Jolly Jumper Sequence

Question

A sequence of n > 0 integers is called a jolly jumper if the absolute values of the difference between successive elements take on all the values 1 through n-1.

So the sequence [4,1,2,4] has absolute differences [3,1,2] which is equivalent to the set [1,2,3] (1 to n-1 where n is the length of original sequence) so it is therefore a jolly jumper.

Sequences have length n>0.

Assume n=1 is a jolly jumper.

Easy mode: Don't worry about stdin/stdout. Just a function that accepts arguments however and returns something that indicates jolly or not

Hard mode: Input on stdin (space separated), and output is "Jolly" / "Not jolly". Capitalization matters.

This is code golf.

EDIT: Sequences can contain negative integers and input on stdin is space separated.

$ jolly 2 -1 0 2
Jolly

$ jolly 19 22 24 25
Jolly

$ jolly 19 22 24 21
Not jolly

Answer 1

Haskell

Easy 4 characters

Returns a list of jolly integers if and only if a list of jolly integers is given as input. This is legal based on "Just a function that accepts arguments however and returns something that indicates jolly or not".

j=id

Alternative easy solution with 61 characters:

Takes in a list and returns the empty list if the sequence is jolly.

import List
j n=zipWith(\x->abs.(x-))n(tail n)\\[1..length n]

Answer 2

Ruby, 92 93 characters

The hard version with input on STDIN.

f=gets.split.each_cons(2).map{|a|eval(a*?-).abs}.sort
$><<(f==[*1..f.size]??J:"Not j")+"olly"

If you start it with -pa (counts as 4) you can save 5 chars:

f=$F.each_cons(2).map{|a|eval(a*?-).abs}.sort
$_=(f==[*1..f.size]??J:"Not j")+"olly"

Answer 3

Java (Hard)

Assumes that input is given through stdin. (not through command line arguments as per example)

Golfed - 325

class JollyJumper {
public static void main(String[] args) {
String[] in = new Scanner(System.in).nextLine().split(" ");
int[] j=new int[in.length-1],k=j.clone();
for(int i=0;i<in.length-1;i++){j[i]=Math.abs(Integer.parseInt(in[i])-Integer.parseInt(in[i+1]));k[i]=i+1;}
Arrays.sort(j);System.out.println(Arrays.equals(j, k)?"Jolly":"Not jolly");
}
}

Un-Golfed

public class JollyJumper {
public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int[] jolly;
    String[] in;

    in = sc.nextLine().split(" ");
    jolly = new int[in.length-1];

    for (int i = 0; i < in.length-1; i++)
        jolly[i] = Math.abs(Integer.parseInt(in[i]) - Integer.parseInt(in[i+1]));

    Arrays.sort(jolly);

    for (int i = 1; i <= in.length-1; i++) {
        if (jolly[i-1] != i) {
            System.out.println("Not jolly");
            return;
        }
    }
    System.out.println("Jolly");
}
}

Answer 4

Scala, easy mode, 123 characters

def j(s:String)={var a=s.sliding(2,1).map(x=>math.abs(x(0)-x(1))).toList
for(c<-1 to a.size)
if(!a.contains(c))false
true}

To run or test on ideone.com:

object Main
{
   def main(args:Array[String])
   {
      def j(s:String):Boolean=
      {
         var a=s.sliding(2,1).map(x=>math.abs(x(0)-x(1))).toList
         for(c<-1 to a.size)
            if(!a.contains(c)) false
         true
      }
      println(j("4124"))
   }
}

Answer 5

Golfscript, easy mode, 21 18 chars

{.@-abs\}*;0]$.,,=

Accepts arguments as an array of ints on the stack, with nothing else on the stack; leaves 1 on the stack if it's jolly and 0 otherwise. To take input on stdin as a space-separated list of ints, prepend

~]

and to output "Jolly" / "Not jolly" (assuming that we're turning this into a program) postpend

"Not jJ"5/="olly"

Answer 6

J (easy), 18

(i.@#-:<:/:])|2-/\

   (i.@#-:<:/:])|2-/\2 _1 0 2
1
   (i.@#-:<:/:])|2-/\19 22 24 25
1
   (i.@#-:<:/:])|2-/\19 22 24 21
0

J (hard), 68

2!:55]1!:2&2'olly',~>('Not j';'J'){~(i.@#-:<:/:])|2-/\".@>2}.ARGV_j_

$ jconsole jumper.ijs 2 -1 0 2
Jolly
$ jconsole jumper.ijs 19 22 24 25
Jolly
$ jconsole jumper.ijs 2 19 22 24 21
Not jolly

Answer 7

Brachylog, 11 bytes (easy)

s₂ᶠ-ᵐȧᵐo~⟦₁

Try it online!

s₂ᶠ-ᵐ - Consecutive differences

ȧᵐ - Absolute values

o - Sort

~⟦₁ - Is the result the range 1 to something?

