Feature #15233
closedSpeeding Up Matrix Powers
Description
I've been looking at SymPy's slow matrix power speed, and noticed that replacing Ruby's matrix power code with the equivalent code from SymPy makes it significantly faster. As this is a recursively defined function, presumably it can be made even faster with a proper iterative version.
Base:
require 'matrix'
Q = Matrix[[0,0,1,0,2,0],
[0,0,0,1,0,0],
[1,0,0,1,0,2],
[0,2,1,0,0,0],
[1,0,0,0,0,0],
[0,0,1,0,0,0]]
r = Vector[1,0,0,0,0,0]
p = (Q**100000000*r).sum
puts p.to_s.size
Output:
time ruby main.rb
35950264
real 1m42.250s
user 1m41.050s
sys 0m1.184s
Modified:
require 'matrix'
def pow(a,num)
if (num==1)
return a
end
if (num%2)==1
return a*pow(a,num-1)
end
ret = pow(a,(num/2))
ret*ret
end
Q = Matrix[[0,0,1,0,2,0],
[0,0,0,1,0,0],
[1,0,0,1,0,2],
[0,2,1,0,0,0],
[1,0,0,0,0,0],
[0,0,1,0,0,0]]
r = Vector[1,0,0,0,0,0]
p = (pow(Q,100000000)*r).sum
puts p.to_s.size
Output:
time ruby main.rb
35950264
real 1m9.489s
user 1m8.661s
sys 0m0.828s
Updated by duerst (Martin Dürst) about 6 years ago
The code doesn't look very Ruby-ish. What about rewriting it as follows:
def pow(a, num)
if num==1
a
elsif num.odd?
a*pow(a, num-1)
else
ret = pow(a, num/2)
ret*ret
end
end
Updated by mame (Yusuke Endoh) about 6 years ago
- Status changed from Open to Assigned
- Assignee set to marcandre (Marc-Andre Lafortune)
Interesting. The current implementation accumulates the value in LSB-to-MSB order, and SymPy's implementation does in MSB-to-LSB because. The former is slower than the latter because the latter multiplies the original matrix (which is smaller than an accumulated matrix for each digit).
@marcandre (Marc-Andre Lafortune), could you review this?
diff --git a/lib/matrix.rb b/lib/matrix.rb
index 62852bdad0..fd9115cfad 100644
--- a/lib/matrix.rb
+++ b/lib/matrix.rb
@@ -1086,12 +1086,12 @@ def **(other)
return self.class.identity(self.column_count) if other == 0
other = -other
end
- z = nil
- loop do
- z = z ? z * x : x if other[0] == 1
- return z if (other >>= 1).zero?
- x *= x
+ z = self
+ (other.bit_length - 2).downto(0) do |i|
+ z *= z
+ z *= self if other[i] == 1
end
+ return z
when Numeric
v, d, v_inv = eigensystem
v * self.class.diagonal(*d.each(:diagonal).map{|e| e ** other}) * v_inv
Updated by CaryInVictoria (Cary Swoveland) about 6 years ago
The O(ln n) method could be written as follows.
def pow1(m, n)
return m if n == 1
p = pow1(m, n/2)
n.even? ? p*p : p*p*m
end
t = Time.now
p = (pow(Q,100000000)*r).sum % 1_000_000
#=> 109376
Time.now-t
#=> 53.2
t = Time.now
p1 = (pow1(Q,100000000)*r).sum % 1_000_000
#=> 109376
Time.now-t
#=> 54.4
Updated by marcandre (Marc-Andre Lafortune) about 6 years ago
Cool. I'll definitely have a look, in a few days probably as I'm travelling right now
Updated by jeremyevans0 (Jeremy Evans) about 5 years ago
- Tracker changed from Bug to Feature
- ruby -v deleted (
ruby 2.5.1p57 (2018-03-29 revision 63029) [x86_64-linux-gnu]) - Backport deleted (
2.3: UNKNOWN, 2.4: UNKNOWN, 2.5: UNKNOWN)
Updated by Anonymous almost 4 years ago
- Status changed from Assigned to Closed
Applied in changeset git|a83a51932dbc31b549e11b9da8967f2f52a8b07c.
[ruby/matrix] Optimize **
Avoiding recursive call would imply iterating bits starting from
most significant, which is not easy to do efficiently.
Any saving would be dwarfed by the multiplications anyways.
[Feature #15233]