Algorithm to sort pairs of numbers - algorithm

Question

I am stuck with a problem and I need some help from bright minds of SO.
I have N pairs of unsigned integerers. I need to sort them. The ending vector of pairs should be sorted nondecreasingly by the first number in each pair and nonincreasingly by the second in each pair. Each pair can have the first and second elements swapped with each other. Sometimes there is no solution, so I need to throw an exception then.
Example:
in pairs:
1 5
7 1
3 8
5 6
out pairs:
1 7 <-- swapped
1 5
6 5 <-- swapped
8 3 <-- swapped
^^ Without swapping pairs it is impossible to build the solution. So we swap pairs (7, 1), (3, 8) and (5, 6) and build the result.
or
in pairs:
1 5
6 9
out:
not possible
One more example that shows how 'sorting pairs' first isn't the solution.
in pairs:
1 4
2 5
out pairs:
1 4
5 2
Thanks

Answer 1

O( n log n ) solution

Answer 2

Let S(n) equals all the valid sort orderings, where n corresponds to pairs included [0,n].
S(n) = []
for each order in S(n-1)
for each combination of n-th pair
if pair can be inserted in order, add the order after insertion to S(n)
else don't include the order in S(n)
A pair can be inserted into an order in maximum of two ways(normal pair and reversed pair).
Maximum orderings = O(2^n)
I'm not very sure about this amortized orderings, but hear me out.
For an order and pair we have four ways of getting sorted orders after insertions
(two orders, one(normal),one(reversed), zero)
No of orderings (Amortized) = (1/4)*2 + (1/4)*1 + (1/4)*1 + (1/4)*0 = 1
Amortized orderings = O(1)
Similarly time complexity will be O(n^2), Again not sure.
Following program finds orderings using a variant of Insertion sort.
debug = False
(LEFT, RIGHT, ERROR) = range(3)
def position(first, second):
""" Returns the position of first pair when compared to second """
x,y = first
a,b = second
if x <= a and b <= y:
return LEFT
if x >= a and b >= y:
return RIGHT
else:
return ERROR
def insert(pair, order):
""" A pair can be inserted in normal order or reversed order
For each order of insertion we will get one solution or none"""
solutions = []
paircombinations = [pair]
if pair[0] != pair[1]: # reverse and normal order are distinct
paircombinations.append(pair[::-1])
for _pair in paircombinations:
insertat = 0
if debug: print "Inserting", _pair,
for i,p in enumerate(order):
pos = position(_pair, p)
if pos == LEFT:
break
elif pos == RIGHT:
insertat += 1
else:
if debug: print "into", order,"is not possible"
insertat = None
break
if insertat != None:
if debug: print "at",insertat,"in", order
solutions.append(order[0:insertat] + [_pair] + order[insertat:])
return solutions
def swapsort(pairs):
"""
Finds all the solutions of pairs such that ending vector
of pairs are be sorted non decreasingly by the first number in
each pair and non increasingly by the second in each pair.
"""
solutions = [ pairs[0:1] ] # Solution first pair
for pair in pairs[1:]:
# Pair that needs to be inserted into solutions
newsolutions = []
for solution in solutions:
sols = insert(pair, solution) # solutions after inserting pair
if sols:
newsolutions.extend(sols)
if newsolutions:
solutions = newsolutions
else:
return None
return solutions
if __name__ == "__main__":
groups = [ [(1,5), (7,1), (3,8), (5,6)],
[(1,5), (2,3), (3,3), (3,4), (2,4)],
[(3,5), (6,6), (7,4)],
[(1,4), (2,5)] ]
for pairs in groups:
print "Solutions for",pairs,":"
solutions = swapsort(pairs)
if solutions:
for sol in solutions:
print sol
else:
print "not possible"
Output:
Solutions for [(1, 5), (7, 1), (3, 8), (5, 6)] :
[(1, 7), (1, 5), (6, 5), (8, 3)]
Solutions for [(1, 5), (2, 3), (3, 3), (3, 4), (2, 4)] :
[(1, 5), (2, 4), (2, 3), (3, 3), (4, 3)]
[(1, 5), (2, 3), (3, 3), (4, 3), (4, 2)]
[(1, 5), (2, 4), (3, 4), (3, 3), (3, 2)]
[(1, 5), (3, 4), (3, 3), (3, 2), (4, 2)]
Solutions for [(3, 5), (6, 6), (7, 4)] :
not possible
Solutions for [(1, 4), (2, 5)] :
[(1, 4), (5, 2)]