Answer 8

J, 30 26 easy mode, 81 76 hard mode

edit: handle lists shorter than 3, fix stdin reading

First line takes care of easy mode, second adds hard mode.

j=:[:*/[:(>:@i.@#=/:~)[:|2-/\]
exit('olly',~[:>('Not j';'J'){~[:j 0".}:)&.stdin''

J reads generally right-to-left:

2-/\ : for every two successive numbers in the list, take the difference

| : absolute value

/:~ : sort in ascending order

>:@i.@# : 1 to n, for a list of n numbers

= : compare the sorted differences with the sequence (using a J "fork")

*/ : multiply all the element-wise booleans; if all the comparisons were 1, their product is 1, so it's jolly

Answer 9

Ruby, 97 102 106 (hard)

Might as well, since everyone else is:

h,*t=gets.split
d=t.map{|i|h,i=i,h;eval(i+?-+h).abs}.sort
$><<(d==[*1..d.size]??J:"Not j")+"olly"

Input taken on stdin.

Answer 10

D

easy (103 83 chars)

returns sum of 1..i.length on Jolly some other number if not (a bit of rules laywering here)

import std.math;auto jolly(I)(I i){int t,l;foreach(r;i){t+=abs(l-r);l=r;}return t;}

hard (142 chars)

input is whitespace delimited and ends on EOF

import std.stdio;import std.math; void main(){int i,j,l,t;while(readf("%d ",&i)>0){t+=abs(l-i);l=i;j++;}write(t==j*++j/2?"J":"Not j","olly");}

Answer 11

Groovy

Easy: 78

j={m=[];it[1..-1].inject(it[0]){p,n->m<<p-n;n};m*.abs().sort()==1..<it.size()}

assert [[2, -1, 0, 2,], [19, 22, 24, 25], [19, 22, 24, 21]].collect { j(it) } == [true, true, false]

Hard: 151

j={m=[];it[1..-1].inject(it[0]){p,n->m<<p-n;n};m*.abs().sort()==1..<it.size()};System.in.eachLine{println "${j(it.split()*.toLong())?'J':'Not j'}olly"}

Answer 12

PowerShell, hard, 117 126

('Not j','J')["$(($a=-split$input)|%{if($x-ne$0){[math]::abs($x-$_)}$x=$_}|sort)"-eq"$(1..($a.Count-1)|sort)"]+'olly'

History:

• 2011-11-18 17:54 (123, −3) – Changed $null to a non-existent variable
• 2011-11-18 18:02 (117, −6) – inlined all variable declarations

Answer 13

Scala

A quick stab - there are probably improvements possible.

Easy: 77

def j(? :Int*)=(?tail,?).zipped.map(_-_).map(math.abs).sorted==(1 to?.size-1)

Hard: 124

val? =args.map(_.toInt)toSeq;print(if((?tail,?).zipped.map(_-_).map(math.abs).sorted==(1 to?.size-1))"Jolly"else"Not jolly")

Answer 14

Q, 64 (hard), 30 (easy)

hard

{$[(1_(!)(#)x)~asc abs 1_(-':)x;(-1"Jolly";);(-1"Not jolly";)];}

easy

{(1_(!)(#)x)~asc abs 1_(-':)x}

Answer 15

J (easy), 19 characters

*/(=i.@#)<:/:~|2-/\

Usage:

    */(=i.@#)<:/:~|2-/\4 2 1 4
1

Vary similar to DCharness's answer, and I would have just added it as a comment but for the fact that he hasn't visited since the 23rd of February.

2-/\ takes the difference between successive pairs of numbers,

| gets the absolute value of each number,

/:~ sorts into ascending order,

<: decrements each number by 1,

(=i.@#) a J hook which generates the sequence of numbers from 0 to the length of the differences list - 1 (i.@#) and compares it with that list =.

