Circular lists

[gamo]

I want to learn an effient way of handle circular lists.

EXAMPLE:

I have a set of @set = qw(a a b b b c c c c);

In a loop...

@list = shuffle(@set);

and I want to know how many diferent circular lists could be generated.

Thanks, best regards



PS: The way I handle this is doing

$s = join '',@list;
$string = $s . $s;

and after that comparing $s with all the previous stored substr
$string,$i,9 for $i (0..9)

BUT this way is very inefficient for a large @set or long loop.

 

[C.DeRykus]

I want to learn an effient way of handle circular lists.

EXAMPLE:

I have a set of @set = qw(a a b b b c c c c);

In a loop...

@list = shuffle(@set);

and I want to know how many diferent circular lists could be generated.

Thanks, best regards

PS: The way I handle this is doing

$s = join '',@list;
$string = $s . $s;

and after that comparing $s with all the previous stored substr
$string,$i,9 for $i (0..9)

BUT this way is very inefficient for a large @set or long loop.

See: perldoc -q permute

 

[gamo]

See: perldoc -q permute

OK, supposing I have all the permutations,
supposing I remove the redundant ones with a hash,
how I detect the circular - identical ones?

Thanks

 

[RedGrittyBrick]

I want to learn an effient way of handle circular lists.

What do you mean by a circular list? AFAIK Perl only knows about plain
lists.

I'd first worry about correctness. I'd only worry about efficiency if
the result is measured to be unacceptably slow.

EXAMPLE:

I have a set of @set = qw(a a b b b c c c c);

That isn't what I'd call a set. To me a set can not contain multiple
identical elements. A set might be (a,b,c).

In a loop...

You can construct a for loop for the range from zero to the length of
@set minus one. Given a set a,b,c - the for loop could be used to yield
a b c
b c a
c a b
This might be the sort of thing you are looking for. I exclude b,c,a
because of the way I have interpreted your mention of "circular list".

@list = shuffle(@set);

and I want to know how many diferent circular lists could be generated.

Assuming you want to rotate an element off the end and prepend it at the
beginning (one interpretation of circularity), I'd maybe use splice.

Read `perldoc -f splice`

Thanks, best regards



PS: The way I handle this is doing

$s = join '',@list;
$string = $s . $s;

see `perlfoc -f push`

and after that comparing $s with all the previous stored substr
$string,$i,9 for $i (0..9)

That seems an unusual approach but unless you show code I'm unsure what
exactly you mean.

BUT this way is very inefficient for a large @set or long loop.

See above.

 

[Jürgen Exner]

I want to learn an effient way of handle circular lists.

EXAMPLE:

I have a set of @set = qw(a a b b b c c c c);

In a loop...

@list = shuffle(@set);

and I want to know how many diferent circular lists could be generated.

Circular lists are an implementation method in programming languages
with pointers (e.g. C or Pascal/Modula), where the end of a linked list
is linked back to its beginning. The advantage is that any algorithm
accessing this list does not need to special code the edge cases for
list beginning and list end, because every element is a middle element.

Yes, you can simulate linked lists (and thus circular linked lists) in
Perl by using references. But why do you want to do that?

And if you use arrays to implement a circular list (which is very easy
to do by always taking the modulo of the index and the length of the
list) then there are exactly lenght of array different representations
of the same circular list because each element could be the first
element of the array.

jue

 

[xhoster]

I want to learn an effient way of handle circular lists.

EXAMPLE:

I have a set of @set = qw(a a b b b c c c c);

In a loop...

@list = shuffle(@set);

Is this the shuffle from List::Util? If so, why use a random method
as part of determining a non-random result? If you want to inspect
every permutation, don't do it randomly.

But I don't know of an permuter which will be efficient in this case.
(By efficient, I mean computing each detectable permutation only once,
rather than computing permutations which are not detectably different
because they just swap the positions of two identical characters.)

and I want to know how many diferent circular lists could be generated.

