5
3
The program has an input of a and b, and outputs the sum of numbers from a to b, inclusive. Score is in bytes. As always, standard loopholes are disallowed.
If your input is in the format [a,b], +3 bytes
Summation from a to b
Answers
24
Stuck, 4 Bytes
I'm amazed Stuck finally is winning something! The inclusive range function checks to see if the top of the stack is a list of length 2, and uses it for the parameters of the range function if so.
t]R+
Input is | separated, as t takes a bunch of inputs separated by |s and places them on the stack. ] just wraps the elements in a list, and lets the range function do it's job.
If that still falls in the penalty, then this works (5 Bytes): ii]R+
Kade
Posted 2015-10-08T20:27:39.633
Reputation: 7 463
3Dammit. I was going to try to make a Stuck solution but the docs on Esolang didn't say that you could use the inclusive range function like that and so I gave up.... – a spaghetto – 2015-10-09T01:28:04.313
2@quartata Sorry, I haven't updated the docs in a while.. :P – Kade – 2015-10-09T02:20:53.513
13
Pyth, 6 4 bytes + 3 = 7 5 bytes
s}vzQ
Avoids the 3-byte penalty imposed by the new input requirements via an extra byte.
Takes input separated from stdin separated by newlines.
kirbyfan64sos
Posted 2015-10-08T20:27:39.633
Reputation: 8 730
11
Python, 26
lambda a,b:(a+b)*(b-a+1)/2
Assumes b>=a. Uses the formula of mean * #summands. The result is a whole number, so it doesn't matter if the / is Python 2's floor division.
Shorter by 2 chars than the direct expression
lambda a,b:sum(range(a,b+1))
xnor
Posted 2015-10-08T20:27:39.633
Reputation: 115 687
3I feel like there should be a char save with ~. Expanding gives the same length (b*b-a*a+a+b)/2. So is -b*~b/2-a*~-a/2. – xnor – 2015-10-08T21:00:00.973
What about (a-a*a-~b*b)/2? – ETHproductions – 2016-10-21T12:55:24.613
8
Matlab/Octave, 14 bytes
@(a,b)sum(a:b)
flawr
Posted 2015-10-08T20:27:39.633
Reputation: 40 560
7
Julia, 15 bytes
s(a,b)=sum(a:b)
This creates a function s that sums the range a:b. This assumes a ≤ b.
Alex A.
Posted 2015-10-08T20:27:39.633
Reputation: 23 761
Upvote for a short answer that is completely legible. – azz – 2015-10-10T04:58:45.317
7
Javascript ES6, 21 bytes
a=>b=>(a+b)*(b-a+1)/2
Used the same approach to that problem that's almost everywhere: "Add up all of the numbers between 1 and 100, inclusive."
Mama Fun Roll
Posted 2015-10-08T20:27:39.633
Reputation: 7 234
1I didn't know you could use lambda's in Javascript! – IEatBagels – 2015-10-09T13:31:19.237
1Oh, it's in the new ES6 specs. – Mama Fun Roll – 2015-10-09T22:49:59.853
You can save a byte with currying (a=>b=> …) – Cyoce – 2016-10-20T21:12:31.027
7
Dyalog APL, 8 7 bytes
+÷2÷1--
This is a dyadic function train which is equivalent to
{(⍺+⍵)÷2÷1-⍺-⍵}
How it works
⍝ Left argument: a, right argument: b
- ⍝ Calculate a-b.
1- ⍝ Subtract the difference from 1 to calculate 1-(a-b) = b-a+1.
2÷ ⍝ Divide 2 by the difference to calculate 2/(b-a+1).
+ ⍝ Calculate a+b.
÷ ⍝ Divide the sum by the quotient to calculate (a+b)/(2/(b-a+1)),
⍝ i.e., (a+b)(b-a+1)/2.
Dennis
Posted 2015-10-08T20:27:39.633
Reputation: 196 637
6
Java, 44 42 bytes
An approach I haven't seen here so far: Via gaussian sums you can derive that the sum of all numbers from a to b is (b-a+1)*(b+a)/2
If you only have to implement a function, you could do it in 42 bytes:
int s(int a,int b){return(b-a+1)*(b+a)/2;}
The conventional approach is 3 bytes longer:
int s(int a, int b){for(;b>a;a+=b--);return a;}
A full program is 142 bytes:
public class S{public static void main(String[]a){int n=Integer.parseInt(a[0]),m=Integer.parseInt(a[1]);System.out.println((m-n+1)*(m+n)/2);}}
flawr
Posted 2015-10-08T20:27:39.633
Reputation: 40 560
3You really should try something other than Java. ;) – kirbyfan64sos – 2015-10-08T20:57:36.863
Would it work to remove the space in int a, int b? Perhaps the space after return as well? – Alex A. – 2015-10-08T21:17:12.397
@AlexA. Thanks=) Do you know about the Java8 lambdas? I haven't installed it yet but I think (int a,int b)->(b-a+1)*(b+a)/2 should work as a lambda? – flawr – 2015-10-08T21:24:17.143
Even without lambdas, public static isn't usually required for functions here. Just int s(...args...) is fine for the signature. – Geobits – 2015-10-09T12:45:30.723
1you could reduce it by 3 characters, if you use long instead int, i'm talking about full program here – user902383 – 2015-10-20T15:41:38.083
I know it's been almost 2 years ago.. But, the conventional approach is 4 bytes longer, not 3 (after you've removed the space at int a,int b. The full class can be 118 bytes: class S{public static void main(String[]a){long n=new Long(a[0]),m=new Long(a[1]);System.out.print((m-n+1)*(m+n)/2);}}. And all three can be lowered if you use Java 8: a->b->(b-a+1)*(b+a)/2 (21 bytes); a->b->{for(;b>a;a+=b--);return a;} (34 bytes); and interface S{static void main(String[]a){long n=new Long(a[0]),m=new Long(a[1]);System.out.print((m-n+1)*(m+n)/2);}} (115 bytes) – Kevin Cruijssen – 2017-09-27T14:12:01.260
6
TI-BASIC, 8 bytes
mean(Ans+Ansmin(ΔList(Ans
Takes input in the form {A,B}.
