Matrix in Range

Question

The Challenge

Given an integer n>0 output a n+1 X n+1 matrix containing all integers from 1 to 2n as shown in the test cases bellow

Test Cases

n=1  

1  2  
2  2

n=2

1   2   4  
2   3   4  
4   4   4

n=5  

1   2   3   4   5   10  
2   3   4   5   6   10  
3   4   5   6   7   10   
4   5   6   7   8   10  
5   6   7   8   9   10  
10  10  10  10  10  10  

n=10  

1   2   3   4   5   6   7   8   9   10  20  
2   3   4   5   6   7   8   9   10  11  20  
3   4   5   6   7   8   9   10  11  12  20  
4   5   6   7   8   9   10  11  12  13  20  
5   6   7   8   9   10  11  12  13  14  20  
6   7   8   9   10  11  12  13  14  15  20  
7   8   9   10  11  12  13  14  15  16  20  
8   9   10  11  12  13  14  15  16  17  20  
9   10  11  12  13  14  15  16  17  18  20  
10  11  12  13  14  15  16  17  18  19  20  
20  20  20  20  20  20  20  20  20  20  20

I think that the pattern is pretty easy, so let's see who will give the shortest answer in bytes.
This is code-golf

Rules

Input must be an integer (1-indexed)

Output can be a matrix (as shown in the test cases) or a list of lists

Giuseppe · Accepted Answer · 2017-08-29 13:08:55Z

10

R, 53 bytes

function(n)rbind(cbind(outer(1:n,1:n,`+`)-1,2*n),2*n)

Uses the outer "product" to generate all sums of the range 1,...,n as a matrix, subtracts 1 from each, then binds 2*n as a column on the right and a row on the bottom, recycling as needed, and returns a matrix as the result.

Try it online!

R, 78 bytes

More naive implementation.

function(n){m=matrix(2*n,n+1,n+1)
for(i in seq(n))m[1:n,i]=(0:(2*n))[1:n+i]
m}

Try it online!

edited Aug 29, 2017 at 13:08

answered Aug 28, 2017 at 21:34

Giuseppe

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\$\begingroup\$ +1 nice, I was thinking about outer but hadn't quite got there \$\endgroup\$
– MickyT
Commented Aug 28, 2017 at 22:00

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Community · Accepted Answer · 2020-06-17 09:04:33Z

7

Mathematica, 61 46 bytes

ArrayFlatten@{{##-1&~Array~{#,#},2#},{2#,2#}}&

thanx @alephalpha for -15 bytes

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Aug 28, 2017 at 21:04

ZaMoC

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\$\begingroup\$ ArrayFlatten@{{Array[+##-1&,{#,#}],2#},{2#,2#}}& \$\endgroup\$
– alephalpha
Commented Aug 29, 2017 at 3:47
\$\begingroup\$ +##-1& can just be ##-1& and you can use infix Array: ArrayFlatten@{{##-1&~Array~{#,#},2#},{2#,2#}}& \$\endgroup\$
– user61980
Commented Aug 29, 2017 at 20:27

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Community · Accepted Answer · 2020-06-17 09:04:33Z

5

MATL, 12 10 bytes

:&+q,!GEYc

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Explanation

:       % Input n (implicit). Push range [1 2 ... n]
&+      % matrix of pairwise additions
q       % Subtract 1
,       % Do twice
  !     %   Transpose
  GE    %   Push 2*n
  Yc    %   Concatenate that value to all rows, thus extending the matrix
        % End (implicit). Display (implicit)

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Aug 28, 2017 at 21:12

Luis Mendo

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jferard · Accepted Answer · 2017-08-28 21:34:37Z

5

Haskell, 48 bytes

f n=[[i..n-1+i]++[2*n]|i<-[1..n]]++[2*n<$[0..n]]

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answered Aug 28, 2017 at 21:34

jferard

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Community · Accepted Answer · 2020-06-17 09:04:33Z

5

Octave, 38 37 35 bytes

@(n)[(x=1:n)+x'-1 z=~x'+2*n;z' 2*n]

Try it online!

or

@(n)~(k=blkdiag((x=1:n)+x'-1,0))*2*n+k

Try it online!

