Question

Converting prefix

Asked by: letharion

I have this expression:
(- (- (+ (+ I19 I02) (- I22 I50)) (* I13  I21)) (* I13 I21))

Which is written with prefix. I'm converting it into a syntax that looks something like:
f += I19 + I02
f += I22 - I50
f -= I13 * I21
f -= I13 * I21

The code I've started working with below looks for the first ) it can find, backtracks to the nearest (, and cuts that expression.
The problem is that once that's been done twice, I've ended up with this:
(- (- (+) (* I13  I21)) (* I13 I21))
inwhich the (+) doesn't make sense. I'm not sure how to proceed. I fear that my solution will turn out to have 25 if-exception rules everyone at it will look horrible.

I'm hoping someone who's done something similar could point me in the right direction.
To make it worse I'm gonna need to handle unary operators too, like sin/cos.

Any general guidelines?

while(true) {
   leftBrace = rightBrace = 0;
   for(int j = 0; j < P.length; j++) {
      if(P[j] == '(') leftBrace = j;
      else if(P[j] == ')') {
         rightBrace = j;
         break;
      }
   }
   if(leftBrace <= 0 || rightBrace <= 0) return;
   else {
       P = P[0..leftBrace - 1] ~ P[rightBrace + 1..P.length];
   }
}

                                  
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Asked On
2009-02-26 at 08:17:40ID24180545
Tags

Prefix Infix

Topics

D Programming Language

,

C / C++ / C# Editors and IDEs

,

Algorithms

Participating Experts
1
Points
500
Comments
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Answers

 

by: NovaDenizenPosted on 2009-02-26 at 09:33:57ID: 23747180

You're trying to compile the expression into a series of operations with only one temporary variable.  You can't do that for expressions in general.  You need a stack or some form of recursion so that you can support an arbitrary number of temporary variables.

 

by: NovaDenizenPosted on 2009-02-26 at 09:43:39ID: 23747301

This is typically done with a recursive function like this:

double eval_expr(expr_node *expr, var_state *vars) {
    switch (expr->type) {
       case EXPR_VAR: return read_variable(var_state, expr->var_name);
       case EXPR_CONST: return expr->const_value;
       case EXPR_OP:  
               switch (expr->op->op_name) {
                     case OP_ADD:  return eval_expr(expr->left, vars) + eval_expr(expr->right, vars);
                     case OP_MUL:  return eval_expr(expr->left, vars) * eval_expr(expr->right, vars);
                    ..... and so on
                }
    }
}

There are also nice ways to do it using objects and polymorphism.

class VarState {
     double var_value(string varname) {.....}
}

abstract class ExprNode {
   abstract double eval(VarState state);
}

class AddNode {
    ExprNode mLeft, mRight;
   AddNode(ExprNode left, ExprNode right) {
     mLeft = left;
     mRight = right;
   }
   double eval(VarState state) {
      return mLeft.eval(state) + mRight.eval(state);
   }
}

 

by: letharionPosted on 2009-02-27 at 02:34:29ID: 23754300

NovaDenizen: Thanks :)
Recursion is probably a good idea.

So I'm lets see if I'm on the right track.
To begin with, such a recurion would start from the outside, as opposed to my approach from the inside.

(- (- (+ (+ I19 I02) (- I22 I50)) (* I13  I21)) (* I13 I21)) Would be evaluated by two function-calls as
(- (+ (+ I19 I02) (- I22 I50)) (* I13  I21))      -          (* I13 I21)          The following function-calls becomes
(+ (+ I19 I02) (- I22 I50))     -           (* I13  I21)
(+ I19 I02)         +    (- I22 I50)

And from there all expressions have been broken down enough to be evaluated, correct so far?
The function would start returning instead, and then returning the new "format" that I wish to acquire?

 

by: letharionPosted on 2009-02-27 at 02:42:48ID: 23754338

Hmm, I'm looking at the code, and thinking that I don't quite understand why you used double as the return value.

Could it be that the expression is actually calculated directly? Because that's not my intention. I need to translate from one programs output to another programs input format so to speak.

 

by: letharionPosted on 2009-02-27 at 07:25:53ID: 23756408

Here's what I've produced so far.
I works well for ECJformat1, but fails with ECJformat2 because there are unary operators nestled in eachother. I guess I need a stack to keep track of those.