Ans ; {A,B}
ΔList( ; {B-A}
min( ; B-A
Ans ; (B-A)*{A,B}
Ans+ ; (B-A+1)*{A,B}
mean( ; (B-A+1)*(A+B)/2
By comparison, the shortest two-variable solution is 11 bytes:
Prompt A,B
sum(randIntNoRep(A,B ;random permutation of integers between, inclusive
lirtosiast
Posted 2015-10-08T20:27:39.633
Reputation: 20 331
5
CJam, 7 bytes
q~),>:+
q~ e# Read and eval input, pushing a and b onto the stack.
), e# Increment b and turn into the range [0 1 2 ... b-1 b].
> e# Discard the first a elements to get [a a+1 ... b-1 b].
:+ e# Reduce + onto the list, computing the sum.
Martin Ender
Posted 2015-10-08T20:27:39.633
Reputation: 184 808
5
Mathematica, 9 bytes
Tr@*Range
This evaluates to an unnamed function taking two integers. I was going to use the built-in Sum, but it's 4 bytes longer:
x~Sum~{x,##}&
Martin Ender
Posted 2015-10-08T20:27:39.633
Reputation: 184 808
5
APL, 9 bytes
-⍨/2!⎕+⍳2
Takes input as a two-element list . Set your index origin to 0 (⎕IO←0) before running.
This is longer than the answer by @Dennis but in my opinion more stylish.
It calculates (B+1 nCr 2)-(A nCr 2) = B(B+1)/2-A(A-1)/2.
lirtosiast
Posted 2015-10-08T20:27:39.633
Reputation: 20 331
Does ⍨ actually take only one byte in some encoding? – Roberto Bonvallet – 2015-10-09T15:06:50.817
@RobertoBonvallet I believe APL uses a special encoding – Cyoce – 2016-01-22T07:25:18.827
5
><>, 14 bytes
This assumes that the stack has already been populated with a and b, so this is basically a function. I'm using this online interpreter which lets you put values onto the stack before the program runs. An extra three bytes for -v (plus a space, for running from a terminal) would put this answer at 17 bytes.
:r:@-1+@+*2,n;
Explanation:
-v ab (initial stack)
: abb duplicates top of stack (b)
r bba reverses stack
: bbaa duplicates top of stack (a)
@ baba rotates top three elements clockwise (baa -> aba)
- ba,(b-a) subtraction (ba -> b-a)
1+ ba,(b-a+1) adds 1 (b-a -> b-a+1)
@ (b-a+1),ba rotates top three elements clockwise
+ (b-a+1),(b+a) addition (ba -> b+a)
* (b-a+1)*(b+a) multiplication
2, (b-a+1)*(b+a)/2 division by 2 (/ is a mirror, so , is used instead)
n; <integer, quit> output top of stack as an integer and quit
This assumes that 0 <= a <= b and uses this closed-form solution (b-a+1)(b+a)/2 to directly calculate the answer.
This code below goes the route of trying to be a full program, but only works for 0 <= a <= b <= 9. That you have to do integer parsing in ><> is annoying. Thanks to Cole, if we assume the input is two single digits, then four bytes can be shaved off. 20 bytes.
ic%:ic%:@$-1+@+*2,n;
The trick is that instead of 68*-, use c% (take the modulus with respect to 12). This will indeed work because 48+x (mod 12) is x (mod 12), which is 0-9 if x is 0-9.
