Is histogram “function” linear?












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I was ask to prove/disprove if for two given images $f_1(x,y)$ and $f_2(x,y)$ the histogram of $f_1(x,y)-f_2(x,y)$ and $h_1-h_2$ are equal, where $h_i$ is the histogram of image i.



Intuitively it seems to be correct, as if we look at the images as matrices, the histogram is a function that take the matrix values and count the number of repetitions ("losing" the indices of each value)



But is there a proper mathematically proof?






statistics image-processing







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    0












    $begingroup$


    I was ask to prove/disprove if for two given images $f_1(x,y)$ and $f_2(x,y)$ the histogram of $f_1(x,y)-f_2(x,y)$ and $h_1-h_2$ are equal, where $h_i$ is the histogram of image i.



    Intuitively it seems to be correct, as if we look at the images as matrices, the histogram is a function that take the matrix values and count the number of repetitions ("losing" the indices of each value)



    But is there a proper mathematically proof?






    statistics image-processing







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I was ask to prove/disprove if for two given images $f_1(x,y)$ and $f_2(x,y)$ the histogram of $f_1(x,y)-f_2(x,y)$ and $h_1-h_2$ are equal, where $h_i$ is the histogram of image i.



      Intuitively it seems to be correct, as if we look at the images as matrices, the histogram is a function that take the matrix values and count the number of repetitions ("losing" the indices of each value)



      But is there a proper mathematically proof?






      statistics image-processing







      share|cite|improve this question











      $endgroup$




      I was ask to prove/disprove if for two given images $f_1(x,y)$ and $f_2(x,y)$ the histogram of $f_1(x,y)-f_2(x,y)$ and $h_1-h_2$ are equal, where $h_i$ is the histogram of image i.



      Intuitively it seems to be correct, as if we look at the images as matrices, the histogram is a function that take the matrix values and count the number of repetitions ("losing" the indices of each value)



      But is there a proper mathematically proof?






      statistics image-processing




      statistics image-processing






      share|cite|improve this question















      share|cite|improve this question













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      edited Jan 17 at 20:10







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      asked Jan 17 at 19:56









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          2 Answers
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          $begingroup$

          Consider one-dimensional images
          $$f_1=[0,0,0,1]$$
          $$f_2=[0,1,0,0]$$
          Then we can say that
          $$h_1=begin{bmatrix}(-1,0)\(0,3)\(1,1)end{bmatrix}$$
          $$h_2=begin{bmatrix}(-1,0)\(0,3)\(1,1)end{bmatrix}$$
          $$h_1-h_2=begin{bmatrix}(-1,0)\(0,0)\(1,0)end{bmatrix}$$
          $$f_1-f_2=[0,-1,0,1]$$
          $$h_{1-2}=begin{bmatrix}(-1,1)\(0,2)\(1,1)end{bmatrix}$$



          In general, if two images are different from each other but have the same histogram, then the histogram of the difference of the images will not be equal to the difference of the histograms.



          Have I misunderstood something? I don't know if there's even a rigorous idea of "histogram subtraction".






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            No, the difference of histograms of images is not equal to the histogram of difference of images.



            Consider the following counter example:





            • $f_1$ image of dimensions $1 times 1$ with $f_1(0,0) = 0$,


            • $f_2$ image of dimensions $1 times 1$ with $f_2(0,0) = 1$.


            Define $f_d(0,0) = f_1(0,0) - f_2(0,0) = 0 - 1 = -1$. Now take the histograms:




            • $h[f_1](x) = begin{cases} 0 & text{for } x = 1,\
              1 & text{for } x = 0,\
              0 & text{for } x = -1,
              end{cases}$

            • $h[f_2](x) = begin{cases} 1 & text{for } x = 1,\
              0 & text{for } x = 0,\
              0 & text{for } x = -1,
              end{cases}$

            • $h[f_1 - f_2](x) = begin{cases} 0 & text{for } x = 1,\
              0 & text{for } x = 0,\
              1 & text{for } x = -1.
              end{cases}$


            The histogram difference results to be




            • $(h[f_1] - h[f_2])(x) = begin{cases} -1 & text{for } x = 1,\
              1 & text{for } x = 0,\
              0 & text{for } x = -1.
              end{cases}$


            Hence we deduce that the histogram operator is not linear:
            $$ h[f_1 - f_2] neq h[f_1] - h[f_2].$$






            share|cite|improve this answer









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              2 Answers
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              $begingroup$

              Consider one-dimensional images
              $$f_1=[0,0,0,1]$$
              $$f_2=[0,1,0,0]$$
              Then we can say that
              $$h_1=begin{bmatrix}(-1,0)\(0,3)\(1,1)end{bmatrix}$$
              $$h_2=begin{bmatrix}(-1,0)\(0,3)\(1,1)end{bmatrix}$$
              $$h_1-h_2=begin{bmatrix}(-1,0)\(0,0)\(1,0)end{bmatrix}$$
              $$f_1-f_2=[0,-1,0,1]$$
              $$h_{1-2}=begin{bmatrix}(-1,1)\(0,2)\(1,1)end{bmatrix}$$



              In general, if two images are different from each other but have the same histogram, then the histogram of the difference of the images will not be equal to the difference of the histograms.



