Yet another Fork question (not for the faint of heart)

[Guest]

Archived from groups: rec.games.trading-cards.magic.rules (More info?)

Alright, take a deep breath. Here goes:

Player A casts Lightning Bolt at Player B.
Player B responds with Fork (Call it Fork1) targetting the Lightning Bolt.
Player B further responds with Fork (Call it Fork2) targetting Fork1.
Player B further responds with Counterspell targetting the Lightning Bolt.

So, the stack (from the top down) looks like this:

Counterspell
Fork2
Fork1
Lightning Bolt

Here's what I think happens - please confirm or correct me:

Counterspell resolves, removing Lightning Bolt from the stack. The stack now
looks like this

Fork2 (targetting Fork1)
Fork1 (targetting Lightning Bolt)

Fork2 resolves, adding a new Fork (Fork3) to the top of the stack. Let's
assume that Player B chooses Fork1 as a target. The stack now looks like:

Fork3 (targetting Fork1)
Fork1 (targetting Lightning Bolt)

Fork3 resolves, adding a new Fork to the stack (Fork4). Fork1 is chosen as a
target. The stack is now:

Fork4 (targetting Fork1)
Fork1 (targetting Lightning Bolt)

Now let's say that both players' hands are empty and they've exhausted all
mana. What the heck happens? Is the game a draw? The stack will always look
like this:

ForkX (targetting Fork1)
Fork1 (targetting Lightning Bolt)


My brain hurts - now it's your turn.

--
KB

 

[Guest]

Archived from groups: rec.games.trading-cards.magic.rules (More info?)

On Mon, 27 Jun 2005 15:53:14 -0400, Briscobar <youcant@sendmespam.com> wrote:
>Alright, take a deep breath. Here goes:

You can't scare me:
http://www.vic.com/~dbd/NFd/exp.analyzed/interrupt.timing.example.RV.Interrupts

>Player A casts Lightning Bolt at Player B.

Okay. LB on the stack.

A then passes. Then:

>Player B responds with Fork (Call it Fork1) targetting the Lightning Bolt.

Okay. Fork1 on the stack on top of LB, targetting LB.

>Player B further responds with Fork (Call it Fork2) targetting Fork1.
>Player B further responds with Counterspell targetting the Lightning Bolt.

Which he can do because he gets priority back each time.

>So, the stack (from the top down) looks like this:
>
>Counterspell
targetting LB
>Fork2
targetting Fork1
>Fork1
targetting LB
>Lightning Bolt
targetting A

>Here's what I think happens - please confirm or correct me:

Well, it gets complex as the Fork2 is resolving, with a possibility that
most players don't see for some reason.

>Counterspell resolves, removing Lightning Bolt from the stack.

And itself of course.

>The stack now looks like this
>
>Fork2 (targetting Fork1)
>Fork1 (targetting Lightning Bolt)
>
>Fork2 resolves, adding a new Fork (Fork3) to the top of the stack. Let's
>assume that Player B chooses Fork1 as a target.

Which is fine. But is NOT THE ONLY LEGAL CHOICE.

>The stack now looks like:
>
>Fork3 (targetting Fork1)
>Fork1 (targetting Lightning Bolt)

Sure; this is a loop, and can go around as many times as B wants.

Is it a mandatory loop? No. Why? Because, each time, the RESOLVING Fork is
ALSO a legal target for the newly-created Fork; it's not off the stack yet,
because it's NOT done resolving. It's about to be, but it isn't yet. So,
each time, "the Fork that created me" is a legal choice also.

Thus, this is a loop with an optional component; B must choose a number of
times to go around it, then after that break it. Which translates to "he
can repeat this N times, where he chooses N, but after that he must choose
to target the Fork that's creating the new Fork, which ends up stopping the
loop because THAT Fork is no longer there when the new Fork tries to resolve".

>Now let's say that both players' hands are empty and they've exhausted all
>mana. What the heck happens? Is the game a draw?

Nope. You've missed one of the options.

IF there were no other options, and nobody broke the loop, then yes, it would
be a draw. But there is another option, each time, and eventually B must take
that option, by our loop rules.

Dave
--
\/David DeLaney posting from dbd@vic.com "It's not the pot that grows the flower
It's not the clock that slows the hour The definition's plain for anyone to see
Love is all it takes to make a family" - R&P. VISUALIZE HAPPYNET VRbeable<BLINK>
http://www.vic.com/~dbd/ - net.legends FAQ & Magic / I WUV you in all CAPS! --K.

 

[Guest]

Archived from groups: rec.games.trading-cards.magic.rules (More info?)

