Pancake Problem

許凱平 & 張傑生

Resources Google

“Bounds for Sorting by Prefix Reversal”(63)

“Pancake Problem” (218) Slides

[S1] CMU’s “Great Theoretical Ideas In Computer Science” course

http://www-2.cs.cmu.edu/~15251/Materials/Lectures/Lecture01/lecture01.ppt

Steven Rudich

Resources Slides

[S2] On the Generalization of the Pancake Problem http://hsu14.cis.nctu.edu.tw/upload/on%20the%20generalization%20of%20the%20pancake%20network.pp

t

Marissa P. Justan, Felix P. Muga II,Ivan Hal Sudborough,

[S3] Pancake problems with restricted prefix reversals and some corresponding Cayley networks

http://hsu14.cis.nctu.edu.tw/upload/pancake.ppt Douglas W. Bass and I. Hal Sudborough

Resources Papers

[P1]Gates W.H., and Papadimitriou, C.H. Bounds for sorting by prefix reversal. Discrete Math. 27 (1979), 47--57.

數學系系圖有 [P2] H. Heydari and I. Hal Sudborough. O

n the Diameter of the Pancake Network. Journal of Algorithms 25, 67-94 (1997).

http://www.atcminc.com/mPublications/EP/EPATCM98/ATCMP003/paper.pdf

History [P1]

n+(1/16)n<= f(n) <=(5n+5)/3 Guess 19n/16 <=f(n)

[P2] N+(1/14)n<=f(n) Disprove “Guess 19n/16 <=f(n)”

Pre-Knowledge Symmetric Group

S2#8 S3#5

Dual Linear Program Skip!

Terms Adjacency

Two neighbor (position) element x,y are said to be adjacent (value)

Iff |x-y| = 1 Free, Singleton

No neighbor of x is adjacent to x x is a singleton

Block Block is a group of adjacent element

Terms Positions Values

P Q

P’ Q’

Q

Q

P

P

Q’

P

P’ Q

P Q

Ideas Algebra Use two-sided pancake

to simulate block in one-sided pancake problem Capital for block Non-Capital for singleton

Notations- - flip one segment operation

-- -- flip two segment operation

P 1...k-1

P’ k-1...1

Q k...n

Q’ n...k

d(x) distance, depth: # of flip convert x to PQ

Lemma P’-P PQ--Q’P’

Operationsx flip y d(x)

PQ 0

PQ’ -- QP’ =d(QP’)++ = 3

P’Q - PQ 1

a P’Q’ -- QP =d(QP)++ = 4

b QP - Q’P =d(Q’P)++=3

QP’ - Q’P’ =d(Q’P’)++ = 2

Q’P -- P’Q =d(P’Q)++ =2

Q’P’ -- PQ 1

State Chart

What it means?

QP

P’Q’

Q’P

PQ

P’Q

PQPQ

PQ

What will be PQR? R’Q’P’

Conjecture 4 is the magic number

Notations adj: # of adjacency blk: # of block Move, flip

t is free, t+o is also free

t is head of string, t is glue t is free

t-1, t+1 is not the second element o =1 or o=-1 adj++; blk++

t is free, t+o is 1st of a block

adj++;

t is in block, t+o is free

Adj++

t is in block, t+o is 1st of blk

Move++; Adj++; Blk--;

t singleton, first element of sigma, connector

s singleton, (t-1)

u singleton, (t+1)

S [...s], block, , [...(t-1)]

S’ [s...]

U [u...]

U’ [...u]

T [t...]

