Solving expressions

jj2007 · December 12, 2010, 10:24:45 PM

Quote from: qWord on December 12, 2010, 09:24:21 PM
Quote from: jj2007 on December 12, 2010, 07:24:45 AMMaybe we could PM each other to exchange ideas and sources?
:thumbu, check your PMs

Thanks, impressive :U

qWord · December 12, 2010, 10:36:26 PM

Quote from: jj2007 on December 12, 2010, 10:24:45 PM
Thanks, impressive :U

thx :bg
however, it is not 100% finish (but usable).

BTW: REAL10-var's also work after calling

Code Select

fSlvSetGlbPrecision <REAL10>
or using the macro fSlv10

RuiLoureiro · December 13, 2010, 04:21:36 PM

Hi Dave,
I dont want to work in twos's complement but in sign-amplitude
we need two variables (not only one): _Sign and _Value
_Value can be any 32 bit value
With two's complement we use only 31 bits for value and 1 bit for signal
and we need only one variable, but i dont want this.

Did you try Calcula12 ? Does it work as expected ?

This is my procedure to solve any expression without ():

Quote
ResolveIt proc

mov esi, offset _TblOperation
xor ebx, ebx

_loop1: mov edi, offset _TblOperand
mov edx, dword ptr [esi + ebx*4]
cmp edx, 3 ; 1=Sub(-) 2=Add(+) 3=Mul(*) 4=Div(/) 5=Rem(\)
jb short _next1
;
push ebx
shl ebx, 4
invoke OperationAB,[edi + ebx + $SIGN1$], [edi + ebx + $OPR1$],
[edi + ebx + $SIGN2$], [edi + ebx + $OPR2$], edx
pop ebx
jc _end

mov dword ptr [esi + ebx*4], 0

call SaveAntRes
call SaveNxtRes
jnc short _next1
;
cmp _NoAnt, 0
je _Ok ; END
;
_next1: add ebx, 1
cmp ebx, dword ptr [esi - 4]
jb short _loop1

;---***---***---***---

xor ebx, ebx

_loop2: mov edi, offset _TblOperand
mov edx, dword ptr [esi + ebx*4]
cmp edx, 0
je short _next2
;
push ebx
shl ebx, 4
invoke OperationAB,[edi + ebx + $SIGN1$], [edi + ebx + $OPR1$],
[edi + ebx + $SIGN2$], [edi + ebx + $OPR2$], edx
pop ebx
jc _end

call SaveNxtRes
jnc short _next2
;
; END
; »»»
_Ok: invoke ToRclCnv, SignW, OperandW, addr ResultZ
clc
ret
;
_next2: add ebx, 1
cmp ebx, dword ptr [esi - 4]
jb short _loop2
; »»»»»»»»»»»»»»»»»»
; It cannot end here
; »»»»»»»»»»»»»»»»»»
call SetError2
stc
_end: ret
ResolveIt endp

qWord · December 13, 2010, 05:31:47 PM

hi RuiLoureiro,

maybe an bug or simply an limit?:

Code Select

(1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1)=ERROR

qWord

RuiLoureiro · December 13, 2010, 05:52:30 PM

Quote from: qWord on December 13, 2010, 05:31:47 PM
hi RuiLoureiro,

maybe an bug or simply an limit?:
Code Select Expand
(1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1)=ERROR
qWord

Hi qWord,
First, thank you
It is a limit of 20 operations if im correct
But i can change the table to operate with 500 or 1000... operations
The tables are defined to use only 20 operations

dd 20
dd 0 <-- current number of operations
_TblOperation dd 20 dup (?)

now works till 200 operations inside brackets

EDIT:
To solve this: -2 + (-2* (12 - 3*5) + 23 ) + 5

the 1ª expression is: -2+(-2*(12-3*5)+23)+5
and we calculate: 12-3*5 (= @ ) @=-3

now, the expression is: -2+(-2*@+23)+5
and we calculate: -2*@+23 (= @ ) @=29

now, the expression is: -2+@+5
at last the result is: 32

When we try to find an operand or an operator we need to test to find @
When we solve an expression with () we replace it with @
This is my linear method :green2

RuiLoureiro · December 14, 2010, 07:30:29 PM

Till now anything OK

the method used:

1- Read expression to _BufferX
2- Copy _BufferX to _BufferY and remove spaces and get
the buffer length Y
If we find @ than ERROR (this is used inside to replace...)

