Add Rainbow Pairs [duplicate]

Question

The rules are as follows:

• Take in a string of integers separated by spaces. Add the first and last numbers, the second and second-last numbers, and so on.
• The program must output the numbers separated by newlines.
• The program must handle negative numbers, but may not handle decimals.
• If an odd amount of integers are inputted, just print the lone integer at the end.
• This is code golf, fewest bytes wins.

Here are some test cases:

> input
output
more output
etc.
-------------------
> 2 2
4

> 1 2 3 4 5 6
7
7
7

> 3 4 2 6
9
6

> 1 2 3 4 5
6
6
3

> -5 3 -8 2
-3
-5

Answer 1

Stax, 9 8 bytes

Thanks to Weijun Zhou for -1 bytes

It actually conforms with the required output format!

Φ┴⌐Öâ9|Ö

Run and debug it

Here's the ungolfed ascii representation of the program.

L   Concatenate all inputs into list
2M  Create two equal-ish size batches, with the bigger piece first
Er  Push the two parts to the stack, and reverse the second one
LM  Put two parts back together as 2-row matrix, then transpose
m|+ For each zipped pair, print the sum on a separate line

Run this one

Answer 2

Perl 5 -a, 23 bytes

#!/usr/bin/perl -a
say$'*s//a/+pop@F for@F

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Perl is unusual in that it is valid to change an array while you are looping over it. So you can have $_ walking from front to back while using pop to effectively walk from back to front while removing array elements which will stop the loop when they meet in the middle:

say$_+pop@F for@F

This however fails for an odd number of elements because in the middle both $_ and pop@F are that middle element so you get twice the value.

So instead I use another odd fact of perl loops: the loop variable is not a copy of the corresponding array position but is an alias for it. If you change $_ then you change the corresponding array element. In this case I want to both get the value of $_ and change it to 0 so that the pop@F returns zero when processing the middle element. The shortest way (without needing ()) to do that is s//a/ which changes e.g. 6 to a6 which evaluates to 0 in a numeric context. The part after the substitution is preserved in $' which will be 6 in the example. And since s/// returns the number of substitutions (1 in this case) the expression s//a/*$' will both set $_ to zero and return the original value.

This however would require a space after say. So finally I use yet another perl oddity, the fact that the expression stack also uses aliases of the values and variables you are processing. Therefore $'*s//a/ also works. When $' is put on the expression stack it is not set yet but when next s//a/ is executed it changes to the old value of $_ in place just before the multiplication is done. This is also the reason why the simpler seeming

say$_+($_=0)+pop@F for@F

does not work. The first $_ gets changed to 0 just before the first addition. You would need to force a copy with something like

say "$_"+($_=0)+pop@F for@F

to make that work

Answer 3

Python 3, 57 bytes, relaxed I/O

def f(a):
 try:x,*a,y=a;return[x+y]+f(a)
 except:return a

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First time to see try..except in golfing.

print version, 57 bytes

def f(a):
 try:x,*a,y=a;print(x+y);f(a)
 except:print(*a)

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printing instead of returning gives the same length.

Python 3, 73 bytes, strict I/O

a=*map(int,input().split()),
while len(a)>1:x,*a,y=a;print(x+y)
print(*a)

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This prints an extra newline at the end if the number of elements is even.

Answer 4

JavaScript ES6, 100 97 bytes

_=>{for(_=_.split(' ').map(Number);0<_.length;)console.log(1<_.length?_.shift()+_.pop():_.pop())}

https://jsbin.com/nufanenehu/edit?js,console

First golfing answer ever. Be gentle.

Answer 5

05AB1E, 7 bytes

#2ä`R+»

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Explanation

#          # split input on spaces
 2ä        # divide into 2 parts
   `       # split separately to stack
    R      # reverse the second one
     +     # add the lists of numbers
      »    # join on newlines

Answer 6

JavaScript (Node.js), 77 58 bytes

a=>g(a.split` `)
g=a=>+a.shift()+~~a.pop()+`
`+(a+a&&g(a))

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Thanks to @Neil for chopping off 17 bytes at once. Also, I found that ~~ works instead of |0, which removes a pair of parens.

