Time complexity of Parallel Reduction Algorithm











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Currently I am studying GPU architecture and its concepts. In parallel Reduction technique, how is the time complexity shown on 29th slide in following NVIDIA guide come O(N/P + log N)? I know that for N threads, it will be O(log N). If we have P threads parallel available then time complexity should be O((N/P)*log P). Right? Where am I wrong here?



Parallel Reduction Techniques






parallel-processing cuda time-complexity gpu-programming reduction







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    up vote
    1
    down vote

    favorite
    1












    Currently I am studying GPU architecture and its concepts. In parallel Reduction technique, how is the time complexity shown on 29th slide in following NVIDIA guide come O(N/P + log N)? I know that for N threads, it will be O(log N). If we have P threads parallel available then time complexity should be O((N/P)*log P). Right? Where am I wrong here?



    Parallel Reduction Techniques






    parallel-processing cuda time-complexity gpu-programming reduction







    share|improve this question
























      up vote
      1
      down vote

      favorite
      1









      up vote
      1
      down vote

      favorite
      1






      1





      Currently I am studying GPU architecture and its concepts. In parallel Reduction technique, how is the time complexity shown on 29th slide in following NVIDIA guide come O(N/P + log N)? I know that for N threads, it will be O(log N). If we have P threads parallel available then time complexity should be O((N/P)*log P). Right? Where am I wrong here?



      Parallel Reduction Techniques






      parallel-processing cuda time-complexity gpu-programming reduction







      share|improve this question













      Currently I am studying GPU architecture and its concepts. In parallel Reduction technique, how is the time complexity shown on 29th slide in following NVIDIA guide come O(N/P + log N)? I know that for N threads, it will be O(log N). If we have P threads parallel available then time complexity should be O((N/P)*log P). Right? Where am I wrong here?



      Parallel Reduction Techniques






      parallel-processing cuda time-complexity gpu-programming reduction




      parallel-processing cuda time-complexity gpu-programming reduction






      share|improve this question













      share|improve this question











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      asked yesterday









      Tapan Modi

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          2 Answers
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          I would like to explain this with an example, consider this array with N=8 elements



          1  2  3  4  5  6  7  8


          The parallel reduction will occur in following steps



          1  2  3  4  5  6  7  8
          
  3    7      11   15
          
    10          26
          
          36


          If you count the number of reduction operations, we have 4,2 and 1 on first, second and third step respectively. So total number of operations we have is 4+2+1=7=N-1 and we do all the reductions in O(N) and we also have log(8)=3 (this is log to base 2) steps so we pay a cost to do these steps which is O(logN). Hence if we used a single thread to reduce in this way we add the two costs as they occur separately of each other and we have O(N+logN). Where O(N) is cost for doing all operations and O(logN) is cost for doing all steps. Now there is no way to parallelize the cost for steps since they have to happen sequentially. However we can use multiple threads to do the operations and divide the O(N) cost to O(N/P). Therefore we have



          Total cost = O(N/P + logN)





          share|improve this answer






























            up vote
            1
            down vote













            I'm not familiar with cuda, but usualy in parallel reductions you do




            • first a local reduction on each processors, which would take O(N/P), and then

            • compute a reduction of the P local results, which takes O(log P) step.


            Hence you get O(N/P + log P).






            share|improve this answer





















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              2 Answers
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              up vote
              0
              down vote



              accepted










              I would like to explain this with an example, consider this array with N=8 elements



              1  2  3  4  5  6  7  8


              The parallel reduction will occur in following steps



              1  2  3  4  5  6  7  8
              
  3    7      11   15
              
    10          26
              
          36


              If you count the number of reduction operations, we have 4,2 and 1 on first, second and third step respectively. So total number of operations we have is 4+2+1=7=N-1 and we do all the reductions in O(N) and we also have log(8)=3 (this is log to base 2) steps so we pay a cost to do these steps which is O(logN). Hence if we used a single thread to reduce in this way we add the two costs as they occur separately of each other and we have O(N+logN). Where O(N) is cost for doing all operations and O(logN) is cost for doing all steps. Now there is no way to parallelize the cost for steps since they have to happen sequentially. However we can use multiple threads to do the operations and divide the O(N) cost to O(N/P). Therefore we have



              Total cost = O(N/P + logN)





              share|improve this answer



























                up vote
                0
                down vote



                accepted










                I would like to explain this with an example, consider this array with N=8 elements



