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 Posted: Fri Aug 17, 2007 12:54 am
I have a problem for you. I suck at math, and suck hard. So, pweez to be figuring this out for me. I'll give you 1,500 gold if you do:
Say in a language, there 29 consonants. Any of these consonants can be aspirated or unaspirated, or made into an ejective. 2 of the consonants can be dentalized in addition to having one of those 3 variations on it, and 6 of them may be retroflexed in addition to those variations.
There are 12 vowels that can be distinguished by being lengthened or not, and any normal or lengthened vowel can be one of two tones.
17 of those 29 aforementioned consonants, if none of the previously mentioned possbile consonant variations are applied to them, may also act as vowels as described above.
2 other consonants stay as they are, and never change.
So, if the syllable structure of this language were to be CV, how many distinguishable syllables are possible? I excpect it to be, how you say, exactly one shitload of 'em.
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Posted: Fri Aug 17, 2007 3:04 am
Okay, this is like the first Maths problem I ever try to solve in English, so I won't guarantee that my answer correct. Here's what I got: Vajra B. Hairava I have a problem for you. I suck at math, and suck hard. So, pweez to be figuring this out for me. I'll give you 1,500 gold if you do: Say in a language, there 29 consonants. Any of these consonants can be aspirated or unaspirated, or made into an ejective. 2 of the consonants can be dentalized in addition to having one of those 3 variations on it, and 6 of them may be retroflexed in addition to those variations. There are 12 vowels that can be distinguished by being lengthened or not, and any normal or lengthened vowel can be one of two tones. 17 of those 29 aforementioned consonants, if none of the previously mentioned possbile consonant variations are applied to them, may also act as vowels as described above. 2 other consonants stay as they are, and never change. So, if the syllable structure of this language were to be CV, how many distinguishable syllables are possible? I excpect it to be, how you say, exactly one shitload of 'em. I added everything in red together. In my calculation I counted the retroflex and dentalized versions as separate phonemes, just to make it easier to count. 29 + 2 + 6 = 37Then you wanted each of them to come in three different variations, right? That makes it 37 x 3 = 113Then you have two extra consonants that don't change, right? 113 + 2 = 115So we have total amount of 115 consontants. Now the vowels: 12 + 17 = 29And then I double it due to the two length variations: 29 x 2 = 58Then I do the same thing again, since there are also two different tones to consider: 58 x 2 = 116So we have 116 different vowels. Now we multiply the vowels with the consontants to find out how many syllables we get if we follow the CV rule: 116 x 115 = 13 340 syllables. I don't know whether I've done it right. sweatdrop But it was fun, I love Maths! And even more when I had to think in English, never done that before. razz Thank you!
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Eccentric Iconoclast Captain
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Posted: Fri Aug 17, 2007 8:56 am
My answer?
A ******** LOT.
Good day to you.
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Posted: Fri Aug 17, 2007 8:58 am
Eccentric Iconoclast My answer? A ******** LOT. Good day to you. Ditto. (I suck at math. That's why I like language.)
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Posted: Fri Aug 17, 2007 9:59 pm
We have 29 consonants. Each of them has the aspirated, unaspirated, and ejective forms. Am I correct in assuming that every consonant must have one of these three forms?
That gives us 29*3 = 87.
Now, in addition, for two consonants we have the dentalized aspirated, the dentalized unaspirated, and the dentalized ejective forms (if I'm reading this correctly). This assume that dentalization and aspiration are independent; in this case, independence means that the consonant can be dentalized or not independent of whether it is aspirated, unaspirated, or ejective.
So, we have 2*3 = 6 more, giving us 93.
In addition addition, we have 6 retroflex. Here was have a bit of a problem: do the retroflexable consonants overlap with the dentalizable consonants?
If not, then we have 6*3 = 18 more, giving us 111 consonants.
If one overlaps, then we have 5*3 + 1*6 = 21 more, giving us 114 total, assuming that retroflexation and dentalization are independent.
If two overlap, i.e. if both dentalizable consonants are retroflexable, and retroflexation and dentalization are independent, then we have 4*3+2*6 = 24 more, giving us 117 total consonants.
We have 12 vowels, which can be either short or long, and either tone a or tone b. Thus we have 12*4 = 48 vowels.
We also have 17 syllabic consonants, which can be turned into any of four vowel forms, giving us 17*4 = 68 more vowels, giving us a total of 116 vowels.
Thus, we get a total of either:
No overlaps (see above): 111*116 = 12876
One overlap: 114*116 = 13224
Two overlaps: 117*116 = 13572
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Posted: Sat Aug 18, 2007 2:31 am
Allright, around 13000. I got it. So as EI said, "a ******** lot". J00 win.
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Posted: Sat Aug 18, 2007 4:12 am
I don't think it's a ******** lot, probably just a metric assload.
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Posted: Sat Aug 18, 2007 1:14 pm
1.57 metric assloads if you do the conversion. But I ain't usin the mertic system, cuz I R ameericaen!
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Posted: Sun Aug 19, 2007 8:20 am
Depends on exceptions; till I figure out what's what I'm not even going to THINK about the numbers.
Guh, I have a math test on Wednesday. First day of class.
Moving this to Help.
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