Finding the maximum frequency in a window in a stream without minimum window size restriction

Problem Description

Problem description:

More specifically:

Define maximum frequency of an item set A = maximum over all windows from the start of the flow until the current state that satisfies a minimum size constraint
(The minimum size constraint is imposed to remove undesirable solutions, e.g., last itemset always have maximum freqeuncy = 1)

Example:

Find the window with the highest possible frequency

Definitions and Notations

Definitions used in Calder's paper:

S = stream
I_k = item sets
count(I, S) = number of occurrence of itemset I in stream S
freq(I, S) = (relative) frequency of itemset I in stream S
freq(I, S) = count(I, S)/|S|

Stream state at time t

Each itemset in the stream numbered with a logical time stamp, starting with 1
S_t = the itemsets that has arrived upto time t

Concatentation of streams:

Suppose:

S1 = < a1, a2, ..., a_m > S2 = < b1, b2, ..., b_n >

Then the concatentation of streams S1 and S2 is:

S1 ⊕ S2 = < a1, a2, ..., a_m, b1, b2, ..., b_n >

Substream:

Suppose:

S = < I1, I2, ..., I_s, I_s+1, ..., I_t, I_t+1, ..., I_m >

Then the substream S[s,t] of stream S is:

S[s,t] = < I_s, I_s+1, ..., I_t >

Last k items in stream:

Suppose:

S = < I1, I2, ..., I_|S|-k+1, I_|S|-1, ..., I_|S| >

Then the last k items last(k,S) of stream S is:

last(k,S) = < I_|S|-k+1, I_|S|-1, ..., I_|S| >

Maximum Frequency:

mfreq^mwl(I, S) = maximum frequency of itemset I in stream S with minimum windows size "mwl"

Formally:

mfreq^mwl(I, S) = max ( freq(I, last(k, S) ) k = mwl .. |S|

NOTES:

Although the maximum frequency is defined as the highest frequency in the last k items in the stream, it is easy to find the maximum frequency in any portion of the stream

All we need to do is record the highest frequency while processing the item sets:

We are currently here is the stream: ....... .............. ......... | | +--------------+ We must find: max. freq. found in this window Algorithm: ....... .............. | | +--------------+ We record: max. freq. found in last(k,S)

Example: computing frequency of an itemset a

S = a b a a a b | | +-----+ freq(a,last 2) = 1/2 (0.5) | | +---------+ freq(a,last 3) = 2/3 (0.66666) | | +-------------+ freq(a,last 4) = 3/4 (0.75) <--------- Max freq !! | | +----------------+ freq(a,last 5) = 3/5 (0.6) | | +---------------------+ freq(a,last 6) = 4/6 = 2/3 (0.6666)

Maximum Window

Maximum Window = the largest window in which the maximum frequency is reached
startmax^mwl(I,S) = the starting index of the maximum window

Example: maximum window

1 2 3 4 5 6 S = a b a a a b | | +-------------+ freq(a,S) = 3/4 | | +---------------------+ freq(s,S) = 4/6 = 2/3 startmax = 3

Important note:

When multiple windows in the stream has the same maximum frequency, then the maximum window is (defined as) the largest window among those with maximum frequency

Naive Algorithm

Naive Algorithm:

1. Buffer all itemset in memory (disk ?) 2. Find the frequency of itemset I in these windows:

I1 I2 ..... ... ... ... ... ... ... ... I_n | | +-------------+ <---- mwl ----> | | +----------------+ <----- mwl+1 ----> .................... | | +---------------------------------------------+

3. Compute the maximum frequency

The naive algorithm is presented to give you an idea what you need to do...

Simplification
- Calders first discuss the solution to a simpler version of the problem.
- We will first consider the maximum frequncy (and maximum window) problem where:

Borders

The naive algorithm check every point in the stream to find the maximum window
Important Fact about the maximum frequency:
A border is a point in the stream that needs to be inspected to find the maximum window

Definition: Border

The time stamp q is a border for itemset A in stream S if:

Graphically:

We are processing the stream S and currently, we are processing item I_n: S: I1 I2 .... I_q ...... I_n
q is a border if there exists a continuation I_n+1 ... I_m such that the element I_q can become the start of a max. window: I1 I2 .... I_q ...... I_n I_n+1 ... I_m | | +-----------------------+ maximum windown (max. freq. is found here)

