OpenEuphoria

2. Re: Better Way to Do This

• Posted by bernie Aug 06, 2009
• 1144 views
• Last edited Aug 07, 2009

function no_atoms( sequence s ) 
integer i = 1 
	for t=1 to length(s) do 
		if atom(i) then return 0 end if 
	end for 
	return 1 
end function 

Would this be faster ?

function no_atoms( sequence s ) 
integer i = length(s) 
while(sequence(s[i]) and i > 0 ) do 
  i -= 1 
end while 
if i = 0 then return 1 else return 0 end if 
end function 

4. Re: Better Way to Do This

• Posted by DerekParnell (admin) Aug 06, 2009
• 1143 views
• Last edited Aug 07, 2009

function no_atoms( sequence s ) 
integer i = 1 
	for t=1 to length(s) do 
		if atom(i) then return 0 end if 
	end for 
	return 1 
end function 

There is a Eu v4 library routine for this. It's in std/types.e

public type sequence_array( object x ) 
    if not sequence(x) then  
        return 0 
    end if 
	 
    for i = 1 to length(x) do 
        if not sequence(x[i]) then  
            return 0 
        end if 
    end for 
    return 1 
end type 

Note that this also allows empty sequences so if you must have at least one element you can use it like this ...

  if length(S) and sequence_array(S) then  
     ... 

By the way, your code above wouldn't work as you don't update the 'i' variable.

6. Re: Better Way to Do This

• Posted by euphoric (admin) Aug 06, 2009
• 1125 views
• Last edited Aug 07, 2009

function no_atoms( sequence s ) 
integer i = length(s) 
while(sequence(s[i]) and i > 0 ) do 
  i -= 1 
end while 
if i = 0 then return 1 else return 0 end if 
end function 

Yeah, that's faster all right! Here's the program I used to test the speeds. Bernie, your suggested code was waaaaaay faster than any other. It should definitely replace our std/lib sequence_array()!

include std/console.e 
 
atom t 
 
function no_atoms_1( sequence s ) 
integer i = 1 
	for t=1 to length(s) do 
		if atom(s[t]) then return 0 end if 
	end for 
	return 1 
end function 
 
function no_atoms_bernie_1( sequence s ) 
integer i = length(s) 
	while sequence(s[i]) and i > 0 do 
		i -= 1 
	end while 
	if i=0 then return 1 else return 0 end if 
end function 
 
function no_atoms_bernie_2( sequence s ) 
integer i = length(s) 
	while sequence(s[i]) and i > 0 do 
		i -= 1 
	end while 
	return not i 
end function 
 
function no_atoms_stdlib( object x ) 
    if not sequence(x) then 
        return 0 
    end if 
 
    for i = 1 to length(x) do 
        if not sequence(x[i]) then 
            return 0 
        end if 
    end for 
    return 1 
end function 
 
sequence a 
a = repeat({},500) 
a &= 1 
 
t = time() 
for x=1 to 100000 do 
	no_atoms_1( a ) 
end for 
?time()-t 
 
t = time() 
for x=1 to 100000 do 
	no_atoms_bernie_1( a ) 
end for 
?time()-t 
 
t = time() 
for x=1 to 100000 do 
	no_atoms_bernie_2( a ) 
end for 
?time()-t 
 
for x=1 to 100000 do 
	no_atoms_stdlib( a ) 
end for 
?time()-t 
 
wait_key() 

The no_atoms_bernie_2() was just my testing of the "not" thingie. Didn't help.

I do have a question. When i gets to zero, why doesn't the test for sequence(s[i]) fail? Or is 0 a valid sequence element?!

8. Re: Better Way to Do This

• Posted by DerekParnell (admin) Aug 06, 2009
• 1128 views
• Last edited Aug 07, 2009

Sorry to burst the bubble, but the test is flawed.

Bernie's routines start from the back of the sequence and yours and mine from the front. Your test data had the atom as the last item so naturally, bernie's would find it first. Plus you forgot to reset the timer for the last test.

