A Conversation for Combinatorial fortytwology (5 * 270 / 32 = 42?) [Work in progress]
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Ratio Rules?
vogonpoet (AViators at A13264670) Started conversation Jun 27, 2001
Hey Argon0 or Iaoth...Are you guys ever going to finish this aricle? Go on, please .
I want to know more concerning having the number one at the start and end of a researcher number...
Are these both allowed? They are rather similar, but I think valid...
(1 x 7)(4 x 3)/ (2 x 1) = 42
((1 x 7)(4 x 3)/ 2) / 1 = 42
Whereas I think these constitue just one way of making 42:
1 x (74 - (32 x 1))
(1 x 74) - (32 x 1)
But then what about
1 x (74 - 32) x 1 ?????
And can I use
1^7 + 42 - (2 x 1)??
(Am not sure about validity of the seventh power)
The whole multiply my 2 by 1 at the end versus divide all by 1 business has rather mamoth effect on my 42 count - I seem to have a ridiculous amount of ways of making 42....
Ratio Rules?
iaoth Posted Jun 27, 2001
To be technically correct, this is actually my Entry. (Does my ego look big in this dress? )
However, the 42 game is Argon0's invention, and I have no idea what he considers valid 42isms. I just accept everything when I construct 42isms in order to calculate 42-count and 42-ratio. On the other hand, I think you're right; some 42isms are so similar they're essentially the same. Some are actually mathematically equivalent.
For example, the first two 42isms in your posting:
(1 x 7)(4 x 3)/ (2 x 1) = 42
((1 x 7)(4 x 3)/ 2) / 1 = 42
I would count these as two different 42isms, but they're actually mathematically equivalent; they're just two different ways of writing the same equation.
X / (Y x Z) = (X / Y) / Z
Note that this is only true when using real values. In the integer domain, you may very well get different results.
I'd say that 1 x (74 - 32) x 1 is the same as the two before it. Essentially, all you're doing in all of them is subtract 32 from 74.
1^7 + 42 - (2 x 1) doesn't equal 42. It's 41. 1^7 = 1, so you're adding 1 to 42 and then subtracting 2.
And I'll have to agree that your Researcher number seems to have a high 42-count. I'll have to check it out when I get my computer back. Apparently, my processor is too hot for some reason...
Ratio Rules?
iaoth Posted Jun 27, 2001
Oh, I see now that it's a typo. You meant 1^7 + 43 - (2 x 1). I guess all this talk about 42 threw you off. Yeah, that'll give you 42 alright.
I wonder why I got that "<winkeye>" in the previous posting. Must've been the parenthesis. Let's see if I can reproduce it.
)
The above line should be
)
without the space.
Ratio Rules?
iaoth Posted Jun 28, 2001
Huh? Sorry, I can't parse that. Note that I write "< winkeye >" (without the spaces), btw.
Ratio Rules?
vogonpoet (AViators at A13264670) Posted Jun 28, 2001
So typing ; - ) without the spaces doesnt work for you? (Sorry, I am a bit slow taking this in )Hmm, seems to work for me (: - / ??)
Ratio Rules?
iaoth Posted Jun 28, 2001
; - ) works, but ( < w i n k e y e > ) doesn't, for some reason. Try writing < winkeye > within parentheses.
Test
iaoth Posted Jun 28, 2001
WEIRD. They must've just fixed the bug or something!
However, it's kind of a cheap workaround. They're just adding a space after the smiley. Look at the previous posting.
Test
vogonpoet (AViators at A13264670) Posted Jun 29, 2001
Surely cheap workarounds are what this guide is all about - with a pioneering spirit and a few rolls of gaffer tape we can mend the world.
Test
vogonpoet (AViators at A13264670) Posted Jul 11, 2001
Well after only having to wait a week for Argon0 to notice my query on 42isms, my front page now includes 97 different combinational 42isms . Now all I need to do is work out how many combinations are possible using / * - + sqrt ^ and ! operators and I will be able to calculate my 42 ratio so far... of course, if you got round round to actually finishing that potential filled article, maybe I wouldnt have to work out the number of combos, which is frankly daunting me
Test
iaoth Posted Jul 11, 2001
No, please, feel free to do all the boring calculations for me. As you can see, I've got it worked out for a simplified 42-game, but the more complex is.. well .. complex.
I think I can do it though, just give me some time and I'll probably get around to it.
