Problem 2592. LocateMembership | Membership of Matrix in another matrix

Created by Vishal in Community

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Let a and b be given by

 a = [ 2 1 1 1   
       1 2 1 1 
       1 1 2 1 
       1 1 1 2
       4 0 0 0 ]

 b = [ 2 1 1
       1 2 1
       1 1 2 ]

then b can be created using a (b is wholly present in a). So output should be location of central element of b in a.

Here output is [3,3] else it should be [0,0].

Solve

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Last solution submitted on Dec 10, 2018

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rifat on 16 Sep 2014

in test case 3, b is not available in a. I might be wrong.. Please Check

Jean-Marie Sainthillier on 16 Sep 2014

Not logic to ask a scalar (if no location), an array (if 1), and a cell (if 2). And why not add the example (difficult) in the tests ?

Vishal on 17 Sep 2014

@rifat : updated, thanks for info

Jan Orwat on 18 Sep 2014

Nice problem. Some solutions (using conv, filter2 etc.) doesn't check conditions properly. Adding a=2*ones(4); b=3*eye(3)+1; or a=0; b=0; or similar can improve the testsuite;

Jean-Marie Sainthillier on 18 Sep 2014

The solution in the example is [2 2;3 3].

Solution Comments

Solution 502674

3 Comments

3 Comments

rifat on 18 Sep 2014

@Jean-Marie, using ismember(a,b) and comparing it with a 2D window of ones works in this particular case. But, ismember does not consider the order of the elements. You can verify this by calling LocateMembership(a,b') for the second test case. You would see that it results [2 4] though it is not correct.

Jean-Marie Sainthillier on 18 Sep 2014

@rifat, you're right. At the beginning, before the first corrections, I thought it was the goal of this problem (b can be created using a in any order) and I kept this approach.By the way @Vishal, it's a very interesting problem !

Vishal on 19 Sep 2014

@Jean - Thanks.

Problem 2592. LocateMembership | Membership of Matrix in another matrix

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Problem 2592. LocateMembership | Membership of Matrix in another matrix

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Last 200 Solutions

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Recent Solvers15

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