Original submission, 77 bytes

a=>g(a.split` `.map(x=>+x))
g=a=>a.shift()+(a.pop()|0)+`
`+(a.length?g(a):'')

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Input as a space-delimited string, output as a newline-delimited string. First converts the input string into an array, then passes to a recursive function that generates the output string.

Answer 7

Brainfuck, 61 bytes

>>+[>,]<[<]>->[[-<<+>>]>[>]<[[<]<<+>>>[>]<-]<[<]>]<<<[<]>[.>]

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This program takes a null-terminated list of base-256 ASCII numbers as input :)

Commented Version

Read input chars until \0
>>+[>,]<
Clear initial 1
[<]>->
[
  Copy two to the left
  [-<<+>>]>
  Head to right side
  [>]<
  Subtract from right side add to left side repeat
  [[<]<< +>>> [>]< - ]
  Start at next spot
  <[<]>
]
Print it all out
<<<
[<]
>[.>]

Satisfying I/O requirements, 81 bytes

>>+[>,>,[-]<]<[<]>->[[-<<+>>]>[>]<[[<]<<+>>>[>]<-]<[<]>]<<<[<]>[.[-]++++++++++.>]

In this version the input characters must be separated and the output is split by newlines.

Answer 8

Jelly, 7 bytes

œs2U2¦S

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Takes a list of integers, such as [1, 2, 3, 4, 5]. Outputs as an array, such as [6, 6, 3].

11 bytes to conform to the strict input specifications.

Answer 9

J, 28 bytes

>.@-:@#({.|:@,:@+/@,:|.@}.)]

ungolfed

>.@-:@# ({. |:@,:@+/@,: |.@}.) ]

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My J is rusty and surely this could be improved -- suggestions encouraged. Almost all the complexity arises from the odd length case and the newline requirement. I had 22 bytes before I noticed the newline requirement:

>.@-:@# ({. +/@,: |.@}.) ]

Answer 10

R, 68 66 bytes

cat(c((n=scan())[y<-1:(x=sum(n|1)/2)],rep(0,x!=x%/%1))+rev(n[-y]))

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Takes input from stdin as in the examples, and prints the output on stdout. Since using scan() is typically shorter than defining a function and taking an array as input, this is the shortest way to take input.

Reads in data, takes the first half (rounded down) of the list, and appends a zero (if necessary), then adds it to the reverse of the second half of the list.

Answer 11

Python 2, 46 bytes

f=lambda x:x[1:]and[x[0]+x[-1]]+f(x[1:-1])or x

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x[1:] is just a shorter way to write len(x)>1

Answer 12

Python 3, 70 bytes

lambda a:[a[i]+a[~i]for i in range(len(a)//2)]+len(a)%2*[a[len(a)//2]]

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+32 bytes for strict input:

Python 3, 102 bytes

a=[*map(int,input().split())]
l=len(a)
for q in[a[i]+a[~i]for i in range(l//2)]+l%2*[a[l//2]]:print(q)

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Answer 13

Ruby + -p, 52 bytes (spec compliant)

a=$_.split.map(&:to_i);loop{puts a.shift+(a.pop||0)}

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A full program, taking a string of space-separated integers and printing the results separated by newlines.

Normally the -p flag forces you to reassign the $_ variable, as it will be implicitly printed at the end of the program. However this code terminates with an NoMethodError when it calls + on nil (the result of shifting the head off an empty array), before that implicit print can happen.

Ruby, 33 bytes (relaxed IO)

->a{puts a.shift+(a.pop||0);redo}

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This lambda skips the integer parsing code, and it is able to save 2 bytes by replacing loop{} with redo since it is already inside of a repeatable block. This also terminates with NoMethodError.