El'endia Starman
Posted 2015-10-08T20:27:39.633
Reputation: 14 504
For your second program, you can shave off some bytes by changing 68*- to c% since we only expect input from 0 to 9. EDIT: also, division is ,, not /. – cole – 2015-10-08T21:53:37.270
1It was not originally clear to me where the number 48 was coming from. I see now that it's because the ASCII codes for the characters 0 through 9 are 48 through 57, but I'm mentioning it in case someone else is mystified. – mathmandan – 2015-10-09T15:50:36.267
5
R - 24 Bytes
f=function(a,b) sum(a:b)
Michal
Posted 2015-10-08T20:27:39.633
Reputation: 209
3You can save yourself 3 by making it an unnamed function and remove the space from it. – MickyT – 2015-10-09T02:12:21.083
You can save 1 byte because you don't really need the space. Also, is it really necessary to declare a function? I see a lot of submissions that doesn't have them so I am getting confused about it. – Mutador – 2015-10-09T04:18:27.443
3@AndréMuta Submissions on this site have to use one of our standard input formats. That means that a and b can be function arguments (as is the case here), they can come from user input like from scan(), or they can be read as command line arguments. – Alex A. – 2015-10-09T05:15:59.130
5
SH, 16 bytes
seq $1 $2|numsum
Where $1 and $2 are the values of two args passed to this program on the command line. numsum is from the num-utils package. Another version that is also 16 bytes is:
seq -s+ $1 $2|bc
deltaray
Posted 2015-10-08T20:27:39.633
Reputation: 151
4
Haskell, 13 bytes
x#y=sum[x..y]
Usage example: 10 # 12 -> 33
nimi
Posted 2015-10-08T20:27:39.633
Reputation: 34 639
4
O, 8 7 bytes
Hjmr]+o
Explanation
H Start an array with a number from input
j Get input as number
mr Range between
]
+ Sum the array
o print it
Try it here
MickyT
Posted 2015-10-08T20:27:39.633
Reputation: 11 735
1I was trying an O answer, but I forgot about mr and ended up with some REALLY long crap. +1 – kirbyfan64sos – 2015-10-08T22:24:05.000
3
PowerShell, 32 31 29 Bytes
$args-join'..'|iex|measure -s
Thanks to TessellatingHeckler for the alternate way to make a range and sum it. Works for any combination of a and b so long as the difference between them is less than 50,000.
Previous version (31):
param($a,$b)($b-$a+1)*($b+$a)/2
Uses the same formula as others, assumes that b>=a.
Previous version (32):
param($a,$b)($a..$b)-join'+'|iex
Pretty trivial. Generates a range of numbers from $a to $b, -joins them with a +, then pipes that to Invoke-Expression which performs the summation. Yay verbosity. Note: This function will break if the two numbers are > 50,000 apart, as that's the (hard-coded) limit of dynamically generated ranges in PowerShell, but it correctly handles b<a, so, y'know...
AdmBorkBork
Posted 2015-10-08T20:27:39.633
Reputation: 41 581
I'll throw in $f={$args-join'..'|iex|measure -s} used like & $f 5 10 -> 45 - measuring just the inner function content as you are, it hits 29 bytes. The rules don't say it must only output the sum. Cool idea with the join/iex, – TessellatingHeckler – 2015-10-09T00:37:53.180
1@TessellatingHeckler Oh, interesting. Note that you don't need the $f={ } if you save it as a ps1 file and execute it as a script from the PS command line. Same with the param() versions. That's why I've only ever measured "just the inner function content" because it's a complete program. – AdmBorkBork – 2015-10-09T12:48:54.977
3
Befunge-93, 18 16 bytes
&:1-*&:1+*\-2/.@
Explanation
& a input as integer (a)
: aa duplicate top of stack (a -> a,a)
1- a,(a-1) subtract 1 from top of stack (a -> a-1)
* a*(a-1) multiply top two values of stack (a,a-1 -> a*(a-1))
& a*(a-1),b input as integer (b)
: a*(a-1),bb duplicate top of stack (b -> b,b)
1+ a*(a-1),b,(b+1) add 1 to top of stack (b -> b+1)
* a*(a-1),b*(b+1) multiply top two values of stack (b,b+1 -> b*(b+1))
\ b*(b+1),a*(a-1) swap top two values of stack (c,d -> d,c)
- b*(b+1)-a*(a-1) subtract top two values of stack (d,c -> d-c)
2/ (b*(b+1),a*(a-1))/2 divide top of stack by 2
.@ <output, exit> output as integer and stop
This uses the following formula:
b(b+1) (a-1)(a-1+1) b(b+1) - a(a-1)
------ - ------------ = ---------------
2 2 2
Unfortunately, the shorter (b-a+1)(b+a)/2 cannot be done in Befunge easily because it requires accessing b and a twice, which is impossible to do when working only with the top two values of the stack. Storing a value in "memory" would take more characters.
El'endia Starman
Posted 2015-10-08T20:27:39.633
Reputation: 14 504
3
C++, 42 bytes
As a function (42 bytes):
int s(int a,int b){return(a+b)*(b-a+1)/2;}
As a full program that reads from STDIN and writes to STDOUT (86 bytes):
#include<iostream>
main(){int a,b;std::cin>>a;std::cin>>b;std::cout<<(a+b)*(b-a+1)/2;}
Both approaches compute the sum using the Gaussian formula.
Ungolfed:
#include <iostream>
int s(int a, int b) {
return (a + b) * (b - a + 1) / 2;
}
int main() {
int x, y;
std::cin >> x;
std::cin >> y;
std::cout << s(x, y) << std::endl;
return 0;
}
Alex A.
