in reply to Re: Re: Better algorithm than bruteforce stack for combinatorial problems? (Take 3 and homage to A::L)
in thread Better algorithm than bruteforce stack for combinatorial problems?
The first argument to NestedLoops is the list of loops so
[ # First loop: all item indices [ 0..$#items ], # @items1 subsequent loops: ( sub { # If we need more items: $sum[@_] < $target # then get more (unique) item indices ? [ ($_+1)..$#items ] # else don't get more items : [] } ) x $#items, ],
becomes the equivalent of
for $_ ( 0..$#items ) { # ... for $_ ( @{ $sum[@_] < $target ? [ ($_+1)..$#items ] : [] } ) { # ... for $_ ( @{ $sum[@_] < $target ? [ ($_+1)..$#items ] : [] } ) { # ... } } }
The sub { ... } in the original code is required to delay the running of the loop computation code instead of running it before NestedLoops is called (at which point $_ and other variables wouldn't contain the rigth values).
The list of items computed by the nested loops is passed to the subs as @_ and the currently innermost loop's variable is also put into $_ so you can use that as shorthand for $_[1].
And this bit
OnlyWhen => sub { # Compute sum of selected items as # sum of @_1 items plus last item: $sum[@_]= $sum[$#_] + $items[$_[1]]; # Return subsets that match desired sum: return $sum[@_] == $target; },
just declares a sub that gets called to determine which lists to return. We'll pretend it is named when() below. And we'll replace the @_ in each $sum[@_] with a hardcoded value to simplify our 'translation' which becomes something close to:
@_= (); for $_ ( 0..$#items ) { push @_, $_; push @return, [ @_ ] if when( @_ ); for $_ ( @{ $sum[1] < $target ? [ ($_+1)..$#items ] : [] } ) { push @_, $_; push @return, [ @_ ] if when( @_ ); for $_ ( @{ $sum[2] < $target ? [ ($_+1)..$#items ] : [] } ) { push @_, $_; push @return, [ @_ ] if when( @_ ); # ... pop @_; } pop @_; } pop @_; }
But instead of pushing each selected list into @return, each call to $iter>() returns the next list that would be pushed.
Note that we loop over indices so we can use ($_+1)..$#items to only loop over indices that we haven't already looped over.
Let's simplify the inner loops. The point of
$sum[@_] < $target ? [ ($_+1)..$#items ] : []
is to avoid looping any deeper if we don't need more items to add up (because we've already reached our desired total). Which can be more clearly written in our translation as
next if $target <= $sum[@_];
(if we do our pops in continue blocks) so we can clarify our example to
@_= (); for $_ ( 0..$#items ) { push @_, $_; push @return, [ @_ ] if when( @_ ); next if $target <= $sum[1]; for $_ ( ($_+1)..$#items ) { push @_, $_; push @return, [ @_ ] if when( @_ ); next if $target <= $sum[2]; for $_ ( ($_+1)..$#items ) { push @_, $_; push @return, [ @_ ] if when( @_ ); # ... } continue { pop @_; } } continue { pop @_; } } continue { pop @_; }
Of course, we can't finish this translation because you can't write loops that nest to some arbitrary depth.
Fiinally, we use the iterator to get each desired set of indices. We use an array slice to convert the list of indices into a list of iitems:
while( my @list= @items[ $iter>() ] ) { warn "$target = sum( @list )\n"; }
I hope that helps explain how this works.
 tye


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Re: Re^3: Better algorithm than bruteforce stack for combinatorial problems? (explain)
by BrowserUk (Patriarch) on May 22, 2004 at 20:53 UTC  
by tye (Sage) on May 22, 2004 at 21:38 UTC 