Archive - Aug 31, 2008

Date

Complexity of remember physics equations

Posted August 31st, 2008 by Mgccl
in

I did some physics problem and now I see something non-desirable with physics:
Too many equations to remember.
Physicist creates so many words and letters to detonate trivial combination. Those increase the amount of stuff to be remembered.
Why do they still do it? Because it help one remember them easier.(Seems like a paradox but not...)

But I still don't like to remember stuff.

From now on
m=mass, t=time, l=length
Pressure. What a useless word.
What if I replace the word pressure with \frac{m}{lt^2}. For example the following questions.

In a system, an object with the area of 14m^2 are under 19pascal pressure, what is the force excreted on this object.

Express elementary way I like.

In a system, an object with the l^2 of 14m^2 are under 19 kg/(m·s^2) \frac{m}{lt^2}, what is the \frac{ml}{t^2} excreted on this object.

I found this much easier.

I don't know how many formulas I have to remember, but I think it's O(n^2)(maybe lower) if there are n physics terms.
I know it improves the speed of computation. It's hash table speed... O(1) for finding relationship between all those placeholders...

I like to reduce it to exactly n formulas. (Lower bound? maybe can be improved, for a lot more computation)
Convert w/e physics term into the 5 basic units. mass, time, length, charge and temperature. Remember that.
The computation is bad...
One have to know should one term multiply or divide another term, which will be consist of a few subtraction and addition because it's purely about cancellation.
This has to be done by trial and error. bring it to O(2^n) computation time. when there is only k terms involved in an equation, the worst number of trials are 2^(n-1).
an simple algorithm
List the exponents in for each basic unit present in the equation on lines. the last line is what we want in the result.

sign m l t
1 -3 1
2 2 1
1 0 1
2 -1 1

add signs so the sum of the rows will result the last row. from trial and error, a few simple techniques can be used to find the answer faster.

sign m l t
+ 1 -3 1
+ 2 2 1
- 1 0 1
2 -1 1

The formula is found, plug in the numbers!

Yes, it is in fact a hand based version of this computer science problem

Given: vectors v_1 to v_n
The solution is unique and exists.

Create an algorithm that can find c_1 to c_{n-1} that are either 1, 0 or -1 so
v_n = \sum _{k=1}^{n-1} c_kv_k

I think there is some improvement to the algorithm. If I can prove the following.

Suppose v_k(x) is the xth element in the vector1.
If the solution of the c's for v_n = \sum _{k=1}^{n-1} c_kv_k is unique.
Then the solution for the c's for v_n(x) = \sum _{k=1}^{n-1} c_kv_k(x) is also unique, for arbitrary x.
If [v_k] is the dimension of a vector2.
Then we expect to reduce to 1/[v_k] of the original running time..


Disproved. Counter example
given<1,a>,<1,b>, want <1,a>

Physics is not my thing. Don't like trivial applied math with experiments.

  1. 1. Is there a specific mathematical notation for that?
  2. 2. Again, any standard notation for that?
Honey Pot that kill bots