Posted 2015-10-08T20:27:39.633
Reputation: 23 761
You can save few bytes if you use lambda instead. Requires changing tag to C++11 though. – Zereges – 2015-10-09T11:35:27.007
@Zereges I don't know much about C++ lambdas. Would it be something like [](int a,int b){return(a+b)*(b-a+1)/2;}? – Alex A. – 2015-10-09T18:48:37.543
yes, exactly like that. Sorry, but I for some reason can address this post to you using @ – Zereges – 2015-10-09T19:50:39.740
@Zereges You don't need to ping me since this is my post; I'll always get notified of a comment on it. And thanks! :) – Alex A. – 2015-10-09T19:53:02.770
You can save 9 bytes from the full program by using std::cin>>a>>b; to get your inout from one line with a space in the middle (eg 1 10) – Noodle9 – 2015-10-11T15:46:49.500
hehe, 42 bytes => +1!! – Stefan – 2016-10-21T12:56:09.507
3
Jelly, 2 bytes
rS
Takes two numbers as command-line arguments. This is non-competing since Jelly was created after this challenge.
r Yields inclusive range [a, b]
S Sum
a spaghetto
Posted 2015-10-08T20:27:39.633
Reputation: 10 647
3
Brain-Flak, 32 30 bytes
([{({}[()])}{}]{()({}[()])}{})
Explanation
This uses the triangulation algorithm to calculate T(a-1) and T(b) then pushes the difference. I actually golfed 2 bytes off the existing world record triangulation algorithm while making this solution.
To push T(a-1) it uses a modified version of the original algorithm. Instead of the simple:
({}[()])({()({}[()])}{})
I removed both the first ({}[()]) and the ().
The reason () was there in the first place was that ({({}[()])}{}) quite conveniently calculates T(n-1) for us so there is no need to decrement if we remove the ()
This first portion is made negative and put next to the second portion which performs standard triangulation inside of the outer push.
([{({}[()])}{}]{()({}[()])}{})
Post Rock Garf Hunter
Posted 2015-10-08T20:27:39.633
Reputation: 55 382
2
Awk, 24 bytes
$0=""($1+$2)*($2-$1+1)/2
Incorporated manatwork's suggestion.
jkabrg
Posted 2015-10-08T20:27:39.633
Reputation: 299
Is shorter if you make it pattern instead of action: $0=""($1+$2)*($2-$1+1)/2. – manatwork – 2015-10-09T09:51:53.973
2
Applescript, 147 137 Bytes
Welp. I'm not winning.
set a to(text returned of(display dialog""default answer""))
set b to(text returned of(display dialog""default answer""))
(b-a+1)*(b+a)/2
Since I'd never seen an Applescript answer before, I decided I'd make one, then immediately realized why it never had been done.
The explanation is fairly straightforward -
set ... to (text returned of (display dialog "" default answer ""))
^
Get the text of ^ the user inputs an answer here
(b-a+1)*(b+a)/2
^ with nothing else below this, Applescript recognizes it as the final
value and prints to console.
Addison Crump
Posted 2015-10-08T20:27:39.633
Reputation: 10 763
Yes, I recognize that on compile, Applescript "helpfully" spaces everything out - this will still compile and run as is, however. – Addison Crump – 2015-10-08T22:51:40.323
2
PHP, 29 bytes
Based on standard stricks to set variables (eg register_globals oldSkool)
while($b>=$a)$r+=$b--;echo$r;
Martijn
Posted 2015-10-08T20:27:39.633
Reputation: 713
2
Rust, 27 bytes
Creates a closure s that takes two unsigned 8-bit integers and returns another.
let s=|m,n|(n-m+1)*(m+n)/2;
jus1in
Posted 2015-10-08T20:27:39.633
Reputation: 81
2
Groovy, 19 Bytes
{a,b->(a..b).sum()}
Magic Octopus Urn
Posted 2015-10-08T20:27:39.633
Reputation: 19 422
1
Lua, 40 Bytes
Used the same method as the Javascript answer, but in Lua.
function y(a,b)return(a+b)*(b-a+1)/2 end
You use this as such
y(3, 6) --3 + 4 + 5 + 6 == 18
boyned
Posted 2015-10-08T20:27:39.633
Reputation: 11
1
Symbolic Python, 49 bytes (46 + 3 penalty)
_=(_[_==_]*-~_[_==_]-_[_>_]*~-_[_>_])/-~(_==_)
Anything involving range, len, or loops would be too difficult in Symbolic Python, and so I decided the easiest way was to use the formula:
$$ \sum_{r=a}^b r = \frac{b(b+1) - a(a-1)}{2} $$
FlipTack
Posted 2015-10-08T20:27:39.633
Reputation: 13 242
1
33, 20 bytes
OcaxcmszOcaxaclaz2do
Explanation
This answer uses the following formula to calculate it: $$a + (a+1) + (a+2) + \cdots + b = {a-a^2+b+b^2 \over 2}$$
O O | Get a and b as two integers delimited by whitespace
ca | (a )
xcm | ( - a^2 )
sz la | ( + ( ))
ca | ( (b ))
xac | ( ( + b^2))
z2d | ( ) / 2
o | Print it
TheOnlyMrCat
Posted 2015-10-08T20:27:39.633
Reputation: 1 079
1
APL(NARS), 8 chars, 16 bytes
{+/⍺..⍵}
test:
f←{+/⍺..⍵}
1 f 2
3
2 f 10
54
This it is 6 chars but I not find a way to assign a name to it
+/↑../
Using as function 2 3
RosLuP
Posted 2015-10-08T20:27:39.633
Reputation: 3 036
1
Reg (a.k.a Unofficial Keg), 17 bytes
Try it online behavior is unknown.