3- Scan _BufferY from right to left to find an expression
inside brackets - try to find first ")" and than "("
If not, exit to 6
4- Copy the expression to a WorkBuffer and than call
the procedure SolveExpress
5- Save the result on _BufferAT (from top to bottom)
Replace the expression by @
Start again until we have not brackets
6- Call SolveExpress to solve the last expression in _BufferY

SolveExpress: build 2 tables: operations and operands
whenever we find @, from bottom to top,
get the values from _BufferAT;
If it is a constant, convert it
When we have the tables, call ResolveIt

Note that the expression is solved from left to right
while we scan _BufferY from right to left;
We move values to _BufferAT from top to bottom
and retrieve from bottom to top.

If _BufferY="-(5-3)+2*((5-3)+2*(3-5))"

1º expression: -(5-3)+2*((5-3)+2*@) _BufferAT = -2;

2º expression: -(5-3)+2*(@+2*@) _BufferAT = -2;+2

3º expression: -(5-3)+2*@ _BufferAT = -2;

4º expression: +@+2*@ _BufferAT = -2; -2

last solution: -6

What do you think about ?
Is there anything wrong ? Tell me

RuiLoureiro · December 15, 2010, 04:18:23 PM

Hi

Quote
The first procedure to solve an expression is this:

If _BufferY is defined like

dd 120
dd 0 <--- number of characters
_BufferY db 121 dup (0)

invoke LookExpress, addr _BufferY to solve the expression in _BufferY

Quote
; _TblOperandAt is an array of 2 dwords to save sign
; and result of each operation (each @)
;
LookExpress proc pRcl:DWORD

mov edx, offset _TblOperandAt
mov dword ptr [edx - 4], 0
;
_start: mov esi, pRcl
mov ecx, dword ptr [esi - 4]
;
cmp ecx, 0
jne short _loop1
;
error1: call SetError
stc
ret

_loop1: sub ecx, 1
movzx eax, byte ptr [esi + ecx]
cmp al, ')'
je short _found1
;
cmp ecx, 0
jne short _loop1
;
; not found -> solve the last expression
; »»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»

invoke SolveExpress, esi
ret
;---
_found1: cmp ecx, 0
je short error1
;
mov _OffsetLst, ecx
;---
_loop2: sub ecx, 1
movzx eax, byte ptr [esi + ecx]
cmp al, '('
je short _found2

cmp al, ')'
je short _found1

cmp ecx, 0
jne short _loop2
je error1
;---
_found2: mov _OffsetFst, ecx

mov ebx, _OffsetLst
sub ebx, ecx
sub ebx, 1

cmp ebx, 0
je short _rem
;
cmp ebx, 1
ja short _found3
;
; Constant
;
mov ebx, _OffsetFst
add ebx, 1
movzx eax, byte ptr [esi + ebx]

cmp al, '@'
je short _rem

cmp al, '0'
jb error1

cmp al, '9'
ja error1
;
; Remove ()
; »»»»»»»»»
_rem: mov ebx, _OffsetFst
mov byte ptr [esi + ebx], 20h
mov ebx, _OffsetLst
mov byte ptr [esi + ebx], 20h
invoke RemSpaces, esi
jmp _start
;---------------------------------------
_found3: mov edi, offset WrkBuffer

push esi

add esi, _OffsetFst
add esi, 1

mov dword ptr [edi - 4], ebx
mov byte ptr [edi + ebx], 0

@@: sub ebx, 1
movzx eax, byte ptr [esi + ebx]
mov byte ptr [edi + ebx], al
jnz short @B

pop esi

; »»»»»»»»»»»»»»»»»»»»»»»»»»»
; Wrk is constant ?
; »»»»»»»»»»»»»»»»»»»»»»»»»»»
invoke VrfOperand, edi, [edi - 4]
jnc _rem ; yes

cmp eax, 1
je error1
;
; Operation
; »»»»»»»»»
invoke SolveExpress, edi
jc error1
; »»»»»»»»»»»»»»»»»»»»»»
; Replace (...) by AT
; »»»»»»»»»»»»»»»»»»»»»»
invoke RemExpress, esi

call AdjustSign

call SaveAt

jmp _start
LookExpress endp

RuiLoureiro · December 15, 2010, 06:50:44 PM

Try Calcula14 and see all expressions one by one

RuiLoureiro · December 16, 2010, 07:03:13 PM

Now, Calcula15 solve expressions like 2^5+2^4+2^3+2^2+2^1+2^0
or 2^3+2*(5-3)+10

RuiLoureiro · December 17, 2010, 09:06:49 PM

The last Calcula15 has a bug: 2^3-2^3 = 16 !!!