Comment of Calder's Algorithm
- Calder's Algorithm is very simple - it is based on some properties of a border and its relationship to the maximum window
- But without knowing these properties of:
  the algorithm is meaningless
- So we must study the properties of a border first

Theorem 1: property of a maximum window

Theorem 1:

Let stream S be:

S = I1 I2 ... I_q-1 I_q ... I_L

If I_q ... I_L is a maximum window for itemset A, then:

This is a max window +-------------+ | | I1 I2 ... I_p ... I_q-1 I_q ... I_r ... I_L | || | +----------++-------+ freq(A, S[p,q-1]) freq(A, S[q,r]) The following property holds: ∀ p ≤ q-1 ∧ r ≥ q: freq(A, S[p,q-1]) < freq(A, S[q,r])

Proof:

Notation used:

maximum window +--------------------+ | | I1 I2 .... I_p ... I_q-1 I_q .... I_r I_r+1 .... I_L | || || | +----------++---------++---------+ B1 B2 B3

Given fact: B2 ⊕ B3 is the maximum window (i.e., has max frequency in the largest window)

Therefore:

freq(A, B1 ⊕ B2 ⊕ B3) < freq(A, B2 ⊕ B3) ..... (1)

Because the window (B1, B2, B3) is larger (in size) than the window (B2, B3):

maximum window +----------------------+ | | I1 I2 .... I_p ... I_q-1 I_q .... I_r I_r+1 .... I_L | || || | +----------++---------++-----------+ B1 B2 B3 | | +----------------------+ freq(A, B2 ⊕ B3) = max. freq | | +----------------------------------+ freq(A, B1 ⊕ B2 ⊕ B3)

so the frequency in window (B1, B2, B3) must be lower than the frequency in window (B2, B3) !!!

It is also true that:

freq(A, B3) ≤ freq(A, B2 ⊕ B3) ..... (2)

Because:

maximum window +----------------------+ | | I1 I2 .... I_p ... I_q-1 I_q .... I_r I_r+1 .... I_L | || || | +----------++---------++-----------+ B1 B2 B3 | | +-----------+ freq(A, B3) | | +----------------------+ freq(A, B2 ⊕ B3) = max. freq

(The = case is included because if two windows have the maximum frequency, the maximum window is the larger of the two).

Denote:

|B1| = k1, |B2| = k2, and, |B3| = k3
count(A, B1) = a1, count(A, B2) = a2, and, count(A, B3) = a3

Then we have:

From Equations (1) and (2), we have:

a1 + a2 + a3 a2 + a3 -------------- < --------- ..... (1) k1 + k2 + k3 k2 + k3 a3 a2 + a3 ---- ≤ --------- ..... (2) k3 k2 + k3

We need to prove this:

a1 a2 ---- < ---- k1 k2

Side note:

a + n n a n ------- < --- <===> --- < --- ........ (H1) b + m m b m Proof: a + n n ------- < --- // Multiply both side by m/n b + m m am + nm <==> ---------- < 1 // Multiply both side by (bn + mn) bn + mn <==> am + nm < bn + mn // Subtract both side by mn <==> am < bn // Multiply both side by 1/bm a n <==> --- < --- // Q.E.D. b m

Continue with the proof... we had:

a1 + a2 + a3 a2 + a3 -------------- < --------- ..... (1) k1 + k2 + k3 k2 + k3 a3 a2 + a3 ---- ≤ --------- ..... (2) k3 k2 + k3

Re-written as:

a1 + (a2 + a3) a2 + a3 -------------- < --------- ..... (1) k1 + (k2 + k3) k2 + k3 a3 a2 + (a3) ---- ≤ ----------- ..... (2) k3 k2 + (k3)

Conclusion:

a1 a2 + a3 ---- < --------- ...... (3) k1 k2 + k3 a2 a3 ---- >= ---- ...... (4) k2 k3

We need one more inequality:

From: a2 a3 ---- >= ---- ...... (4) k2 k3 or re-written in a different order: a3 a2 ---- <= ---- k3 k2 Using (H1) above, we get: a3 + a2 a2 --------- <= ---- k3 + k2 k2 (The <= sign will be preserved instead of the < sign) Or re-written in a different order: a2 + a3 a2 --------- <= ---- ....... (5) k2 + k3 k2