                1  2  3  4  5  6  7  8


                The parallel reduction will occur in following steps



                1  2  3  4  5  6  7  8
                
  3    7      11   15
                
    10          26
                
          36


                If you count the number of reduction operations, we have 4,2 and 1 on first, second and third step respectively. So total number of operations we have is 4+2+1=7=N-1 and we do all the reductions in O(N) and we also have log(8)=3 (this is log to base 2) steps so we pay a cost to do these steps which is O(logN). Hence if we used a single thread to reduce in this way we add the two costs as they occur separately of each other and we have O(N+logN). Where O(N) is cost for doing all operations and O(logN) is cost for doing all steps. Now there is no way to parallelize the cost for steps since they have to happen sequentially. However we can use multiple threads to do the operations and divide the O(N) cost to O(N/P). Therefore we have



                Total cost = O(N/P + logN)





                share|improve this answer

























                  up vote
                  0
                  down vote



                  accepted







                  up vote
                  0
                  down vote



                  accepted






                  I would like to explain this with an example, consider this array with N=8 elements



                  1  2  3  4  5  6  7  8


                  The parallel reduction will occur in following steps



                  1  2  3  4  5  6  7  8
                  
  3    7      11   15
                  
    10          26
                  
          36


                  If you count the number of reduction operations, we have 4,2 and 1 on first, second and third step respectively. So total number of operations we have is 4+2+1=7=N-1 and we do all the reductions in O(N) and we also have log(8)=3 (this is log to base 2) steps so we pay a cost to do these steps which is O(logN). Hence if we used a single thread to reduce in this way we add the two costs as they occur separately of each other and we have O(N+logN). Where O(N) is cost for doing all operations and O(logN) is cost for doing all steps. Now there is no way to parallelize the cost for steps since they have to happen sequentially. However we can use multiple threads to do the operations and divide the O(N) cost to O(N/P). Therefore we have



                  Total cost = O(N/P + logN)





                  share|improve this answer














                  I would like to explain this with an example, consider this array with N=8 elements



                  1  2  3  4  5  6  7  8


                  The parallel reduction will occur in following steps



                  1  2  3  4  5  6  7  8
                  
  3    7      11   15
                  
    10          26
                  
          36


                  If you count the number of reduction operations, we have 4,2 and 1 on first, second and third step respectively. So total number of operations we have is 4+2+1=7=N-1 and we do all the reductions in O(N) and we also have log(8)=3 (this is log to base 2) steps so we pay a cost to do these steps which is O(logN). Hence if we used a single thread to reduce in this way we add the two costs as they occur separately of each other and we have O(N+logN). Where O(N) is cost for doing all operations and O(logN) is cost for doing all steps. Now there is no way to parallelize the cost for steps since they have to happen sequentially. However we can use multiple threads to do the operations and divide the O(N) cost to O(N/P). Therefore we have



                  Total cost = O(N/P + logN)






                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited yesterday

























                  answered yesterday









                  Nirvedh Meshram

                  17010




                  17010
























                      up vote
                      1
                      down vote













                      I'm not familiar with cuda, but usualy in parallel reductions you do




                      • first a local reduction on each processors, which would take O(N/P), and then

                      • compute a reduction of the P local results, which takes O(log P) step.


                      Hence you get O(N/P + log P).






                      share|improve this answer

























                        up vote
                        1
                        down vote













                        I'm not familiar with cuda, but usualy in parallel reductions you do




                        • first a local reduction on each processors, which would take O(N/P), and then

                        • compute a reduction of the P local results, which takes O(log P) step.


                        Hence you get O(N/P + log P).






                        share|improve this answer























                          up vote
                          1
                          down vote










                          up vote
                          1
                          down vote









                          I'm not familiar with cuda, but usualy in parallel reductions you do




                          • first a local reduction on each processors, which would take O(N/P), and then

                          • compute a reduction of the P local results, which takes O(log P) step.


                          Hence you get O(N/P + log P).






                          share|improve this answer












                          I'm not familiar with cuda, but usualy in parallel reductions you do




                          • first a local reduction on each processors, which would take O(N/P), and then

                          • compute a reduction of the P local results, which takes O(log P) step.


                          Hence you get O(N/P + log P).







                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered yesterday









                          Julien

                          794




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