Answer 3

This is a fun problem. I came up with Tom's solution independently, here's my Python code:
class UnableToAddPair:
pass
def rcmp(i,j):
c = cmp(i[0],j[0])
if c == 0:
return -cmp(i[1],j[1])
return c
def order(pairs):
pairs = [list(x) for x in pairs]
for x in pairs:
x.sort()
pairs.sort(rcmp)
top, bottom = [], []
for p in pairs:
if len(top) == 0 or p[1] <= top[-1][1]:
top += [p]
elif len(bottom) == 0 or p[1] <= bottom[-1][1]:
bottom += [p]
else:
raise UnableToAddPair
bottom = [[x[1],x[0]] for x in bottom]
bottom.reverse()
print top + bottom
One important point not mentioned in Tom's solution is that in the sorting stage, if the lesser values of any two pairs are the same, you have to sort by decreasing value of the greater element.
It took me a long time to figure out why a failure must indicate that there's no solution; my original code had backtracking.

Answer 4

Below is a simple recursive depth-first search algorithm in Python:
import sys
def try_sort(seq, minx, maxy, partial):
if len(seq) == 0: return partial
for i, (x, y) in enumerate(seq):
if x >= minx and y <= maxy:
ret = try_sort(seq[:i] + seq[i+1:], x, y, partial + [(x, y)])
if ret is not None: return ret
if y >= minx and x <= maxy:
ret = try_sort(seq[:i] + seq[i+1:], y, x, partial + [(y, x)])
if ret is not None: return ret
return None
def do_sort(seq):
ret = try_sort(seq, -sys.maxint-1, sys.maxint, [])
print ret if ret is not None else "not possible"
do_sort([(1,5), (7,1), (3,8), (5,6)])
do_sort([(1,5), (2,9)])
do_sort([(3,5), (6,6), (7,4)])
It maintains a sorted subsequence (partial) and tries to append every remaining pair to it both in the original and in the reversed order, without violating the conditions of the sort.
If desired, the algorithm can be easily changed to find all valid sort orders.
Edit: I suspect that the algorithm can be substantially improved by maintaining two partially-sorted sequences (a prefix and a suffix). I think that this would allow the next element can be chosen deterministically instead of trying all possible elements. Unfortunately, I don't have time right now to think this through.

Answer 5

Update: this answer is no longer valid since question was changed
Split vector of pairs into buckets by first number. Do descending sort on each bucket. Merge buckets in ascending order of first numbers and keep track of second number of last pair. If it's greater than current one there is no solution. Otherwise you will get solution after merge is done.
If you have stable sorting algorithm you can do descending sort by second number and then ascending sort by first number. After that check if second numbers are still in descending order.

