Transcript Optimality of an Algorithm Solving the k

On Optimal Solutions for the
Bottleneck Tower of Hanoi Problem
Yefim Dinitz
Shay Solomon
Dept. of Comp. Sci., Ben-Gurion University
ISRAEL
Contents of the lecture
1. Background on the Tower of Hanoi.
2. The Bottleneck Tower of Hanoi (BTH)
problem.
3. The family of all optimal solutions to BTH.
4. Investigation of the configuration graph of
BTH: its diameter and average distance.
1. Background
• The first version of the puzzle was marketed by Edouard
Lucas in 1883, under the name "Tower of Hanoi“:
- Given are 3 pegs, denoted by A,B and C, and n disks of
distinct sizes placed on peg A in decreasing order, largest
at the bottom to the smallest at the top.
- The goal of the puzzle is to transfer all disks to peg C,
placed in the same order, by the minimum possible
number of moves.
- The rules are to move a single disk from (the top of) one
peg to (the top of) another one at each step, following the
"divine rule": to never have a larger disk above a smaller
one.
Move number
Example for general n
1
2
…
…
…
n-1
n
A
B
C
Move number
Cn-1
The optimal transfer for general n
1
2
…
…
…
n-1
n
A
B
C
Move number
Cn-1 + 1
The optimal transfer for general n
1
2
…
…
…
n-1
A
B
n
C
Move number
2 Cn-1 + 1
The optimal transfer for general n
1
2
…
…
…
n-1
n
A
B
C
The recurrence formula is
cn = 2cn-1 + 1, c1 = 1.
Hence, cn = 2^n -1.
This classical Tower of Hanoi problem with
three pegs is a basic problem, taught as a
basic example of a recursion program.
There is a great proliferation of
new ToH puzzles posed over the
last twenty years.
There is a bibliography made by Paul
Stockmeyer, which covers over 350
papers on the subject.
2. The Bottleneck ToH problem
In 1981, D. Wood posed the following
generalization of ToH, which we call the
“Bottleneck Tower of Hanoi Problem, BTHn= BTHn,k”,
depending on an integer parameter k, k>0.
The k-relaxed Placement Rule:
Besides the usual placement of smaller
disks above bigger ones, a bigger disk j
may be placed above a smaller disk i, if
their size difference j - i < k.
i+k
….
….
i
Definitions
• A placement of the n disks [1..n] on a
single peg is called a tower of n disks.
• The decreasing bottom-top placement
of the n disks [1..n] on a peg is called
the perfect tower.
1
2
3
…
…
n-2
n-1
n
Description of BTHn:
• There are 3 pegs: A,B, and C.
• The goal is to transfer the perfect tower on
peg A to the perfect tower on peg C.
• The placement rule is k-relaxed.
(Note that if k=1, we arrive at the classical
problem.)
Example:
k=4, n=7
We note that there are many legal
towers for a given k.
1
2
3
As k grows, there are more legal
towers.
7
6
5
4
Legal
(The two extreme) examples:
k=1, one legal tower (perfect tower).
k≥n, n! possible towers.
• D.Poole [1992] suggested the sequence of
moves αn (described in the next slide),
which solves BTHn. He showed its
optimality within just one simple strategy.
• Beneditkis, Berend, and Safro [1998]
proved its optimality for the case k=2.
• We proved optimality of αn for the general
case [ISAAC’2006].
• Remark: another proof of optimality of αn for
BTHn, by different techniques, was suggested
independently by Xiaomin Chen et al.
(unpublished).
αn(A,C) (not explicitly)
1.(*) Move disks [1..(n-1)] from A,
by the minimum possible number of moves,
to any (legal) state on B.
2. Move disk n from A to C.
3.(*) Move disks [1..(n-1)] from B to the
perfect tower on C, by the minimum
possible number of moves.
1
2
3
4
….
….
….
….
….
n-1
n
A
B
C
[1..n-1]
n
A
B
C
[1..n-1]
n
A
B
C
1
2
3
4
….
….
….
….
….
n-1
n
A
B
C
“Somehow” tower-move(*): To move a
tower of m disks on one peg, to a tower
of m disks on another peg, in any order.
• D. Poole and BBS used, for the
“somehow” problem, the following optimal
sequence of moves, βm:
βm(source,target):
(for the explanation, assume that the initial configuration is perfect)
• If m ≤ k, move all disks from source to
target one by one.