So for this purpose, (a,b,c) and (b,c,a) are not different, as they close
to the same circle? You could canonicalize by insisting that a circular
list be represented by that equivalent linear list which is alphabetically
first. One consequences is that one of the 'a' would be fixed in the first
position, and would no longer need to be permuted. (but tested against the
other 'a' to see if it is first or second)

Thanks, best regards

PS: The way I handle this is doing

$s = join '',@list;
$string = $s . $s;

and after that comparing $s with all the previous stored substr
$string,$i,9 for $i (0..9)

BUT this way is very inefficient for a large @set

How to make it more efficient is highly dependent on what you want to do
once you detect the sameness. You haven't really spelled that out. It
will also depend on the nature of @set, i.e. how redundant the elements of
it are. Since the @set you show us isn't "large", it is hard to
extrapolate from your example up to your actual use case.

or long loop.

What loop is hypothetically long?

Xho

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[Jim Gibson]

I want to learn an effient way of handle circular lists.

What is your definition of a "circular list"?

EXAMPLE:

I have a set of @set = qw(a a b b b c c c c);

In a loop...

@list = shuffle(@set);

and I want to know how many diferent circular lists could be generated.

I think you want to know how many unique permutations may be generated
that are invariant under rotation (moving the last element to the front
of the list any number of times).

You can use the advice given in 'perldoc -q permute' to generate all
possible permutations. Because the elements in your example are not
unique, you will generate some permutations which are also not unique.
If this is actually the case, then you want an efficient way of
ignoring redundant permutations.

One way would be to attach a unique key to each member of your list,
e.g. qw( a1 a2 b3 b4 b5 c6 c7 c8 c9 ). Then, you only accept a
permutation that has each repeated element in order. For example ( a1,
a2, ... ) would be accepted but ( a2, a1, ... ) would be rejected. You
apply this test for all the a's, b's, and c's in the permutation. If
any are out of order, reject the permutation. To use the list, strip
off the keys.

For the circularity problem, each permutation can be rotated to produce
all of the equivalent cases. For N elements (9 in your sample case),
there will be N equivalent cases. To avoid generating the redundant
cases, select one element (e.g. a1 from your list), place it in the
first location, and generate all possible permutations of the remaining
N-1 elements, using the method described above if you have non-unique
elements. The result should be all possible circular lists (if I
understand your definition correctly).

No hashes needed!

 

[gamo]

What do you mean by a circular list? AFAIK Perl only knows about plain lists.
....
Given a set a,b,c - the for loop could be used to yield
a b c
b c a
c a b
This might be the sort of thing you are looking for. I exclude b,c,a because
of the way I have interpreted your mention of "circular list".

If a,b,c is a list
b,c,a is the same list rotated b,c,a,b,c,a (note the a,b,c)
c,a,b is the same too, because c,a,b,c,a,b (note the a,b,c)

a,c,b is another list because a,c,b,a,c,b is new

I expect this clarifies the problem.

Best regards

 

[gamo]

Is this the shuffle from List::Util? If so, why use a random method
as part of determining a non-random result? If you want to inspect
every permutation, don't do it randomly.

I don't want to inspect every permutation because the number of
permutations is n! = n*(n-1)*(n-2)...*1 and a problem of 20! is
intractable.

But I don't know of an permuter which will be efficient in this case.
(By efficient, I mean computing each detectable permutation only once,
rather than computing permutations which are not detectably different
because they just swap the positions of two identical characters.)

Suppose that I have the huge list of permutations in memory.
Wich are the circular rotations of another?

So for this purpose, (a,b,c) and (b,c,a) are not different, as they close
to the same circle? You could canonicalize by insisting that a circular
list be represented by that equivalent linear list which is alphabetically
first. One consequences is that one of the 'a' would be fixed in the first
position, and would no longer need to be permuted. (but tested against the
other 'a' to see if it is first or second)

That's right. If I have a list with only one member of type 'a',
I could stick that as the beginning of the chain, to compare with,
and to store.

...

How to make it more efficient is highly dependent on what you want to do
once you detect the sameness. You haven't really spelled that out. It
will also depend on the nature of @set, i.e. how redundant the elements of
it are. Since the @set you show us isn't "large", it is hard to
extrapolate from your example up to your actual use case.



What loop is hypothetically long?