*/ multiples the list of 1s and 0s generated by the previous verb.

Answer 16

Scala easy:138 153, 170 (was errornous, improved later)

def j(i:String)={
def a(s:Seq[Int])=(s zip s.tail).map(x=>(x._2-x._1))
a(a(i.split(" ").map(_.toInt)).map(math.abs).sorted).toSet.size==1}

ungolfed:

def jolly (input: String) = { 
      val list = input.split (" ").map (_.toInt)

      def stepsize (s: Seq[Int]) = 
        (s zip s.tail).map (x=> (x._2 - x._1))

      val first = stepsize (input.split (" ").map (_.toInt))
      val pos = first.map (math.abs)
      val unique = stepsize (pos.sorted).toSet
      (unique.size) == 1
}

The idea is, that we build the second derivation:

Original: 4 1 2 4
Stepsize:  -3 1 2 (first)
     abs:   3 1 2
  sorted:   1 2 3 
Stepsize:     1 1 
  to Set:       1 
    size:       1

Scala hard 172 182, 205 (was errornous/improved):

def j{
def a(s:Seq[Int])=(s zip s.tail).map(x=>(x._2-x._1))
println((if(a(a(readLine.split(" ").map(_.toInt)).map(math.abs).sorted).toSet.size==1)"J"else"Not j")+"olly")}
j

more or less the same as above.

Answer 17

PHP, easy, 129

For a given array $s of integers:

for($i=1;$i<count($s);$i++)$a[abs($s[$i]-$s[$i-1])]=1;
for($i=1;$i<count($s);$i++)if(!isset($a[$i]))die('Not Jolly');echo 'Jolly';

The ungolfed version:

for( $i=1; $i<count( $s ); $i++ )
    $a[ abs( $s[$i] - $s[$i-1] ) ] = 1;

for( $i=1; $i < count($s); $i++ )
    if( !isset( $a[$i] ) )
        die( 'Not Jolly' );

echo "Jolly";        

Answer 18

Jelly, 7 6 bytes (easy)

IAṢ⁼J$

Try it online!

IAṢ⁼J$    jolly function on N:
IAṢ       the increment list: get all the Increments, take their Absolute values, and Ṣort them
   ⁼      compare that to...
    J$    range from 1 to len(N) -- this has an extra number, but that's fine because...
          ...the increment list is one shorter, and ⁼ will only compare that many values

Takes input as comma-separated numbers in the first argument. Returns 1 if the sequence is jolly, and 0 if it isn't!

7-byte solution:

LRṖḟIA$

Try it online!

Takes input as comma-separated numbers in the first argument. Returns nothing if the list is a jolly jumper sequence, and something if it isn't.

Adding this line makes it work with the hard spec:

Jelly, 27 22 bytes (hard, feedback welcome!)

ɠḲVIAṢ⁼J$ị“¢⁼D“¡KṀȥƘạ»

Try it online!

ɠḲVIAṢ⁼J$ị“¢⁼D“¡KṀȥƘạ»
ɠḲV                     read a line, split on spaces and eValuate the numbers
   IAṢ⁼J$               jolly function: see above!
         ị              ịndex the result into (remember Jelly is one-indexed, so 0 wraps around to the back):
          “¢⁼D“          "Jolly" compressed if true,
              ¡KṀȥƘạ»   or, "Not jolly" compressed if false!

27-byte (hard) solution:

LRṖḟIA$
ɠḲVÇ“¡KṀȥƘạ»“¢⁼D»L?

Try it online!

Takes space-separated numbers on stdin, and outputs "Jolly" or "Not jolly".

Explanation:

LRṖḟIA$               jolly function:
LRP                   make a range (R) from 1 to the input length (L), popping off (P) the last number to make it 1 to N-1.
   ḟ                  reverse filter: remove all the elements from that range that are members of...
    IA$               the increment list: get all the increments, take their absolute values (expressed as one monad via '$').
ɠḲVÇ“¡KṀȥƘạ»“¢⁼D»L?    i/o main function:
ɠḲV                   read a line from stdin, split it on spaces and have Python parse each number (handling negative signs)
   Ç             ?    run the above, and use the result on the following conditional:
                L?    if the length of the result is truthy (non-empty):
    “¡KṀȥƘạ»          then, return Jelly compressed string "Not jolly",
            “¢⁼D»     else, return Jelly compressed string "Jolly".

