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By motty (Mon Apr 10, 2006 at 05:14:19 PM EST) fun loving cowwqas., rtfo, mp3s, fights, girls, drinks, maths fun challenge (all tags)
A Maths Fun Challenge! Which is bugging me.

Also, a stupid bloody weekend, some new music, and other things. Maybe. How should I know?

I haven't even written it yet.

Right. Maths Fun Challenge. This is beginning to bug me now, though it may well turn out to be trivial.

From a nameless message-board elsewhere which you may or may not have already seen:

If you say that A = 1 and B = 2 and C = 3 and so on and use this to add up the letters in numbers then there are only two numbers which equal themselves TWO HUNDRED AND FIFTY ONE, and TWO HUNDRED AND FIFTY NINE.

I ended up perpetrating this excrescence of a Perl script showing this conjecture to hold for numbers between 0 and 9999.

The following was then suggested - in essence, after showing by brute force that the conjecture held for numbers from 0 to 9999, 'every time you add a zero on, you'd add maybe another two words to the total, coming to substantially less than ten times the alphabetic total, but the number itself has gone up by 10 times and so the growth of the number will always be higher than the growth of the alphabetical total.'


Not quite.

After all, at some point you run out of words for numbers. For arbitrarily large numbers, the words that might be used to represent them will eventually also have to become arbitrarily long. You might argue that new words would be of analagous length to existing words, but this is not the case as you will simply run out of letter combinations after a while, and at some point, new words for numbers would always have to be longer than previous new words for numbers.

Since it is not reasonable to suggest that such new words would necessarily always be of minimal length, they themselves can be of arbitrary length. In which case, the original conjecture falls, though we cannot say which numbers it falls to.

I'm so glad to have cleared that up. Unless I'm talking bollocks, which is possible. Hence the Maths Fun Challenge bit, to wit:

Am I talking bollocks?

It was a stupid weekend, in which I played three gigs in less than twenty-four hours on Friday night and Saturday during the day.

I really shouldn't have booked myself two gigs on the Friday. I should have turned down the second one. Mind you, I attempted to, but after I told the guy from the Australian band, 'well, no, I am actually playing earlier on that day' he replied that it was ok as he was booking a late set anyway, so I'd be able to come down after my first gig. My mistake was not to immediately reply 'fuck off', so I found myself roped into it.

So, unwillingly, and late, after the first gig on Friday night (to which the cello player failed to turn up in the end, being taken unwell) I tore myself away from the interesting band who had followed the interesting 20's / 30's / 40's jazz singer I'd stayed to see and buggered off to the 12-bar in Denmark Street.

Turned out that S. the new drummer, who was also running the night, had overbooked himself, and was trying to fit no less than eight bands into the evening, including us.

See if you can guess what his favourite drug is. Go on, guess.


We went on and played three songs, which was short and fun, and then I hit the bar heavily.

Which was when it started getting silly. Some random Italian guy who wanted to make everyone think he was Russian kept buying me shots of vodka, which it would have been as impolite to turn down as it would have been impolite not to down in one then whack the bar with the empty shot glass. But I should have guessed that it would all come to no good.

Pretty soon I was at the opposite end of the bar to random Italian guy for no particular reason, and found myself being flirted with outrageously by some random and cute Canadian girl. A., the singer from my band was in the vicinity, and then... bam. Canadian girl disappears. Oh no, there she is, flirting outrageously with A.

I'll have another drink then. Oh, hello cute Canadian girl. Why, you wish to flirt outrageously with me some more. That's fine by me. How are you doing anyway?

So she oscillated between the two of us for some time... at which point my memory becomes hazy.

Next thing I know I'm walking along Oxford Street yelling at A. down the phone.

Then I'm in Cafe Helen on Edgware Road eating shwarma.

Then I'm waking up at home next day in a panic, realising that no, it's ok, I have time to get to my piano gig that afternoon, so long as I get up and leave immediately.

And I don't know what happened.

I rang A. after the piano thing apologising and asking if I'd taken a swing at him or anything. He told me no, it was all cool and he too had been very drunk, I'd just upped and left, then sent him an angry text message, following which he'd rung.

I don't know. I'm kind of embarrassed. I have no idea why I am telling you all this.

Anyway. The other rock band I'm playing with have some new mp3s if you want to hear some loud rock type things.

And I have now been procrastinating from things I ought to be doing for even longer than before.

I'll stop now.

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MaFC, Silly Weekend, Stuff | 12 comments (12 topical, 0 hidden)
Possible but not necessary by onethreeseven (4.00 / 1) #1 Mon Apr 10, 2006 at 07:04:03 PM EST
As you point out it all depends on the behavior for number words beyond your standard set (say, out past a decillion).

It's trivially possible to concoct such words so the conjecture fails.  (Let the word for 1036 consist of exactly 1036 A's.)

It's likewise possible to concoct a system where the conjecture provably holds.  (Let the word for 103n, n > 11 be n Z's followed by "illion", and apply a little elbow grease to prove a logarithmic bound on the letter sum for N with sufficiently low constant.)

