Problem 319. Biggest Value in the (Neighbor)Hood

Created by Hans

Solve Later

{"relationships":[{"relationshipType":"http://schemas.mathworks.com/matlab/code/2013/relationships/document","targetMode":"","relationshipId":"rId1","target":"/matlab/document.xml"},{"relationshipType":"http://schemas.mathworks.com/matlab/code/2013/relationships/output","targetMode":"","relationshipId":"rId2","target":"/matlab/output.xml"}],"parts":[{"partUri":"/matlab/document.xml","relationship":[],"contentType":"application/vnd.mathworks.matlab.code.document+xml","content":"<?xml version=\"1.0\" encoding=\"UTF-8\"?>\n<w:document xmlns:w=\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\"><w:body><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>For this challenge you get two inputs: a matrix A and an integer value n. Your function should return a Matrix B of the same size as A with integer values. Thereby, the entry B(i,j) counts the occurrence of lower values in the neighborhood A(i-n:i+n,j) (column wise), while symmetric boundary conditions (A(i&lt;1,j) == A(-i+1,j) should be used (compare to padarray('symmetric')). For example,</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>assume n to be 2 and a matrix A containing double values, e.g.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>A =</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"code\"/></w:pPr><w:r><w:t><![CDATA[ 0.3147 -0.4025 -0.3424 -0.3581 0.1557 0.2577 0.2060\n 0.4058 -0.2215 0.4706 -0.0782 -0.4643 0.2431 -0.4682\n -0.3730 0.0469 0.4572 0.4157 0.3491 -0.1078 -0.2231\n 0.4134 0.4575 -0.0146 0.2922 0.4340 0.1555 -0.4538\n 0.1324 0.4649 0.3003 0.4595 0.1787 -0.3288 -0.4029]]></w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>then, your function should return</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>B =</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"code\"/></w:pPr><w:r><w:t><![CDATA[ 2 1 1 1 3 4 4\n 3 2 4 2 0 2 0\n 0 2 3 3 3 1 3\n 4 2 0 1 4 3 1\n 2 4 3 4 1 1 3]]></w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>Explanation: Consider the value -0.3730 in the first column of A. The two entries above and below are all bigger and thus, the corresponding entry in B is 0. However, the entry 0.4134 is bigger than its neighbors (symmetric boundary condition) and thus, the corresponding entry in B is 4. You can assume that n&lt;=size(A,1)</w:t></w:r></w:p></w:body></w:document>"},{"partUri":"/matlab/output.xml","contentType":"text/xml","content":"<?xml version=\"1.0\" encoding=\"UTF-8\" standalone=\"no\" ?><embeddedOutputs><metaData><evaluationState>manual</evaluationState><layoutState>code</layoutState><outputStatus>ready</outputStatus></metaData><outputArray type=\"array\"/><regionArray type=\"array\"/></embeddedOutputs>"}]}

For this challenge you get two inputs: a matrix A and an integer value n.
Your function should return a Matrix B of the same size as A with integer values.
Thereby, the entry B(i,j) counts the occurrence of lower values in the neighborhood 
A(i-n:i+n,j) (column wise), while symmetric boundary conditions (A(i&lt;1,j) == A(-i+1,j)
should be used (compare to padarray('symmetric')). For example,assume n to be 2 and a matrix A containing double values, e.g.A =<pre> 0.3147 -0.4025 -0.3424 -0.3581 0.1557 0.2577 0.2060
 0.4058 -0.2215 0.4706 -0.0782 -0.4643 0.2431 -0.4682
 -0.3730 0.0469 0.4572 0.4157 0.3491 -0.1078 -0.2231
 0.4134 0.4575 -0.0146 0.2922 0.4340 0.1555 -0.4538
 0.1324 0.4649 0.3003 0.4595 0.1787 -0.3288 -0.4029</pre>then, your function should returnB =<pre> 2 1 1 1 3 4 4
 3 2 4 2 0 2 0
 0 2 3 3 3 1 3
 4 2 0 1 4 3 1
 2 4 3 4 1 1 3</pre>Explanation: Consider the value -0.3730 in the first column of A. The two
entries above and below are all bigger and thus, the corresponding entry in B is 0. 
However, the entry 0.4134 is bigger than its neighbors (symmetric boundary condition)
and thus, the corresponding entry in B is 4.
You can assume that n&lt;=size(A,1)

For this challenge you get two inputs: a matrix A and an integer value n.
Your function should return a Matrix B of the same size as A with integer values.
Thereby, the entry B(i,j) counts the occurrence of lower values in the neighborhood 
A(i-n:i+n,j) (column wise), while symmetric boundary conditions (A(i<1,j) == A(-i+1,j)
should be used (compare to padarray('symmetric')). For example,

assume n to be 2 and a matrix A containing double values, e.g.

A =

0.3147   -0.4025   -0.3424   -0.3581    0.1557    0.2577    0.2060
    0.4058   -0.2215    0.4706   -0.0782   -0.4643    0.2431   -0.4682
   -0.3730    0.0469    0.4572    0.4157    0.3491   -0.1078   -0.2231
    0.4134    0.4575   -0.0146    0.2922    0.4340    0.1555   -0.4538
    0.1324    0.4649    0.3003    0.4595    0.1787   -0.3288   -0.4029

then, your function should return

B =

2     1     1     1     3     4     4
     3     2     4     2     0     2     0
     0     2     3     3     3     1     3
     4     2     0     1     4     3     1
     2     4     3     4     1     1     3

Explanation: Consider the value -0.3730 in the first column of A. The two
entries above and below are all bigger and thus, the corresponding entry in B is 0. 
However, the entry 0.4134 is bigger than its neighbors (symmetric boundary condition)
and thus, the corresponding entry in B is 4.
You can assume that n<=size(A,1)

Solve

Solution Stats

91 Solutions
23 Solvers

Last Solution submitted on Nov 16, 2021

Last 200 Solutions

Problem Comments

4 Comments

4 Comments

Show 1 older comment

Venu Lolla on 14 Feb 2012

Is the B matrix described above correct?

Alfonso Nieto-Castanon on 15 Feb 2012

no, I believe the first and last rows should read [1 0 0 0 2 3 3] and [1 3 2 3 0 0 2], respectively

rifat on 5 Feb 2015

Is the test suite correct now? An old problem though...

Jan Orwat on 5 Feb 2015

Hello Rifat, I've solved this one recently, and it seems to be ok.

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Problem Tags

matrix neighborhood operation

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