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Aug 29, 2017 at 4:17

rahnema1

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Leaky Nun · Accepted Answer · 2017-08-28 21:01:52Z

4

Jelly, 11 bytes

+þḶ;€Ḥ;ḤzḤG

Try it online!

answered Aug 28, 2017 at 21:01

Leaky Nun

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\$\begingroup\$ Was just about to post byte for byte without the G which is superfluous to requirements \$\endgroup\$
– Jonathan Allan
Commented Aug 28, 2017 at 21:04
\$\begingroup\$ Nice use of outer product \$\endgroup\$
– Adalynn
Commented Aug 28, 2017 at 21:05
\$\begingroup\$ @JonathanAllan another solution uses ;€ḤZ;€Ḥ... \$\endgroup\$
– Leaky Nun
Commented Aug 28, 2017 at 21:13
\$\begingroup\$ ...and yet another is Ḷ;Ḥ©µ+þ‘«® ;p \$\endgroup\$
– Jonathan Allan
Commented Aug 28, 2017 at 21:48
1

\$\begingroup\$ @JonathanAllan and another uses +€Ḷ;Ḥṁ€;€Ḥ :p \$\endgroup\$
– Erik the Outgolfer
Commented Aug 29, 2017 at 10:16

| Show 1 more comment

Neil · Accepted Answer · 2017-08-28 21:13:24Z

4

JavaScript (ES6), 64 bytes

f=
n=>[...Array(n+1)].map((_,i,a)=>a.map((_,j)=>n-i&&n-j?i-~j:n+n))

<input type=number min=0 oninput="t.innerHTML=f(+this.value).map(a=>`<tr>${a.map(b=>`<td>${b}</td>`).join``}</tr>`).join``"><table id=t>

answered Aug 28, 2017 at 21:13

Neil

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Jakob · Accepted Answer · 2017-09-01 18:44:33Z

4

Java 8, 99 bytes

Lambda from Integer to int[][] (e.g. Function<Integer, int[][]>). Surprisingly resistant to golfing.

n->{int p=n+1,o[][]=new int[p][p],i=0,r,c;while(i<p*p)o[r=i/p][c=i++%p]=r<n&c<n?r-~c:2*n;return o;}

Try It Online

Ungolfed lambda

n -> {
    int
        p = n + 1,
        o[][] = new int[p][p],
        i = 0,
        r, c
    ;
    while (i < p * p)
        o[r = i / p][c = i++ % p] =
            r < n & c < n ?
                r - ~c
                : 2 * n
        ;
    return o;
}

Acknowledgments

-1 byte thanks to Kevin Cruijssen

edited Sep 1, 2017 at 18:44

answered Aug 29, 2017 at 1:01

Jakob

2,7691 gold badge13 silver badges24 bronze badges

\$\begingroup\$ You can golf a byte by starting i=0 and putting the ++ at [c=i++%p]. \$\endgroup\$
– Kevin Cruijssen
Commented Aug 29, 2017 at 11:52

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totallyhuman · Accepted Answer · 2017-08-28 21:21:32Z

3

Python 2, 64 62 61 bytes

-3 bytes thanks to Mr. Xcoder.

lambda n:[range(i+1,i-~n)+[n*2]for i in range(n)]+[[n*2]*-~n]

Try it online!

I'm probably missing a key pattern though.

Python 2, 76 bytes

lambda n:[[[n*2,i-~j][n-i and n-j>0]for j in range(n+1)]for i in range(n+1)]

Try it online!

edited Aug 28, 2017 at 21:21

answered Aug 28, 2017 at 20:51

totallyhuman

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1

\$\begingroup\$ As usual, *(n+1) is *-~n. \$\endgroup\$
– Mr. Xcoder
Commented Aug 28, 2017 at 20:54
1

\$\begingroup\$ Some more obfuscation for 61 bytes \$\endgroup\$
– Mr. Xcoder
Commented Aug 28, 2017 at 20:58
\$\begingroup\$ If you want to adopt the Python 3 version too, the shortest I could get is 64 bytes \$\endgroup\$
– Mr. Xcoder
Commented Aug 28, 2017 at 21:15
\$\begingroup\$ The 76 bytes version can be reduced to 72 bytes \$\endgroup\$
– Halvard Hummel
Commented Aug 29, 2017 at 8:14

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Community · Accepted Answer · 2020-06-17 09:04:33Z

3

Pyth, 18 bytes

+m+rd+QdyQSQ]*]yQh

Maybe I am missing an obvious pattern (cc @totallyhuman)...

