# Efficiency/runtime trouble - may not be Ruby-specific

Hi all - apologies in advance since this issue is not necessarily a
Ruby issue, but I’ve written the program in Ruby and the community
here has been very helpful/friendly in my experience.

I’m working on another problem from Project Euler (
http://www.projecteuler.net
). This one is problem 73 (I’m not actually that smart, I just jump
around ):

“Consider the fraction, n/d, where n and d are positive integers. If
nd and HCF(n,d)=1, it is called a reduced proper fraction. How many
fractions lie between 1/3 and 1/2 in the sorted set of reduced proper
fractions for d 10,000?”

So I set up an hcf(n,d) function that prime-factor-izes n and d to see
if n/d is reduced and proper:

Primes_under_100 = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41,
43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
def hcf(n,d)
Primes_under_100.each do |prime|
if (prime < Math.sqrt(d) &&
n % prime == 0 &&
d % prime == 0) ||
d % n == 0
return 2
end
end
return 1
end

Then I ran that through a loop that builds an array of unique
instances of fractions between 1/2 and 1/3:

red_prop_fract = []
for d in 2…10_000 do
for n in 1…d do
if hcf(n,d) == 1 && n.to_f/d > 1.0/3 && n.to_f/d < 0.5
red_prop_fract.push(n.to_f/d) unless red_prop_fract.include?
(n.to_f/d)
end
end
end

p red_prop_fract.length

Now, I’m fairly certain that the algorithm is right, because I ran it
on their example of d < 9 and got the correct answer. However, this
setup is taking ridiculously long and I can’t determine why. Running
on a computer with a 2.4GHz CPU and 256 Mb of RAM, I started it at
4:30 yesterday and it hasn’t terminated!

I know that a nested ‘for’ loop that goes to n is O(n^2). And I know
that the array I’m building is rather big, and growing it gradually is
going to cause a lot of allocations and GCs. But I can’t figure out
why it’s being so ridiculous.

My first rule as a programmer is “Assume all problems are my fault.”

Insights?

TIA,
Andrew

On 7/25/07, Kaldrenon [email protected] wrote:

fractions lie between 1/3 and 1/2 in the sorted set of reduced proper
n % prime == 0 &&

that the array I’m building is rather big, and growing it gradually is
going to cause a lot of allocations and GCs. But I can’t figure out
why it’s being so ridiculous.

My first rule as a programmer is “Assume all problems are my fault.”

Insights?

TIA,
Andrew

Hi,

I don’t want to spoil the fun, so just few hints:

hcf(14,35) == 1 which is wrong.

I don’t think you have a problem with GC. Your algorithm is too slow.
You can verify this using profiler.

There are two ways of enhancing the speed:

1. enhacing the algorithm itself, i.e. lowering the O(n^2)and
2. tweaking the implementation (i.e. the constants)

Install the profiler, gem install ruby-prof, and run it
ruby-prof -p graph_html yourprog.rb > profile.html

Have a look at the output, and find the parts that take the most time.

Suggested changes:

1. Move things out of cycle if it doesn’t depend on the inner variable.

2. Cache hard-to-compute expressions, especially those involving floats.

3. Use Set instead of array.

4. Move easy-to-compute conditions (<, >) before the hard ones (hcf)

5. Optimize the ranges of your loops

6. Try to avoid sqrt and floats as much as possible.

Some stats (core2duo T7200 2ghz/1gb ram/4mb cache, ruby 1.8.5, wxpsp2)

d <= 500 (12687)
with change #3: 6.981
with #3 and #4: 1.484
with #3,4,2: 1.203

d <= 700 (24836)
with #3,4,2: 2.515
with #3,4,2,5: 2.141

d <= 1000 (50695)
with #3,4,2,5: 5.094
fixing hcf: 2.547

(I’m not sure about the results, I suspect at least 1/2 is included
there somehow, if not more false positives)

J.

