How to make this faster?

[jzakiya]

I have this code snippet below:

p=11
while p <= sqrtN
return false if
n%(p) == 0 or n%(p+2) ==0 or n%(p+6) == 0 or n%(p+8) ==0 or
n%(p+12) == 0 or n%(p+18) ==0 or n%(p+20) == 0 or n%(p+26) ==0 or
n%(p+30) == 0 or n%(p+32) ==0 or n%(p+36) == 0 or n%(p+42) ==0 or
n%(p+48) == 0 or n%(p+50) ==0 or n%(p+56) == 0 or n%(p+60) ==0 or
n%(p+62) == 0 or n%(p+68) ==0 or n%(p+72) == 0 or n%(p+78) ==0 or
n%(p+86) == 0 or n%(p+90) ==0 or n%(p+92) == 0 or n%(p+96) ==0 or
n%(p+98) == 0 or n%(p+102)==0 or n%(p+110)== 0 or n%(p+116)==0 or
n%(p+120)== 0 or n%(p+126)==0 or n%(p+128)== 0 or n%(p+132)==0 or
n%(p+138)== 0 or n%(p+140)==0 or n%(p+146)== 0 or n%(p+152)==0 or
n%(p+156)== 0 or n%(p+158)==0 or n%(p+162)== 0 or n%(p+168)==0 or
n%(p+170)== 0 or n%(p+176)==0 or n%(p+180)== 0 or n%(p+182)==0 or
n%(p+186)== 0 or n%(p+188)==0 or n%(p+198)== 0 or n%(p+200)==0
p += mod
end
return true

that I can replace with this:

modk,r=0,1; p=11
while (p+modk) <= sqrtN
return false if ( s=false; res[1..-1].each {|r| s ||= n%(r+modk)==0}; s)
modk +=mod
end
return true

The second snippet is much shorter than the first, but slower.

Is there a way to make the line below faster?

return false if ( s=false; res[1..-1].each {|r| s ||= n%(r+modk)==0}; s)

What I want to do is take an element from res and do -- n%(r+modk)==0.
If it becomes true I want to return false (immediately if possible). If it's always false for every array element I go through the loop again until the end
condition is met.

 

[Robert Klemme]

I have this code snippet below:

p=11
while p <= sqrtN
return false if
n%(p) == 0 or n%(p+2) ==0 or n%(p+6) == 0 or n%(p+8) ==0 or
n%(p+12) == 0 or n%(p+18) ==0 or n%(p+20) == 0 or n%(p+26) ==0 or
n%(p+30) == 0 or n%(p+32) ==0 or n%(p+36) == 0 or n%(p+42) ==0 or
n%(p+48) == 0 or n%(p+50) ==0 or n%(p+56) == 0 or n%(p+60) ==0 or
n%(p+62) == 0 or n%(p+68) ==0 or n%(p+72) == 0 or n%(p+78) ==0 or
n%(p+86) == 0 or n%(p+90) ==0 or n%(p+92) == 0 or n%(p+96) ==0 or
n%(p+98) == 0 or n%(p+102)==0 or n%(p+110)== 0 or n%(p+116)==0 or
n%(p+120)== 0 or n%(p+126)==0 or n%(p+128)== 0 or n%(p+132)==0 or
n%(p+138)== 0 or n%(p+140)==0 or n%(p+146)== 0 or n%(p+152)==0 or
n%(p+156)== 0 or n%(p+158)==0 or n%(p+162)== 0 or n%(p+168)==0 or
n%(p+170)== 0 or n%(p+176)==0 or n%(p+180)== 0 or n%(p+182)==0 or
n%(p+186)== 0 or n%(p+188)==0 or n%(p+198)== 0 or n%(p+200)==0
p += mod
end
return true

that I can replace with this:

modk,r=0,1; p=11
while (p+modk) <= sqrtN
return false if ( s=false; res[1..-1].each {|r| s ||= n%(r+modk)==0}; s)
modk +=mod
end
return true

The second snippet is much shorter than the first, but slower.

Is there a way to make the line below faster?

return false if ( s=false; res[1..-1].each {|r| s ||= n%(r+modk)==0}; s)

What I want to do is take an element from res and do -- n%(r+modk)==0.
If it becomes true I want to return false (immediately if possible). If it's always false for every array element I go through the loop again until the end
condition is met.

Here's another approach for your benchmarking:

OFFSETS = [0, 2, 6, 8, ...]

....

p = 11

while p <= sqrtN
return false if OFFSETS.any? {|off| n % (p + off) == 0}
p += mod
end

return true

Cheers

robert