Answer 6

The swapping in your case is just a sort of a 2-element array.
so you can
tuple[] = (4,6),(1,5),(7,1),(8,6), ...
for each tuple -> sort internal list
=> (4,6),(1,5),(1,7),(6,8)
sort tuple by 1st asc
=> (1,5),(1,7),(4,6),(6,8)
sort tuple by 1nd desc
=> (1,7),(1,5),(4,6),(6,8)

Answer 7

The first thing I notice is that there is no solution if both values in one tuple are larger than both values in any other tuple.
The next thing I notice is that tuples with a small difference become sorted towards the middle, and tupples with large differences become sorted towards the ends.
With these two pieces of information you should be able to figure out a reasonable solution.
Phase 1: Sort each tuple moving the smaller value first.
Phase 2: Sort the list of tuples; first in descending order of the difference between the two values of each tuple, then sort each grouping of equal difference in ascending order of the first member of each tuple. (Eg. (1,6),(2,7),(3,8),(4,4),(5,5).)
Phase 3: Check for exceptions. 1: Look for a pair of tuples where both elements of one tuple are larger than both elements of the other tuple. (Eg. (4,4),(5,5).) 2: If there are four or more tuples, then look within each group of tuples with the same difference for three or more variations (Eg. (1,6),(2,7),(3,8).)
Phase 4: Rearrange tuples. Starting at the back end (tuples with smallest difference), the second variation within each grouping of tuples with equal difference must have their elements swapped and the tuples appended to the back of the list. (Eg. (1,6),(2,7),(5,5) => (2,7),(5,5),(6,1).)
I think this should cover it.