We have all the necessary inequalities (inequalties (3) and (5)) to prove the theorem:

a1 a2 + a3 ---- < --------- ...... (3) k1 k2 + k3 a2 + a3 a2 --------- <= ---- ....... (5) k2 + k3 k3
It follows that: a1 a2 ---- < ---- k1 k2

Translate back using the notations:

maximum window +----------------------+ | | I1 I2 .... I_p ... I_q-1 I_q .... I_r I_r+1 .... I_L | || || | +----------++---------++-----------+ B1 B2 B3 ^ ^ ^ | | | v v v frequencies: a1/k1 a2/k2 a3/k3
a1 a2 ---- < ---- k1 k2 means: I1 I2 ... I_p ... I_q-1 I_q ... I_r ... I_L | || | +----------++-------+ freq(A, S[p,q-1]) freq(A, S[p,r]) ∀ p ≤ q-1 ∧ r ≥ q: freq(A, S[p,q-1]) < freq(A, S[p,r])

Q.E.D.

Corollary 1: property of a border

Corollary 1:

Let stream S be:
(Remember the definition of a border: it is a point that can potentially become the start of the maximum window - with an appropriate continuation)

Position q is a border for itemset A, if an only if:

I1 I2 ... I_j ... I_q-1 I_q ... I_k ... I_L | || | +----------++-------+ freq(A, S[j,q-1]) freq(A, S[p,k]) ∀ j ≤ q-1 ∧ k ≥ q: freq(A, S[j,q-1]) < freq(A, S[q,k])

Proof: part 1 (only if, easy)

We must prove:

border

itemset A

then

continuation

starting position

maximum window

Graphically:

maximum window +---------------+ | | I1 I2 ... I_j ... I_q-1 I_q ... I_k ... I_L | || | +----------++--------+ freq(A, S[j,q-1]) freq(A, S[p,k]) ∀ j ≤ q-1 ∧ k ≥ q: freq(A, S[j,q-1]) < freq(A, S[q,k])

If q is a border, then - using the appropriate continuation - q will become the start location of a maximum window:

I1 I2 ... I_j ... I_q-1 I_q ... I_k ... I_L a b c d .... | | +------------------------------+ maximum window

And because [q, ...] is a maximum window, according to Theorem 1:

maximum window +----------------------------+ | | I1 I2 ... I_p ... I_q-1 I_q ... I_r ... I_L a b c d .... | || | +----------++--------+ freq(A, S[p,q-1]) freq(A, S[q,r]) ∀ p ≤ q-1 ∧ r ≥ q: freq(A, S[p,q-1]) < freq(A, S[q,r])

Proof: part 2 (if, hard...)

We must prove:

I1 I2 ... I_j ... I_q-1 I_q ... I_k ... I_L | || | +----------++--------+ freq(A, S[j,q-1]) freq(A, S[p,k]) ∀ j ≤ q-1 ∧ k ≥ q: freq(A, S[j,q-1]) < freq(A, S[q,k])

then position q is a border for itemset A

The proof is divided into 2 cases:

Case 1: q is the right most border

If q is the right most border, then q can be easily shown to be the start of a maximum window:

I1 I2 ... I_j ... I_q-1 I_q ... I_k ... I_L | || | +----------++---------------+ freq(A, S[j,q-1]) freq(A, S[p,L]) ∀ j ≤ q-1: freq(A, S[j,q-1]) < freq(A, S[q,L])

With a different notation:

(a1+a2)/(k1+k2) +---------------------------+ | | I1 I2 ... I_j ... I_q-1 I_q ... I_k ... I_L | || | +----------++---------------+ a1/k1 a2/k2 ∀ j ≤ q-1: a1/k1 < a2/k2

The frequency in any larger window is:

a1 + a2 a2 --------- < ---- k1 + k2 k2

which is always less than the frequency in S[q,L].

