Scratch

Actually, I am revising my statement.  I just realized that it doesn't matter where we put the multi-letter ops, because they happen first anyway.  So, they're not really ops, they're more like numbers. 

amcerbu wrote:

scimonster wrote:

@amcerbu: Is "1" the first thing? According to OOO, it goes "PEMDAS" - Parentheses, exponents, multiplication/division, addition/subtraction. I want to put our multi-letter ops with ^s.

Sorry for being unclear.  The system I described ranks operations of higher precedence (things that happen first) with higher numbers. 

amcerbu wrote:

Code:

Operator:   ^ * / + - %
Precedence: 4 2 2 1 1 3
Operators with higher precedence happen first - exponentiation, then modulo, multiplication/division, and addition/subtraction

Since % (shortcut for modulo) has precedence 3, it will execute after exponentiation, but before multiplication.  That means that its arguments are only the numbers immediately before and after.  5 + 3 % 4 * 2 will evaluate 5 + ( 3 % 4 ) * 2.  However, 3 % 4 ^ 2 will evaluate 3 % ( 4 ^ 2 ).  I would like to say again that the user must format multi-letter operations like this: sqrt(x), sin(x).  That way, the value of the multi-letter operation is taken care of before moving on.  Think about this:

Code:

//In this representation, position up or down indicates bracket hierarchy.  
//The highest line is the innermost bracket set.  
//In our project, the program will evaluate the diagram "top to bottom."  
                                    3      
               4 + 5           log(   )    
      2 / sin(       ) + sqrt(          )
4 + (                                     )

Together, this makes:
4 + ( 2 / sin( 4 + 5 ) + sqrt( log( 3 ) ) )

I listed multi-letter operations with a precedence of 2 because, in fact, they will execute before the rest of the "line" on which the appear.  Consider 2 / sin( 4 + 5 ).  sin( 4 + 5 ) will be calculated before performing the 2 / ___ part. 

Hope that clears up what I said earlier.

EDIT:  Finished exams!  I can start helping out on this project.  I'll be free for about the next 3 weeks.

Last edited by amcerbu (2011-05-28 03:06:59)

I'm planning on using nested if-else blocks and a variable. Then I will repeatedly check if the character (insert variable here) is a parenthesis and the character after that is a parenthesis (or bracket). I will just be checking for implied multiplication. Should I use an item of a list instead of a variable?

Have I quit, or am i just inactive on here?

ssss wrote:

Have I quit, or am i just inactive on here?

And, when you claimed to quit...
I think that's when you were labeled as inactive and your job given to scratcher7_13.

@Hardmath123:

Your code previously was correct, that the program will scan the sub-expression from left to right and evaluate as it goes.  However, there is one problem.

Exponents have right-to-left association, meaning that

a^b^c = a^(b^c) ≠ (a^b)^c

So, for the exponents only, the program will have to scan right-to-left, not left-to-right.

Last edited by amcerbu (2011-05-29 18:13:27)

So, the correct form is

a^(b^c)

I just wanted to make sure that was made clear before someone started on the computation part.

amcerbu wrote:

So, the correct form is

Code:

a^(b^c)

I just wanted to make sure that was made clear before someone started on the computation part.

Okay, I want to help make the computation part, but the issue is, where is the base?

bbbeb wrote:

amcerbu wrote:

So, the correct form is

Code:

a^(b^c)

I just wanted to make sure that was made clear before someone started on the computation part.

Okay, I want to help make the computation part, but the issue is, where is the base?

Well, technically, the base is b, and then a.  The evaluator would do the following for something like 2^3^4

2^3^4
Start at end, scan until find ^
     Check terms before and after ^ 
          Return 3, 4 
     Perform 3^4
          Return 81
          Replace 3^4 in 2^3^4 with 81
     
Start at end, scan until find ^
     Check terms before and after ^
          Return 2, 81
     Perform 2^81
          Return 2.417e24
          Replace 2^81 with 2.417e24

Move on...

Better?

EDIT: For a sub-expression not containing any brackets (so, lowest possible level), the number of necessary repetitions for any given operator is the number of that operator.  Consider the following:

1 + 4 * 3 / 2 ^ 2

Since it contains no sub-expressions (brackets), we can evaluate it immediately. 

First, we scan right to left and find there to be 1 exponentiation.  So, we repeat the exponentiation sequence above 1 time. 

1 + 4 * 3 / 4

Then, we scan for multiplication or division; there are two.  We scan left to right this time, first simplifying 4 * 3 to 12.  On the second repetition, we perform 12 / 4 and simplify the expression further.

1 + 3

Now, there is only one addition or subtraction operation, so we complete the simplification process, and are left with...

4

Last edited by amcerbu (2011-05-31 00:56:47)

That could work for after we get all the special forms.
Do you program Squeak? XD
When you wrote "t" did you mean "t1"?

By the way, the script above could easily be simplified, but I'm too lazy.

Or, amcerbu's free, he can do it.
amcerbu, job for you!!!

scratcher7_13 wrote:

I'm sorry, but I won't be able to do anything with Scratch for the next month. I'll be too busy. Maybe ssss can do this job anyway.

So, what were you attempting to do?