Answer 8

This is a very interesting question. Here is my solution to it in VB.NET.
Module Module1
Sub Main()
Dim input = {Tuple.Create(1, 5),
Tuple.Create(2, 3),
Tuple.Create(3, 3),
Tuple.Create(3, 4),
Tuple.Create(2, 4)}.ToList
Console.WriteLine(Solve(input))
Console.ReadLine()
End Sub
Private Function Solve(ByVal input As List(Of Tuple(Of Integer, Integer))) As String
Dim splitItems As New List(Of Tuple(Of Integer, Integer))
Dim removedSplits As New List(Of Tuple(Of Integer, Integer))
Dim output As New List(Of Tuple(Of Integer, Integer))
Dim otherPair = Function(indexToFind As Integer, startPos As Integer) splitItems.FindIndex(startPos, Function(x) x.Item2 = indexToFind)
Dim otherPairBackwards = Function(indexToFind As Integer, endPos As Integer) splitItems.FindLastIndex(endPos, Function(x) x.Item2 = indexToFind)
'split the input while preserving their indices in the Item2 property
For i = 0 To input.Count - 1
splitItems.Add(Tuple.Create(input(i).Item1, i))
splitItems.Add(Tuple.Create(input(i).Item2, i))
Next
'then sort the split input ascending order
splitItems.Sort(Function(x, y) x.Item1.CompareTo(y.Item1))
'find the distinct values in the input (which is pre-sorted)
Dim distincts = splitItems.Select(Function(x) x.Item1).Distinct
Dim dIndex = 0
Dim lastX = -1, lastY = -1
'go through the distinct values one by one
Do While dIndex < distincts.Count
Dim d = distincts(dIndex)
'temporary list to store the output for the current distinct number
Dim temOutput As New List(Of Tuple(Of Integer, Integer))
'go through each of the split items and look for the current distinct number
Dim curIndex = 0, endIndex = splitItems.Count - 1
Do While curIndex <= endIndex
If splitItems(curIndex).Item1 = d Then
'find the pair of the item
Dim pairIndex = otherPair(splitItems(curIndex).Item2, curIndex + 1)
If pairIndex = -1 Then pairIndex = otherPairBackwards(splitItems(curIndex).Item2, curIndex - 1)
'create a pair and add it to the temporary output list
temOutput.Add(Tuple.Create(splitItems(curIndex).Item1, splitItems(pairIndex).Item1))
'push the items onto the temporary storage and remove it from the split list
removedSplits.Add(splitItems(curIndex))
removedSplits.Add(splitItems(pairIndex))
If curIndex > pairIndex Then
splitItems.RemoveAt(curIndex)
splitItems.RemoveAt(pairIndex)
Else
splitItems.RemoveAt(pairIndex)
splitItems.RemoveAt(curIndex)
End If
endIndex -= 2
Else
'increment the index or exit the iteration as appropriate
If splitItems(curIndex).Item1 <= d Then curIndex += 1 Else Exit Do
End If
Loop
'sort temporary output by the second item and add to the main output
output.AddRange(From r In temOutput Order By r.Item2 Descending)
'ensure that the entire list is properly ordered
'start at the first item that was added from the temporary output
For i = output.Count - temOutput.Count To output.Count - 1
Dim r = output(i)
If lastX = -1 Then
lastX = r.Item1
ElseIf lastX > r.Item1 Then
'!+ It appears this section of the if statement is unnecessary
'sorting on the first column is out of order so remove the temporary list
'and send the items in the temporary list back to the split items list
output.RemoveRange(output.Count - temOutput.Count, temOutput.Count)
splitItems.AddRange(removedSplits)
splitItems.Sort(Function(x, y) x.Item1.CompareTo(y.Item1))
dIndex += 1
Exit For
End If
If lastY = -1 Then
lastY = r.Item2
ElseIf lastY < r.Item2 Then
'sorting on the second column is out of order so remove the temporary list
'and send the items in the temporary list back to the split items list
output.RemoveRange(output.Count - temOutput.Count, temOutput.Count)
splitItems.AddRange(removedSplits)
splitItems.Sort(Function(x, y) x.Item1.CompareTo(y.Item1))
dIndex += 1
Exit For
End If
Next
removedSplits.Clear()
Loop
If splitItems.Count = 0 Then
Dim result As New Text.StringBuilder()
For Each r In output
result.AppendLine(r.Item1 & " " & r.Item2)
Next
Return result.ToString
Else
Return "Not Possible"
End If
End Function
<DebuggerStepThrough()> _
Public Class Tuple(Of T1, T2)
Implements IEqualityComparer(Of Tuple(Of T1, T2))
Public Property Item1() As T1
Get
Return _first
End Get
Private Set(ByVal value As T1)
_first = value
End Set
End Property
Private _first As T1
Public Property Item2() As T2
Get
Return _second
End Get
Private Set(ByVal value As T2)
_second = value
End Set
End Property
Private _second As T2
Public Sub New(ByVal item1 As T1, ByVal item2 As T2)
_first = item1
_second = item2
End Sub
Public Overloads Function Equals(ByVal x As Tuple(Of T1, T2), ByVal y As Tuple(Of T1, T2)) As Boolean Implements IEqualityComparer(Of Tuple(Of T1, T2)).Equals
Return EqualityComparer(Of T1).[Default].Equals(x.Item1, y.Item1) AndAlso EqualityComparer(Of T2).[Default].Equals(x.Item2, y.Item2)
End Function
Public Overrides Function Equals(ByVal obj As Object) As Boolean
Return TypeOf obj Is Tuple(Of T1, T2) AndAlso Equals(Me, DirectCast(obj, Tuple(Of T1, T2)))
End Function
Public Overloads Function GetHashCode(ByVal obj As Tuple(Of T1, T2)) As Integer Implements IEqualityComparer(Of Tuple(Of T1, T2)).GetHashCode
Return EqualityComparer(Of T1).[Default].GetHashCode(Item1) Xor EqualityComparer(Of T2).[Default].GetHashCode(Item2)
End Function
End Class
Public MustInherit Class Tuple
<DebuggerStepThrough()> _
Public Shared Function Create(Of T1, T2)(ByVal first As T1, ByVal second As T2) As Tuple(Of T1, T2)
Return New Tuple(Of T1, T2)(first, second)
End Function
End Class
End Module
The input
1 5
2 3
3 3
3 4
2 4
Produces the output
1 5
2 4
2 3
3 4
3 3
And
3 5
6 6
7 4
Outputs
Not Nossible
Comments
I found this problem quite challenging. It took me some 15 minutes to come up with with a solution and an hour or so to write and debug it. The code is littered with comments so that anyone can follow it.