26
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For each row and then column of a matrix, we can add an extra entry with the sum of the last two entries in that row or column. For example with the following input matrix:

[ 1 1 1 ]
[ 2 3 4 ]

The resulting matrix would be:

[ 1 1 1 2 ]
[ 2 3 4 7 ]
[ 3 4 5 9 ]

Given an input of an integer N and an [X,Y] matrix of size at least 2x2, perform the above expansion N times and output the result. The resulting matrix will always be of size [X+N,Y+N].

Examples:

Input:                     Output:

2, [ 0 0 ]                 [ 0 0 0 0 ]
   [ 0 0 ]                 [ 0 0 0 0 ]
                           [ 0 0 0 0 ]
                           [ 0 0 0 0 ]


3, [ 1 1 1 ]               [ 1  1  1  2  3  5 ]
   [ 2 3 4 ]               [ 2  3  4  7 11 18 ]
                           [ 3  4  5  9 14 23 ]
                           [ 5  7  9 16 25 41 ]
                           [ 8 11 14 25 39 64 ]
\$\endgroup\$

14 Answers 14

8
\$\begingroup\$

MATL, 13 14 15 16 20 21 bytes

2*:"!tP2:Y)sv

Thanks @Zgarb for removing 1 byte!

Try it online!

2*         % implicitly input number N and multiply by 2
:          % create vector [1,2,...,2*N]
"          % for loop: do this 2*N times
  !        %   transpose. Implicitly input matrix in the first iteration
  tP       %   duplicate and flip vertically
  2:       %   vector [1,2]
  Y)       %   pick submatrix formed by the first two rows
  s        %   sum of each column
  v        %   append as a new row
           % end for
           % implicit display
\$\endgroup\$
4
  • 1
    \$\begingroup\$ I don't know MATL, but wouldn't it be shorter to loop 2N times than to loop twice N times? \$\endgroup\$
    – Zgarb
    Feb 5 '16 at 0:25
  • \$\begingroup\$ @Zgarb Of course! How did I miss that? Thanks!! \$\endgroup\$
    – Luis Mendo
    Feb 5 '16 at 0:42
  • \$\begingroup\$ Does MATL have a built-in for doubling a number? \$\endgroup\$
    – Zgarb
    Feb 5 '16 at 0:46
  • \$\begingroup\$ @Zgarb No. You need 2* (postfix notation). Maybe it should have a one-character built-in, it's used often. Also 2^ (square). But I'm running out of code space :-) \$\endgroup\$
    – Luis Mendo
    Feb 5 '16 at 0:49
6
\$\begingroup\$

J, 19 bytes

(v"1@v=.,[+&{:}:)^:

This defines an adverb, which takes the number on its left, and produces a verb taking the matrix on its right. For the second example, it gives

  3 ((v"1@v=.,[+&{:}:)^:) 2 3 $ 1 1 1 2 3 4
1  1  1  2  3  5
2  3  4  7 11 18
3  4  5  9 14 23
5  7  9 16 25 41
8 11 14 25 39 64

Explanation

(v"1@v=.,[+&{:}:)^:  Left argument x, right argument y
(               )^:  Repeat x times:
     v=.               Bind the following verb to v, and apply to y:
         [    }:         y and y-without-last-item
          +&{:           Sum of their last items
        ,                Append that to y
                       (v automatically threads to rows)
 v"1@                  then apply v to columns
\$\endgroup\$
0
3
\$\begingroup\$

K, 23 bytes

{x(2({x,+/-2#x}'+)/)/y}

In action:

  {x(2({x,+/-2#x}'+)/)/y}[3;(1 1 1;2 3 4)]
(1 1 1 2 3 5
 2 3 4 7 11 18
 3 4 5 9 14 23
 5 7 9 16 25 41
 8 11 14 25 39 64)

Try it here.

\$\endgroup\$
1
  • \$\begingroup\$ it still works if you remove the leading {x and trailing y} \$\endgroup\$
    – ngn
    Feb 6 '18 at 1:50
3
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Jelly, 15 13 12 bytes

-1 byte by @Dennis

ṫ-S;@"Z
ÇḤ}¡

Like @LuisMendo's MATL answer, this transposes the array before doing the transformation along one axis. Therefore, we need to call the function 2*n times.

ṫ-S;@"Z       Helper link. Input: x (2D array)
 -              Numeric literal: -1
ṫ               Get x[-1:], i.e. last two rows in x
  S             Sum
   ;@"          Append each to x. " is 'zipWith'; @ switches argument order.
      Z         Transpose the array.
ÇḤ}¡          Main link. Input: a, n
Ç               Call the last link on a
 Ḥ}             2n
   ¡            times.
  

Try it here.