Answer 14

Haskell, 93 80 bytes

^{Saved 13 bytes thanks to Laikoni!}

f s|x<-read<$>words s=mapM print$zipWith(+)(take(div(1+length x)2)x)$reverse$0:x

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Answer 15

Clojure, 124 bytes

#(loop[l(map bigint(.split %" "))](println(+(first l)(or(last(rest l))0)))(if(>(count l)2)(recur(concat(butlast(rest l))))))

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Ungolfed:

#(loop [l (map bigint (.split % " "))]
   (println (+ (first l) (or (last (rest l)) 0)))
   (if (> (count l) 2)
     (recur (concat (butlast (rest l))))))

An anonymous function that takes the input string and splits it. loop is rerun with every iteration with the "middle" items of the changing list if there are enough items left. Combining butlast and rest creates the "middle" list.
Interesting is the part where the sum is printed. Simply adding (first l) and (last l) only works for an even number of items. When an uneven number of items is passed, first and last will hit the same item. Using rest will remove the first item, in this case the only one, and return an empty list. Using last will return nil which is converted into zero by the or function.

Answer 16

Haskell, 78 72 70 67 bytes

mapM print.f.map read.words
f(x:r@(_:_))=(x+last r):f(init r)
f e=e

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Edits:

• -6 bytes by using mapM print instead of unlines.map show from GolfWolf's Haskell answer.
• -2 bytes by removing +0 which I though was necessary for type inference but wasn't. Thanks to GolfWolf for pointing out.
• -3 bytes by adding r@(_:_) to make the f[x]=f[x] pattern superfluous. Thanks again to GolfWolf.

Answer 17

Perl 6 -n,  58  42 bytes

.&{$_,{S:s/(\S+) (.*?) (\S+)?$ {say $0+($2//0)}/$1/}...''}

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.say for .=words[^*/2]Z+ |.[$_-1...$_/2],0

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---

If the input/output restrictions are relaxed 40 bytes

*.[^*/2,{$_-1...$_/2}].&roundrobin».sum

Test it

Answer 18

Java (OpenJDK 9), 76 bytes

a->{for(int i=0,l=a.length;i<l--;)System.out.println(i<l?a[i++]+a[l]:a[i]);}

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With strict input

Java (JDK 10), 115 bytes

s->{var a=s.split(" ");for(int i=0,l=a.length;i<l--;)System.out.println(i<l?new Long(a[i++])+new Long(a[l]):a[i]);}

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Credits

• -10 bytes on strict input thanks to Kevin Cruijssen

Answer 19

JavaScript (Node.js), 56 bytes

f=x=>([a,m='',b]=x.split(/ (.*) | /),b?a- -b+`
`+f(m):a)

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Split the string by all the middle numbers and capture them all as a group, add the ends (coercing strings to numbers via a- -b, which is shorter than any alternatives I know of), and concatenate the result with a line break and f(all captured middle numbers) where available.

Answer 20

Python 3, 108 bytes

i=[*map(int,input().split())]
l=len(i)
for a,b in zip(i,i[:int(l/2-.5):-1]):print(a+b)
if l&1:print(i[l//2])

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Non competitive, just thought it would be nice to share this zip approach :)
Too bad slicing and reversing in python is a bit finicky

Answer 21

Python 2, 111 107 103 bytes

i=map(int,input().split())
for y in[i[x]+i[-(x+1)]for x in range(len(i)/2)]+[i[x+1]]*(len(i)%2):print y

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Takes input in "strict" format (space separated string of numbers) and outputs a trailing newline if the count is even. Works because in Python 2, list comprehensions leak the last value of the variable x so it can be used later to identify the middle number.

Python 2, 79 75 69 bytes

lambda i:[i[x]+i[-(x+1)]for x in range(len(i)/2)]+[i[x+1]]*(len(i)%2)

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Shorter version with input and output as lists. Adds an empty string to the output for even numbers.