So, S[q,L] is the maximum window

Case 2: q is not the right most border

If q is not the right most border, then we have another border x to the right of q:

border border | | v v I1 I2 .... I_j ... I_q-1 I_q ... I_k ... I_x-1 I_x ... I_y ... I_L | | | || | +----------+ +-----------------++--------------------+
Facts: (1) Because q is a border: freq(A, S[i,q-1]) < freq(A, S[q, k]) (2) Because x is a border: freq(A, S[k,x-1]) < freq(A, S[x, y])

We show that:
(So that q will become the right most border and we have case 1 again).
By consecutively applying this procedure, any "inner" border can become the right most border - (and therefore, become the starting position of the maximum window)

Consider the two right most border:

border border | | v v I1 I2 ... I_q-1 I_q .... I_q'-1 I_q' .... I_L | || | +-------------++------------+ freq(A, S[q, q'-1]) freq(A, S[q', L]) ^ ^ | | v v a1/k1 a2/k2
From the fact that q' is a border, we have that: freq(A, S[q, q'-1]) < freq(A, S[q', L]) or: a1 a2 ---- ≤ ---- k1 k2 (NOTE: a2/k2 can be equal to a1/k1, because if the frequency are the same, then the larger window is the maximum window)
Claim: We can always find 2 integers x and y, such that: a1 a2 + x ---- = -------- k1 k2 + y

Example:

3 7 3 9 --- ≤ --- --- = ---- 4 8 4 16 3 7 + 2 --- = ------- 4 8 + 8

Now, consider the following continuation:

border border Cont = x A's and y O's | | +------------------------+ v v | | I1 I2 ... I_q-1 I_q .... I_q'-1 I_q' .... I_L A A ... A O O ... O | || | +-------------++------------+ freq(A, S[q, q'-1]) freq(A, S[q', L]) ^ ^ | | v v a1/k1 a2/k2 Then: freq(A, S[q, q'-1]) = freq(A, S[q', L+Cont]) And: q' is no longer a border with this continuation !

We have effectively made q the new right-most border with this specific continuation
So we can always make an "inner" border into a right-most border with a certain continuation and reduce it to case 1 - and that concludes the proof

Summary data structure to find the maximum window

Based on the above theorem and corollary, Calders designed the following summary data structure that contains all the information on the border points that will help us find the maximum window (and the maximum frequency)

Supose the border locations for the item set A are:

p₁
p₂
...
p_r

and define:

a₁ = count(A, S[p₁, p₂ - 1])
a₂ = count(A, S[p₂, p₃ - 1])
...
a_r = count(A, S[p_r, p_t])

Graphically:

S: I1 I2 .... I_p₁-1 I_p₁ ... I_p₂-1 I_p₂ ... I_p₃-1 .... I_{p_r} ... I_t | || | | | +-----------++------------+ +---------+ a1 a2 a_r

The summary of the stream for item set A consists of this array:

+----+----+------+----+ | p₁ | p₂ | ..... | p_r | +----+----+------+----+ | a₁ | a₂ | ..... | a_r | +----+----+------+----+

Example:

Suppose the border positions are: V V V S = b a a a b a a b a b b a a a a b a 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

Summary of the stream:

+----+----+----+ | 2 | 12 | 17 | +----+----+----+ | 6 | 4 | 1 | +----+----+----+ (There are 6 a's from pos 2 to 11) (There are 4 a's from pos 12 to 16) (There are 1 a's from pos 17 to ...)

Computing frequencies at borders using the summary:

S: I1 I2 .... I_p₁-1 I_p₁ ... I_p₂-1 I_p₂ ... I_p₃-1 .... I_{p_r-1} ... I_{p_r-1} I_{p_r} ... I_t | || | | | | | +-----------++------------+ +-------------+ +---------+ a1 a2 a_r-1 a_r | | +---------+ a_r freq(A, S[p_r,t]) = ------------ t - p_r + 1 | | +-------------------------+ a_r-1 + a_r freq(A, S[p_r,t]) = ------------ t - p_r-1 + 1

Example:

Suppose the border positions are: V V V S = b a a a b a a b a b b a a a a b a 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

Frequencies at the border locations:

+----+----+----+ | 2 | 12 | 17 | +----+----+----+ | 6 | 4 | 1 | +----+----+----+ t = 17 freq(a, S[17,17]) = 1/1 freq(a, S[12, 17]) = (4+1)/6 = 5/6 freq(a, S[2, 17]) = (6+4+1)/16 = 11/16

An invariant of the frequencies at the border locations

Property:

Let S_t be the stream with border locations p₁, p₂,..., p_r.