On Jul 25, 2007, at 11:45 AM, Kaldrenon wrote:

nd and HCF(n,d)=1, it is called a reduced proper fraction. How many
if (prime < Math.sqrt(d) &&
instances of fractions between 1/2 and 1/3:

I know that a nested ‘for’ loop that goes to n is O(n^2). And I know
that the array I’m building is rather big, and growing it gradually is
going to cause a lot of allocations and GCs. But I can’t figure out
why it’s being so ridiculous.

My first rule as a programmer is “Assume all problems are my fault.”

Insights?

cents. Basically, you need a better algorithm. However, even if you
were to stick with a brute-force approach, it could be speeded up.
For one thing, you could just count the reduced fractions as you go
along rather than accumulating a list of their values and counting
that. Second, you could use the mathn standard library, which adds a
gcd method to Integer (gcd is another name for HCF). Third, you can
avoid converting integers to floats and doing float arithmetic. When
I made these changes to your algorithm, I got the following results:

~ mg: time /Users/mg/Desktop/test.rb
Number of reduced fractions is 5066251

real 7m52.685s
user 7m22.083s
sys 0m2.275s

This is on a 2-GHz iMac. It’s still very slow, but a lot faster than
what you were experiencing. And it doesn’t use much memory. I’m not
showing my code here for fear of being a spoiler. I will post it if

Regards, Morton

On Jul 25, 2:42 pm, “Jano S.” [email protected] wrote:

1. Move things out of cycle if it doesn’t depend on the inner variable.

I try to do this in general, but am I correct in my thinking that it’s
not applicable to this script?

1. Cache hard-to-compute expressions, especially those involving floats.

I’m still very much a novice - could you explain to me in simple terms
what you mean, or point me to a document/resource on this topic?

1. Use Set instead of array.

Had a BIG impact on run-time, thank you!

1. Move easy-to-compute conditions (<, >) before the hard ones (hcf)

This is something I feel dumb for not remembering myself - although
some of these other things are new material for me, this is something
I’ve dealt with before. Thanks for the reminder. Also had a big impact
on time.

1. Optimize the ranges of your loops

As with tip #1, is there a way this technique can improve this script?
I know it’s wise in general.

1. Try to avoid sqrt and floats as much as possible.

Like #1 and #5, I think - Can I really avoid floats, since I need to
get a number that’s between two floats?

Some stats (core2duo T7200 2ghz/1gb ram/4mb cache, ruby 1.8.5, wxpsp2)
d <= 500 (12687)
with change #3: 6.981
with #3 and #4: 1.484
with #3,4,2: 1.203

On my system (2.4ghz/256mb ram/(don’t know cache), ruby 1.8.6, wxpsp2)
(taking into account a number of other programs were running at the
time, it’s much slower than it was for you but the rate of change is
comparable.)
d <= 500
with change #3: ~38 seconds
with changes #3 and #4: ~8 seconds.

I also took Morton’s advice and dropped my hcf method in favor of
Mathn’s gcd2(), along with dropping the array/set for a counter. I was
wary of this at first until I realized that with /reduced/ proper
fractions, there’s no chance of a duplicate value. I had been using
the “push unless include?” style in order to avoid duplication, but
avoiding duplication is inherent.

My full script is now:

p Time.now
require ‘mathn’
red_prop_fract = 0
for d in 2…10_000 do
for n in 1…d do
if n.to_f/d > 1.0/3 && n.to_f/d < 0.5 && n.gcd2(d) == 1
red_prop_fract += 1
end
end
end

p red_prop_fract.length
p Time.now

Many thanks to both of you! I always love it when people who know
their stuff are genuinely willing to help newbies like me.

-Andrew

On 7/25/07, Kaldrenon [email protected] wrote:

On Jul 25, 2:42 pm, “Jano S.” [email protected] wrote:

1. Move things out of cycle if it doesn’t depend on the inner variable.

I try to do this in general, but am I correct in my thinking that it’s
not applicable to this script?

In your script Math.sqrt was computed inside an each loop. Now it’s
not there anymore, so it’s fixed.

1. Cache hard-to-compute expressions, especially those involving floats.

I’m still very much a novice - could you explain to me in simple terms
what you mean, or point me to a document/resource on this topic?

For this I meant n.to_f/d that was there three times. I’m not sure if
ruby reuses the value or computes it three times.