\$\endgroup\$
0
2
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ES6, 134 bytes

(n,a)=>[...a.map(b=>[...b,...Array(n)].map(c=>(c<1/0?0:c=a+d,d=a,a=c))),...Array(n)].map(b=>(b?0:b=[...a].map((c,j)=>c+d[j]),d=a,a=b))

Explanation:

(n,a)=> // arguments n is number to expand, a is original array
    [...
        a.map(b=> // for each row in a
            [...b,...Array(n)] // append n elements to the row
            .map(c=>(c<1/0?0:c=a+d,d=a,a=c))) // scan the elements and fill the new ones by summing the previous two
        ,...Array(n)] // append n rows
    .map(b=>(b?0:b=[...a].map((c,j)=>c+d[j]),d=a,a=b)) // scan the rows and fill the new rows by summing the previous two rows
\$\endgroup\$
2
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Haskell, 67 bytes

o%m=m++[o(+)(last m)$last$init m]
(!!).iterate(map(id%).(zipWith%))

Usage example:

*Main> ( (!!).iterate(map(id%).(zipWith%)) ) [[1,1,1],[2,3,4]] 3
[[1,1,1,2,3,5],[2,3,4,7,11,18],[3,4,5,9,14,23],[5,7,9,16,25,41],[8,11,14,25,39,64]]

How it works:

(!!).iterate(    ...         )  -- repeatedly apply ... to the first agrument and
                                -- pick the iteration defined by the second arg
                   (zipWith%)   -- for one iteration add a new row and
          map(id%)              -- then a new element at the end of each each row

o%m                             -- add row or element at the end of a row resp.
                                -- argument o is a "modify function"
                                --          m the whole matrix or a row
 m++[    (last m)(last$init m)] -- take m and append the result of combining the
                                -- last and 2nd last element of m
     o(+)                       -- with a modified version of (+)
                                -- modification is none (aka. id) when adding an
                                -- element to the end of a row and
                                -- zipping elementwise (zipWith) when adding a row
\$\endgroup\$
5
  • \$\begingroup\$ I'm a haskell novice. I've got as far as sudo apt-get install haskell-platform and am running the ghci REPL which gives me a Prelude> prompt. When I paste in o%m=m++[o(+)(last m)$last$init m] I get <interactive>:2:4: parse error on input '='. Can you give me a little primer either running this from a source file or in the REPL? \$\endgroup\$ Feb 5 '16 at 16:59
  • \$\begingroup\$ @DigitalTrauma: either put the o%m=... line (and only this line) in a file called, let's say fib-matrix.hs. Then you can use the :l fib-matrix.hs command in ghci to load the definitions and call the main function just as described in my usage example. -- Or use let o%m=... in ( (!!). ... ) [[1,1,1]...] 3. \$\endgroup\$
    – nimi
    Feb 5 '16 at 17:16
  • 1
    \$\begingroup\$ @DigitalTrauma: oh, there's a 3rd way: give the main function a name, e.g. add a f= in front of the second line: f=(!!).iterate..., save both lines in a file and load it via l: <filename.hs>. Then you can call f [[1,1,1],[2,3,4]] 3, etc. \$\endgroup\$
    – nimi
    Feb 5 '16 at 17:20
  • \$\begingroup\$ I'm not sure I would accept this as valid haskell, the top line is a function definition and needs modification for use in the REPL, but the 2nd line can be can only be used in the REPL. \$\endgroup\$ Feb 7 '16 at 3:25
  • \$\begingroup\$ @DanielHill: there's a topic on meta which allows unnamed functions that depend on global helper functions. \$\endgroup\$
    – nimi
    Feb 7 '16 at 8:59
2
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CJam, 17 16 bytes

q~2*{~_2$.+]z}*p

Input format is the matrix first (as a CJam-style 2D array) and the number of iterations afterwards.

Test it here.

Explanation

Turns out this is the same solution as everyone else's:

q~      e# Read and evaluate input.
2*      e# Double the iteration count.
{       e# Run this block that many times...
  ~     e#   Dump all rows on the stack.
  _     e#   Copy the last row.
  2$    e#   Copy the penultimate row.
  .+    e#   Vectorised addition.
  ]     e#   Wrap all rows in a new array.
  z     e#   Transpose such that the next iteration processes the other dimension.
}*
p       e#   Pretty-print.
\$\endgroup\$
1
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Seriously, 20 bytes

,,τ"┬`;d@d@X+@q`M"£n

Takes input the matrix (as a 2D list), then N. Outputs a 2D list.

This version doesn't work on the online interpreter for some reason, but does work with this pre-challenge commit.

A version that works online, for 23 bytes:

,τ",┬`;d@d@X+@q`M"nkΣ£ƒ

Takes input in the opposite order (N, then matrix).

Try it online!

I will add an explanation after I sleep for a little while. Working around interpreter bugs is never fun.

\$\endgroup\$
1
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Pyth, 13 12 bytes

u+Rs>2dCGyEQ

Try it online. Test suite.

