# Bouncing in an array

### Introduction

Arrays can also be seen as a field for a bouncing ball. This of course sounds very vague, so here is an example of an input:

[1, 2, 3, 4, 5, 6, 7, 8, 9]
[9, 8, 7, 6, 5, 4, 3, 2, 1]
[1, 2, 3, 4, 5, 6, 7, 8, 9]


The challenge is to output the bounced arrays. These are made from diagonal patterns which bounce at the edges of the field. This path is pointed upwards. The path for the first bounced array (in which the path is directly bounced off the egde), is:

[1, -, -, -, 5, -, -, -, 9]
[-, 8, -, 6, -, 4, -, 2, -]
[-, -, 3, -, -, -, 7, -, -]


From left to right, this would result in [1, 8, 3, 6, 5, 4, 7, 2, 9]. This is our first bounced array. The path for the second bounced array:

[-, 2, -, -, -, 6, -, -, -]
[9, -, 7, -, 5, -, 3, -, 1]
[-, -, -, 4, -, -, -, 8, -]


This results in [9, 2, 7, 4, 5, 6, 3, 8, 1]. The path for the third bounced array is:

[-, -, 3, -, -, -, 7, -, -]
[-, 8, -, 6, -, 4, -, 2, -]
[1, -, -, -, 5, -, -, -, 9]


This results in [1, 8, 3, 6, 5, 4, 7, 2, 9]. So the three bounced arrays are:

[1, 8, 3, 6, 5, 4, 7, 2, 9]
[9, 2, 7, 4, 5, 6, 3, 8, 1]
[1, 8, 3, 6, 5, 4, 7, 2, 9]


Given at least 1 array containing only non-negative integers, with all the arrays having the same length, output all the bounced arrays.

### Test cases

Test case 1:

Input:                       Output:
[1, 2, 3, 4, 5]              [1, 7, 3, 9, 5]
[6, 7, 8, 9, 0]              [6, 2, 8, 4, 0]

Input:                       Output:
[1, 2, 3, 4, 5]              [1, 2, 3, 4, 5]

Input:                       Output:
[0, 0, 0, 0, 0, 0, 0, 0]     [0, 9, 0, 9, 0, 9, 0, 100]
[9, 9, 9, 9, 9, 9, 9, 100]   [9, 0, 9, 0, 9, 0, 9, 0]
[0, 0, 0, 0, 0, 0, 0, 0]     [0, 9, 0, 9, 0, 9, 0, 100]

Input:                       Output:
[0, 1, 2, 3, 4, 5]           [0, 7, 14, 9, 4, 11]
[6, 7, 8, 9, 10, 11]         [6, 1, 8, 15, 10, 5]
[12, 13, 14, 15, 16, 17]     [12, 7, 2, 9, 16, 11]

Input:                       Output:
[0, 0, 0, 0, 0, 0]           [0, 2, 2, 6, 2, 6]
[1, 2, 3, 4, 5, 6]           [1, 0, 3, 2, 5, 2]
[2, 2, 2, 2, 2, 2]           [2, 2, 0, 4, 2, 4]
[9, 8, 7, 6, 5, 4]           [9, 2, 3, 0, 5, 2]


This is , so the submission with the least amount of bytes wins!

• Jelly wins this. – lirtosiast Jan 22 '16 at 18:22
• Can you please add a three-array testcase where the last array is different then the first, and a four-array testcase? – ETHproductions Jan 22 '16 at 19:11
• I can't find the problem description. Can anyone tell me where it is? – feersum Jan 22 '16 at 19:14
• Yes. I can't find a description of the task. – feersum Jan 22 '16 at 19:20
• @LuisMendo it's the only way to bounce, at line 0 it can't go up more – edc65 Jan 22 '16 at 22:15

## Pyth, 17 bytes

>lQC.e.>bkC+_PtQQ


Explanation:

                      implicit: Q=input
>                     First
l Q                   len(Q) elements of the
C .e                  Transpose of enumerated map lambda b,k:
.>                 Rotate the kth element rightwards by
b
k               k.
C +              Transpose of: Q concatenated to
_ P t Q        itself reversed without first and last elements.
Q


Try it here.

# JavaScript (ES6), 70

a=>a.map((r,k)=>r.map((e,j)=>(a[k-=d]||a[d=-d,k-=d+d]||r)[j],d=1,++k))


TEST

F = a=>a.map((r,k)=>r.map((e,j)=>(a[k-=d]||a[d=-d,k-=d+d]||r)[j],d=1,++k))

test = [{
I: [
[1, 2, 3, 4, 5],
[6, 7, 8, 9, 0]
],
O: [
[1, 7, 3, 9, 5],
[6, 2, 8, 4, 0]
]
}, {
I: [
[1, 2, 3, 4, 5]
],
O: [
[1, 2, 3, 4, 5]
]
}, {
I: [
[0, 0, 0, 0, 0, 0, 0, 0],
[9, 9, 9, 9, 9, 9, 9, 100],
[0, 0, 0, 0, 0, 0, 0, 0]
],
O: [
[0, 9, 0, 9, 0, 9, 0, 100],
[9, 0, 9, 0, 9, 0, 9, 0],
[0, 9, 0, 9, 0, 9, 0, 100]
]
}, {
I: [
[0, 1, 2, 3, 4, 5],
[6, 7, 8, 9, 10, 11],
[12, 13, 14, 15, 16, 17]
],
O: [
[0, 7, 14, 9, 4, 11],
[6, 1, 8, 15, 10, 5],
[12, 7, 2, 9, 16, 11]
]
}, {
I: [
[0, 0, 0, 0, 0, 0],
[1, 2, 3, 4, 5, 6],
[2, 2, 2, 2, 2, 2],
[9, 8, 7, 6, 5, 4]
],
O: [
[0, 2, 2, 6, 2, 6],
[1, 0, 3, 2, 5, 2],
[2, 2, 0, 4, 2, 4],
[9, 2, 3, 0, 5, 2]
]
}];