On 27 Jun 2005, David DeLaney wrote:

[snip]
>>The stack now looks like:
>>
>>Fork3 (targetting Fork1)
>>Fork1 (targetting Lightning Bolt)
>
>Sure; this is a loop, and can go around as many times as B wants.
>
>Is it a mandatory loop? No. Why? Because, each time, the RESOLVING Fork is
>ALSO a legal target for the newly-created Fork; it's not off the stack yet,
>because it's NOT done resolving. It's about to be, but it isn't yet. So,
>each time, "the Fork that created me" is a legal choice also.
[snip]

This is only true if the topmost fork remains on the stack while it
resolves...

At what point during resolution does an item leave the stack?

i.e. Could you cite the rule number?

Thanks,
Gene P.
Slidell LA


--
Alcore Nilth - The Mad Alchemist of Gevbeck
alcore@uurth.com

 

[Guest]

Archived from groups: rec.games.trading-cards.magic.rules (More info?)

On Mon, 27 Jun 2005, Briscobar wrote:

> Alright, take a deep breath. Here goes:
>
> Player A casts Lightning Bolt at Player B.
> Player B responds with Fork (Call it Fork1) targetting the Lightning Bolt.
> Player B further responds with Fork (Call it Fork2) targetting Fork1.
> Player B further responds with Counterspell targetting the Lightning Bolt.
>
> So, the stack (from the top down) looks like this:
>
> Counterspell
> Fork2
> Fork1
> Lightning Bolt
>
> Here's what I think happens - please confirm or correct me:
>
> Counterspell resolves, removing Lightning Bolt from the stack. The stack now
> looks like this
>
> Fork2 (targetting Fork1)
> Fork1 (targetting Lightning Bolt)
>
> Fork2 resolves, adding a new Fork (Fork3) to the top of the stack. Let's
> assume that Player B chooses Fork1 as a target. The stack now looks like:
>
> Fork3 (targetting Fork1)
> Fork1 (targetting Lightning Bolt)
>
> Fork3 resolves, adding a new Fork to the stack (Fork4). Fork1 is chosen as a
> target. The stack is now:
>
> Fork4 (targetting Fork1)
> Fork1 (targetting Lightning Bolt)
>
> Now let's say that both players' hands are empty and they've exhausted all
> mana. What the heck happens? Is the game a draw? The stack will always look
> like this:
>
> ForkX (targetting Fork1)
> Fork1 (targetting Lightning Bolt)
>
>
> My brain hurts - now it's your turn.

Ok, let's have a look at Fork:

Fork
{R}{R}
Instant
Put a copy of target instant or sorcery spell onto the stack, except
that it copies ~this~'s color and you MAY choose new targets for the
copy.

So player B *may* choose new targets for the copy but he doesn't have
to. So player B has to choose a number, and will choose a new target
for the copy that many times. After that he can't choose a new target
so the copy will be targetting `the Lightning Bolt'. Since this is an
illegal target it's countered on resolution.

--
David

 

[Guest]

Archived from groups: rec.games.trading-cards.magic.rules (More info?)

In newsine.GSO.4.63.0506272211040.19645@blade016.cs.vu.nl,
David de Kloet <dskloet@few.vu.nl> rambled:
>
> Ok, let's have a look at Fork:
>
> Fork
> {R}{R}
> Instant
> Put a copy of target instant or sorcery spell onto the stack, except
> that it copies ~this~'s color and you MAY choose new targets for the
> copy.
>
> So player B *may* choose new targets for the copy but he doesn't have
> to.

Right. Nothing requires him to choose a new target.

> So player B has to choose a number, and will choose a new target
> for the copy that many times.

I don't understand. Do you mean that because this is a loop, he'll have to
choose a number (say, 10,000) instead of going through the motions 10,000
separate times? Alright, that's fine.

> After that he can't choose a new target
> so the copy will be targetting `the Lightning Bolt'. Since this is an
> illegal target it's countered on resolution.

After the 10,000th Fork (Fork10,000) resolves, it creates a new copy of Fork
(Fork10,001). This Fork must have a target, and the only legal target is
Fork1. He can't choose the Lightning Bolt, because it has long since been
removed from the stack by a Counterspell, so it's not there to copy. The
only legal target is Fork1 (the first one cast, that targetted the Lightning
Bolt - it still hasn't resolved because all the Forks keep getting added to
the top of the stack), so I think Player B *has* to choose Fork1.

Maybe I'm misunderstanding what you're saying - care to run it by me again?

--
KB

 

[Guest]

Archived from groups: rec.games.trading-cards.magic.rules (More info?)