¿¿ï_'(:|_)(+).
This takes 2 inputs separated by newlines.
1
Ruby, 22 bytes
->a,b{eval [*a..b]*?+}
Just as long as using the expression in @xnor's Python answer:
->a,b{(a+b)*(b-a+1)/2}
Test it:
->a,b{eval [*a..b]*?+}[10,20] #=> 165
daniero
Posted 2015-10-08T20:27:39.633
Reputation: 17 193
4Not everyone is posting incomplete snippets. When not otherwise specified, one should assume that answers must use one of our default input methods. – Alex A. – 2015-10-08T21:01:28.193
@AlexA. yup, but I think the challenge is poorly worded. I'll check back later and update if it's been rephrased. Gotta run. – daniero – 2015-10-08T21:06:40.093
1
Math++, 23 bytes
?>a
?>b
(b-a+1)*(b+a)/2
SuperJedi224
Posted 2015-10-08T20:27:39.633
Reputation: 11 342
1
Squirrel, 21 bytes
@(a,b)(b-a+1)*(b+a)/2
Creates an anonymous function.
a spaghetto
Posted 2015-10-08T20:27:39.633
Reputation: 10 647
1
J, 19 bytes
f=:4 :'+/x+i.1+y-x'
1 f 1
1
1 f 2
3
5 f 10
45
It would be longer if it calculated the answer directly, because it would need some parentheses. I can't quickly find a builtin way to "range from a to b", so it does:
# The parameters are x on the left, y on the right
f=:4 : ' # f is a dyad verb defined as...
+/ # The sum of ...
x + # x added to the list of ... (turns 0 1.. to x x+1..)
i. # the integers from 0 to ...
1+y-x # 1 + y -x (sequence length)
' # end definition string.
5 f 10 ->
1 + 10 - 5 =
6
i. 6 =
0 1 2 3 4 5
5 + 0 1 2 3 4 5 =
5 6 7 8 9 10
+/ 5 6 7 8 9 10 =
5 + 6 + 7 + 8 + 9 + 10 =
45
TessellatingHeckler
Posted 2015-10-08T20:27:39.633
Reputation: 2 412
Have you tried the algorithms from Dennis' and my APL answers? – lirtosiast – 2015-10-09T00:29:39.140
@ThomasKwa Not yet, I can't even follow them. (Your APL answer has one character which my browser can't render and can't copy into the TryAPL site, one that I can't follow the TryAPL cheat sheet explanation, and I don't know what an index origin is. Dennis's answer - sounds like a train might become a J hook or fork, but I haven't tried to follow what it's doing). – TessellatingHeckler – 2015-10-09T00:51:48.207
you don't need to count f=: so this is really 16 bytes – Cyoce – 2017-09-28T01:32:45.653
1
PHP, 39 bytes
As usual not even close to the shortest answer:
<?=array_sum(range($argv[1],$argv[2]));
Runs from command line like:
php sum.php 10 20
insertusernamehere
Posted 2015-10-08T20:27:39.633
Reputation: 4 551
Not a wonderfull solution, but +1 for doing the same thing I thought of – Martijn – 2015-10-09T08:19:08.453
@Martijn Thanks. I couldn't think of anything shorter even using the calculation ($a+$b)*($b-$a+1)/2 isn't shorter than using the long function names. – insertusernamehere – 2015-10-09T10:45:52.467
I base my php answers on register_globals in the old settings, so $_GET['a'] would be accessable via $a, saving bytes – Martijn – 2015-10-09T10:59:29.853
@Martijn Yeah, but you can't post any answer that runs in PHP 5.4 or better, right?
– insertusernamehere – 2015-10-09T11:00:59.0401Nope. 'luckily' I've got a 5.2 test environment so I can apply this. Then again, I cant use the funky new possibilities – Martijn – 2015-10-09T11:03:21.623
1
Swift 2, 27 bytes
Swift seems to be able to infer types much more easily these days...
let f={($0+$1)*($1-$0+1)/2}
Called with f(a, b)
GoatInTheMachine
Posted 2015-10-08T20:27:39.633
Reputation: 463
1
R, 18 Bytes
sum(scan():scan())
Mutador
Posted 2015-10-08T20:27:39.633
Reputation: 1 361
1
C++11 lambda, 37 bytes
[&b](int a,int&b){b=(a+b)*(b-a+1)/2;}
Invocation:
int main()
{
int a,b;
std::cin >> a; std::cin >> b;
std::cout << ([&b](int a,int&b){b=(a+b)*(b-a+1)/2;}(a,b),b) << std::endl;
return 0;
}
renonsz
Posted 2015-10-08T20:27:39.633
Reputation: 121
Do you really need [&b]? You're still taking it as a reference in the argument list. – kirbyfan64sos – 2015-10-14T18:15:04.223
1
Perl 6 (14 chars / bytes)
The only obvious solution is to create a list from a to b, and then to reduce it with the addition operator.
[+] a..b
That would be a snippet, so let's place it in a pointy block.