console.log = x => O.textContent += x + '\n';

test.forEach(t => {
var r = F(t.I),
ok = r.join \n == t.O.join \n
console.log((ok ? 'OK' : 'KO') + '\nInput\n' + t.I.join \n + '\nOutput\n' + r.join \n + '\n')
})
<pre id=O></pre>

## CJam, 31 30 bytes

q~__W%1>W<+_z,_@*<zee::m>z\,<p


Input and output as a list of CJam-style arrays.

Test it here.

Most definitely golfable...

# Jelly, 2925 24 bytes

ṖḊm-³;Z
,ZL€R_€/‘ż€Çị/€€


Thanks for @Sp3000 for helping me golf off 5 bytes!

Try it online!

## Ruby (2.2.2p95), 124 bytes

->*a{b=a.length;b<2?a:b.times.map{|i|d=i>0?-1:1;c=i;a[0].length.times.map{|j|r=a[c][j];c+=d;d*=-1if c==0||c==a.length-1;r}}}


This could probably be a lot better. I'll figure out how later!

# Japt, 554941 39 bytes

Wow, that was both really tricky and insanely fun.

C=2*Nl -2Nw £YoY+Ul)£NgLmX%CC-X%C)gY} ·


Test it online!

Outputs in reverse order to the examples. This will break slightly on inputs of more than 100 arrays; hopefully this doesn't make too much difference.

           // Implicit: N = array of input arrays, U = first input array, J = -1, L = 100
// Let's use the first example. 3 input arrays, each of length 9.
C=2*Nl -2  // Set variable C to 2*N.length - 2. In the example, C is 4.
Nw      }  // Reverse N and map each index Y to:
YoY+Ul)    //  Create the range [Y...U.length+Y).
//  The first time through (Y=0), we have   [0, 1, 2, 3, 4, 5, 6, 7, 8]
£       }  //  Map each item X and index Y to:
X%C        //   Take X mod C.                          [0, 1, 2, 3, 0, 1, 2, 3, 0]
C-X%C      //   Take C - (X mod C).                    [4, 3, 2, 1, 4, 3, 2, 1, 4]
Lm         //   Take the minimum of these and 100.     [0, 1, 2, 1, 0, 1, 2, 1, 0]
Ng         //   Get the array at the resulting index.
gY         //   Get the item at index Y in this array. [1, 8, 3, 6, 5, 4, 2, 7, 9]
·          // Join the result with newlines. I guess this isn't necessary, but it helps with readability.


### Non-competing version, 36 bytes

C=J+Nl)òC±C ®óUl)£NgLmX%CC-X%C)gY} ·


I had implemented these two number functions prior to the challenge:

• ò - same as o, but returns [X..Y] instead of [X..Y)
• ó - same as o, but returns [X..X+Y) instead of [X..Y)

But due to a misplaced 0, they were buggy and always returned empty arrays. This has now been fixed.

Test it online!

# Python 2, 107106108105 104 bytes

(Dropped some extra parens) (Wrong starting location (ب_ب) ) (already had a list of that length)

def b(a):
r,e=len(a)-1,enumerate
print[[a[abs((i-o-r)%(r*2or 1)-r)][i]for i,_ in e(q)]for o,q in e(a)]


It's legal to have the input as an argument to a function, right? This is my first time submitting my code golf answer.

• Yes it is legal :) – Adnan Jan 24 '16 at 2:16

## APL, 33 chars

{(⍳≢⍵){0⌷(⍺-⍳≢⍉⍵)⊖⍵⍪1↓¯1↓⊖⍵}¨⊂↑⍵}


Assume ⎕IO←0. The idea is that the bouncing movement can be obtained by simple shift upward of a matrix, if the original matrix is augmented along the first dimension with the matrix reversed with its first and last row shaved. Graphically:

1 - - - - - 1 - - - -
- 2 - - - 2 - 2 - - -
- - 3 - 3 - - - 3 - 3
- - - 4 - - - - - 4 -


from

1 - - - - - 1 - - - -
- 2 - - - - - 2 - - -
- - 3 - - - - - 3 - -
- - - 4 - - - - - 4 -
- - - - 3 - - - - - 3
- - - - - 2 - - - - -


In APL reverse and upward rotate are the same symbol: ⊖.

• USE THE FIRST STEFANO. ⊃ instead of 0⌷. – Adalynn Jul 31 '17 at 22:34
• I am afraid "first" and "0⌷" give two very different results when applied to a nested array. Try it for yourself. First vs. 0⌷ – lstefano Aug 2 '17 at 9:30

## Clojure, 125 bytes

Wow, this accumulated characters quite fast.

(fn[v](let[H(count v)r range R(r H)](for[i R](map #((v %2)%)(r(count(v 0)))(drop i(cycle(concat R(reverse(r 1(dec H))))))))))


Just trying to save bytes by let-defining frequently used values.

# Jelly*, 15 bytes

JṚŒḄṖṙ’Ʋị⁸Jị"\$€


Try it online!