Ace, King and Queen
Place a Ace (A), a Queen (Q) and a King (K) on three slots in a row like AQK (A on extreme left, K on extreme right). Reverse the position of the cards (ie, KQA) in the least possible number of moves. In a valid move, a card can be moved either left or right into an empty slot or placed onto a card of a higher rank. (eg, a Q can be placed on a K, but not vice-versa). Also, only the top card of the stack can be moved. Now, what is the least number of moves required if:
(a) there are only three slots?
(b) there is an empty slot to the left of the Ace?
(c) there is an empty slot to the right of the King?
(a) there are only three slots?
(b) there is an empty slot to the left of the Ace?
(c) there is an empty slot to the right of the King?
Labels: logic, mathemagic
posted by Rajesh Lal at
12:45 PM
23 Comments:
It says...
"In a valid move, a card can be ...
....placed onto a card of a higher rank. (eg, a A can be placed on a K, but not vice-versa)."
A is not higher than a K is it?
This seems to contradict itself.
Only 3 slots? No!
It looks like it's describing 5 slots, if theres and empty one left of the A and another right of the K.
You only need 4 slots.
Put the king in the empty slot beside the ace, move the ace where the king was, then place the king where the ace was.
This is a preliminary guess before I go to a group meeting, but the way I see it, regardless of slots to the left or right, I can do it in four moves:
Move K on top of Q.
Move A on top of K/Q.
Move A to right slot.
Move K to left slot.
But, I'm assuming since you asked the last two questions, one of the implied rules was that you cannot have all three cards stacked. Will work more later.
Actually, I thought of another way:
Turn all the cards around, then walk to the other side of the table.
Zero moves! Or three, if you'd like. :P
Thhe following seems to be ridiculously long (6 moves) but:
-AQK-
--(A/Q)K-
--Q(A/K)-
--QKA
--(K/Q)-A
-KQ-A
-KQA-
I'm assuming you have to end up in the three middle slots.
I see that's what Ragknot said, except I inverted A and K.
Anonymous 4:51 seems to have a reasonable interpretation.
Is the 5 slots a red-herring?
aaarrgh. Not sure which way to read highest card rule - but that just corresponds to a mirror solution.
Okay there are only three slots.
So, you would have to do:
AQK
-(A/Q)K
-Q(A/K)
Q-(A/K)
QAK
(A/Q)-K
(A/Q)K-
A(Q/K)-
AKQ
-(A/K)Q
-K(A/Q)
K-(A/Q)
KAQ
(A/K)-Q
(A/K)Q-
K(A/Q)-
KQA
16 moves.
Then if there is a space next to the Ace,
-AQK
A-QK
AQ-K
AQK-
-(A/Q)K-
-Q(A/K)-
-QKA
Q-KA
QK-A
-(Q/K)-A
-KQA
10 moves
Then, if there's a space next to the King,
AQK-
-(A/Q)K-
-Q(A/K)-
-QKA
Q-KA
QK-A
-(Q/K)-A
-KQA
7 moves
I assume there's an error.
And possibly it is way I understand ... or misunderstand the Trick.
The best answer I think... Is one move... walk your self around to the other side of the table.
That was cool.
Can improve Anonymous's scenario 2.
-AQK
A-QK
A-(K/Q)-
AKQ-
-(A/K)Q-
-K(A/Q)-
-KQA
that's only 6 moves rather than 10.
I'm beginning to believe that there aren't some more cunning solutions.
Also Scenario 3
AQK-
-(A/Q)K-
-Q(A/K)-
-QKA
-(K/Q)-A
KQ-A
KQA-
6 rather than 7 moves.
But for scenario 1, there seem to be no options (without simply going backwards).
Ragknot your first comment is a valid point, my bad . I have changed that now to Q can be placed on K.
Hi Rajesh. We've already finished the problem, having recognised and overridden the conflict in the original rules, but with the reversed rule. We'd only be repeating ourselves, unless there really is something we've missed.
I think it can be solved with 3 slots in 20 moves
AQK
.(QA)K
.Q(KA)
Q.(KA)
QAK
(QA).K
(QA)K.
Q(KA).
QKA
.(KQ)A
.(KQA).
A(KQ).
AKQ
.(KA)Q
.K(QA)
K.(QA)
KAQ
(KA).Q
(KA)Q.
K(QA).
KQA
I.scenario 1:
(11 movements) from AQK:
A_[K/Q]
_A[K/Q]
_[A/Q]K
QAK
Q[A/K]_
_[A/K/Q]_
KAQ
K[A/Q]_
[K/Q]A_
[K/Q]_A
KQA
II.Scenario 2:
(8 movements) from _AQK:
_[A/Q]_K
QA_K
Q_AK
Q_[A/K]_
QKA_
QK_A
_[K/Q]_A
_KQA
III.Scenario 3:
(8 movements) from AQK_:
A_[K/Q]_
_A[K/Q]_
_AKQ
_[A/K]_Q
KA_Q
K_AQ
K_[A/Q]_
KQA_
ionuts.. your scenario 1 is actually 12 moves.. This is what I had come up with before looking at the comments..
You went from _[A/K/Q]_ to KAQ.. that's 2 moves. moving the Q right and the K left.
12 moves is correct when following the rule that a card can only be stacked onto a higher ranking card.
A_(Q/K)
_A(Q/K)
_(A/Q)K
QAK
Q(K/A)_
_(Q/K/A)_
_(K/A)Q
KAQ
K(Q/A)
(K/Q)A_
(K/Q)_A
KQA
(A/Q)_K
(A/Q)K_
Q(A/K)_
QKA
bah!.. those last 4 lines of my previous post is garbage i forgot to delete...
What you guys up to? You've got to beat 16, 6 and 6. Why show longer sets of moves? I'm assuming that you can't stack 3 cards together, else you've got to beat 4,4,4 moves.
Chris, the rules don't say you can't stack 3 cards, as long as they are "placed onto a card of a higher rank". Gratz to ionutz and EB, 12 is indeed the least possible number of moves,(as far as i can tell).
This post has been removed by the author.
Hi mo. I've already acknowledged the 3 stack move (it was given in the third post) in both my second and my last post. It only uses 4 moves in all three scenarios. I don't think that can be beaten.
Hi mo. The original question wasn't written correctly. Up to the point the Rajesh changed it to it's current form, we took it that a high card could be stacked on a low card and not the other way. That understanding was then reversed by Rajesh. I jumped to a false conclusion that reversing the priority didn't significantly change the problem. That was wrong.
Because with the original rule and allowing 3 cards to be stacked, all scenarios could be done using the same 4 moves, I'd considered that version of the problem to be almost trivial. Hence my assumption (not ruling) was that you couldn't stack 3.
I found ionutz's notation confusing. A-[K/Q] suggests King over Queen (as everyone else had done - except Kurt, who lost me). I think ionutz meant Q over King in the example. EB a typo on his 3rd and 10th move.
So AQK -> A-(Q/K) -> -A(Q/K) -> -(Q/A)K -> QAK -> Q(K/A)-
-> -(Q/K/A)- -> -(K/A)Q -> KAQ -> K(Q/A)- -> (Q/K)A- -> (Q/K)-A
-> KQA => 12 moves - credit to EB for that. I can't see a way to do scenario (a) without using a 3 card stack or in less moves.
I haven't re-examined the other scenarios.
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