Test
vogonpoet (AViators at A13264670) Posted Jul 11, 2001
I would try to do it now, but the whole minus sign operator thang is kind of holding me back...
Test
vogonpoet (AViators at A13264670) Posted Jul 11, 2001
It was indeed a bad bad confusing footnote - so confusing that it banished all math comprehension from my soul . Luckily I have now recovered enough to not be confused, alhtough still not feeling desperate to work it all out... for a general equation, I guess operators that require one parameter will require a different term from the ones requiring two, ! has one parameter, and sqrt, I think leaving out trig functions for now is reasonable, not just because I havent used any ...
Test
iaoth Posted Jul 11, 2001
Hmm, that sounds like a simplification (what an ugly word that is). To calculate how many combinations there are, you have to count how many choices you have at each point. For example, you could put a '-' or a '+' in front of the first digit. The same goes for all of the following digits, except you can use multiplication, division and concatenation there as well. So you get 2 x 5^(n - 1), where n is the number of digits. (Now that I think about it, I suspect that I made a mistake in the Entry, I'll have to check that.)
Now, if you want to add '!' and 'sqrt' in there, it gets much more complicated. In that case, we can have any combination of '!' and 'sqrt' for each digit (for example, sqrt(sqrt(x!!)!)!!). In fact, we'll have to add '-' to that list as well: -sqrt(sqrt(-(-x!)!)!!). Since we can decide in which order to perform these operators using parentheses, we can think of this as a string where 's' means 'square root': "-!-!s!!s-" for the most recent example.
Suddenly, the concept of 42-ratio loses all meaning. Let's see why, shall we? Of course, the string mentioned above can be of ~any~ length. We can add '!' or 'sqrt' for as long as we like, which immediately means that there are an infinite number of ways to combine a number of digits using these rules. However, if we limit the length of the string in some way, we can reach some kind of result. Say that the string can only be m characters long. Then the number of combinations is the sum 3^1 + ... + 3^(m - 1) + 3^m, which is 3^(m + 1) - 1. We have this number of choices for every digit besides the five choices of operators mentioned earlier, which means we now have 2 x 5^(n - 1) x (3^(m + 1) - 1)^n.
And we haven't even begun talking about 'to the power of y'. We'd have to limit that one as well, because we clearly have an infinite number of possible values of y there...
So basically, in the real 42-game, everyone has a 42-ratio of x divided by infinity, which is a number that doesn't exist. We have to set up some kind of limits to have a number of possible combinations other that infinity. That kinda sucks, but I think it's rather reasonable.
Test
iaoth Posted Jul 11, 2001
"3^1 + ... + 3^(m - 1) + 3^m, which is 3^(m + 1) - 1"
I'm not so sure about this part, on second thought. The sum is correct, but I don't think it adds up to 3^(m + 1) - 1.
Test
vogonpoet (AViators at A13264670) Posted Jul 13, 2001
Ive been thinking about it - I tried not to, as as I already said, its too complicated for me to work on, especially after too much alchohol, however, I might think properly tomorrow, as I have to stay sober tonight, I really do
I might even do it when I am not tired if I am lucky - it is about 20 days since I was last not tired, but I have great plans for sleep tonight
Key: Complain about this post
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Ratio Rules?
- 1: vogonpoet (AViators at A13264670) (Jun 27, 2001)
- 2: iaoth (Jun 27, 2001)
- 3: iaoth (Jun 27, 2001)
- 4: iaoth (Jun 27, 2001)
- 5: vogonpoet (AViators at A13264670) (Jun 28, 2001)
- 6: iaoth (Jun 28, 2001)
- 7: vogonpoet (AViators at A13264670) (Jun 28, 2001)
- 8: iaoth (Jun 28, 2001)
- 9: iaoth (Jun 28, 2001)
- 10: iaoth (Jun 28, 2001)
- 11: vogonpoet (AViators at A13264670) (Jun 29, 2001)
- 12: iaoth (Jun 29, 2001)
- 13: vogonpoet (AViators at A13264670) (Jul 11, 2001)
- 14: iaoth (Jul 11, 2001)
- 15: vogonpoet (AViators at A13264670) (Jul 11, 2001)
- 16: iaoth (Jul 11, 2001)
- 17: vogonpoet (AViators at A13264670) (Jul 11, 2001)
- 18: iaoth (Jul 11, 2001)
- 19: iaoth (Jul 11, 2001)
- 20: vogonpoet (AViators at A13264670) (Jul 13, 2001)
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