Date: 1 June 1982 2222-EDT (Tuesday)
From: Dan Hoey at CMU-10A
To: Cube-Lovers at MIT-MC
Subject: Lower bounds for the 4x4x4 (Long)
Message-Id: <01Jun82 222230 DH51@CMU-10A>
Lower bounds for solving the 4^3 puzzle
In order to prevent counting positions which arise from whole-cube
moves, I do not move the DLB corner. So the generators I use are
U1 U1' R1 R1' F1 F1'
U2 U2' R2 R2' F2 F2'
U3 U3' R3 R3' F3 F3'
where the digit indicates how many layers are being moved and the
prime indicates a counterclockwise quarter-twist.
Let us break up any process into "syllables", where a syllable is a
maximal nonempty string of generators with the same letter. Note
that the order of the generators within a syllable is irrelevant.
We may make a syllable canonical by simplifying so that
1. A generator and its inverse do not both appear,
2. Clockwise generators appear at most twice,
3. Counterclockwise generators appear at most once, and
4. Generators appear in numerical order.
For each letter there are 63 canonical syllables. Omitting the
letter, they are:
1,1',2,2',3,3' Six of length one
11,12,12',13,13',1'2,1'2',1'3,1'3',22,
23,23',2'3,2'3',33 Fifteen of length two
112,112',113,113',122,123,123',12'3,
12'3',133,1'22,1'23,1'23',1'2'3,1'2'3',
1'33,223,223',233,2'33 Twenty of length three
1122,1123,1123',112'3,112'3',1133,1223,
1223',1233,12'33,1'223,1'223',1'233,
1'2'33,2233 Fifteen of length four
11223,11223',11233,112'33,12233,1'2233 Six of length five
112233 One of length six
(Exercise for the reader: There is a reason these are binomial
coefficients. Why did we skip a row?)
The number of canonical syllable strings containing N generators is
P(-N) = 0 if -N < 0,
P(0) = 1,
P(N) = 6 C(N-1) P(N-1) + 15 C(N-2) P(N-2) + 20 C(N-3) P(N-3)
+ 15 C(N-4) P(N-4) + 6 C(N-5) P(N-5) + C(N-6) P(N-6)
if N > 0,
where C(N) is the number of ways of choosing a new letter after N
generators: C(-N) = 0 if -N < 0, C(0) = 3, C(N) = 2 if N > 0.
Evaluating this recurrence yields
P( 0)= 1 P(13)<1.338E15 P(26)<1.404E30 P(39)<1.472E45
P( 1)= 18 P(14)<1.914E16 P(27)<2.007E31 P(40)<2.105E46
P( 2)= 261 P(15)<2.737E17 P(28)<2.871E32 P(41)<3.011E47
P( 3)= 3732 P(16)<3.915E18 P(29)<4.106E33 P(42)<4.307E48
P( 4)= 53379 P(17)<5.599E19 P(30)<5.873E34 P(43)<6.160E49
P( 5)= 763506 P(18)<8.009E20 P(31)<8.400E35 P(44)<8.811E50
P( 6)=10920771 P(19)<1.146E22 P(32)<1.202E37 P(45)<1.261E52
P( 7)<1.563E 8 P(20)<1.639E23 P(33)<1.719E38 P(46)<1.803E53
P( 8)<2.235E 9 P(21)<2.344E24 P(34)<2.459E39 P(47)<2.579E54
P( 9)<3.196E10 P(22)<3.353E25 P(35)<3.516E40 P(48)<3.688E55
P(10)<4.572E11 P(23)<4.795E26 P(36)<5.029E41 P(49)<5.275E56
P(11)<6.539E12 P(24)<6.858E27 P(37)<7.194E42 P(50)<7.545E57
P(12)<9.352E13 P(25)<9.810E28 P(38)<1.029E44 P(51)<1.080E59
The number of positions exactly N qtw from SOLVED is at most P(N),
so the number of positions within N qtw of SOLVED is at most
P(0)+P(1)+...+P(N). Since there are 7.07E53 marked and 7.40E45
colored positions, this would give us lower bounds of 40 qtw for
the marked cube and 47 qtw for the colored cube.
But we can improve these lower bounds by one each, using the parity
hack I described some time ago. Every qtw is an odd permutation of
the corner cubies, so the number of positions with even corner
permutations within 2N+1 qtw of SOLVED is at most
P(0)+P(2)+...+P(2N) and the number of positions with odd corner
permutations within 2N+2 qtw of solved is at most
P(1)+P(3)+...+P(2N+1).
There are 3.70E45 colored positions with odd corner permutations,
and fewer than 1.583E45 odd-length canonical processes with length
at most 39. So some odd colored positions require at least 41 qtw
to solve.
There are 3.53E53 marked positions with even corner permutations,
and fewer than 1.939E53 even-length canonical processes with length
at most 46. So some even marked positions require at least 48 qtw
to solve.
Directions for future hacks
Note that we could also distinguish between positions with even or
odd edge permutations. The recurrence gets hairier, but my
analysis of that problem indicates that the numbers get very close
very quick, so no luck.
We could divide the positions into buckets based on other quotients
of the group. For instance, count the number of positions with
each of the 729 possible corner orientations. This should be
feasible, but probably no help for the 4^3 puzzle. For the 3^3,
where there are 2187 corner buckets, some of which stay empty for a
fair number of moves (I seem to recall 8), and our current lower
bound is small, I think there is a chance of improvement. Any
God's numerologist willing to hack the good hack?