Test Suite.

"Pretty print" Test Suite.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Aug 28, 2017 at 20:52

Mr. Xcoder

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Mr. Xcoder · Accepted Answer · 2017-08-28 21:29:19Z

2

Proton, 48 44 bytes

-4 bytes thanks to @totallyhuman!

n=>[(i+1..i-~n)+[n*2]for i:0..n]+[[n*2]*-~n]

Try it online!

edited Aug 28, 2017 at 21:29

answered Aug 28, 2017 at 21:03

Mr. Xcoder

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\$\begingroup\$ 44 bytes. \$\endgroup\$
– totallyhuman
Commented Aug 28, 2017 at 21:05
\$\begingroup\$ @totallyhuman thanks. \$\endgroup\$
– Mr. Xcoder
Commented Aug 28, 2017 at 21:05

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MickyT · Accepted Answer · 2017-08-28 22:06:15Z

2

R, 54 63 67 bytes

function(n)cbind(rbind(sapply(1:n-1,'+',1:n),2*n),2*n)

Try it online!

Thanks to @Guiseppe for the pointer for sapply and the 9 bytes

edited Aug 28, 2017 at 22:06

answered Aug 28, 2017 at 21:49

MickyT

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\$\begingroup\$ Actually, this approach can be improved as well: sapply(1:n-1,'+',1:n) but then it's a mere 1 byte more than using outer (the quotes are obviously backticks) \$\endgroup\$
– Giuseppe
Commented Aug 28, 2017 at 22:00
1

\$\begingroup\$ \` works for escaping backticks in a code block delimited by backticks @Giuseppe \$\endgroup\$
– Cyoce
Commented Aug 29, 2017 at 3:07

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Oliver · Accepted Answer · 2017-08-29 13:55:42Z

2

Japt, 17 16 bytes

õ_óU pU*2ÃpUô@*2

Try it online!

edited Aug 29, 2017 at 13:55

answered Aug 28, 2017 at 21:20

Oliver

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0xffcourse · Accepted Answer · 2017-09-07 18:54:34Z

2

Recursiva, 50 bytes

Only 10 bytes shorter than python and thus it is official, Recursiva is not a golfing-language at all... It is an esolang though. :D

|{Ba+++++'PJ"	"W+Z~}B+~}'Va'+}'Va'AD'VaJ"	"W*;aADa

Try it online!

edited Sep 7, 2017 at 18:54

answered Aug 29, 2017 at 6:22

0xffcourse

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cleblanc · Accepted Answer · 2017-08-28 21:29:35Z

1

C (gcc), 119 116 115 107 bytes

i,j;f(n){for(;j<n;j++,printf("%d\n",2*n))for(i=0;i++<n;printf("%d\t",j+i));for(;j=i--;printf("%d\t",n*2));}

Try it online!

edited Aug 28, 2017 at 21:29

answered Aug 28, 2017 at 21:18

cleblanc

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\$\begingroup\$ i<n+1 is i<=n (117 bytes). \$\endgroup\$
– Mr. Xcoder
Commented Aug 28, 2017 at 21:20
\$\begingroup\$ @Mr.Xcoder Thanks ! \$\endgroup\$
– cleblanc
Commented Aug 28, 2017 at 21:22
\$\begingroup\$ 101 bytes \$\endgroup\$
– ceilingcat
Commented Aug 5, 2020 at 20:48

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fergusq · Accepted Answer · 2017-08-29 16:42:30Z

Röda, 44 bytes

f n{seq 1,n|[[seq(_,_1+n-1)]+2*n];[[2*n]*n]}

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Explanation:

f n{seq 1,n|[[seq(_,_1+n-1)]+2*n];[[2*n]*n]}
f n{                                       } /* Function f(n)         */
    seq 1,n                                  /* Sequence 1..n         */
           |                                 /* For each _1:          */
              seq(_,_1+n-1)                  /*   Sequence _1.._1+n-1 */
             [             ]                 /*   As list             */
                            +2*n             /*   Append 2*n          */
            [                   ]            /*   Push to the stream  */
                                   [2*n]     /* List [2*n]            */
                                        *n   /* Multiplied by n       */
                                  [       ]  /* Push to the stream    */

Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Dyalog APL, 29 bytes

Requires ⎕IO←0

{(S,⍨1+¯1 ¯1↓∘.+⍨⍳⍵+1)⍪S←2×⍵}

Try it online!