Any suggestions on making the code "easier on the eyes" would be mostly appreciated.
Opening this in a month (when for example If's are gonna be added) is likely to yield a lot of confusion :P

import std.stdio;
 
int main() {
   string ECJformat1 = "(- (- (+ (+ I19 I02) (- I22 I50)) (* I13 I21)) (* I13 I21))";
   string ECJformat2 = "(- (- (exp (+ I52 -0.32028294)) (cos (sin I29))) (+ (log 4.7557163) (cos (exp I05))))";
 
   string ELformat = eval_expr(ECJformat2);
   writefln();
   writefln(ECJformat2);
   writefln(ELformat);
   return 0;
}
 
string eval_expr(string expr, char lastOp = '+') {
   expr = expr[1..expr.length - 1];    //Strip outer ()
   string op1, op2;
   bool binary = false, lastBinary = false;
   char operator = expr[0];
   if(lastOp == '+' || lastOp == '-' || lastOp == '*' || lastOp == '/') lastBinary = true;
   if(opCode == '+' || opCode == '-' || opCode == '*' || opCode == '/') {
      expr = expr[2..expr.length];     //Remove binary opCode from expr
      binary = true; 
   } else expr = expr[4..expr.length]; //Remove unary opCode
 
   if(expr[0] == '(') {    //There is an inner expression, split it
      int closes = 0;
      foreach(pos, ch; expr) {
         if(ch == '(') closes++;
         else if(ch == ')') { 
            if(--closes == 0) {
               /*if(binary) writefln("OP1: " ~ expr[0..pos+1] ~ "\tOP2: " ~ expr[pos+2..expr.length]);
               else writefln("OP1: " ~ expr);
               writefln();*/
               op1 = eval_expr(expr[0..pos+1], operator);                        //Operand 1/Left
               if(binary) op2 = eval_expr(expr[pos+2..expr.length], operator);   //Operand 2/Right
               break;
            }
         }
      }   
   } else {             //No inner expr, convert to ELformat
      string ELexpr; 
      if(lastBinary) {
         foreach(pos, ch; expr) {
            if(ch == ' ') ELexpr = "f[0] " ~ lastOp ~ "= " ~ expr[0..pos] ~ operator ~ expr[pos+1..expr.length] ~ "\n";
         }
      } else ELexpr = "f[0] " ~ operator ~ "= " ~ lastOp ~ "(" ~expr ~ ")\n";
 
      return ELexpr;
   }
   return op1 ~ op2;
}

                                              
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by: NovaDenizenPosted on 2009-02-27 at 09:12:32ID: 23757569

I don't think you have a sound algorithm yet.  Does your code handle this?
(+ (sin a1) (cos a2))
To handle this, there is no way of getting around that fact that you need two temporary variables.  It looks like your method is limited to one temporary variable.

From looking at your code, I'm guessing that f is an array that could be used as a stack for evaluating expressions.

So the expression above could be converted to:
f[0] = sin(a1)
f[1] = cos(a2)
f[0] += f[1]

Am I right so far?

 

by: letharionPosted on 2009-03-11 at 08:28:39ID: 23858430

You're perfectly correct :)

I decided that it might be easier to attack the problem at another stage, which is before it's converted to somewhat-human-readable-format.

At that stage it looks like
105 205 4 204 204 2 .....
Below is the code that parses this, and it works well. However, it's no longer in D. I might rewrite in D as an interesting exercise, but right now I've got other things to attend to :-)

The code is somewhat ugly/hard to read, so any suggestions on improving it would be appreciated.
I'll credit your second post with points since you that's the idea/code that got me started :)

void translate(vector<float> PFProgram) {
   vector<float>::iterator step = PFProgram.begin();
   int nextF = 201, del = 0;
   ofstream m_outputFile;
   m_outputFile.open("out2.stat", fstream::app);
   ostringstream opL, opR, op, IFProgram;
   while(*step != 5) {
      if(*step == 1) *step++;
      else if(*step < 100) {
         op << OpToStr(*step);
         IFProgram << OpToStr(nextF) << " = ";
         //The ERCs cause some problems. If the left OP is ERC, we need to move one extra back, but if the right OP is ERC, we need to move one back AND move the left check one back.
         if(*(step - 2) == 1) {
            opR << *(step - 1);
            del = 1;
            if(*(step - 4) == 1) {
               del = 2;
               opL << *(step - 3);
            } else opL << OpToStr(*(step - 3));
         } else {
            opR << OpToStr(*(step - 1));
            if(*(step - 3) == 1) {
               opL << *(step - 2);
               del = 1;
            } else opL << OpToStr(*(step - 2));
         }
         IFProgram << opL.str() << " " << op.str() << " " << opR.str() << ";" << endl;
         *(step) = nextF++;
         PFProgram.erase(step - (2 + del), step);
         del = 0;
         opL.str("");
         opR.str("");
         op.str("");
         step = (PFProgram.begin() - 1);
      }
      *step++;
   } IFProgram << "f[0] = " << OpToStr(--nextF) << ";";
   m_outputFile << IFProgram.str() << endl << endl << "-------------------------" << endl << endl;
}
                                              
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by: letharionPosted on 2009-03-11 at 08:29:33ID: 31551688

Thanks a lot for showing me a correct direction :)

 

by: NovaDenizenPosted on 2009-03-11 at 10:29:54ID: 23859880

Again, I really can't follow your code but here's how I'd structure it.

 

by: NovaDenizenPosted on 2009-03-11 at 11:31:03ID: 23860579

So numbers < 100 are operators, what are the other numbers?  Is it a postfix representation?

20120131-EE-VQP-002

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