Here is some revised code.

include std/console.e 
 
atom t 
 
function no_atoms_1( sequence s ) 
integer i = 1 
	for t=1 to length(s) do 
		if atom(s[t]) then return 0 end if 
	end for 
	return 1 
end function 
 
function no_atoms_bernie_1( sequence s ) 
integer i = length(s) 
	while sequence(s[i]) and i > 0 do 
		i -= 1 
	end while 
	if i=0 then return 1 else return 0 end if 
end function 
 
function no_atoms_bernie_2( sequence s ) 
integer i = length(s) 
	while sequence(s[i]) and i > 0 do 
		i -= 1 
	end while 
	return not i 
end function 
 
function no_atoms_stdlib( sequence x ) 
    for i = 1 to length(x) do 
        if not sequence(x[i]) then 
            return 0 
        end if 
    end for 
    return 1 
end function 
 
sequence a 
-- Put atom in the middle. 
a = repeat({},250) & 1 & repeat({},250) 
 
t = time() 
for x=1 to 100000 do 
	no_atoms_1( a ) 
end for 
?time()-t 
 
t = time() 
for x=1 to 100000 do 
	no_atoms_bernie_1( a ) 
end for 
?time()-t 
 
t = time() 
for x=1 to 100000 do 
	no_atoms_bernie_2( a ) 
end for 
?time()-t 
t = time() 
for x=1 to 100000 do 
	no_atoms_stdlib( a ) 
end for 
?time()-t 
 
wait_key() 

I get these results ...

c:\temp>eui test 
0.5 
0.641 
0.656 
0.469 

Good question ... don't know the answer yet.

10. Re: Better Way to Do This

• Posted by euphoric (admin) Aug 06, 2009
• 1149 views
• Last edited Aug 07, 2009

Thank you Derek and Fernando for your diligent coder brains and hawk-like eyes.

I could understand if the test for i was first, because it would short circuit, right? Hmmmmm.

Take your time. You'll figure it out.

12. Re: Better Way to Do This

• Posted by LarryMiller Aug 06, 2009
• 1122 views
• Last edited Aug 07, 2009

include std/console.e 
 
atom t 
 
function no_atoms_1( sequence s ) 
integer i = 1 
	for t=1 to length(s) do 
		if atom(s[t]) then return 0 end if 
	end for 
	return 1 
end function 
 
function no_atoms_bernie_1( sequence s ) 
integer i = length(s) 
	while sequence(s[i]) and i > 0 do 
		i -= 1 
	end while 
	if i=0 then return 1 else return 0 end if 
end function 
 
function no_atoms_bernie_2( sequence s ) 
integer i = length(s) 
	while sequence(s[i]) and i > 0 do 
		i -= 1 
	end while 
	return not i 
end function 
 
function no_atoms_stdlib( sequence x ) 
    for i = 1 to length(x) do 
        if not sequence(x[i]) then 
            return 0 
        end if 
    end for 
    return 1 
end function 
 
sequence a 
-- Put atom in the middle. 
a = repeat({},250) & 1 & repeat({},250) 
 
t = time() 
for x=1 to 100000 do 
	no_atoms_1( a ) 
end for 
?time()-t 
 
t = time() 
for x=1 to 100000 do 
	no_atoms_bernie_1( a ) 
end for 
?time()-t 
 
t = time() 
for x=1 to 100000 do 
	no_atoms_bernie_2( a ) 
end for 
?time()-t 
t = time() 
for x=1 to 100000 do 
	no_atoms_stdlib( a ) 
end for 
?time()-t 
 
wait_key() 

c:\temp>eui test 
0.5 
0.641 
0.656 
0.469 

The answer is simple. i never reaches 0, the loop exits when i is 1. I discovered this when I changed the data to include only sequences. i then reached 0 and produced the usual error.

14. Re: Better Way to Do This

• Posted by bernie Aug 07, 2009
• 1146 views

function no_atoms( sequence s ) 
integer i = 1 
	for t=1 to length(s) do 
		if atom(i) then return 0 end if 
	end for 
	return 1 
end function 

function no_atoms( sequence s ) 
integer i = length(s) 
while(sequence(s[i]) and i > 0 ) do 
  i -= 1 
end while 
if i = 0 then return 1 else return 0 end if 
end function 

Sorry I went to bed and realized I put the short-circuit wrong.

It should of been this:

function no_atoms( sequence s ) 
integer i = length(s) 
while( i > 0 and sequence(s[i]) ) do -- <----<< 
  i -= 1 
end while 
if i = 0 then return 1 else return 0 end if 
end function 

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