( One of the ways to create a lambda expression. )
-> \a, \b { [+] a..b } # 22 chars
->\a,\b{[+] a..b} # 17 chars
Or better yet use a placeholder parameterized block.
{[+] $^a..$^b} # 14 chars
Examples of it's use:
say {[+] $^a..$^b}(4,7) # 22
# store it as a subroutine and in a scalar
my &s = my $s = {[+] $^a..$^b};
say s 4,8; # 30
say $s.(3,8); # 33
say ( {[+] $^a..$^b} for 4,7, 4,8, 3,8, ); # (22 30 33)
If you wanted to be cheap, you could modify the parser by adding a list summation operator. Which technically could be considered a new language at that point. Although that would be a made up language so wouldn't be a valid answer anyway.
( This is actually how it might be written in the Rakudo implementation if it was added )
sub prefix:<∑> (+@a) is looser(&[,]) {
[+] @a
}
### new language starts here ###
{∑$^a..$^b} # only 11 chars / 12 bytes in UTF8
( Scoring it with bytes feels wrong because Perl 6 and thus this new anonymous language only deal with graphemes in strings. )
Brad Gilbert b2gills
Posted 2015-10-08T20:27:39.633
Reputation: 12 713
1
Minkolang 0.8, 14 bytes (INVALID)
Minkolang was created after this challenge was posted, so I post this answer only in the sense of contributing to a catalog.
nndr-[d1-]$+N.
Explanation
n input as integer (stack: a)
n input as integer (stack: a,b)
d duplicate top of stack (stack: a,b,b)
r reverses stack (stack: b,b,a)
- subtraction (stack: b,b-a)
[ starts a For loop that will run b-a times
d duplicates top of stack
1- subtracts 1
] closes For loop (stack: b,b-1,b-2,...,a+1,a)
$+ sums the stack
N. outputs as integer and stops
The formula-based solutions like what I used in my Befunge and ><> answers are 2 and 3 bytes longer, respectively:
nd1-*nd1+*r-2:N.
ndnd3R-1+1R+*2:N.
El'endia Starman
Posted 2015-10-08T20:27:39.633
Reputation: 14 504
1
jq, 25 characters
(24 characters code + 1 character command line option)
[range(.[0];.[1]+1)]|add
Sample run:
bash-4.3$ jq -s '[range(.[0];.[1]+1)]|add' <<< '10 15'
75
jq, 27 characters
(24 characters code + 3 character penalty)
[range(.[0];.[1]+1)]|add
Sample run:
bash-4.3$ jq '[range(.[0];.[1]+1)]|add' <<< '[10,15]'
75
Not specifically interesting, just the -s option which comes handy here: it interprets raw input as array. (Yes, the code itself is identical in both solutions.)
manatwork
Posted 2015-10-08T20:27:39.633
Reputation: 17 865
1
Cyoce
Posted 2015-10-08T20:27:39.633
Reputation: 2 690
1
05AB1E, 2 bytes
ŸO
Explanation:
# Implicit input
# Implicit input
Ÿ # Push [a...b]
O # Sum list
# Implicit print
Oliver Ni
Posted 2015-10-08T20:27:39.633
Reputation: 9 650
0
Stephen
Posted 2015-10-08T20:27:39.633
Reputation: 12 293
0
TI-Basic, 14 bytes
Input :.5(YY+Y+X-XX
Bonus: no lists or list commands.
Timtech
Posted 2015-10-08T20:27:39.633
Reputation: 12 038
The X-value should be smaller. – Timtech – 2017-09-14T18:13:34.867
0
Xcali
Posted 2015-10-08T20:27:39.633
Reputation: 7 671
0
Excel, 20 bytes
=(B1^2+B1-A1^2+A1)/2
Wernisch
Posted 2015-10-08T20:27:39.633
Reputation: 2 534
0
Husk, 2 bytes
Σ…
Explanation
-- implicit inputs: A B | 2 7
… -- inclusive range [A,B] | [2,3,4,5,6,7]
Σ -- sum | 2+3+4+5+6+7 == 27
ბიმო
Posted 2015-10-08T20:27:39.633
Reputation: 15 345
0
K (oK), 14 12 bytes
Solution:
+/^/!:'|0 1+
Examples:
> +/^/!:'|0 1+2 7
27
> +/^/!:'|0 1+5 10
45
Explanation:
Slightly different to doing .5*(x+y)*1+y-x. Instead, calculate ranges, take largest except smallest and then sum up:
+/^/!:'|0 1+ / solution
0 1+ / vectorised addition of 0 1 / 2 7 -> 2 8
| / reverse: / 2 8 -> 8 2
!:' / til (!:) each (') / (8;2) -> (0 1 2 3 4 5 6 7;0 1)
^/ / except (^) over (/) / (0 1 2 3 4 5 6 7;0 1) -> 2 3 4 5 6 7
+/ / add (+) over (/) aka sum / 2 3 4 5 6 7 -> 27
streetster
Posted 2015-10-08T20:27:39.633
Reputation: 3 635
0
ATaco
Posted 2015-10-08T20:27:39.633
Reputation: 7 898
0
Casio Basic, 20 bytes
sum(seq(x,x,a,b))
Simple function. 17 bytes for the program, 3 bytes to enter a,b as arguments.
