# Sum of primes between given range

Write the shortest code for finding the sum of primes between $$\a\$$ and $$\b\$$ (inclusive).

Input

1. $$\a\$$ and $$\b\$$ can be taken from command line or stdin (space seperated)
2. Assume $$\1 \le a \le b \le 10^8\$$

Output Just print the sum with a newline character.

Bonus Points

1. If the program accepts multiple ranges (print one sum on each line), you get extra points. :)
• The upper limit is too big to allow many interesting solutions (if they have to complete in reasonable time, at least). – hallvabo Jan 28 '11 at 9:13
• @hallvabo You find inefficient solutions interesting? – Matthew Read Jan 28 '11 at 9:25
• @hallvabo, That's ok. I don't think anyone minds an ineffcient solution. If other's object, i'll be more than happy to lower the limit – st0le Jan 28 '11 at 9:38
• Just made and ran a not very optimised or concise version of the program in C#, using 1 to 10^8. Assuming my algorithm's correct, it ran in under 1m30s, and didn't overflow from a long. Seems like a fine upper limit to me! – Nellius Jan 28 '11 at 12:08
• A quick easy check: sum of primes between 1 and 100 = 1060. – Nellius Jan 28 '11 at 12:50

# Python - 194

File fsoe-sum.py:

S,E=input()
L={}
n=2
s=0
while n<=E:
try:
P=L[n];del L[n]
except:
P=[n]
if S<=n: s+=n
for p in P:
m=n+p
try:
if p not in L[m]:L[m].append(p)
except:
L[m] = [p]
n+=1
print s


Filesize is 194 bytes when using tabs to indent and no final newline.

Not the shortest pythonish solution but do you see the enbedded sieve? ;-)

Run:

$python fsoe-sum.py 1,1000000 37550402023$ python fsoe-sum.py
1,2000000
142913828922
$python fsoe-sum.py 1000001,2000000 105363426899$ python -c 'print 142913828922-37550402023'
105363426899


# Perl, 94

my$s;map{my($a,$b)=($_,0);for(2..$a-1){$a%$_==0&&$b++}$b or$s+=$a}($ARGV[0]..$ARGV[1]);print$s


This takes input from the command line. It doesn't use regex.

# Perl, 62 with bonus

use ntheory":all";while(<>){say vecsum(@{primes(split/\s+/)})}


Takes lines with two whitespace separated numbers and prints the sum of primes within the range. Exits when it sees EOF.

47 for the simple case we assume the input magically arrives in $a and$b like a few other solutions:

use ntheory":all";forprimes{$s+=$_}$a,$b;say$s  or use ntheory":all";say vecsum(@{primes($a,$b)})  With a newer module version that can be 39 characters: use ntheory":all";say sum_primes($a,$b)  # Julia, 69 bytes a,b=int(split(readline()));println(sum(setdiff(primes(b),primes(a))))  This reads a space-delimited pair of integers from STDIN and prints the result to STDOUT. The only thing I've really golfed here is whitespace; otherwise this is probably how I would go about it in a non-golfing context. Ungolfed + explanation: # Read a line from STDIN, split it into an array on the space, convert # the elements to integers, and assign the first element to a and the # second to b a, b = int(split(readline())) # Get the primes between a and b inclusive. primes(x) returns the primes # <= x, so the set difference of primes(b) and primes(a) will get us only # those between a and b d = setdiff(primes(b), primes(a)) # Print the sum to STDOUT println(sum(d))  Just realized how old this challenge is. ## Ruby, 60 bytes require'prime';p=->a,b{eval Prime.each(b).reject{|x|x<a}*?+}  ## Usage puts p[*gets.split.map(&:to_i)]  ### Inputs & Outputs 2 10 #=> 10 3 10 #=> 8 100 1000 #=> 75067  # Java 7, 239 237 bytes class M{public static void main(String[]a){c(new Long(a[0]),new Long(a[1]));}static void c(long a,long b){long r=0,i=a-1;for(;++i<=b;r+=p(i)?i:0);System.out.print(r);}static boolean p(long n){int i=2;while(i<n)n=n%i++<1?0:n;return n>1;}}  Ungolfed & test case: class M{ static void c(long a,long b){ long r = 0, i = a-1; for(; ++i <= b; r += p(i) ? i : 0); System.out.print(r); } public static void main(String[] a){ c(new Long(a[0]), new Long(a[1])); } static boolean p(long n){ int i = 2; while(i < n){ n = n % i++ < 1 ? 0 : n; } return n > 1; } }  Usage: java -jar M.jar 5 17 Output: 53 # Casio-Basic, 42 bytes sum(seq(piecewise(isPrime(x),x,0),x,a,b  Uses a hybrid/piecewise function that returns the number if it's prime, otherwise return 0. seq runs this over the range a to b, then sum adds it all up. 39 bytes for the function, 3 bytes to enter a,b as parameters. # Haskell, 47 44 bytes a!b=sum[x|x<-[a..b],all((<)0.mod x)[2..x-1]]  SEJPM helped me save 3 bytes! • 44 bytes: a!b=sum[x|x<-[a..b],all((<)0.mod x)[2..x-1]] (using all instead of and) – SEJPM Oct 11 '17 at 12:00 • Fails for 1!3 (gives 6 instead of 5).. – ბიმო Aug 9 '18 at 0:16 • a?b=sum[n|n<-[a..b],n>1,all((>0).mod n)[2..n-1]], works for 1?3 – Benji yesterday # APL(NARS), 11 chars, 22 bytes {+/⍵/⍨0π¨⍵}  test one range: {+/⍵/⍨0π¨⍵} 1..23 100 {+/⍵/⍨0π¨⍵} 1..22 77 {+/⍵/⍨0π¨⍵}1..1 0  test two ranges: {+/⍵/⍨0π¨⍵}¨(1..22)(1..23) 77 100  # Desmos, 124 bytes I know this is a long time ago, but I'm bored :). f(a,b)=\sum_{n=a}^b\left\{\prod_{k=2}^{n-1}\left\{\operatorname{mod}(n,k)=0:0,1\right\}-\left\{n<2:1,0\right\}=1:n,0\right\}  Try it on Desmos! I had to waste 23 bytes just for the edge case of 0 and 1(-\left\{n<2:1,0\right\}). I feel like this could definitely be golfed further, but I'm not that skilled at Desmos. I especially think that the edge cases 0 and 1 could be handled a lot better than what I'm doing currently. # Scala, 260 bytes object P extends App{ def c(M:Int)={val p=(false::false::true::List.range(3,M+1).map(_%2!=0)).toArray for(i<-(3 to M) if p(i)) {var j=i*i while(j<M){p(j)=false j+=i}} p} val l=args.map(_.toInt) val p=c(l(1)) println((l(0)to l(1)).filter(p).map(_.toLong).sum)}  A self-written primes-sieve. time scala P 3900000 4000000 25811704341 real 0m8.288s user 0m6.968s sys 0m0.456s  # PowerShell, 94 bytes $a,$b=$args[0,1]
(.{$p=2..$b
while($p){$p[0];$p=@($p|?{$_%$p[0]})}}|
?{$_-gt$a}|
measure -s).sum


Python 3: 99 chars


l,h=map(int,input().split())
print(sum(p for p in range(l,h+1) if all(p%i for i in range(2,p)))-1)


• This is not inclusive. – jamylak Jul 30 '12 at 6:23
• As noted by jamylak, this is not an inclusive range (try inputs 2 23), and is therefore invalid per the challenge spec. As so, I’ve flagged your answer for deletion. – caird coinheringaahing Oct 16 at 1:52
• I am removing this answer as per our policy on handling invalid submissions. Please feel free to edit this solution so it's valid and flag it for undeletion. – HyperNeutrino Oct 16 at 3:19
• Whoever reflagged this for deletion, it was undeleted after being edited. I'm marking as "Looks OK", let me know if it's still invalid. – Redwolf Programs Oct 17 at 23:19

Python 3: 259 characters This is a longer solution, but it's more efficient than the one-liner I first posted.

import itertools as i
l,h=map(int,input().split())
F=lambda p:lambda x:x%p
def S(s):
while 1:
yield (p:=next(s),s:=filter(F(p),s) if p*p<=h else s)[0]
P=lambda:(yield from S(i.count(2)))
print(sum(i.takewhile(lambda e:e<=h,i.dropwhile(lambda e:e<l,P()))))