So no bollocks here, as far as I can tell; the answer is just "it depends".

exactly as i was going to say by 256 (4.00 / 1) #12 Tue Apr 11, 2006 at 06:18:14 AM EST
although i would also draw attention to the three key descriptors of the namespaces in this exercise.
  1. the letters are a larger counting base than the decimal numbers.
  2. letters are combined into words for numbers in a manner that is generally efficient but less than ideally so because:
    2a. letters represent sounds and thus can only be combined in ways that don't violate the rules of the spoken language

    2b. words describe more than just numbers and many of the most efficient letter combinations are used for non-number concepts

3. there is also that you are combining the values of the letters in a simple conflation whereas the numbers of course are being combined into a positional numeral system, arguably the most efficient way of using tokens to describe data.

Given these factors, it is hardly surprising that both of your collisions happen very close to each other.

I would expect to see the combined letter values start off slightly low (as the lowest ordinals are very old and common concepts and thus have short names) and then increase quickly somewhere shortly after ten. After this initial acceleration the letter values will climb in a steady step function, briefly dropping or leveling out at intervals such as "forty" or "one thousand" where combinatorial naming fails.

The number valueson the other hand will start lower but climb at a constant (and faster) rate.

It would be entirely reasonable for there to be a chance overlap in the very early numbers (consider how close the meaning of "a" is to "one") or at one of these brief dips at the break points. But the chance of an overlap at a breakpoint only exists when the values of the combined letters are higher than those of the numbers.

and once the numbers values exceed the letter values, only a conscious attempt by number namers could bring it back. the best way to assure yourself of this would be to have your perl script print a graph of the values, i expect you would see something like this:

and just looking at the shapes of the lines should make it clear that it would require an extremely large anomaly in the number names (much much larger than that that would ever be required) to bring the two lines back to intersection.

an interesting experiment would be to try the same thing using number names from different languages, I predict that your results would be very similar across all alphabet-based human languages.
I don't think anyone's ever really died from smoking. --ni

[ Parent ]
am I talking bollocks? by martingale (4.00 / 2) #2 Mon Apr 10, 2006 at 10:57:40 PM EST
I think that's fundamentally wrong, since bollocks are plural. In fact, thinking about it you may well have meant to say "Are I talking bollocks?", which of course is just as ungrammatical for a completely different reason, even though it now accords with the multiplicity of the object. Perhaps it should therefore be properly written as "Are we talking bollocks?", which now makes grammatical sense but has the unfortunate side effect of encompassing more than yourself, unless you happen to be called Liz in RL. Besides, everyone knows bollocks can't talk, as they don't have a larynx.

Which brings us to my final conjecture: the expression "am I talking bollocks" cannot be properly expressed in the English language as intended.
$E(X_t|F_s) = X_s,\quad t > s$

-1, doesn't mention Gödel [nt] by gazbo (4.00 / 1) #4 Tue Apr 11, 2006 at 12:58:11 AM EST

I recommend always assuming 7th normal form where items in a text column are not allowed to rhyme.

[ Parent ]
sqrt(-1) by martingale (4.00 / 1) #6 Tue Apr 11, 2006 at 01:15:22 AM EST
You can't bring in Goedel, he's in a consistent completion.
$E(X_t|F_s) = X_s,\quad t > s$
[ Parent ]
Also, your conjecture is not 100% true by gazbo (4.00 / 1) #5 Tue Apr 11, 2006 at 01:10:55 AM EST

the expression "am I talking bollocks" cannot be properly expressed in the English language as intended except by the queen.

I recommend always assuming 7th normal form where items in a text column are not allowed to rhyme.

[ Parent ]
dammit by martingale (4.00 / 1) #7 Tue Apr 11, 2006 at 01:17:06 AM EST
I'd forgotten about Camilla.
$E(X_t|F_s) = X_s,\quad t > s$
[ Parent ]
When I go out on the piss by nebbish (4.00 / 2) #3 Tue Apr 11, 2006 at 12:50:11 AM EST
I stick a Post-It note to my eye that says "No internet, email or text messaging after 5 pints." It's the only way.

It's political correctness gone mad!

the growth rate argument works. by rmg (4.00 / 1) #8 Tue Apr 11, 2006 at 04:07:22 AM EST
obviously, if there's not a word for a number, it cannot provide a counterexample to your conjecture.

[t]rolling retards conversation, period.
pronunciation of numbers by sasquatchan (4.00 / 2) #9 Tue Apr 11, 2006 at 04:48:26 AM EST
OK, I figure I might as well ask here..

Several grade school teachers hammered into our heads that using "AND" while giving a whole number (in words) is incorrect. "AND" should only be used to indicate a decimal point (eg cents or tenths/hundredths etc)

Thus, 259 would be "two hundred fifty nine" -- no "AND" in there.

Yet, I hear national broadcasters get that wrong quite regularly. I'm pretty sure my Strunk and White doesn't cover this anywhere...

Americanism by motty (4.00 / 1) #10 Tue Apr 11, 2006 at 04:56:57 AM EST
Americanism americanism americanism americanism americanism. Also, and childishly, la la la I can't hear you.

That's up there with omitting 'and' from newspaper headlines.

In English English, though, we say 'two hundred and fifty-nine'.

I amd itn ecaptiaghle of drinking sthis d dar - Dr T

[ Parent ]
(Comment Deleted) by yicky yacky (4.00 / 2) #11 Tue Apr 11, 2006 at 05:15:04 AM EST

This comment has been deleted by yicky yacky

[ Parent ]
MaFC, Silly Weekend, Stuff | 12 comments (12 topical, 0 hidden)