numbermaniac
Posted 2015-10-08T20:27:39.633
Reputation: 639
0
J-uby, 9 Bytes
:!~|:/&:+
Explanation:
:!~ generate range a..b
| then
:/&:+ sum it together
Expansion:
:!~|:/&:+
->(a,b){ (:/&:+).(a!~b) }
->(a,b){ :+ / (a !~ b) }
->(a,b){ (a..b).reduce(:+) }
Cyoce
Posted 2015-10-08T20:27:39.633
Reputation: 2 690
0
LOLCODE, 184 bytes
HAI 1.3
I HAS A B
I HAS A J
GIMMEH J
I HAS A Q
GIMMEH Q
MAEK Q A NUMBR
IM IN YR L UPPIN YR i TIL BOTH SAEM DIFF OF Q AN SUM OF i AN J AN -1
VISIBLE SUM OF i AN J
IM OUTTA YR L
KTHXBYE
FantaC
Posted 2015-10-08T20:27:39.633
Reputation: 1 425
In some interpreters, you can exclude the "HAI" and "KTHXBYE" – MilkyWay90 – 2018-11-17T03:33:46.730
0
MilkyWay90
Posted 2015-10-08T20:27:39.633
Reputation: 2 264
0
Japt -x, 3 bytes
Nrò
Explanation
I use N to avoid the +3 input penalty.
I would normally format the input as \$[a,b]\$ and my code would be rò.
N reads the entire input.
r reduces that by: ò, which creates the range from \$a\$ to \$b\$.
-x sums the output.
Oliver
Posted 2015-10-08T20:27:39.633
Reputation: 7 160
0
Alchemist, 135 bytes
_+a->
0_+0m+a+b->u+x+v+y
0m+0a+b->v+y
0a+0b+0m+u+v->m
0a+0b+0m+0u+v->m
m+u+v->a+b+m
m+0u+v->b+m
m+0u+0v->
0a+0b+x+y->
0a+0b+0x+y->Out_I
Output in unary, try it online or view the remaining \$\texttt{y}\$-atoms here!
Explanation
Pretty similar to this answer, uses that \$a \leq b\$: Basically we ensure with 0_ on the second rule that the first rule must always be applied first (initially the universe always contains one
\$\texttt{_}\$-atom), removing one \$\texttt{a}\$-atom and then summing both \$\texttt{#x} = [1\dotsc \texttt{#a}]\$ and \$\texttt{#y} = [1\dotsc \texttt{#b}]\$ in parallel, once there are no \$\texttt{a}\$- or \$\texttt{b}\$-atoms left, we subtract them and finally print the remaining \$\texttt{y}\$-atoms.
Ungolfed
# Remove one a-atom:
_+a->
# Sum [1..#a] (to out_a) and [1..#b] (to out_b) in parallel:
0_+0mov+ a+b->u+out_a+v+out_b
0mov+0a+b->v+out_b
0a+0b+0mov+ u+v->mov
0a+0b+0mov+0u+v->mov
mov + u+ v->a+b+mov
mov +0u+ v->b+mov
mov +0u+0v->
# If we're done with that:
0a+0b+ out_a+out_b-> # difference: out_b - out_b
0a+0b+0out_a+out_b->Out_I # output rest in unary
ბიმო
Posted 2015-10-08T20:27:39.633
Reputation: 15 345
0
JavaScript, 21 bytes
Input is taken in currying syntax: f(a)(b)
a=>b=>b*b-a*a-(a+b)*2
Oliver
Posted 2015-10-08T20:27:39.633
Reputation: 7 160
0
recursive
Posted 2015-10-08T20:27:39.633
Reputation: 8 616
0
Malbolge, 57283 bytes
TryItOnline link didn't fit in the answer. What a shame!
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Krzysztof Szewczyk
Posted 2015-10-08T20:27:39.633
Reputation: 3 819
0
C preprocessor, 40 bytes
#define S(a,b) (((a)+(b))*((b)-(a)+1)/2)
renonsz
Posted 2015-10-08T20:27:39.633
Reputation: 121
1This submission doesn't add anything particularly useful, and is similar to multiple submissions already here. What makes your post different? Perhaps adding that would make it a more popular solution. – Conor O'Brien – 2015-10-09T16:23:14.143
Is codegolf useful? C/C++ cannot compete with interpreted languages in that way, so avoiding typenames and return keyword was the deal. – renonsz – 2015-10-12T06:15:43.033
0
Bash/GNU, 21 Bytes
expr `seq -s" + " $@`
called with arguments a and b
seqgenerates a sequence fromatob, separated by " + "exprevaluates and prints the resulting expression
Fabian Schmengler
Posted 2015-10-08T20:27:39.633
Reputation: 1 972
0
gs2, 5 bytes
WÉ'Od 57 90 27 4f 64
Read two numbers, dump them on the stack, increment the top one, take the range between them, sum it.
Lynn
Posted 2015-10-08T20:27:39.633
Reputation: 55 648
0
Burlesque - 6 bytes
per@++ usage:
$ echo "10 20" | blsq --stdin "per@++"
165
mroman
Posted 2015-10-08T20:27:39.633
Reputation: 1 382
0
DUP, 15 bytes
[^^+^3ø-1+*2/.]
