# 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 <= a <= b <= 108

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 3: 86 chars

a,b=map(int,input().split())
P=k=1
s=0
while k<=b:s+=P%k*k*(k>=a);P*=k*k;k+=1
print(s)


Uses the factorial trick with Wilson's Theorem to check whether k is prime. P%k is 1 if k is prime and 0 otherwise. If it is prime, k is added to the running sum s.

# 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

# 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

### Scala: 260

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


# 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.

# Jelly, 3 bytes

æRS


Try it online!

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

L,d@ßrÞP¦+


Try it online!

## How it works

D,f,@@,		; Define a dyadic function, f
; Example arguments:	[2 23]
d	; Duplicate;	STACK = [2 23 23]
@	; Reverse;	STACK = [23 23 2]
ßr	; Range;	STACK = [23 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22]
Þ	; Filter on:
P	;  Primality	STACK = [23 2 3 5 7 11 13 17 19]
¦+	; Sum;		STACK = [100]


# 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