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Lonelygirl15
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hagbard celine
Casual Observer
Joined: 23 Oct 2006
Posts: 91
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 Posted: Tue Dec 12, 2006 5:23 pm Post subject: |
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| hagbard celine wrote: | 0/0 = 0
0/1 = 1
0/2 = 2
0/5 = 5
0/6 = 6
1/4 = 5
1/5 = 6
1/6 = 7
2/4 = 6
2/5 = 7
3/3 = 6
3/4 = 7
3/6 = 9
which if we were to turn this into hex...
we would have to use 5 6 or 7 to get a valid letter...
so we have 55 6666 777
for the second number we can use 0129 and i suppse we could use up a 5 6 7 also...
so we have
50 P
51 Q
52 R
55 U
56 V
57 W
59 Y
60 `
61 a
62 b
65 e
66 f
67 g
69 i
70 p
71 q
72 r
75 u
76 v
77 w
79 y
plus using ccf from the drop:
5C \
5F _
6C l
6F o
7C |
7F DEL
Others:
20 = space
21 = !
25 = %
26 = &
27 = '
29 = )
2C = ,
2F = /
8 letters. two 5's four 6's three 7's two C's one F one 0 one 1 one 2 one 9 |
OK, one f...
i choose 6F = o
two 5's three 6's three 7's two C's one 0 one 1 one 2 one 9
two C's? l (6C) and , (2C)
---
o , l
two 5's two 6's three 7's one 0 (P`p) one 1 (Qaq!) one 9 (Yiy)) |
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ziz-lite
Suspiciously Absent
Joined: 18 Nov 2006
Posts: 14
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| Posted: Wed Dec 13, 2006 12:32 am Post subject: C C F:3 3 6 |
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Someone may have posted this elsewhere, but C is the 3rd letter and F is the 6th letter of the alphabet. Does this suggests a way to read the dominoes?
Also, have we tried overlapping the stencil letters like the text does for a split second in the last Tachyon vid? Perhaps that will generate another letter or number. |
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sparkyb
Casual Observer
Joined: 07 Dec 2006
Posts: 35
Location: Somerville, MA
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| Posted: Wed Dec 13, 2006 6:31 am Post subject: |
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| hagbard celine wrote: | 0/0 = 0
0/1 = 1
0/2 = 2
0/5 = 5
0/6 = 6
1/4 = 5
1/5 = 6
1/6 = 7
2/4 = 6
2/5 = 7
3/3 = 6
3/4 = 7
3/6 = 9
which if we were to turn this into hex...
we would have to use 5 6 or 7 to get a valid letter...
so we have 55 6666 777
for the second number we can use 0129 and i suppse we could use up a 5 6 7 also...
so we have
50 P
51 Q
52 R
55 U
56 V
57 W
59 Y
60 `
61 a
62 b
65 e
66 f
67 g
69 i
70 p
71 q
72 r
75 u
76 v
77 w
79 y
plus using ccf from the drop:
5C \
5F _
6C l
6F o
7C |
7F DEL
Others:
20 = space
21 = !
25 = %
26 = &
27 = '
29 = )
2C = ,
2F = /
8 letters. two 5's four 6's three 7's two C's one F one 0 one 1 one 2 one 9 |
I'm not sure if I think the strategy of adding up the digits on the dice and forming 8 character ascii out of that (plus the ccf) is right. However, here are my thoughts on it just in case.
Let's assume that letters are really all we are going to use. The C's and F can only go after a 6, which gives us l, l, o. and 556777 left for first digits. the 5's and 7's are the same letters but just lowercase or uppercase. the 0 must go with one of them giving us a P/p. The 1 probably goes with the 6 because 'a' is a lot better to have than Q/q. The 2 becomes an R/r, the 9 becomes a Y/y, and the last letter is either a W/w or U/u. We can figure out capitalization later, but we can anagram lloparyw or lloparyu.
In case I'm wrong and the 1 is a Q/q. Well then we probably want a u (75). so now we have a 6 and a 5/7 (depending on the case of the Q/q and P/p) left for the leftmost digit and 2 and 9 for the right. which is either b and Y/y or i and R/r. So in this case we can anagram llopquby or llopquir.
If I'm wrong about the u going wit the Q/q, then we could have W/w instead. So we can anagram llopqbyw or llopqirw. And also instead of a 5 or 7 going on the right of another 5 or 7, they could go on the right of the 6 so we can anagram llopqyre or llopqyrg. Finally the 6 could go on the right so we can anagram llopqyrv.