              Have I misunderstood something? I don't know if there's even a rigorous idea of "histogram subtraction".






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Consider one-dimensional images
                $$f_1=[0,0,0,1]$$
                $$f_2=[0,1,0,0]$$
                Then we can say that
                $$h_1=begin{bmatrix}(-1,0)\(0,3)\(1,1)end{bmatrix}$$
                $$h_2=begin{bmatrix}(-1,0)\(0,3)\(1,1)end{bmatrix}$$
                $$h_1-h_2=begin{bmatrix}(-1,0)\(0,0)\(1,0)end{bmatrix}$$
                $$f_1-f_2=[0,-1,0,1]$$
                $$h_{1-2}=begin{bmatrix}(-1,1)\(0,2)\(1,1)end{bmatrix}$$



                In general, if two images are different from each other but have the same histogram, then the histogram of the difference of the images will not be equal to the difference of the histograms.



                Have I misunderstood something? I don't know if there's even a rigorous idea of "histogram subtraction".






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Consider one-dimensional images
                  $$f_1=[0,0,0,1]$$
                  $$f_2=[0,1,0,0]$$
                  Then we can say that
                  $$h_1=begin{bmatrix}(-1,0)\(0,3)\(1,1)end{bmatrix}$$
                  $$h_2=begin{bmatrix}(-1,0)\(0,3)\(1,1)end{bmatrix}$$
                  $$h_1-h_2=begin{bmatrix}(-1,0)\(0,0)\(1,0)end{bmatrix}$$
                  $$f_1-f_2=[0,-1,0,1]$$
                  $$h_{1-2}=begin{bmatrix}(-1,1)\(0,2)\(1,1)end{bmatrix}$$



                  In general, if two images are different from each other but have the same histogram, then the histogram of the difference of the images will not be equal to the difference of the histograms.



                  Have I misunderstood something? I don't know if there's even a rigorous idea of "histogram subtraction".






                  share|cite|improve this answer









                  $endgroup$



                  Consider one-dimensional images
                  $$f_1=[0,0,0,1]$$
                  $$f_2=[0,1,0,0]$$
                  Then we can say that
                  $$h_1=begin{bmatrix}(-1,0)\(0,3)\(1,1)end{bmatrix}$$
                  $$h_2=begin{bmatrix}(-1,0)\(0,3)\(1,1)end{bmatrix}$$
                  $$h_1-h_2=begin{bmatrix}(-1,0)\(0,0)\(1,0)end{bmatrix}$$
                  $$f_1-f_2=[0,-1,0,1]$$
                  $$h_{1-2}=begin{bmatrix}(-1,1)\(0,2)\(1,1)end{bmatrix}$$



                  In general, if two images are different from each other but have the same histogram, then the histogram of the difference of the images will not be equal to the difference of the histograms.



                  Have I misunderstood something? I don't know if there's even a rigorous idea of "histogram subtraction".







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 17 at 20:18









                  ShapeOfMatterShapeOfMatter

                  1794




                  1794























                      0












                      $begingroup$

                      No, the difference of histograms of images is not equal to the histogram of difference of images.



                      Consider the following counter example:





                      • $f_1$ image of dimensions $1 times 1$ with $f_1(0,0) = 0$,


                      • $f_2$ image of dimensions $1 times 1$ with $f_2(0,0) = 1$.


                      Define $f_d(0,0) = f_1(0,0) - f_2(0,0) = 0 - 1 = -1$. Now take the histograms:




                      • $h[f_1](x) = begin{cases} 0 & text{for } x = 1,\
                        1 & text{for } x = 0,\
                        0 & text{for } x = -1,
                        end{cases}$

                      • $h[f_2](x) = begin{cases} 1 & text{for } x = 1,\
                        0 & text{for } x = 0,\
                        0 & text{for } x = -1,
                        end{cases}$

                      • $h[f_1 - f_2](x) = begin{cases} 0 & text{for } x = 1,\
                        0 & text{for } x = 0,\
                        1 & text{for } x = -1.
                        end{cases}$


                      The histogram difference results to be




                      • $(h[f_1] - h[f_2])(x) = begin{cases} -1 & text{for } x = 1,\
                        1 & text{for } x = 0,\
                        0 & text{for } x = -1.
                        end{cases}$


                      Hence we deduce that the histogram operator is not linear:
                      $$ h[f_1 - f_2] neq h[f_1] - h[f_2].$$






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        No, the difference of histograms of images is not equal to the histogram of difference of images.