Uses the same algorithm to most answers. Takes as input the matrix as a 2D array on the first line and n on the second line.

Explanation

u        yEQ     do 2*N times, starting with input matrix:
       CG          transpose
 +R                append to each row:
   s                 sum of
    >2d              last 2 elements of row
\$\endgroup\$
1
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Matlab, 60 bytes

I was first messing about with Matlab's fancy indexing methods (i.e., A(end+1,:)=sum...) before I figured that in this rare case, simple concatenation is actually cheaper in Matlab. Too bad I had to convert this into an actual function. Should work with Octave as well.

function A=f(A,n)
for i=1:2*n
A=[A;sum(A(end-1:end,:))]';end

I suppose this is a prime example of how not to make algorithms. For A=2x2, n=1000 this algorithm already takes 5 seconds on my laptop, n=2000 it's almost 50 seconds! (or approximately 30s if A is a gpuArray thanks to my trusty Quadro 1000M)

\$\endgroup\$
4
  • \$\begingroup\$ I don't have a copy of Matlab. Can I run this under GNU octave? If so can you give instructions? \$\endgroup\$ Feb 5 '16 at 17:06
  • 1
    \$\begingroup\$ Yes I called it Matlab because it doesn't use any Octave specific functions. Simply put it in a file called f.m and run as e.g. f([0,1;2,3],1000) \$\endgroup\$
    – Sanchises
    Feb 5 '16 at 17:26
  • \$\begingroup\$ I see. 1) save as f.m. 2) Start octave. 3) Paste load f.m; f([1,1,1;2,3,4],3) into the REPL prompt - works for me. \$\endgroup\$ Feb 5 '16 at 17:39
  • \$\begingroup\$ If you say so! I only use the octave online website so no idea how it should work otherwise. I'll see if I can permalink from there \$\endgroup\$
    – Sanchises
    Feb 5 '16 at 18:19
1
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Java, 2179 bytes

Just Worked it out:- This code is on Java Language.

import java.util.Scanner;

public class FebonnaciMatrix {
        static Scanner scan=new Scanner(System.in);

        public static void main(String[] args) {
        
        int x,y;
        System.out.println("For the Array to Work Upon:- ");
        
        System.out.println("Enter the Row:- ");
        int row=scan.nextInt();
        System.out.println("Enter the Column:- ");
        int col=scan.nextInt();
        
        int inpArr[][]=new int[row][col];
        
        System.out.println("Enter the values");
        inpArr=inpValues(row,col);
        
        System.out.println("The Input Array is:- ");
        display(inpArr,row,col);
        
        System.out.println("Input the Array size of Febonacci Array ");
        
        System.out.println("Enter the Row");
        int frow=scan.nextInt();
        System.out.println("Enter the Column");
        int fcol=scan.nextInt();
        
        int febArr[][]=new int[frow][fcol];
        febArr=copyValue(inpArr,febArr,row,col);
        
        for(x=0;x<row;x++)
        {
            for(y=col;y<fcol;y++)
                febArr[x][y]=febArr[x][y-2]+febArr[x][y-1];
        }
        
        for(x=row;x<frow;x++)
        {
            for(y=0;y<fcol;y++)
                febArr[x][y]=febArr[x-2][y]+febArr[x-1][y];
        }
        
        System.out.println();
        System.out.println("The Febonacci Array:-");
        display(febArr,frow,fcol);
    }
    
    static void display(int[][] arr,int row,int col)
    {
        int x,y;
        for(x=0;x<row;x++)
        {
            for(y=0;y<col;y++)
                System.out.print(arr[x][y]+"\t");
            System.out.println();
        }
    }
    
    static int[][] inpValues(int row,int col)
    {
        int arr[][]=new int[row][col];
        int x,y;
        for(x=0;x<row;x++)
        {
            for(y=0;y<col;y++)
            {
                System.out.print("Enter the value:- ");
                arr[x][y]=scan.nextInt();
            }
        }
        return arr;
    }
    
    static int[][] copyValue(int[][] old, int[][] ne, int row,int col)
    {
        int x,y;    
        for(x=0;x<row;x++)
        {
            for(y=0;y<col;y++)
                ne[x][y]=old[x][y];
            