The final case I will consider is that 20, a space, is a useful character (although rare in passwords). so after llo and space, we have 556777 for first digits and just 19 for sure on the right. if the 1 goes with the 6 we've got 'a' and Y/y and either 2 W/w's, 2 U/u's, or 1 of each. (there are some restrictions on case but that isn't important yet. So we can anagram lloayww, lloayuu, or lloaywu knowing we need to find at least 2 words. If the 9 goes with the 6 we've got 'i' and Q/q and again the same combinations of W/w's and U/u's (although now u's are more likely), so we can anagram lloiquw, lloiquu, or lloiqww. If 5 goes with the 6 we get Q/q, Y/y, e, and either W, w, or u. so we can anagram lloqyeu or lloqyew. If 7 goes with 6 we get Q/q, Y/y, g, and either W/w or U/u. So we can anagram lloqygu and lloqygw. Finally if the 6 is on the right we have Q/q, Y/y, V/v, and W/w or U/u. So we can anagram lloqyvu or lloqyvw.
Summary of possible letter combination (required caps in parens):
lloparyw (2 of PRYW)
lloparyu (1 of PRYU)
llopquby (1 of PQY)
llopquir (1 of PQR)
llopqbyw (2 of PQYW)
llopqirw (2 of PQRW)
llopqyre (1 of QYR)
llopqyrg (2 of QYR)
llopqyrv (2 of QYRV)
with a space:
lloayww (2 of YWW)
lloayuu (none)
lloaywu (1 of YW)
lloiquw (1 of QW)
lloiquu (none)
lloiqww (2 of QWW)
lloqyeu (none)
lloqyew (1 of YW)
lloqygu (1 of QY)
lloqygw (2 of QYW)
lloqyvu (1 of QY)
lloqyvw (2 of YQW)
I did the math and there are 1,229,760 total possible combinations based on this (ignoring the fact that a capital is likely to be the first letter and other things that would cause a human to regect certain combinations). Obviously if you felt like just plugging these 21 lines into an anagram solver it might make things much easier. A lot of them probably have no anagrams. I tried a few but lost patience so if someone else wants to give it a try, go ahead. Just for the hell of it I wrote a brute force checker that will check all 1.2 million in about 16 hours. |
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sparkyb
Casual Observer
Joined: 07 Dec 2006
Posts: 35
Location: Somerville, MA
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| Posted: Wed Dec 13, 2006 6:41 am Post subject: Another possible way to combine the dice |
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What if, instead of 13 numbers (the addition) the dice represented 26 digits. And instead of putting 2 digits together to form ascii characters in hex, we were to put them together to form numbers that stood for letters of the alphabet (a=01, b=02... z=26). Stencils are hollow letters that need to be filled it, so those could be saying that 3 of the 13 letters formed will be C, C, and F. (03, 03, and 06). That leaves us with:
3- 0's
4- 1's
3- 2's
2- 3's
3- 4's
3- 5's
2- 6's
10 more letters need to be formed and the first digits must be 0, 1, or 2 which uses up all the 0's, 1's, and 2's. Which would give us some combination of the letters c,d,e,f,m,n,o,p,w,x,y,z.
This assumes only letters and doesn't specify case (all lowercase?). And yet, it still doesn't seem to be enough locked in info (like we had with the XX/Xy puzzle) to figure out what it should be, so I don't know if it is right. It was just something I thought of. |
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diastres
Lonely Fan
Joined: 28 Nov 2006
Posts: 151
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| Posted: Wed Dec 13, 2006 7:19 am Post subject: Re: Another possible way to combine the dice |
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| sparkyb wrote: | What if, instead of 13 numbers (the addition) the dice represented 26 digits. And instead of putting 2 digits together to form ascii characters in hex, we were to put them together to form numbers that stood for letters of the alphabet (a=01, b=02... z=26). Stencils are hollow letters that need to be filled it, so those could be saying that 3 of the 13 letters formed will be C, C, and F. (03, 03, and 06). That leaves us with:
3- 0's
4- 1's
3- 2's
2- 3's
3- 4's
3- 5's
2- 6's
10 more letters need to be formed and the first digits must be 0, 1, or 2 which uses up all the 0's, 1's, and 2's. Which would give us some combination of the letters c,d,e,f,m,n,o,p,w,x,y,z.
This assumes only letters and doesn't specify case (all lowercase?). And yet, it still doesn't seem to be enough locked in info (like we had with the XX/Xy puzzle) to figure out what it should be, so I don't know if it is right. It was just something I thought of. |
-points towards XX/Xy thread.-
http://lonelygirl15.com/forum/viewtopic.php?p=98673#98673
Password: WarPylol |
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