• Otherwise:
1. Recursively perform βm-k(source,auxiliary).
2. Move k disks [(m-k+1)..m] one by one from
source to target.
3. Recursively perform βm-k(auxiliary,target).
K=4, m=7
1
2
3
4
5
6
7
source
auxiliary
target
4
5
3
6
2
7
1
source
auxiliary
target
7
source
3
6
2
5
1
4
auxiliary
target
1
2
3
7
6
5
4
source
auxiliary
target
K=4, m=11
1
2
3
4
5
6
7
8
9
10
11
source
auxiliary
target
1
2
3
8
7
9
6
10
5
11
4
source
auxiliary
target
1
2
3
7
11
6
10
5
9
8
4
source
auxiliary
target
1
2
3
4
5
6
7
11
10
9
8
source
auxiliary
target
(**) We remark that, if we start with the
perfect tower on A, and perform βm(A,B)
followed by βm(B,C), the result is the
perfect tower on C.
αn(A,C), reformulated
1. perform βn-1(A,B).
2. Move disk n from A to C.
3. perform βn-1(B,C).
By earlier remark (**), αn solves BTHn.
1
2
3
4
….
….
….
….
….
n-1
n
A
B
C
1
2
3
….
….
n-k-1
n-1
….
n-k+1
n-k
n
A
B
C
1
2
3
….
….
n-k-1
n-1
….
n-k+1
n-k
n
A
B
C
1
2
3
4
….
….
….
….
….
n-1
n
A
B
C
Notation and numerical results
• bm = The number of moves in βm
• an = The number of moves in αn
• bm = 2bm-k + k ; b1 = 1,…, bk = k.
bm = r (2^q) + k (2^q -1),
where m = qk +r.
• an = 2bn-1 + 1 ; a1 = 1.
an = r (2^(q+1)) + k (2^(q+1) -2) + 1,
where n-1=qk + r.
3. Family of all optimal solutions to
BTHn
We prove that, in any optimal solution to
BTHn, disk n makes only one move, from
peg A to C.
Before that move, all disks 1,…,n-1 have
been moved from peg A to peg B.
After that move, all disks 1,..,n-1 will move
from peg B to peg C.
Is αn the unique optimal solution?
In the case k=1, the answer is yes (and is
well known): there exists a unique solution
for the classical problem.
For the case k ≥ 2, the answer is
No!
Let us see a different sequence of moves,
of the same length as that of αn:
Example n=10, k=3
1
2
3
4
5
6
7
8
9
10
A
B
C
4
5
6
7
8
3
9
2
10
1
A
B
C
7
8
3
6
9
2
5
10
1
4
A
B
C
1
2
7
3
8
6
9
5
10
A
4
B
C
1
2
3
10
A
9
6
8
5
7
4
B
C
Until this point, the movesequence is the same as
αn
3
2
9
6
1
8
5
10
7
4
A
B
C
3
6
2
9
1
8
5
10
7
4
A
B
C
2
5
3
6
9
1
8
10
A
7
4
B
C
1
4
2
5
3
6
9
8
10
A
7
B
C
1
4
2
5
3
6
9
8
10
7
A
B
C
3
6
5
4
A
9
2
8
1
7
10
B
C
1
2
3
6
9
5
8
4
7
A
10
B
C
1
2
3
7
6
8
5
9
4
10
A
B
C
6
3
5
2
4
1
7
8
9
10
A
B
C
4
5
6
7
3
8
2
9
1
10
A
B
C
1
2
3
4
5
6
7
8
9
10
A
B
C
Tower-Move
• tower-move – A sequence of moves that
transfers a tower on one peg to a tower on
another peg.
• For example:
- αn is a shortest perfect-to-perfect
tower-move of n disks.
- βm is a shortest “somehow” tower-move
of m disks.
General structure of an optimal
tower-move
• It should be similar to the structure of the
(recursive) classic Tower of Hanoi solution,
but the transfer unit is a block of k disks
(as in the example).
• Main Proposition 1: The only possible
change, in an optimal solution, may be to
interleave the two upper blocks.
• Main Observation 2: The ways of
interleaving, each time, are independent.
How many optimal solutions exist?
• Any op
timal solution (i.e., a shortest p.t.p.
tower move of [1..n]) contains 2^(┌(n-1)/k┐-1)
tower-moves of the union of the two upper blocks.
