The main idea of the article is to prove that there exist a tester of monotonicity with query and time complexity.

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Presentation transcript:

The main idea of the article is to prove that there exist a tester of monotonicity with query and time complexity

Consider the task of checking monotonicity of functions defined over partially ordered set S. Suppose that for some distribution on pairs with and for every function where C defends on S only. Then for every and every function for pairs selected according to the same distribution

For each 2 functions - the fraction of instances On which - the minimum distance between function and any other monotone function - the probability that a pair selected according to witnesses that is not monotone.

Incrementally transform into a monotone function, while insuring that for each repaired violated edge, the value of the function changed only in a few points.

- arbitrary monotone function at distance from

Claim: Proof: by the definition of CLEAR by the definition of MON

 Interval of a violated edge with respect to function -  two intervals cross if they intersect in more than one point. example: [2,3], [4,6] [1,6]

Lemma: The function has the following properties: has no violated edges whose intervals cross. 3. The interval of a violated edge with respect to is contained in the interval of this edge with respect to.

Define Note: 1.is monotone and takes values from We will check the 4 possibilities for : 1. - not possible. Why? 2.- agree on is violated by and. Proves (1) & (3). If cross Contradiction to the monotonicity of CLEAR definition

3. - is violated Therefore intersects in one point only -. This proves (2) In case (1) & (3) follows. If not then (1) & (3) follows symmetric to case 3.

Lemma: given define: Those functions have the following properties:

1.The SQUASH operator never adds new violated edges

2.Note:

3.Note: Why? the distance from to the set of monotone functions is at most the distance to a particular monotone function :

We will prove by induction on that for every function the following hypothesis: Base case : In the theorem we assumed - By the definition of detect we get the hypothesis.

Lets assume the hypothesis holds for and prove it for :