Any feedback much appreciated!

Answer 19

Haskell, 59 57 bytes

f n=all(`elem`map abs(zipWith(-)n$tail n))[1..length n-1]

Easy mode, returns jollyness as a boolean. Thanks to @Laikoni for two bytes.

Try it online!

Answer 20

Japt, 32 30 bytes Hard

-2 Bytes from @Shaggy

`not jo¥y`s4*Näa n äa e¥1
hUÎu

Try it online!

Answer 21

Python 3, 117 (hard)

l=[*map(int,input().split())]
print(["Not j","J"][{abs(a-b)for a,b in zip(l[1:],l[:-1])}=={*range(1,len(l))}]+"olly")

Try it online!

Answer 22

JavaScript: 105 (easy mode)

Golfed:

function a(l){for(r=i=1;i<(m=l.length);i++){for(j=t=0;j+1<m;)t+=(d=l[j]-l[++j])*d==i*i;t||(r=0)}return r}

Un-golfed:

function isJolly(list){
    //Iterate over i to list.length-1
    for(r=i=1;i<(length=list.length);i++){
        //Check the differences between all consecutive elements squared minus i squared.  Set t to true if one was found.
        for(j=t=0;j+1<length;)t+=(diff=list[j]-list[++j])*diff==i*i;

        //if t is not true, return value is 0
        t||(r=0)
    }
    return r
}

Answer 23

Perl, 89 (hard)

86 characters of code + 3 for running with the -p option

@a=0;$a[abs($1-$2)]=1while s/(\S+) (\S+)/$2/;$_='Jolly';(grep{!defined}@a)&&s/J/Not j/

Answer 24

Javascript (hard): 138

a=prompt().split(" ")
i=0;b=[];c=[]
while(b[i]=Math.abs(a[i]-a[++i]),c[i-1]=i,i<a.length-1);b.sort()
alert(b+""==c+""?"Jolly":"Not jolly")

Answer 25

R, Easy, 110

f=function(s){p=NULL;l=length;for (i in 2:l(s))p=c(p,abs(s[i]-s[i-1]));ifelse(all(sort(p)==(1:(l(s)-1))),1,0)}

Usage:

f(c(2, -1, 0, 2))
[1] 1
f(c(19, 22, 24, 25))
[1] 1
f(c(19, 22, 24, 21))
[1] 0

Answer 26

Python, 72 (easy), 114 (hard)

Easy:

def f(a):return len(set(map(lambda x,y:abs(x-y),a[1:],a[:-1])))>len(a)-2

Hard:

a=map(int,raw_input().split())
print('Not j','J')[len(set(map(lambda x,y:abs(x-y),a[1:],a[:-1])))>len(a)-2]+'olly'

Answer 27

Python, 255 characters

r=[19,22,24,25]
i=0
k=[ i+1 for i in range(len(r)-1)]
def jolly(a):
    p=[]
    i=0
    while i<len(a)-1: 
       p.append(abs(a[i+1]-a[i]))
       i+=1
    p.sort() 
    if p==k:
       return 'jolly'
    else:
       return 'Not jolly'

print(jolly(r))

Answer 28

C, 119(hard), 97(easy)

b,c,a[];main(k){while(~scanf("%d",a+c))k=c++;for(c=k;b<c*c;)k-abs(a[b%c]-a[b++%c+1])?:k--;puts(k?"Not jolly":"Jolly");}

The easy solution reads the input from the arguments and returns a 0 as exit code if the input is a jolly jumper sequence:

i,k;main(int c,char**a){for(k=c-=2,a++;i<c*c;)k-abs(atoi(a[i%c])-atoi(a[i++%c+1]))?:k--;exit(k);}

Answer 29

APL (50 49 47, hard)

'Not jolly' 'Jolly'[1+K[⍋K←¯1↓|Z-1⌽Z]≡¯1↓⍳⍴Z←⎕]

Easy (24):

{K[⍋K←¯1↓|⍵-1⌽⍵]≡¯1↓⍳⍴⍵}

The function takes an array and returns 0 or 1.

Answer 30

Japt, 11 10 bytes

Êo1 eUäa n

Run it online

Saved 1 byte thanks to Shaggy