S: I1 I2 .... I_p₁-1 I_p₁ ... I_p₂-1 I_p₂ ... I_p₃-1 .... I_{p_r} ... I_t | || | | | +-----------++------------+ +---------+ a1 a2 a_r

Let Summary(S_t) be the summary of the stream S_t:

+----+----+------+----+ | p₁ | p₂ | ..... | p_r | (p_i is a border) +----+----+------+----+ | a₁ | a₂ | ..... | a_r | +----+----+------+----+

Then:

a₁ a₂ a_r-1 a_r --------- < --------- < .... < --------- < ------------- p₂ - p₁ p₃ - p₂ p_r - p_r-1 t - p_r + 1

Proof:

From Corollary 1 (see: click here ):

border

itemset A

if an only if

I1 I2 ... I_j ... I_q-1 I_q ... I_k ... I_L | || | +----------++--------+ freq(A, S[j,q-1]) freq(A, S[p,k]) ∀ j ≤ q-1 ∧ k ≥ q: freq(A, S[j,q-1]) < freq(A, S[q,k])

Since p₂ is a borders location, we have:

I1 I2 ... I_j ... I_p₂-1 I_p₂ ... I_k ... I_L | | | | +----------+ +--------+ freq(A, S[j,p₂-1]) freq(A, S[p₂,k]) ∀ j ≤ p₂-1 ∧ k ≥ p₂: freq(A, S[j,p₂-1]) < freq(A, S[p₂,k])

This is also true for j = p₁ and k = p₃ - 1:

freq(A, S[ p₁,p₂-1]) < freq(A, S[p₂, p₃-1])

which translates to

S: I1 I2 .... I_p₁-1 I_p₁ ... I_p₂-1 I_p₂ ... I_p₃-1 .... I_{p_r} ... I_t | || | | | +-----------++------------+ +---------+ a1 a2 a_r a₁ freq(A, S[ p₁,p₂-1]) = --------- p₂ - p₁ a₂ freq(A, S[ p₂,p₃-1]) = --------- p₃ - p₂

By applying Corollary 1 to each border position, we obtain the result:

a₁ a₂ a_r-1 a_r --------- < --------- < .... < --------- < ------------- p₂ - p₁ p₃ - p₂ p_r - p_r-1 t - p_r + 1

Obtaining the maximum frequency from the summary when mwl = 1

Because of the property:

a₁ a₂ a_r-1 a_r --------- < --------- < .... < --------- < ------------- p₂ - p₁ p₃ - p₂ p_r - p_r-1 t - p_r + 1

and the fact that:

Only border positions can be a starting position of a maximum window

We can conclude that when mwl = 1, that:

the maximum frequency of an item set A is:

a_r mfreq^mwl=1(A, S) = ------------- t - p_r + 1

The starting position of the maximum window is:

startmax(A, S) = p_r

Algorithm to maintain the summary information

The algorithm is presented in a simplified manner first:
The algorithm will be generalized later

The algorithm to maintain the summary information is as follows:

// A = the item for which the max. frequency is sought // Input consists of item sets, hence the test A ⊆ I S_t = ∅; while ( not end of stream ) do { I = next item set; // S_t = current summary // S_t+1 = summary after processing the next item set S_t+1 = ∅; // New summary information if ( S_t == ∅ ) { if ( A ⊆ I ) { S_t+1 = [(t+1, 1)]; // One border } } else { if ( A ⊆ I ) { /* ----------------------------------------- An addition of A will either: 1. extend the last border region 2. create a new border It will not remove existing border ----------------------------------------- */ if ( a_r == t - p_r + 1 ) { // Situation: .... A A A A A S_t+1 = S_t; Update: a_r = a_r + 1; // extend last border } else { // Situation: .... A A A A O S_t+1 = S_t ⊕ [(t+1), 1]; // create new border } } else { /* ----------------------------------------- An addition of not-A may cause a border to disappear... ----------------------------------------- */ S_t+1 = S_t; // Original set of border... // Remove borders that do not satisfy the property (click here) i = r; while ( i > 1 ) { a_i a_i-1 + a_i if ( ------------- ≤ ------------- ) t - p_i + 1 t - p_i-1 + 1 { a_i-1 = a_i-1 + a_i; // Merge the counts in 2 consecutive segments delete entry (p_i, p_i) from S; i--; } else { break; } } } } } /* ------------------- Maximum window ------------------- */ Max. window = [p_r..t]; a_r Max. frequency = ----------- t - r + 1

Rational of the algorithm:

