, 1) = possibles(j)
If possibles(j) = mid(input(1..n), i, 1) (and not "_")
{decrease points(1..n) by 1
If points(1..n) = 0: remove from input(1..n) as above
If points(1..n) = len(guess) - i: force frame for remaining characters
If points(1..n) > len(guess) - i: error
If input(m) = input(l) and points(m) <> points(l): error}}}
Return all frame(x, y) where not in error
INPUT
etcher 1
etches 1
ethnic 3
_tatic 2
e_otic 2
OUTPUT
e_ani_
e_an_c
_toni_
_ton_c
__cnic
__hhic
1
e?????:
Xtcher 0
Xtches 0
Xthnic 2
Xtatic 2
X_otic 2
>>
X_hnic 2
X_atic 2
X_otic 1
11
e_????:
XXhnic 2
XXatic 2
XXotic 1
111
e_h???:
XXXnic 1
XXXtic 2
XXXtic 1
<err>
112
e_a???:
XXXnic 2
XXXtic 1
XXXtic 1
1121
e_an??:
XXXXic 1
XXXXic 1
XXXXic 1
>>
XXXXic 1
11211
e_ani?:
XXXXXc 0
112111
e_ani_
11212
e_an_?:
XXXXXc 1
112121
e_an_c
1112
e_at??:
XXXXic 2
XXXXic 0
XXXXic 0
<err>
1113
e_a_??:
XXXXic 2
XXXXic 1
XXXXic 1
<err>
113
e_o???:
XXXnic 2
XXXtic 2
XXXtic 0
<err>
114
e__???:
XXXnic 2
XXXtic 2
XXXtic 1
<err>
2
_?????:
Xtcher 1
Xtches 1
Xthnic 3
Xtatic 2
X_otic 2
21
_t????:
XXcher 0
XXches 0
XXhnic 2
XXatic 1
XXotic 3
>>
XXhnic 2
XXatic 1
XXotic 2
211
_th???:
XXXnic 1
XXXtic 1
XXXtic 2
<err>
212
_ta???:
XXXnic 2
XXXtic 0
XXXtic 2
<err>
213
_to???:
XXXnic 2
XXXtic 1
XXXtic 1
2131:
_ton??:
XXXXic 1
XXXXic 1
XXXXic 1
>>
XXXXic 1
21311:
_toni?:
XXXXXc 0
213111:
_toni_
21312:
_ton_?:
XXXXXc 1
213121:
_ton_c
214
_t_???:
XXXnic 2
XXXtic 1
XXXtic 2
<err>
22
__????:
XXcher 1
XXches 1
XXhnic 3
XXatic 2
XXotic 2
221
__c???:
XXXher 0
XXXhes 0
XXXnic 3
XXXtic 2
XXXtic 2
>>
XXXnic 3
XXXtic 2
>>
221111
__cnic
222
__h???:
XXXher 1
XXXhes 1
XXXnic 2
XXXtic 2
XXXtic 2
2221
__hh??:
XXXXer 0
XXXXes 0
XXXXic 2
XXXXic 2
XXXXic 2
>>
XXXXic 2
>>
222111
__hhic
2222
__hn??:
XXXXer 1
XXXXes 1
XXXXic 1
XXXXic 2
XXXXic 2
<err>
2223
__ht??:
XXXXer 1
XXXXes 1
XXXXic 2
XXXXic 1
XXXXic 1
<err>
2224
__h_??:
XXXXer 1
XXXXes 1
XXXXic 2
XXXXic 2
XXXXic 2
>>
__h_ic
<err>
223
__a???:
XXXher 1
XXXhes 1
XXXnic 3
XXXtic 1
XXXtic 2
<err>
224
__o???:
XXXher 1
XXXhes 1
XXXnic 3
XXXtic 2
XXXtic 1
<err>
225
___???:
XXXher 1
XXXhes 1
XXXnic 3
XXXtic 2
XXXtic 2
>>
XXXher 1
XXXhes 1
XXXnic 3
XXXtic 2
>>
___nic 3
<err>
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kentn
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response 490 of 1103:
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May 10 02:29 UTC 2002 |
evanid 6 (aruba) ---> Ding! Ding! Ding! Winner!
(Just to be complete. Definition has already been provided in #484.)
Your word, Mark!
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aruba
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response 491 of 1103:
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May 10 03:13 UTC 2002 |
I'll come up with a word soon.
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aruba
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response 492 of 1103:
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May 12 23:19 UTC 2002 |
OK, I'm thinking of a 6-letter word.
evanid 0 (lastword)
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kentn
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response 493 of 1103:
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May 13 10:41 UTC 2002 |
unease
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aruba
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response 494 of 1103:
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May 13 17:02 UTC 2002 |
unease 0 (kentn)
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gelinas
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response 495 of 1103:
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May 18 04:27 UTC 2002 |
doubts
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aruba
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response 496 of 1103:
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May 18 04:44 UTC 2002 |
doubts 1 (gelinas)
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kentn
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response 497 of 1103:
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May 18 12:15 UTC 2002 |
series
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aruba
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response 498 of 1103:
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May 18 14:02 UTC 2002 |
series 1 (kentn)
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gelinas
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response 499 of 1103:
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May 21 04:45 UTC 2002 |
serial
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