How?

1+¯1 ¯1↓∘.+⍨⍳⍵+1 the upper-left portion of the array
(S,⍨...)⍪S←2×⍵ the corner

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Aug 28, 2017 at 21:15

Adalynn

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Sasha · Accepted Answer · 2017-08-30 19:43:31Z

1

><>, 84+2 Bytes

+2 for -v flag

Prints with tabs between values, and newlines between rows. Also prints a trailing tab on the last line.

Try it online

1:r:&r&)?\0:r:&r&(?\~$:@2*nao1+!
a0./:r:0~<.17+1o9\ \$:@$:@+n9o1+
   \&r&)?;$:@2*n /

Pre-golfing

1>:r:&r&)?\0>    :r:&r&(?\~$:@2*nao1+\
            \+1o9n+@:$@:$/
 \                                   /
          \~0>:r:&r&)?;$:@2*n9o1+\
             \                   /

answered Aug 30, 2017 at 19:43

Sasha

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Community · Accepted Answer · 2020-06-17 09:04:33Z

0

SOGL V0.12, 22 bytes

∫HA.∫a+}.«¹}¹.I.«r∙rο+

Try it Here!
leaves output on the stack, which you can view in the console

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Aug 28, 2017 at 20:53

dzaima

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Community · Accepted Answer · 2020-06-17 09:04:33Z

0

Jelly, 14 bytes

R+’r@R;€Ḥz0;€Ḥ

Try it online!

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Aug 28, 2017 at 21:06

Adalynn

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Xcali · Accepted Answer · 2017-08-29 05:36:43Z

0

Perl 5, 56 bytes

$,=$";say($_..$_+$"-1,$"*2)for 1..($"=<>);say(($"*2)x$")

Try it online!

answered Aug 29, 2017 at 5:36

Xcali

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Valentin Michalak · Accepted Answer · 2017-08-29 08:59:44Z

0

Kotlin, 59 bytes

Array(n+1){c->Array(n+1){r->when(n){c,r->n*2;else->1+c+r}}}

Try it online.

answered Aug 29, 2017 at 8:59

Valentin Michalak

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Joshua · Accepted Answer · 2017-08-30 11:49:23Z

Clojure, 153 135 bytes

List of lists? Yay, Lisp

(fn[n](loop[r[] i 0 d (* 2 n)](if(= i n)(conj r(conj(repeat n d)d))(recur(conj r(conj(vec(map #(+ i %)(range 1(inc n))))d))(inc i)d))))

Ungolfed:

(defn a[n]
  (loop [r[] i 0 d (* 2 n)]
    (if(= i n)
      (conj r(conj(repeat n d)d))
      (recur
        (conj r
            (conj (vec (map #(+ i %)(range 1(inc n)))) d))
        (inc i)
        d))))

Anonymous function that takes the input as argument and returns a list of lists.

Output of n=5:

[[1 2 3 4 5 10] [2 3 4 5 6 10] [3 4 5 6 7 10] [4 5 6 7 8 10] [5 6 7 8 9 10] (10 10 10 10 10 10)]

Emigna · Accepted Answer · 2017-08-30 12:44:22Z

0

05AB1E, 17 bytes

FLN+I·¸«ˆ}·¸I>.×ˆ

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Explanation

F                   # for N in [0 ... input-1]
 L                  # push range [1 ... input]
  N+                # add N to each
    I·¸«            # append input*2
        ˆ           # add to global list
         }          # end loop
          ·¸        # push [input*2]
            I>.×    # repeat it input+1 times
                ˆ   # add to global list
                    # implicitly output global list

edited Aug 30, 2017 at 12:44

answered Aug 30, 2017 at 12:38

Emigna

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Community · Accepted Answer · 2020-06-17 09:04:33Z