The following property must hold for all the border positions:

a₁ a₂ a_r-1 a_r --------- < --------- < .... < --------- < ------------- p₂ - p₁ p₃ - p₂ p_r - p_r-1 t - p_r + 1

When item set I contains A, the frequency of the last term changes as follows:

from: to: a_r a_r + 1 ------------- --------------- t - p_r + 1 t - p_r + 1 + 1

And because:

a_r a_r + 1 ------------- ≤ --------------- t - p_r + 1 t - p_r + 1 + 1

We will maintain the property:

a₁ a₂ a_r-1 a_r + 1 --------- < --------- < .... < --------- < ---------------- p₂ - p₁ p₃ - p₂ p_r - p_r-1 t - p_r + 1 + 1

So, no old border will be deleted

But when item set I does not contain A, the frequency of the last term changes as follows:

from: to: a_r a_r + 0 ------------- --------------- t - p_r + 1 t - p_r + 1 + 1

And because:

a_r a_r ------------- > --------------- t - p_r + 1 t - p_r + 1 + 1

We may violate the property:

a₁ a₂ a_r-1 a_r --------- < --------- < .... < --------- ?? ---------------- p₂ - p₁ p₃ - p₂ p_r - p_r-1 t - p_r + 1 + 1

If the border property is indeed violated, the border position is removed:

Why a border is deleted when the "border property" no longer applies

Rationale of the removal of a border position:

Border property:

Given the stream:

S: I1 I2 .... I_p₁-1 I_p₁ ... I_p₂-1 I_p₂ ... I_p₃-1 .... I_{p_r} ... I_t | || | | | +-----------++------------+ +---------+ a1 a2 a_r

with the summary:

+----+----+------+----+ | p₁ | p₂ | ..... | p_r | +----+----+------+----+ | a₁ | a₂ | ..... | a_r | +----+----+------+----+

Then the borders always satisfy this property:

a₁ a₂ a_r-1 a_r --------- < --------- < .... < --------- < ------------- p₂ - p₁ p₃ - p₂ p_r - p_r-1 t - p_r + 1

The border p_r is deleted when:

..................... I_{p_r-1} ........ I_{p_r-1} ...... I_t | || | +------------++-------------+ freq(A, S[p_r-1, p_r-1]) ≥ freq(A, S[p_r, p_t])

The reason is that:

This position cannot be the starting position of a maximum window !!!

The underlying reason that the deleted position cannot be the starting position of a maximum window is that:

no continuation

maximum window

A continuation will create one of these 2 outcomes:

If the continuation contains only a few itemset A, we will get outcome 1: ..................... I_{p_r-1} ........ I_{p_r-1} ...... I_t ......... I_t' | || | +------------++-------------+ | | | | +----------------------------+ | | +------------------------------------------+ freq(A, S[p_r-1, t']) ≥ freq(A, S[p_r, t']) in outcome 1, the window S[p_r-1, t'] has a higher frequency than the window S[p_r, t']
If the contiuation contains sufficiently large number of itemset A, the frequency of A in window S[p_r, t'] will exceed that in window S[p_r-1, t']. However, look at the frequency in the continuation in outcome 2: ..................... I_{p_r-1} ........ I_{p_r-1} ...... I_t ......... I_t' | || | +------------++-------------+ | | +----------------------------+ | | +------------+ freq(A, S[t+1, t']) ≥ freq(A, S[p_r, t']) The frequency in the continuation must be higher than that in window S[p_r, t']

Example Calder's Algorithm

Input Stream:

S = b a a a b a a b a b b a a a a b a

Partial Algorithm result:

1 2 3 4 5 6 7 8 9 10 S = b a a a b a a b a b b a a a a b a | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | V | | | | | | | | | | | | | | | | [] | | | | | | | V V V | V | | +-++-++-+ | +-+-+ | | |2||2||2| | |2|6| | | +-++-++-+ | +-+-+ | | |1||2||3| | |3|1| | | +-++-++-+ | +-+-+ | | | | | V V | +-+ +-+-+ | |2| |2|6| | +-+ +-+-+ | |3| |3|2| | +-+ +-+-+ | | V 2 2+3 ----- ? ----- 8-6+1 8-2+1 2 5 --- ? --- 3 7 2 5 --- ≤ --- 3 7 Delete border ! New summary: +-+ |2| +-+ |5| +-+

Worked out example (from paper)