1. Optimize the ranges of your loops

As with tip #1, is there a way this technique can improve this script?
I know it’s wise in general.

require ‘mathn’
MAX_D = 1000
red_prop_fract = 0
for n in 2…(MAX_D/2).ceil do
lower = n2+1
upper = [n
3-1,MAX_D].min
for d in lower…upper do
if n.gcd2(d) == 1
red_prop_fract += 1
end
end
end

1. Try to avoid sqrt and floats as much as possible.

Like #1 and #5, I think - Can I really avoid floats, since I need to
get a number that’s between two floats?

As morton said, instead of checking a/b > c/d you can do ad > bc.
And even in case of storing the fractions you can encode them into a
Fixnum: (n<<16 + d)

You can optimize sqrt when you know you are iterating over a range:

sqrt =1
sqr = 1
for n in 1…10000
if n > sqr
sqr += 2*sqrt + 1
sqrt += 1
end

# now sqrt == Math.sqrt(n).ceil

#… do whatever you need with sqrt
end

When looking at the source of Mathn.gcd2, (and after profiling) we
can see that most of the script is spent inside Prime.succ, so another
way to get a little gain is to copy the code, and replace

def prime_division
raise ZeroDivisionError if self == 0
ps = Prime.new
value = self
pv = []
for prime in ps

with

def prime_division
value = self
pv = []
for prime in Primes_under_100

(we can skip the zero check, as we do not send zero)

But for me gcd is still faster (there we can get a little gain as well
if we drop the .abs calls)

The last thing I could do is to skip over when both n and d are even -
it’s a fast check that avoids expensive gcd check

Inlining the function also helps a bit.
This is my version:
t = Time.now
MAX_D = 1000
red_prop_fract = 0
for n in 2…(MAX_D/2).ceil do
lower = n2+1
upper = [n
3-1,MAX_D].min
for d in lower…upper do
next if (d|n) & 1 == 0
min = n
max = d
while min > 1
tmp = min
min = max % min
max = tmp
end
if min == 1
red_prop_fract += 1
end
end
end
p red_prop_fract
p Time.now-t

On Jul 26, 3:32 am, “Jano S.” [email protected] wrote:

For this I meant n.to_f/d that was there three times. I’m not sure if
ruby reuses the value or computes it three times.

In other words, if the exact same calculation is needed in more than
one place, store it instead of recalculating? Gotta remember that
one…

end
end

…I can’t believe I didn’t even think to narrow the loop down to d’s
where d is less than three but more than two times larger than n.

min = n
end
p red_prop_fract
p Time.now-t

That’s a good-looking solution.

I’m still getting much longer times than the two of you have cited,
but I’m chalking that up to the fact that my work machine is a low-
cost, out-of-the-box Dell and has worse specs.

Thanks again!
Andrew

On 7/26/07, Kaldrenon [email protected] wrote:

On Jul 26, 3:32 am, “Jano S.” [email protected] wrote:

For this I meant n.to_f/d that was there three times. I’m not sure if
ruby reuses the value or computes it three times.

In other words, if the exact same calculation is needed in more than
one place, store it instead of recalculating? Gotta remember that
one…

Right, but it depends on calculation’s complexity. sometimes creating
a variable for it could be longer that the calculation itself. For
best results, always use profile and Benchmark class.

For me profiler says, most of the time is spent in gcd calculation.
So, further optimizations should go into either avoiding gcd as much
as possible, speeding it up and/or caching its results somehow
(sometimes you can trade memory complexity for space). I haven’t found
any such optimizations.

Final note: this was an exercise in optimizing as much as possible. In
normal code one does not do such things, as this would complicate the
maintainability of the code. It’s important to weight speed vs.
complexity of the code…

J.

On Jul 25, 2007, at 5:29 PM, Kaldrenon wrote:

end
end

p red_prop_fract.length
p Time.now

That’s very close to what I did. However, the conditional clause of
the if-statement can be improved a little:

`````` if 3 * n > d && 2 * n < d && n.gcd(d) == 1
``````

Staying with integers is faster, and gcd is a little faster than gcd2.

Regards, Morton