Anonymous lambda. Usage:
1 10[^^+^3ø-1+*2/.]!
Explanation
Uses the formula (a+b)*(b-a+1)/2.
Mama Fun Roll
Posted 2015-10-08T20:27:39.633
Reputation: 7 234
0
Elixir, 37 bytes
fn(a,b)->Enum.reduce(a..b,0,&+/2)end
poster child example for reduce, i guess. allows b > a.
srecnig
Posted 2015-10-08T20:27:39.633
Reputation: 171
0
Mego
Posted 2015-10-08T20:27:39.633
Reputation: 32 998
0
Pylons, 3 bytes
iks
How it works:
i # Take command line input.
k # Push the inclusive range(stack[-2]..stack[-1])
s # Sum the stack.
Morgan Thrapp
Posted 2015-10-08T20:27:39.633
Reputation: 3 574
0
Forth, 27 bytes
0
: f 1+ SWAP DO I + LOOP ;
The parameters much be pushed smallest first. Run like this:
1 4 f .
mbomb007
Posted 2015-10-08T20:27:39.633
Reputation: 21 944
0
J, 11 bytes
-:@<:@-*-@+
This is a tacit verb. Example usage:
v =: -:@<:@-*-@+
5 v 10
45
This verb is a fork of three verbs: -:@<:@-, *, and -@+. A fork x (f g h) y is equivalent to (x f y) g (x h y). Thus, x (-:@<:@-*-@+) y = (x(-:@<:@-)y) * (x (-@+) y). Likewise, the composition of two verbs x (f @ g) y equals f (x g y). Using this rule, given that <: decrements, and -: divides its argument by two, x(-:@<:@-)y = (x - y - 1)/2 and x(-@+)y = -(x+y). Therefore x(-:@<:@-*-@+)y = (x - y - 1) / 2 * -(x+y), and by moving the sign from -(x+y) to (x-y-1), we get (y - x + 1) / 2 * (x + y).
Cyoce
Posted 2015-10-08T20:27:39.633
Reputation: 2 690
Dennis's APL solution translated into J for 7 bytes: +%2%1-- – Jonah – 2017-09-28T02:21:05.290
0
QBIC, 15 bytes
::[a,b|d=d+c]?d
:: takes two command line parameters and names them a and b. Then, we create a FOR loop (using iterator c) from a to b. We keep adding c to d, then print d and exit.
steenbergh
Posted 2015-10-08T20:27:39.633
Reputation: 7 772
What is QBIC? Where can I find out about it? A google search brings up this answer. – Post Rock Garf Hunter – 2016-10-21T05:43:13.163
@WheatWizard This is basically the most complete repository on what QBIC is and where to get it.
– steenbergh – 2016-10-21T06:09:38.3370
Detour, 11 bytes
{+(}-!>)*d.
This uses the (a+b)*(b-a+1)/2 formula.
+ encodes (a+b), -!> encodes (b-a+1) (really -(a-b)+1), and *d. multiplies (*) the two together, halves it (d), and outputs it (.). {…} makes two copies of a and b and (…) manipulate where the two copies go so that the result of (a+b) doesn't get fed into the (b-a+1) section.
Cyoce
Posted 2015-10-08T20:27:39.633
Reputation: 2 690
6Can
bbe smaller thana, and if so, what should the output be? – Zgarb – 2015-10-08T20:39:23.6673If
bis greater than or equal toaThere's a closed-form solution to this problem:(b-a+1)*(b+a)/2– Blckknght – 2015-10-08T20:55:39.5402Can we take input as a two-element list? – lirtosiast – 2015-10-08T20:57:03.290
254 hours, 25 answers and 46 answer votes after posting, you change it so that b>a must error? Ugh. Is that an allowed codegolf thing? (I guess that's a wrong change as well - or is it intended to be a countdown now?) – TessellatingHeckler – 2015-10-09T01:27:52.727
15You are penalizing people for taking input in the format of [a,b]? Why? How does this improve the question? – MickyT – 2015-10-09T01:41:02.757
1B/c [a,b] uses 2 more bytes than a b – None – 2015-10-09T01:42:10.307
25
The overwhelming consensus regarding defaults for reading several pieces of input is that two integers can be read in list from with no penalty. You can obviously override those defaults in your challenge, but it would be preferable to do so when posting the question, not 5 hours later. In case you didn't know, we have a Sandbox where you can get feedback from the community and iron out the details before posting the actual challenge.
– Dennis – 2015-10-09T02:16:35.7206@Hurricane996
[a,b]in concept doesn't always use 2 more bytes thana b. What's meant is a list/vector/pair/whatever you want to call it, which doesn't have to look like[a,b]. Such a collection is a more convenient means of input in some languages rather than two separate inputs. – Alex A. – 2015-10-09T02:41:24.590If I were to make a program in brainf***, how would I take input? – 0WJYxW9FMN – 2016-10-20T19:49:49.480
2-1 because of numerous modifications. – Erik the Outgolfer – 2016-10-21T12:46:42.070