• For any such even-numbered tower-move, except
for the last tower-move, its final configuration can
be chosen arbitrarily from a certain set (**) of
C(k+r,k)-C(k+r,k+1) configurations.
• These are the only choices, for an optimal solution.
How many such configurations exist?
An auxiliary problem equivalent to (**):
Let n ≤ t+p.
What is the number of integer solutions that
satisfy the two following conditions:
• ∑xi =n,
• ∑xi ≤ j+p,
i=[0..t], xi≥0.
0≤j ≤n-p, i=[0..j].
The number of different solutions to (**)
is:
C(n+t,t) – C(n+t,t+p+1)
The number of all optimal
solutions to BTHn is:
k=1:
A unique optimal solution
k ≥ 2:
(C(k+r,k)-C(k+r,k+1))^[┌2^(┌(n-1)/k┐-2)┐ -1],
where r= (n-1) mod k.
4.1 Diameter of the configuration
graph
Questions :
1. Find a pair of disk arrangements over A,B and
C, requiring the longest sequence of moves in
order to get from one to the other.
2. Find the length of that sequence (later referred
as “diameter”).
Let us translate these questions to the language of
graphs:
• For BTHn , the corresponding
configuration graph, G=(V,E) is the
(undirected) graph s.t.:
• V is the set of all legal arrangements of the
n disks over A, B and C.
• E = {e=(u,v)| u,v in V, s.t. u and v are
reachable one from the other by a single
move}.
The configuration
graph for the case
n=2,k=2
(),(2),(1)
(),(2,1),()
(),(),(1,2)
(1,2),(),()
(2),(),(1)
(2,1),(),()
(),(),(2,1)
(2),(1),()
(1),(),(2)
(),(1),(2)
(),(1,2),()
(1),(2),()
The diameter is 4
(),(2),(1)
(),(2,1),()
(),(),(1,2)
(1,2),(),()
(2),(),(1)
(2,1),(),()
(),(),(2,1)
(2),(1),()
(1),(),(2)
(),(1),(2)
(),(1,2),()
(1),(2),()
We found the answer for the second
question, i.e. the diameter, D(n,k),
up to a constant factor.
D(n,k) = Θ((k^2)∙2^(n/k))
This is k times the shortest length of
a perfect-to-perfect tower-move.
Reminder: an = Θ(k ∙2^(n/k))
Tighter bounds for D(n,k):
• Let n ≤ k
D(n,k) = Θ(n∙logn).
• Let k ≤ n ≤ 2k
D(n,k) = Θ(n∙logn + (n-k)^2).
• For n>2k:
D(n,k) = Θ((k^2)∙2^(n/k)).
We believe (based on similar open questions
raised by D. Knuth) that both aforementioned questions:
1. Finding a pair of disk arrangements
requiring a longest sequence of moves.
2. Finding the length of that sequence (i.e,
the previously mentioned diameter) exactly.
should be difficult even for the (degenerate)
case n≤k.
4.2 Average distance
We find the average number of moves
required to get from one configuration to
another, taken over all pairs of
configurations, up to a constant factor.
In other words, we bound the average
distance between nodes in the
configuration graph of BTHn.
We prove that the average, Avg(n,k), is
asymptotically the same as the aforementioned diameter.
For n ≥ 3k:
Avg(n,k) = Θ(D(n,k)) = Θ((k^2)∙2^(n/k))
This is k times the shortest length of a
p.t.p. tower-move.
Avg(n,k) for some values
of n vs. k:
• Let n ≤ k:
Avg(n,k) = Θ(D(n,k)) = Θ(nlogn).
• For n>3k:
Avg(n,k) = Θ(D(n,k)) = Θ((k^2)∙2^(n/k)).
In our paper at ISAAC’06, we considered
optimal algorithms for ToH problems
with generalized placement rules.
We appreciate your attention.
The way of proof (for diameter)
Upper bound: We show an explicit way to
transfer any tower of n disks in
O((k^2)∙2^(n/k)) moves.
Lower bound
• Observation. When any disk out of the k biggest
disks [(n-k+1)..n] moves, all n-2k disks in [1..(n2k)] should be on the spare peg.
• We show two configurations of the k biggest
disks, such that any tower-move between them
requires Ω(k) changes of the spare peg.
• As a consequence, for those configurations and
the case n>2k, Ω(k) tower-moves of the n-2k
smallest disks are required.
As we know, each one of those tower-moves
requires Ω(k∙2^((n-2k)/k)) moves.