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Finding Frequent Elements in a Data Stream

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• Notations
  • Notations used in the paper:

    Σ = alphabet with n symbols x = ( x1, x2, ... xN) x is a sequence of values. Each xi is an element from the alphabet Σ ( x is the input stream) N = number of elements in the input stream fx(a) = number of occurrences of a in x θ = threshold value           (0 ≤ θ ≤ 1) Heavy elements: I(x, θ) = {   a ∈ Σ   |   fx(a) > θ × N   }

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• Maximum number of frequent elements
  • Lemma 2.1:

    The number of elements that has frequency greater than θN is at most 1/θ I.e.: |   I(x, θ)   |   ≤  1/θ

  • Proof:

    If there are more than 1/θ elements with frequency θN or more, then there will be more than (1/θ) × θN = N elements in the stream But there at only N elements in the stream.

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• Algorithm to find frequent elements in a data stream
  • Proposed Algorithm:

    Variable utilization: K = set of potentially heavy elements count[e] = adjusted count of element e ∈ K K = ∅; // Initialize while ( not end of stream ) { x = next item in stream; /* ------------------------------------- Insert phase ------------------------------------- */ if ( x ∈ K ) { count[x]++; } else { insert x into K; count[x] = 1; } /* ---------------------------------- Delete phase ---------------------------------- */ if ( |K| > 1/θ ) { for ( each e in K ) { count[e]--; if ( count[e] == 0 ) { delete e from K; } } } } Output: K contains all elements with frequency > θN

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• Theorem 2.3a: |K| ≤ 1/θ
  • Theorem 2.3a:

    The size of the output set K is at most equal to 1/θ elements

    Proof:

    The final phase of the algorithm is the delete phase The delete phase is triggered when |K| > 1/θ It will delete at least one element (Specifically, the newly inserted element that caused K to become larger than 1/θ will be deleted along with other elements whose count has become 1) Therefore, when the algorithm terminates, |K| is at most equal to 1/θ

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• Theorem 2.3b: I(x, θ) ⊆ K
  • Theorem 2.3b:

    The heavy elements with count ≥ θN are included in the output K of the algorithm

    Proof:

    To proof that A &sube B, we can show that: ∀ x ∈ A:   x ∈ B In this case, it is easier to show the reverse: A &sube B   ⇔   ∀ x ∉ B:   x ∉ A See: Let a ¬in K be an arbitrary element that was not included in K at the end of the algorithm Let fx(a) be the (actual) number of occurence of a in the input stream (fx(a) > 0, otherwise the proof is trivial...) Therefore, a was delete by the delete phase The delete phase will decrement the count of each element in the current set K when the |K| > 1/θ Therefore, each occurence of a was deleted together with 1/θ -1 other symbols So the total number of symbols deleted due to the removal of a is: fx(a) × (1 (a)   +   (1/θ - 1) (the other symbols)) Or:     fx(a) × 1/θ There are at most N symbols removed, therefore: fx(a) × 1/θ   <   N Or:     fx(a)   <   θN Since fx(a) < θN, it follows that: x ∉ I(x,θ) In other words: ∀ x ∉ K:   x ∉ I(x,θ) Therefore: I(x,θ) &sube K