Sorting with Cranes

1
Sorting with Cranes

• Mike Atkinson
• University of Otago
• Joint work with Michael Albert

2
Question

• Why is a stack limited to pushing or popping one
  item at a time?

• Algorithmically to implement last in-first out
  behaviour.
• Implementationally no reason at all.
• In fact, block pops may be more efficient than
  single element pops in a linked list
  implementation.

3
Enter the forkstack

• Storage as for a stack
• Push a sequence of arbitrary length (but first
  element winds up on top)
• Pop a sequence of arbitrary length.
• Mental image a stack of boxes being manipulated
  by a forklift or a crane.

4
Moving from input to output
Input
Forkstack
Output
5
Questions

• Which rearrangements of an input sequence are
  possible?
• Alternatively, which input sequences can be
  sorted?

6
Sorting 236415
Input
Output
Forkstack
7
Sorting 236415
Input
Output
Forkstack
8
Sorting 236415
Input
Output
Forkstack
9
Sorting 236415
Input
Output
Forkstack
10
Sorting 236415
Input
Output
Forkstack
11
Sorting 236415
Input
Output
Forkstack
12
Sorting 236415
Input
Output
Forkstack
13
Sorting 236415
Input
Output
Forkstack
14
Sorting 236415
Input
Output
Forkstack
15
Sorting 236415
Sorted means increasing from top to bottom
Input
Output
Forkstack
16
Which sequences are sortable?

• Concrete approach seek an algorithm that sorts
  input sequences whenever this is possible.
• Abstract approach seek a characterisation of
  obstructions to sortability (bad configurations)
• Or in between

17
The dreaded 13

• If, in the stack, an element sits directly on top
  of a larger, but not next larger, element, then
  we cannot recover.
• First abstract characterisation
• A sequence is unsortable if any schedule of
  moves that does not produce bad output,
  eventually produces the dreaded 13.
• Not tremendously useful!

18
Or is it?

• Use the idea of avoiding 13s at all costs as the
  primary component of an algorithm for sorting.
• Only extra ingredient doing output when you can
  is never harmful.

19
Sorting 236415

• The 23 at the top need never be split.
• In fact, they should never be split.
• We cant move it with the 6 because of the
  dreaded 13 (36 in this case)

20
Sorting 326415

• The 32 at the stop still need to be moved before
  the 6.
• But they should be moved one element at a time,
  to reduce the chances of a later dreaded 13.

21
Almost decreasing?

• A sequence is almost decreasing if it is
  decreasing, except for some steps of 1.
• Eg 14 12 8 9 5 2 3 4 1
• To avoid the dreaded 13 the maximal initial
  almost decreasing subsequence of the input needs
  to be moved before the next element.
• As a block if non-consecutive.
• To make it increasing if consecutive.

22
The algorithm

• repeat
• As much output as possible
• Move the a.d. head as above
• until input is empty
• if stack is empty
• Hurray!
• else
• Waah!

Linear time test for forkstack sortability
23
Modifications

• We may wish to consider the case where either the
  push, the pop, or both operations are limited in
  size (dishwasher and stower)
• No real need to change the algorithm, it may just
  fail earlier when a move is required which
  overloads an operations capacity.

24
Being obstructive

• Any unsortable sequence contains a minimal
  unsortable subsequence.
• The algorithm implies that there are (to within
  renaming) a finite number of such
• examine the stack just before a crash
• determine what went wrong and why
• In fact, they are 35142, together with 45 of
  length six, and 6 of length 7.

25
Counting

• We wish to count the number of sortable
  permutations of each length.
• This allows us to compare the powers of various
  forkstack models,
• And to compare the sorting power of forkstacks
  with other data structures.

26
The forkstack in the kitchen

• The awful truth
• Sometimes the kitchen has no automatic dishwasher
• Washing dishes has to be done by people!

27
Washing, drying, and stowing
Forkstack with one-element pushes, multi-element
pops
28
Washing, drying, and stowing
When dish t is washed the drying stack is empty
When dish t-1 is washed the drying stack contains
t only
2
1
The intermediate dish segments are sortable
segments on ranges of values
29
Washing, drying, and stowing
a0 lt a1 lt at
30
Washing, drying, and stowing
This recursive decomposition allows the counting
problem to be solved in this case. Let cn be the
number of sortable permutations if the push
operation is of single items and the pop
operation is unlimited. Define
Geometric series
Then
31
Employ weaker stowers?

• If the stower can move just one dish at a time,
  thats sorting on an ordinary stack.
• Powerful stowers discussed above.
• In between? The two cases above correspond to the
  first term, and entire sum of the geometric
  series. The intermediate cases impose a limit on
  the size of t and are exactly the partial sums.
• Should we pay for a powerful stower?

32
Employ weaker stowers?

• The equations tell us that a stower who can lift
  up to k dishes allows about akn dish sequences of
  length n to be stowed in order
• a14 a24.730 a34.919 a44.976
• a54.993 a64.998 .. a? 5
• Local improvements in the manipulation of data
  rapidly lose their effectiveness.

33
Other parameters

• We know about the sorting power of a forkstack if
  pushes of s1 items and pops of any length t are
  allowed.
• What about other values of s and t?
• The only solved case is s2, t2. Here we can
  sort about 5.412n sequences of length n (so st2
  is more powerful than s1, t ? )
• If s ?, t ? we can sort at most 9n sequences of
  length n

34
Open question

• Solve the counting problem for the case s ?, t
  ?