Take this as an example (not tested)

#!/usr/local/bin/perl -w

use List::Util qw(shuffle);
@a = qw(a a a a a r r g g n);
for (1..10_000_000){
@set = shuffle(@a);
$s = join '',@set;
$two = $s . $s;
next if ($two =~ /gg/);
for $i (0..9){
$r = substr $two,$i,10;
$hash{$r}++;
if ($hash{$r}==1){
$counter++;
$p[$counter]=$r;
}
}
$ok=1;
for $j (1..$counter-10){
if ($s eq $p[$j]){
$ok=0;
last;
}
}
if ($ok==1){
$exito++;
print "$exito\n";
}
}
print "El número de permutaciones circulares es $exito\n";


__END__
I know that the solution by a previous version is 588.
But the meomory is eaten by the program.
Best regards,

 

[gamo]

Take this as an example (not tested)

#!/usr/local/bin/perl -w

use List::Util qw(shuffle);
@a = qw(a a a a a r r g g n);
for (1..10_000_000){
@set = shuffle(@a);
$s = join '',@set;
$two = $s . $s;
next if ($two =~ /gg/); $precounter=$counter;
for $i (0..9){
$r = substr $two,$i,10;
$hash{$r}++;
if ($hash{$r}==1){
$counter++;
$p[$counter]=$r;
}
}
$ok=1; for $j (1..$precounter){
if ($s eq $p[$j]){
$ok=0;
last;
}
}
if ($ok==1){
$exito++;
print "$exito\n";
}
}
print "El número de permutaciones circulares es $exito\n";


__END__
I know that the solution by a previous version is 588.

 

[xhoster]

I think you want to know how many unique permutations may be generated
that are invariant under rotation (moving the last element to the front
of the list any number of times).

You can use the advice given in 'perldoc -q permute' to generate all
possible permutations. Because the elements in your example are not
unique, you will generate some permutations which are also not unique.
If this is actually the case, then you want an efficient way of
ignoring redundant permutations.

One way would be to attach a unique key to each member of your list,
e.g. qw( a1 a2 b3 b4 b5 c6 c7 c8 c9 ). Then, you only accept a
permutation that has each repeated element in order. For example ( a1,
a2, ... ) would be accepted but ( a2, a1, ... ) would be rejected. You
apply this test for all the a's, b's, and c's in the permutation. If
any are out of order, reject the permutation. To use the list, strip
off the keys.

It might be possible to integrate that technique into the Fischer-Kause
algorithm so that only the unique ones are generated, rather than having
to generate, test, and discard. Now that would be some subtle code.

For the circularity problem, each permutation can be rotated to produce
all of the equivalent cases. For N elements (9 in your sample case),
there will be N equivalent cases. To avoid generating the redundant
cases, select one element (e.g. a1 from your list), place it in the
first location, and generate all possible permutations of the remaining
N-1 elements, using the method described above if you have non-unique
elements. The result should be all possible circular lists (if I
understand your definition correctly).

I think this is a good start, but it still produces redundant results.

For example, if @set = qw/A A C/; # and we number the elements as you did
above, then this method will return both:

A1 A2 C1
A1 C1 A2

Which are circularly the same once the digits are removed.

Xho

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[gamo]

It might be possible to integrate that technique into the Fischer-Kause
algorithm so that only the unique ones are generated, rather than having
to generate, test, and discard. Now that would be some subtle code.


I think this is a good start, but it still produces redundant results.

For example, if @set = qw/A A C/; # and we number the elements as you did
above, then this method will return both:

A1 A2 C1
A1 C1 A2

Which are circularly the same once the digits are removed.

Xho

Thanks to all the responders, but I think now that the problem is
unsolvable. Since you need to compare a new candidate with those of
the past, the problem is more or less O(n^2). No matter if the algo-
rithm is random or deterministic in the enumeration.

Best regards,

 

[Jürgen Exner]

If a,b,c is a list
b,c,a is the same list rotated b,c,a,b,c,a (note the a,b,c)
c,a,b is the same too, because c,a,b,c,a,b (note the a,b,c)

a,c,b is another list because a,c,b,a,c,b is new

I expect this clarifies the problem.