One of the voices in my head - or was it Gene P.? - just said...
>
> On 27 Jun 2005, David DeLaney wrote:
>
> [snip]
> >>The stack now looks like:
> >>
> >>Fork3 (targetting Fork1)
> >>Fork1 (targetting Lightning Bolt)
> >
> >Sure; this is a loop, and can go around as many times as B wants.
> >
> >Is it a mandatory loop? No. Why? Because, each time, the RESOLVING Fork is
> >ALSO a legal target for the newly-created Fork; it's not off the stack yet,
> >because it's NOT done resolving. It's about to be, but it isn't yet. So,
> >each time, "the Fork that created me" is a legal choice also.
> [snip]
>
> This is only true if the topmost fork remains on the stack while it
> resolves...
>
> At what point during resolution does an item leave the stack?
>
> i.e. Could you cite the rule number?

413.2h A spell is put into play from the stack under the control of the
spell's controller (for permanents) or is put into its owner's graveyard
from the stack (for instants and sorceries) AS THE FINAL STEP OF THE
SPELL'S RESOLUTION.

(Emphasis mine)

 

[Guest]

Archived from groups: rec.games.trading-cards.magic.rules (More info?)

Gene P. <alcore@uurth.com> wrote:
>On 27 Jun 2005, David DeLaney wrote:
>>Sure; this is a loop, and can go around as many times as B wants.
>>
>>Is it a mandatory loop? No. Why? Because, each time, the RESOLVING Fork is
>>ALSO a legal target for the newly-created Fork; it's not off the stack yet,
>>because it's NOT done resolving. It's about to be, but it isn't yet. So,
>>each time, "the Fork that created me" is a legal choice also.
>[snip]
>
>This is only true if the topmost fork remains on the stack while it
>resolves...

....Okay, haven't had to say this in a while, but: Please don't post here
AND email me the exact same thing. I have to answer posts here so everyone
can see the answer, but I read email -first-, and have already typed you out
a reply which I sent to you alone. And now I have to retype it all, and I
hate hate hate having to do that. I'm perfectly happy to answer email
questions, don't get me wrong - but I'd really rather not turn around and
ALSO have to answer here.

Correct, that is what occurs.

>At what point during resolution does an item leave the stack?

After it is finished having its effect. See 212.5b, 212.7b, 217.4a, 401.7, and
413.2h . The 'effect' of a creature, enchantment, or artifact spell is
simply "put the card into play as a creature / enchantment / artifact".

So yes, the Fork that's creating the copy is still on the stack at the
time it creates the copy ... because it's not done with its effect yet and
hasn't finished resolving. So it's a legal target if the copy is also a Fork,
though it's very soon going to stop being one.

>i.e. Could you cite the rule number?

As above.

Dave
--
\/David DeLaney posting from dbd@vic.com "It's not the pot that grows the flower
It's not the clock that slows the hour The definition's plain for anyone to see
Love is all it takes to make a family" - R&P. VISUALIZE HAPPYNET VRbeable<BLINK>
http://www.vic.com/~dbd/ - net.legends FAQ & Magic / I WUV you in all CAPS! --K.

 

[Guest]

Archived from groups: rec.games.trading-cards.magic.rules (More info?)

Briscobar <youcant@sendmespam.com> wrote:
>> After that he can't choose a new target
>> so the copy will be targetting `the Lightning Bolt'. Since this is an
>> illegal target it's countered on resolution.
>
>After the 10,000th Fork (Fork10,000) resolves, it creates a new copy of Fork
>(Fork10,001). This Fork must have a target,

Yep. Well ... nope. WHILE Fork10,000 resolves, it creates Fork10,001. It
does not wait until it's done resolving, THEN create Fork10,001 after it's
fully finished resolving by "time delay" somehow.

>and the only legal target is Fork1.

Nope. Fork10000 is still on the stack; it's not done resolving yet. It _is_
a legal target, though very soon it will stop being one.

>Maybe I'm misunderstanding what you're saying - care to run it by me again?

No - you're just overlooking the other possibility. (Don't feel bad - nearly
everyone does, even if they -know- the rules about resolving spells inside and
out, for some reason.)

Dave
--
\/David DeLaney posting from dbd@vic.com "It's not the pot that grows the flower
It's not the clock that slows the hour The definition's plain for anyone to see
Love is all it takes to make a family" - R&P. VISUALIZE HAPPYNET VRbeable<BLINK>
http://www.vic.com/~dbd/ - net.legends FAQ & Magic / I WUV you in all CAPS! --K.

 

[Guest]

Archived from groups: rec.games.trading-cards.magic.rules (More info?)