                        Consider the following counter example:





                        • $f_1$ image of dimensions $1 times 1$ with $f_1(0,0) = 0$,


                        • $f_2$ image of dimensions $1 times 1$ with $f_2(0,0) = 1$.


                        Define $f_d(0,0) = f_1(0,0) - f_2(0,0) = 0 - 1 = -1$. Now take the histograms:




                        • $h[f_1](x) = begin{cases} 0 & text{for } x = 1,\
                          1 & text{for } x = 0,\
                          0 & text{for } x = -1,
                          end{cases}$

                        • $h[f_2](x) = begin{cases} 1 & text{for } x = 1,\
                          0 & text{for } x = 0,\
                          0 & text{for } x = -1,
                          end{cases}$

                        • $h[f_1 - f_2](x) = begin{cases} 0 & text{for } x = 1,\
                          0 & text{for } x = 0,\
                          1 & text{for } x = -1.
                          end{cases}$


                        The histogram difference results to be




                        • $(h[f_1] - h[f_2])(x) = begin{cases} -1 & text{for } x = 1,\
                          1 & text{for } x = 0,\
                          0 & text{for } x = -1.
                          end{cases}$


                        Hence we deduce that the histogram operator is not linear:
                        $$ h[f_1 - f_2] neq h[f_1] - h[f_2].$$






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          No, the difference of histograms of images is not equal to the histogram of difference of images.



                          Consider the following counter example:





                          • $f_1$ image of dimensions $1 times 1$ with $f_1(0,0) = 0$,


                          • $f_2$ image of dimensions $1 times 1$ with $f_2(0,0) = 1$.


                          Define $f_d(0,0) = f_1(0,0) - f_2(0,0) = 0 - 1 = -1$. Now take the histograms:




                          • $h[f_1](x) = begin{cases} 0 & text{for } x = 1,\
                            1 & text{for } x = 0,\
                            0 & text{for } x = -1,
                            end{cases}$

                          • $h[f_2](x) = begin{cases} 1 & text{for } x = 1,\
                            0 & text{for } x = 0,\
                            0 & text{for } x = -1,
                            end{cases}$

                          • $h[f_1 - f_2](x) = begin{cases} 0 & text{for } x = 1,\
                            0 & text{for } x = 0,\
                            1 & text{for } x = -1.
                            end{cases}$


                          The histogram difference results to be




                          • $(h[f_1] - h[f_2])(x) = begin{cases} -1 & text{for } x = 1,\
                            1 & text{for } x = 0,\
                            0 & text{for } x = -1.
                            end{cases}$


                          Hence we deduce that the histogram operator is not linear:
                          $$ h[f_1 - f_2] neq h[f_1] - h[f_2].$$






                          share|cite|improve this answer









                          $endgroup$



                          No, the difference of histograms of images is not equal to the histogram of difference of images.



                          Consider the following counter example:





                          • $f_1$ image of dimensions $1 times 1$ with $f_1(0,0) = 0$,


                          • $f_2$ image of dimensions $1 times 1$ with $f_2(0,0) = 1$.


                          Define $f_d(0,0) = f_1(0,0) - f_2(0,0) = 0 - 1 = -1$. Now take the histograms:




                          • $h[f_1](x) = begin{cases} 0 & text{for } x = 1,\
                            1 & text{for } x = 0,\
                            0 & text{for } x = -1,
                            end{cases}$

                          • $h[f_2](x) = begin{cases} 1 & text{for } x = 1,\
                            0 & text{for } x = 0,\
                            0 & text{for } x = -1,
                            end{cases}$

                          • $h[f_1 - f_2](x) = begin{cases} 0 & text{for } x = 1,\
                            0 & text{for } x = 0,\
                            1 & text{for } x = -1.
                            end{cases}$


                          The histogram difference results to be




                          • $(h[f_1] - h[f_2])(x) = begin{cases} -1 & text{for } x = 1,\
                            1 & text{for } x = 0,\
                            0 & text{for } x = -1.
                            end{cases}$


                          Hence we deduce that the histogram operator is not linear:
                          $$ h[f_1 - f_2] neq h[f_1] - h[f_2].$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Feb 15 at 21:26









                          simonetsimonet

                          1668




                          1668






























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