0

J, 29 bytes

(}:@(][\1+i.@+:),]#+:),.>:#+:

ungolfed

(}:@(] [\ 1+i.@+:) , ]#+:) ,. >:#+:

explanation

(}:@(] [\ 1+i.@+:)                   NB. make the inner part of the matrix
          1+i.@+:                      NB. 1..2*n, where n is the input
    (] [\ 1+i.@+:)                     NB. fork: infixes (sliding window) of length n, over 1..2*n
(}:@                                   NB. remove last element
                   , ]#+:)           NB. add a row of 2*n to the end
                           ,. >:#+:  NB. add a column of 2*n to entire result above

Try it online!

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Aug 29, 2017 at 8:44

Jonah

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\$\begingroup\$ Heh, my APL answer and your answer have the same byte count! Can you add an explanation please? \$\endgroup\$
– Adalynn
Commented Aug 30, 2017 at 2:06
\$\begingroup\$ @Zacharý, updated. fwiw, this could probably be golfed at least a bit further by someone like myself, and perhaps by additional 10+ bytes by a J expert. \$\endgroup\$
– Jonah
Commented Aug 30, 2017 at 17:29

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user45941user45941 · Accepted Answer · 2017-09-07 20:41:47Z

Actually, 23 bytes

;;Rnkp@;r♀+@;τ;(♀q)@α@q

Try it online!

Explanation:

;;Rnkp@;r♀+@;τ;(♀q)@α@q
;;                       two copies of input
  R                      range(1, input+1)
   n                     copy input times
    kp@                  push stack to list, remove first element
       ;r                push range(input)
         ♀+              pairwise addition (add the value in the range to each value in the corresponding list)
           @;            duplicate input again
             τ;          input*2, duplicate that
               (♀q)      append input*2 to each list
                   @α@q  append a row of input*2

Chris · Accepted Answer · 2017-09-07 22:48:37Z

Clojure v1.8, 97 bytes

#(conj(mapv(fn[i](conj(vec(range i(+ % i)))(* 2 %)))(range 1(inc %)))(vec(repeat(inc %)(* 2 %))))

Try it online!

Explanation

(range 1(inc %))                           Numbers from 1 to 'n'
(mapv ... (range 1(inc %)))                For each one of these numbers
(fn[i](conj(vec(range i(+ % i)))(* 2 %)))  Create the numbers from 'i' to (n+i-1), convert to vector and insert '2*n' to the vector
#(conj ... (vec(repeat(inc %)(* 2 %))))    Insert to the previous vector a vector of repeated '2*n's

Zongor · Accepted Answer · 2017-09-08 01:05:38Z

0

Fortran (GFortran), 93, 82 bytes

SUBROUTINE M(n)
PRINT*,(CHAR(10),(i-1+j,i=1,n),2*n,j=1,n)
PRINT*,(2*n,j=1,n+1)
END

Try it online!

edited Sep 8, 2017 at 1:05

answered Sep 8, 2017 at 0:47

Zongor

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Matrix in Range

The Challenge

Test Cases

Rules

28 Answers 28

R, 53 bytes

R, 78 bytes

Mathematica, 61 46 bytes

MATL, 12 10 bytes

Explanation

Haskell, 48 bytes

Octave, 38 37 35 bytes

Jelly, 11 bytes

JavaScript (ES6), 64 bytes

Java 8, 99 bytes

Ungolfed lambda

Acknowledgments

Python 2, 64 62 61 bytes

Python 2, 76 bytes

Pyth, 18 bytes

Proton, 48 44 bytes

R, 54 63 67 bytes

Japt, 17 16 bytes

Recursiva, 50 bytes

C (gcc), 119 116 115 107 bytes

Röda, 44 bytes

Dyalog APL, 29 bytes

How?

><>, 84+2 Bytes

Pre-golfing

SOGL V0.12, 22 bytes

Jelly, 14 bytes

Perl 5, 56 bytes

Kotlin, 59 bytes

Clojure, 153 135 bytes

05AB1E, 17 bytes

J, 29 bytes

ungolfed

explanation

Actually, 23 bytes

Clojure v1.8, 97 bytes

Explanation

Fortran (GFortran), 93, 82 bytes

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