Now, Calcula15 solve expressions like: 2^5+2^4+2^3+2^2+2^1+2^0
or 2^3+2*(5-3)+10 or (((2^3)))+1 or -(-2^3+1)-5 -2^3+2^3 ...

It is under test. Did you find another bug ?

RuiLoureiro · December 18, 2010, 07:32:02 PM

The last Calcula15 has a bug: a procedure VrfOperand proc pRcl:DWORD, Len:DWORD
exits with ret 4 instead of ret 8

Calcula16 does (1-2*3)^3 but not (2^3)^3 (the base cannot be an exponential)
and (-5)^3 or (-5)^2 -5^3 (the exponent should be 0 to +31 )

it does any integer expression with
Sub (-), Add (+), Mul (*), Div (/), Remainder (\), Exp(^)
i hope !

New Calcula16 because ---3^3 doesnt give error

RuiLoureiro · December 19, 2010, 05:41:15 PM

I am working now in Calcula16...meanwhile

I decided to show all Calcula14 project (it doesnt make exponentials)
Calcula14.zip is = .asm, .inc, .glb & .exe
You can assemble it: Use Console Assemble & Link
Express your opinion or comments if you want

RuiLoureiro · December 20, 2010, 06:20:58 PM

the linear method used in Calcula16:

1- Read expression to _BufferX

2- Copy _BufferX to _BufferY and remove spaces and get
the buffer length Y

Remove some types of expressions (to have less problems after):
this (x+y..z) is = x+y...z
this (((x+y)))+z is = (x+y)+z

Test some types of errors (to have less problems after)
... +-(x+y) is ERROR
... -+-x+y is ERROR

Now, START

invoke LookExponent:
3- Scan _BufferY from right to left to find the operator ^
If not, exit to 5

4- For each operator ^, find exponent and base ( base^exponent )

If base is an expression, copy it to BufferB
and «invoke LookExpress, addr BufferB» to get the base

Save the result on _TableEX (from top to bottom) and replace
the expression "base^exponent" by #
Go on (goto 3)

invoke LookExpress:
5- Scan _BufferY from right to left to find an expression
inside brackets - try to find first ")" and than "("
If not, exit to 8

6- Copy the expression to a WorkBuffer and than call
the procedure SolveExpress

7- Save the result on _TableAT (from top to bottom)
Replace the expression by @
Start again until we have not brackets (goto 5)

8- Call SolveExpress to solve the last expression in _BufferY

9- The result: the value is in OperandW and sign in SignW

10- Convert it to string _ResultZ

11- Show string _ResultZ

SolveExpress: build 2 tables: operations and operands
whenever we find @, from bottom to top,
get the values from _TableAT;
whenever we find #, from bottom to top,
get the values from _TableEX;

If it is a constant, convert it
When we have the tables, call ResolveIt

ResolveIt gives the result in OperandW and SignW

We can exit from a procedure with 3 types of results:
a) correct (exit clc)
b) error (exit stc EAX=0)
c) overflow (exit stc EAX=1)

Note that each expression is solved by SolveExpress
from left to right
while we scan _BufferY from right to left;

We move values to _TableAT or _TableEX from
top to bottom and retrieve them from bottom to top.

_TableAt and _TableEx work like a stack: when we
scan we PUSH; when we want the values, we POP

Anyone has any other idea to add.
Do you want help us
EDIT:
The procedures doesnt solve expressions with exponentials or powers inside
brackets. The result cannot be correct

oex · December 21, 2010, 10:45:11 AM

This link might be of use to you

http://www.php.net/manual/en/language.operators.precedence.php

RuiLoureiro · December 21, 2010, 04:42:26 PM

Thanks oex,
That link say nothing to me

Calcula16 has a problem of brackets precedence
If the expression has not brackets, the result is correct
One simple example is this: 2^4 + (2*2^3) + 2^2 = 36 (Calcula16 gives 44)
I am working on Calcula16 to solve this problem

News:

Solving expressions