        }
        return ne;
    }

}
\$\endgroup\$
5
  • 1
    \$\begingroup\$ Welcome to Programming Puzzles and Code Golf! The question is tagged code-golf which means that answers are competing to be written in the shortest amount of code possible (in bytes). Your answer may well solve the problem, but I see little attempt to "golf" the code (i.e. make it as short as possible). There are plenty of trivial opportunities to do this with your code, e.g. variables with 1-char names, and removal of unnecessary whitespace. Beyond that you can read these tips, specifically for java \$\endgroup\$ Feb 5 '16 at 21:11
  • \$\begingroup\$ ... Have a look at the tag-wiki for code-golf, especially the How should I answer a code golf? Any hints? section. Also note that java is notoriously hard to golf down to short code, in comparison with many other languages. This shouldn't dissuade you though - if you have a well-golfed java answer, it is likely to be quite popular, even if it is longer than all the other answers. Don't be put off by all the mind-bendingly short esolang answers - this community tends to be good at taking language handicaps into account. \$\endgroup\$ Feb 5 '16 at 21:18
  • \$\begingroup\$ @DigitalTrauma- Thanks... for helping me out as a newbee to this... I'll surely go through the links and come up with a new code... \$\endgroup\$ Feb 5 '16 at 21:21
  • \$\begingroup\$ Since you're a new user, I took the liberty of editing your answer for better formatting. In particular a) a clear title stating language and number of bytes, b) code formatting of your code. On all stackexchange sites, code formatting is easy - simply prefix all your code lines with 4 spaces. In fact even easier - in the edit box, select your code, then click the {} at the top of the edit box - this will automatically do this prefixing. \$\endgroup\$ Feb 5 '16 at 21:22
  • \$\begingroup\$ Okay... I will just check it out... \$\endgroup\$ Feb 5 '16 at 21:27
1
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Python, 103 105 bytes

f=lambda n,L:f(n-1,[l+[sum(l[-2:])]for l in L])if n else L
lambda n,L:zip(*f(n,map(list,zip(*f(n,L)))))

Anonymous function takes list of list and passes to recursive function f. Output is transposed and then passed to f again, then the output of the second go is re-transposed. Output is a list of tuples

Saved two bytes thanks to bakuriu

\$\endgroup\$
1
  • 1
    \$\begingroup\$ n>0 could simply be n, since you start with a positive n and when you reach 0 its value is false. \$\endgroup\$
    – Bakuriu
    Feb 5 '16 at 21:44
1
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APL (Dyalog Classic), 17 bytes

{⍉⍵,⊢/2+/⍵}⍣2⍣⎕⊢⎕

Try it online!

\$\endgroup\$
0
\$\begingroup\$

Perl 6,  87 73  71 bytes

->\c,\m{for ^c {.[+*]=[+] .[*X-1,2]for m;m.push: [map {[+] m[*X-1,2;$_]},m[0].keys]};m}
->\c,\m{for ^c {.[+*]=[+] .[*X-1,2]for m;m[+*]=[m[*-2;*] »+«m[*-1]]};m}
->\c,\m{for ^c {.[+*]=[+] .[*X-1,2]for m;m[+*]=[m[*-2;*]Z+m[*-1;*]]};m}
-> \c, \m {
  for ^c { # 0 ..^ c

    # each row
    .[+*]                            # new column at the end of row ($_)
          = [+] .[ * X- 1,2 ]        # add up the last two entries in row ($_)
                              for m; # for every row

    # too bad this was longer than the above code
    # m[*;+*]=map *+*,m[*;*-2,*-1]

    # each column
    m[ +* ]                 # add new row
            = [             # make it an Array rather than a List
                m[ *-2; * ] # the second to last row
                »+«         # added columnwise with
                m[ *-1 ]    # the last row
              ]
  };

  m # return the result
}

Usage:

use v6.c;
# give it a lexical name
my &code = ->\c,\m{ … }

my @return = code 3,[[1,1,1],[2,3,4]];

put '[ ', $_».fmt('%2d'), ' ]' for @return;

put '';

put @return.perl ~~ {S:g/' '//};
[  1  1  1  2  3  5 ]
[  2  3  4  7 11 18 ]
[  3  4  5  9 14 23 ]
[  5  7  9 16 25 41 ]
[  8 11 14 25 39 64 ]

[[1,1,1,2,3,5],[2,3,4,7,11,18],[3,4,5,9,14,23],[5,7,9,16,25,41],[8,11,14,25,39,64]]
\$\endgroup\$
3
  • \$\begingroup\$ Pasting this into perl6 gives me some errors. I'm a perl novice - what am I doing wrong? \$\endgroup\$ Feb 5 '16 at 17:04
  • \$\begingroup\$ @DigitalTrauma I'm sorry I should have written the usage as my &code = ->\c,\m{ … } to make it clear that the ->\c,\m{ … } needs to be replaced with the code above. I usually use an implicit $_ or @_ or explicit placeholder parameters $^a because they tend to be shorter. I just didn't think about it. Also make sure that you are using a new enough version ( $*PERL.compiler.version !before 2015.12 ) \$\endgroup\$ Feb 5 '16 at 22:00
  • \$\begingroup\$ @DigitalTrauma You can also go on the #perl6 channel on freenode.net and use camelia (like this) to run the code (precede lines with m: and a space) You can also msg camelia directly \$\endgroup\$ Feb 5 '16 at 22:44

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