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+--- Thread: From addition to multiplication? (/thread-3124.html)
RE: From addition to multiplication? - greatgamer34 - 04-18-2014
Why, just why,,,,,, didnt i think of that?!!!
NICE JOB AFTYPOO!
RE: From addition to multiplication? - Iceglade - 04-18-2014
(04-18-2014, 10:45 PM)AFtExploision Wrote: I got it. You have two numbers, m and n. You do (10^m * 10^n) [Ten to the m times ten to the n]. Now, log base 10 of the answer to that equation will be m+n. (x^a * x^b = x^a+b), as algebra taught us. The base can be any number, I picked 10 because you could do powers of 10 on paper, and it is simple to tell if a number is a nice log of 10, (a log that returns a whole number), so this method is very friendly to use without a computer. (But on paper, you could draw a line n length, a line m length so that lines n and m form a longer line, and find the length of the longer line to get m + n, but my log way is computer-friendly, you'd have to add to find the endpoints so you could calculate the lines if you tried to implement it in code). IDK why none of us thought of this, its pretty simple actually. Now give me a prize!
...so much for vector rotation :p
RE: From addition to multiplication? - AFtExploision - 04-18-2014
(04-18-2014, 11:02 PM)Iceglade Wrote:[17:00:49] Aft-pan: I wrote like 500 words on using vector rotation to figure it out(04-18-2014, 10:45 PM)AFtExploision Wrote: I got it. You have two numbers, m and n. You do (10^m * 10^n) [Ten to the m times ten to the n]. Now, log base 10 of the answer to that equation will be m+n. (x^a * x^b = x^a+b), as algebra taught us. The base can be any number, I picked 10 because you could do powers of 10 on paper, and it is simple to tell if a number is a nice log of 10, (a log that returns a whole number), so this method is very friendly to use without a computer. (But on paper, you could draw a line n length, a line m length so that lines n and m form a longer line, and find the length of the longer line to get m + n, but my log way is computer-friendly, you'd have to add to find the endpoints so you could calculate the lines if you tried to implement it in code). IDK why none of us thought of this, its pretty simple actually. Now give me a prize!
...so much for vector rotation :p
[17:01:00] Aft-pan: then I realized that vector rotation uses addition
[17:01:08] Alex Jarvis: Haha!
[17:01:17] Aft-pan: Then I tried cross-products
[17:01:21] Aft-pan: Couldnt get it to work
[17:01:35] Aft-pan: tried translation, realised pc's used addition to translate
RE: From addition to multiplication? - David - 04-19-2014
Whay didn't I think of that? What's your adress aft? I'll send you a cake.
RE: From addition to multiplication? - AFtExploision - 04-19-2014
Your closet
RE: From addition to multiplication? - David - 04-19-2014
(04-19-2014, 04:32 PM)AFtExploision Wrote: Your closet
Couldn't find you. I left the cake there, though.
RE: From addition to multiplication? - greatgamer34 - 04-19-2014
When I checked my closet.... There was a cake in there.... Creepy. I'll send it to Aft!
RE: From addition to multiplication? - Konstacon - 04-20-2014
Oh, you can be cheap and do the determinant of:
|m -n|
|1 1|
which is equal to m-(-n) or m+n... pretty cheap i guess.
RE: From addition to multiplication? - Iceglade - 04-20-2014
(04-20-2014, 12:43 AM)snugglycreeper9 Wrote: Oh, you can be cheap and do the determinant of:
|m -n|
|1 1|
which is equal to m-(-n) or m+n... pretty cheap i guess.
I don't think you can calculate a determinant without addition/subtraction.
RE: From addition to multiplication? - AFtExploision - 04-20-2014
(04-20-2014, 01:03 AM)Iceglade Wrote:(04-20-2014, 12:43 AM)snugglycreeper9 Wrote: Oh, you can be cheap and do the determinant of:
|m -n|
|1 1|
which is equal to m-(-n) or m+n... pretty cheap i guess.
I don't think you can calculate a determinant without addition/subtraction.
Which is why I ditched the vector rotation, as far as I know to calculate the coordinates after rotation you need to add / subtract sines and cosines