In news:slrndc0nda.dak.dbd@gatekeeper.vic.com,
David DeLaney <dbd@gatekeeper.vic.com> rambled:
>
<snip thourough explanation>

Thanks, David. I did indeed overlook the possibility of targetting the
"parent Fork", if you will. I like to think I'm good on the rules, but this
one slipped by me. I'm aware that the spell is on the stack until it's put
in the graveyard (or into play), since that's the last step of resolving
spells. I just overlooked the possibility of targetting a nearly-resolved
spell. Oh well. Now I know.

Thanks.

--
KB

 

[Guest]

Archived from groups: rec.games.trading-cards.magic.rules (More info?)

On Mon, 27 Jun 2005, Briscobar wrote:

> In newsine.GSO.4.63.0506272211040.19645@blade016.cs.vu.nl,
> David de Kloet <dskloet@few.vu.nl> rambled:
>>
>> Ok, let's have a look at Fork:
>>
>> Fork
>> {R}{R}
>> Instant
>> Put a copy of target instant or sorcery spell onto the stack, except
>> that it copies ~this~'s color and you MAY choose new targets for the
>> copy.
>>
>> So player B *may* choose new targets for the copy but he doesn't have
>> to.
>
> Right. Nothing requires him to choose a new target.
>
>> So player B has to choose a number, and will choose a new target
>> for the copy that many times.
>
> I don't understand. Do you mean that because this is a loop, he'll have to
> choose a number (say, 10,000) instead of going through the motions 10,000
> separate times? Alright, that's fine.
>
>> After that he can't choose a new target
>> so the copy will be targetting `the Lightning Bolt'. Since this is an
>> illegal target it's countered on resolution.
>
> After the 10,000th Fork (Fork10,000) resolves, it creates a new copy of Fork
> (Fork10,001). This Fork must have a target, and the only legal target is
> Fork1. He can't choose the Lightning Bolt, because it has long since been
> removed from the stack by a Counterspell, so it's not there to copy.

He doesn't *choose* Lighting Bolt. He just doens't choose a new
target. As you say above:
"Nothing requires him to choose a new target."
So the copy will have the same target as the original which is the
illegal Lightning Bolt.

As David mentioned there's also the possibility to target the
resolving Fork which I overlooked. (And it seems he is overlooking the
possibility I mentioned ;-) ).

--
David

 

[Guest]

Archived from groups: rec.games.trading-cards.magic.rules (More info?)

David de Kloet <dskloet@few.vu.nl> wrote:
>He doesn't *choose* Lighting Bolt. He just doens't choose a new
>target. As you say above:
>"Nothing requires him to choose a new target."
>So the copy will have the same target as the original which is the
>illegal Lightning Bolt.
>
>As David mentioned there's also the possibility to target the
>resolving Fork which I overlooked. (And it seems he is overlooking the
>possibility I mentioned ;-) ).

Oh, since the original Fork is still patiently targetting the long-gone
Lightning Bolt - sure, yes. (I pointed that all of these three were there
out to Scott on the judgelist not too long ago, and now I'm caught in the
same overlooking trap...) Three! Three lovely possible targets! Ah, ah, ah!

Dave
--
\/David DeLaney posting from dbd@vic.com "It's not the pot that grows the flower
It's not the clock that slows the hour The definition's plain for anyone to see
Love is all it takes to make a family" - R&P. VISUALIZE HAPPYNET VRbeable<BLINK>
http://www.vic.com/~dbd/ - net.legends FAQ & Magic / I WUV you in all CAPS! --K.

 

[Guest]

Archived from groups: rec.games.trading-cards.magic.rules (More info?)

Briscobar <youcant@sendmespam.com> wrote:
>Thanks, David. I did indeed overlook the possibility of targetting the
>"parent Fork", if you will. I like to think I'm good on the rules, but this
>one slipped by me. I'm aware that the spell is on the stack until it's put
>in the graveyard (or into play), since that's the last step of resolving
>spells. I just overlooked the possibility of targetting a nearly-resolved
>spell. Oh well. Now I know.

This is also the basis of a trick that uses a Deflection or Misdirection or
the like to counter a counterspell: use the Deflection to change the
Counterspell's target to? The Deflection itself...

Dave
--
\/David DeLaney posting from dbd@vic.com "It's not the pot that grows the flower
It's not the clock that slows the hour The definition's plain for anyone to see
Love is all it takes to make a family" - R&P. VISUALIZE HAPPYNET VRbeable<BLINK>
http://www.vic.com/~dbd/ - net.legends FAQ & Magic / I WUV you in all CAPS! --K.