Answer 22

FORTH 76 Bytes

: A BEGIN DEPTH 1 - ROLL + CR . DEPTH 2 <= UNTIL DEPTH 1 = IF CR . THEN ;

Output:

1 2 3 4 5 a
6
6
3  ok

2 2 A
4  ok

-5 3 -8 2 A
-3
-5 ok

Answer 23

Python 2, 112 bytes

def f(s):
 s=[int(i)for i in s.split(' ')];s.insert(len(s)//2,0)
 for i in range(len(s)//2):print s[i]+s[-i-1]

Answer 24

Perl 6/Rakudo 45 bytes

Perl6 -ne

{say .shift+(.pop||0) while $_}(.words.Array)

Super simple, shifts and pops, Perl makes the empty list (with odd count) easy with .pop||0.

Answer 25

Red, 105 bytes

func[s][b: to-block s
l: length? b
repeat n l / 2[print b/(n) + b/(l - n + 1)]if odd? l[print b/(n + 1)]]

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Red, 91 bytes (Relaxed IO - a list ot integers)

func[b][l: length? b
repeat n l / 2[print b/(n) + b/(l - n + 1)]if odd? l[print b/(n + 1)]]

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Answer 26

Python 3, 136 bytes

def f(x):
 x,o=[int(i) for i in x.split()],[];y=len(x)
 for i in range(int(y/2)):o+=[x[i]+x[-i-1]]
 if y%2!=0:o+=[x[int(y/2)]]
 return o

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Answer 27

Ruby, 67 bytes

This can probably be improved, but I'm not sure how...

def x(a);puts a.shift.to_i+a.pop.to_i;return if a.length==0;x a;end

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Answer 28

C, 128 bytes

This version is more in keeping with the terms of the challenge, but has a hard-coded maximum of 9 numbers that it can accept.

i[9],j,k;f(char*s){for(s=strtok(s," ");j++,s;i[j]=atoi(s),s=strtok(0," "));while(--j>=++k)printf("%d\n",i[j]+(j==k?0:i[k]));}

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I let C do the tokenizing for me, which allows me to send each string directly to atoi(). Afterwards, I just add the mirror indexes until the start>end.

As I couldn't get a array length and strtok() destroys the original string, making backtracking difficult, and I didn't want to have to realloc() to resize arrays, I hard-coded an array with a maximum of 9 values.

C, 107 bytes (uses argv)

i,j,*k;f(char**s){k=(int*)s;while(s[++i])k[i]=atoi(s[i]);while(++j<=--i)printf("%d\n",k[i]+(i==j?0:k[j]));}

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This version makes use of the fact that argv already has been tokenized; as a bonus, it doesn't have a hard-coded limit on the number of numbers! It does, however, assume that sizeof(int)==sizeof(char*), which is truly evil and non-portable.

Answer 29

PHP, 107 bytes

New to this, so please forgive my lack of correct formatting.

for($i=0,$a=explode(" ",$argv[1]),$c=count($a);$i<$c/2;$i++)echo$i==$c-$i-1?$a[$i]:$a[$i]+$a[$c-$i-1]."\n";

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Answer 30

Java (OpenJDK 8), 163 bytes

a->{String[] b=a.split(" ");int c=b.length;a="";for(int d=0;d<c/2;d++){a+=(Integer.parseInt(b[d])+Integer.parseInt(b[c-1-d]))+"\n";}if(c%2>0){a+=b[c/2];}return a;}

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Well It’s sure not going to win me the bounty, but all in all I’m pretty happy with my work here.

Explanation:

a->{String[] b=a.split(" ");    //Create String array to get rid of the pesky spaces
int c=b.length;    
a="";    //Since we now only need the array, we can reuse the original string to save a few bytes
for(int d=0;d<c/2;d++){    //Loop half(even) or half-1(odd) of the array length
  a+=(Integer.parseInt(b[d])+    
   Integer.parseInt(b[c-1-d]))+    //Add the ‘rainbow pairs’ to the output string
  "\n";}    //Add a newline to each sum and end the loop
if(c%2>0){a+=b[c/2];}    //If the number of input numbers is odd, print the middle digit
return a;}    //the Output string is the Input string