T’ [...t]

# don’t care charactors

x input pattern

y output pattern

Goal ts, tu, stu, tS, StU, tU,U’tS’, T’s, T’u, #T’S, #T’U

t is free, t+o is also free

t#s--#ts, o=-1 t#u--#tu, o=1

t is free, t+o is 1st of a block

t#S' -- #tS, o=-1 t#U -- #tU, o=1

Case 3: t is free, S, U’ For o=-1 Steps

t#S#U’#--- S’#t#U’ #-- #St#U’ #----- U#tS’##-- #U’tS’##

Observation Move+=4 Adj+=2 Blk--;

Case3: t is free, S, U’ For o=-1 Steps

t1S2U’3 --- S’1’t2U’3-- 1St2U’3 ----- U2’tS’1’3-- 2U’tS’1’3

New Notation Case1, fig2(a)

t#s--#ts t#s-(2)-#ts

New Notation Case3, fig2(c)

t#S#U’#-(3)-S’#t#U’ #-(2)-#St#U’ #-(5)-U#tS’##-(2)-#U’tS’##

Briefly t#S#U’#-(3,2,5,2)-#U’tS’##

1’3 Chances ：無心插柳柳成蔭

case Description 1’ 3 adj blk

31non-adjacent 2 -1

32 form a new block a a+1 2+1 -1+1

33merge a block with a

singleton(...a) a+1 2+1 -1

34 merge two block (...a) (a+1...) 2+1 -1-1

Case 6, fig2(f) or 2(g) 2(f)

T#u#S#-(3,2,5,2)-#STu## 2(g)

T#U’#s#-(3,2,5,2)-#sTU## Move+=4;Adj+=2;Blk--; Case 6 share the same operation

property of Case 3, ground…

Table2

Action 1 2 31 32 33 34 4 5 7

flip 1 1 4 4 4 4 1 2 1

adj 1 1 2 3 3 3 1 1 1

blk 1 0 -1 0 -1 -2 0 -1 -1

Bill’s Typos or I’m wrong Case 5 rename to Case7, and

Case7 rename to Case5 Case 7, 2(e) Case 5, 2(h) or 2(k)

3 basic equation When string is sorted

move = Sum( move(i) * #case(i) ) Let x(i) = case(i)

Adj = n-1 Blk =1

Total move: Use Table2

Action 1 2 3 4 5 7

flip 1 1 4 1 2 1

z=X(1)+X(2)+4X(3)+X(4)+2X(5)+X(7)

adj: Use Table2

a: initial adjacency a+X(1)+X(2)+2X(31)+3X(32)+3X(33)+3X(34)+X(4)+X(5)

+X(7) Eq(1)

n-1=a+X(1)+X(2)+2X(31)+3X(32)+3X(33)+3X(34)+X(4)+X(5)+X(7)

Action 1 2 31 32 33 34 4 5 7

adj 1 1 2 3 3 3 1 1 1

blk : Use Table2

Action 1 2 31 32 33 34 4 5 7

blk 1 0 -1 0 -1 -2 0 -1 -1

b: initial blk B+x(1)-x(31)-x(33)-2x(34)-x(5)-x(7) Eq(2)

1=B+x(1)-x(31)-x(33)-2x(34)-x(5)-x(7)

(1)(3), b<=a Eq(1) Eq(3)

n-1=a+X(1)+X(2)+2X(31)+3X(32)+3X(33)+3X(34)+X(4)+X(5)+X(7)

b<=a N-

1>=b+X(1)+X(2)+2X(31)+3X(32)+3X(33)+3X(34)+X(4)+X(5)+X(7)

Eq(3)

Upper bound 1=B+x(1)-x(31)-x(33)-2x(34)-x(5)-x(7) N-1>=

b+X(1)+X(2)+2X(31)+3X(32)+3X(33)+3X(34)+X(4)+X(5)+X(7)

z=X(1)+X(2)+4X(3)+X(4)+2X(5)+X(7) Goal: Max(z)

X1=(n+1)/3, x2=x4=x5=x7=b=0 X3=x31=(n-2)/3 Z = (5n-7)/3 Upper bound = z+4=(5n+5)/3

Dual Linear Program Duality Theorem of Von Neumna, Kuh

n and Tucker, Gale, and Dantzig I did not study this…

Amazing Network