Improved the dot product between two vectors and a sparse matrix

[albertomercurio]

The changed function is a little bit faster than the current one.

N1 = 1024*10
N2 = 1024*20
T = Float64

x = Array(sprand(T, N1, 0.1))
y = Array(sprand(T, N2, 0.1))
A = sprand(T, N1, N2, 1/N1)

@benchmark dot($x, $A, $y)

BenchmarkTools.Trial: 10000 samples with 1 evaluation.
 Range (min … max):  45.800 μs … 287.700 μs  ┊ GC (min … max): 0.00% … 0.00%
 Time  (median):     57.800 μs               ┊ GC (median):    0.00%
 Time  (mean ± σ):   58.845 μs ±  10.107 μs  ┊ GC (mean ± σ):  0.00% ± 0.00%

               ▁█                                               
  ▁▁▂▃▄▅▅▃▃▄▃▃▄██▆▄▃▂▃▂▂▂▂▁▁▂▁▂▂▂▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁ ▂
  45.8 μs         Histogram: frequency by time         96.6 μs <

 Memory estimate: 0 bytes, allocs estimate: 0.



@benchmark mydot($x, $A, $y)

BenchmarkTools.Trial: 10000 samples with 1 evaluation.
 Range (min … max):  37.200 μs … 896.900 μs  ┊ GC (min … max): 0.00% … 0.00%
 Time  (median):     50.900 μs               ┊ GC (median):    0.00%
 Time  (mean ± σ):   49.762 μs ±  14.435 μs  ┊ GC (mean ± σ):  0.00% ± 0.00%

   ▂▁▃             ▆█▃                                          
  ▆████▅▄▄▃▃▃▂▂▃▃▅▄███▇▅▃▃▂▂▂▂▂▂▂▂▁▁▁▁▁▁▂▂▂▂▁▂▂▁▁▂▂▂▁▁▁▁▁▁▁▁▁▁ ▂
  37.2 μs         Histogram: frequency by time         84.9 μs <

 Memory estimate: 0 bytes, allocs estimate: 0.

And with vectors without zeros:

x = rand(T, N1)
y = rand(T, N2)
A = sprand(T, N1, N2, 1/N1)

@benchmark dot($x, $A, $y)

BenchmarkTools.Trial: 10000 samples with 1 evaluation.
 Range (min … max):  166.900 μs …  1.750 ms  ┊ GC (min … max): 0.00% … 0.00%
 Time  (median):     214.800 μs              ┊ GC (median):    0.00%
 Time  (mean ± σ):   224.317 μs ± 37.141 μs  ┊ GC (mean ± σ):  0.00% ± 0.00%

                 ▆█                                             
  ▁▁▁▁▁▁▂▂▂▂▆▅▇▅▅███▆▆▄▆█▆▄▃▂▂▃▃▂▂▂▂▂▂▂▂▂▁▂▂▂▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁ ▂
  167 μs          Histogram: frequency by time          331 μs <

 Memory estimate: 0 bytes, allocs estimate: 0.


@benchmark mydot($x, $A, $y)

BenchmarkTools.Trial: 10000 samples with 1 evaluation.
 Range (min … max):  146.700 μs … 991.600 μs  ┊ GC (min … max): 0.00% … 0.00%
 Time  (median):     178.900 μs               ┊ GC (median):    0.00%
 Time  (mean ± σ):   184.942 μs ±  25.418 μs  ┊ GC (mean ± σ):  0.00% ± 0.00%

       ▁   ▃▃▅██▆▃▁ ▁▂                                           
  ▂▁▂▄▅█▄▃▃███████████▇▆█▅█▄▄▃▃▂▂▂▂▂▂▂▁▁▁▁▁▂▂▂▂▂▁▁▁▁▂▂▂▁▁▁▁▁▁▁▁ ▃
  147 μs           Histogram: frequency by time          277 μs <

 Memory estimate: 0 bytes, allocs estimate: 0.

[albertomercurio]

For the moment I left the check iszero for the elements of the y vector. However, this slows down a lot the calculation. I think that, if the user knows that the vector contains several zeros, he/she can convert it to a sparsevector and execute the specific dot product for this type, leaving the maximum performances for the operations with really dense vectors.

[dkarrasch]

Nice. A few comments.

[dkarrasch]

I think that, if the user knows that the vector contains several zeros, he/she can convert it to a sparsevector and execute the specific dot product for this type,

Good point. Let's assume that for AbstractVectors generic elements are not zero, and skip that test. The main optimization comes from the known sparsity of A.

[dkarrasch]

Would be interesting to see the timings now without the zero-test.

[albertomercurio]

N1 = 1024*20
N2 = 1024*40
T = Float64
x = rand(T, N1)
y = rand(T, N2)


@benchmark dot($x, $A, $y)

BenchmarkTools.Trial: 10000 samples with 1 evaluation.
 Range (min … max):  261.300 μs …  1.293 ms  ┊ GC (min … max): 0.00% … 0.00%
 Time  (median):     293.500 μs              ┊ GC (median):    0.00%
 Time  (mean ± σ):   322.014 μs ± 70.194 μs  ┊ GC (mean ± σ):  0.00% ± 0.00%

  ▂▂██▆▅▆▅▅▄▄▄▃▄▃▃▂▂▂▂▂▂▂▂▂▂▂▂▁▁ ▁▁▁▁▁  ▁▁                     ▂
  ██████████████████████████████████████████▇█████▇▇█▇▆▇▇▇▆▆▅▆ █
  261 μs        Histogram: log(frequency) by time       575 μs <

 Memory estimate: 0 bytes, allocs estimate: 0.



@benchmark mydot($x, $A, $y)

BenchmarkTools.Trial: 10000 samples with 1 evaluation.
 Range (min … max):  213.700 μs … 743.500 μs  ┊ GC (min … max): 0.00% … 0.00%
 Time  (median):     233.200 μs               ┊ GC (median):    0.00%
 Time  (mean ± σ):   245.754 μs ±  40.873 μs  ┊ GC (mean ± σ):  0.00% ± 0.00%

  ▄▆▅███▇▇▅▅▄▃▃▂▁▂▂▁  ▁▁                                        ▂
  ████████████████████████████████▇▇████▇██▇▆▇▆▆▆▇▇▆▆▆▆▆▆▅▅▆▅▅▄ █
  214 μs        Histogram: log(frequency) by time        428 μs <

 Memory estimate: 0 bytes, allocs estimate: 0.

[codecov]

Codecov Report

Merging #410 (3d62748) into main (54f4b39) will increase coverage by 0.78%.
Report is 3 commits behind head on main.
The diff coverage is 100.00%.

@@            Coverage Diff             @@
##             main     #410      +/-   ##
==========================================
+ Coverage   92.42%   93.20%   +0.78%     
==========================================
  Files          12       12              
  Lines        7666     7666              
==========================================
+ Hits         7085     7145      +60     
+ Misses        581      521      -60     

Files Changed	Coverage Δ
src/linalg.jl	91.55% <100.00%> (+4.54%)	⬆️

... and 3 files with indirect coverage changes

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[albertomercurio]

I think it is finished. The codecov is decreased because the amount of lines is less.

[dkarrasch]

One last detail I noticed, but afterwards we should really finally merge this.