Are you confirming Red's understanding of the problem or are you
correcting his understanding? To me both versions, his and yours, seem
to be the same.

If they are, then generating all your "circular lists" is trivial, as is
computing their number.

Obviously there are exactly as many circular lists as there are elements
in the original list, because each element can become the first element
in a result list.

And you can generate them by using a running $index (0..$#list) and for
each result list concatenate the elements from $index to $#list with the
elements from 0 to $index-1.

jue

 

[sln]

And you can generate them by using a running $index (0..$#list) and for
each result list concatenate the elements from $index to $#list with the
elements from 0 to $index-1.

jue

Haven't seen any code yet. I think this is easier for you to write the
manipulation of ring buffers in English than to actually do it.

Anybody can profess how it can be done, unless you actually do it of course.
That separates the men from the boys.

sln

 

[sln]

Haven't seen any code yet. I think this is easier for you to write the
manipulation of ring buffers in English than to actually do it.

Anybody can profess how it can be done, unless you actually do it of course.
That separates the men from the boys.

sln

No problem bro, i'm writing a class that does it all

sln

 

[Tim Greer]

No problem bro, i'm writing a class that does it all

sln

You're replying to yourself again? Just so you know, you've replied to
your previous post saying you've not seen the code yet, and just
replied saying "no problem" and you're writing a class that does it
all?

 

[Rom]

You're replying to yourself again? Just so you know, you've replied
to your previous post saying you've not seen the code yet, and just
replied saying "no problem" and you're writing a class that does it
all?

Maybe it's a multiple identify disorder? That may sound like a joke but it
is in fact a real human ailment. I mean all the trolls over the years in
various groups likely suffer from one mental illness or another. I must
admit, in all my years, I cannot recall seeing someone reply to themselves
in this way. Of course it is possible he just was tired and made an error
when replying to someone else; it happens.

 

[Tim Greer]

Maybe it's a multiple identify disorder? That may sound like a joke
but it is in fact a real human ailment. I mean all the trolls over the
years in various groups likely suffer from one mental illness or
another. I must admit, in all my years, I cannot recall seeing someone
reply to themselves in this way. Of course it is possible he just was
tired and made an error when replying to someone else; it happens.

Actually, he replies to himself all of the time. I really don't mind
it, it just seems weird.

 

[xhoster]

I don't want to inspect every permutation because the number of
permutations is n! =3D n*(n-1)*(n-2)...*1 and a problem of 20! is
intractable.

There only that many permutations if your "set" has only unique letters,
which in your example it does not. Anyway, may gut feeling is that to get
an accurate count based on stochastic sampling, your will need to be about
as large as the underlying domain, anyway. But I could be wrong.

Suppose that I have the huge list of permutations in memory.
Wich are the circular rotations of another?=20

canonicalize! Or solve the problem analytically. If your set is such that
no permutation can have a self-rotation, then every circular list will
have exactly N linear representations.

Take this as an example (not tested)

#!/usr/local/bin/perl -w

use List::Util qw(shuffle);
@a =3D qw(a a a a a r r g g n);

I has able to solve this analytically, without any programming, because
of the uniqueness of 'n'. compute the permutations of 9 letters with
5,2,2 degeneracy, then treat gg as a single unit, computing perutations of
8 letters with 5,2,1 degeneracy, and subtract.

for (1..10_000_000){
@set =3D shuffle(@a);
$s =3D join '',@set;
$two =3D $s . $s;
next if ($two =3D~ /gg/);

Now here is a wrinkle you haven't shown us before. It will be devastating
to certain approaches to the problem.

snip of code which I don't understand the logic behind, and which doesn't
seem to work.

__END__
I know that the solution by a previous version is 588.

That's what I got analytically, too.

Xho

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[xhoster]

Thanks to all the responders, but I think now that the problem is
unsolvable.

For some input sizes and structures, clearly it is intractable. But the
same is true for nearly all problems.

Since you need to compare a new candidate with those of
the past, the problem is more or less O(n^2).

Where n is what, the size of @set, or the factorial of that size?

Xho

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