Matrix in "slash" order

Question

Given two positive numbers N >= 2 and N <= 100 create a matrix which follows the following rules:

• First Number starts at position [0,0]
• Second Number starts at position [0,1]
• Third number goes below First Number (position [1,0])
• Following numbers goes in "slash" direction
• Range of numbers used is [1, N1 * N2]. So, numbers goes from starting 1 to the result of the multiplication of both inputs.

Input

• Two numbers N >= 2 and N <= 100. First number is the amount of rows, Second number the amount of columns.

Output

• Matrix. (Can be outputted as a multidimensional array or a string with line breaks)

Example:

Given numbers 3 and 5 output:

1   2   4   7   10
3   5   8   11  13
6   9   12  14  15

Given numbers 2 and 2

1   2
3   4

Given Numbers 5 and 5

1   2   4   7   11
3   5   8   12  16
6   9   13  17  20
10  14  18  21  23
15  19  22  24  25

The shortest code in bytes wins.

Answer 1

Jelly, 6 5 bytes

pSÞỤs

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How it works

pSÞỤs  Main link. Left argument: n. Right argument: k

p      Take the Cartesian product of [1, ..., n] and [1, ..., k], yielding
       [[1, 1], [1, 2], ..., [n, k-1], [n, k]].
 SÞ    Sort the pairs by their sums.
       Note that index sums are constant on antidiagonals.
   Ụ   Grade up, sorting the indices of the sorted array of pairs by their values.
    s  Split the result into chunks of length k.

Answer 2

Python 3, 91 bytes

def f(n,k):M=[(t//k+t%k,t)for t in range(n*k)];return zip(*k*[map([M,*sorted(M)].index,M)])

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Answer 3

R, 101 60 54 bytes

function(M,N)matrix(rank(outer(1:M,1:N,"+"),,"l"),M,N)

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Thanks to @nwellnhof for the suggestion of rank

Ports Dennis' Jelly answer.

Old answer, 101 bytes:

function(M,N)matrix(unsplit(lapply(split(1:(M*N),unlist(split(x,x))),rev),x<-outer(1:M,1:N,"+")),M,N)

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split is doing most of the work here; possibly there's a golfier algorithm but this definitely works.

Explanation:

function(M,N){
x <- outer(1:M,1:N,"+")			# create matrix with distinct indices for the antidiagonals
idx <- split(x,x)			# split into factor groups
items <- split(1:(M*N),unlist(idx))	# now split 1:(M*N) into factor groups using the groupings from idx
items <- lapply(items,rev)		# except that the factor groups are
					# $`2`:1, $`3`:2,3, (etc.) but we need
                                        # $`2`:1, $`3`:3,2, so we reverse each sublist
matrix(unsplit(items,x),M,N)		# now unsplit to rearrange the vector to the right order
					# and construct a matrix, returning the value
}

Try it online! -- you can use wrap a print around any of the right-hand sides of the assignments <- to see the intermediate results without changing the final outcome, as print returns its input.

Answer 4

Java 10, 121 120 109 105 bytes

m->n->{var R=new int[m][n];for(int i=0,j,v=0;i<m+n;)for(j=++i<n?0:i-n;j<i&j<m;)R[j][i-++j]=++v;return R;}

-11 bytes thanks to @OlivierGrégoire.
-4 bytes thanks to @ceilingcat.

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Explanation:

m->n->{                // Method with two integer parameters and integer-matrix return-type
  var R=new int[m][n]; //  Result-matrix of size `m` by `n`
  for(int i=0,j,       //  Index integers, starting at 0
          v=0;         //  Count integer, starting at 0
      i<m+n;)          //  Loop as long as `i` is smaller than `m+n`
    for(j=++i<n?0      //   Set `j` to 0 if `i+1` is smaller than `n`
               :i-n;   //   or to the difference between `i` and `n` otherwise
        j<i&j<m;)      //   Inner loop `j` until it's equal to either `i` or `m`,
                       //   so basically check if it's still within bounds:
      R[j][i-++j]=++v; //    Add the current number to cell `j, i-(j+1)`
  return R;}           //  Return the result-matrix

Answer 5

J, 15 bytes

$1(+/:@;)</.@i.

-4 more bytes for this solution by miles. Thanks!

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J, 22 19 bytes

-3 bytes thanks to FrownyFrog!

,$[:>:@/:@/:@,+/&i.

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An implementation of Dennis' fantastic Jelly solution in J.

Explanation:

Dyadic verb, takes left and right argument (m f n)

+/&i. creates lists 0..m-1 and 0..n-1 and makes an addition table for them:

   3 +/&i. 5
0 1 2 3 4
1 2 3 4 5
2 3 4 5 6

[:>:@/:@/:@, flattens the table and grades the list twice and adds 1 to it:

   3 ([:>:@/:@/:@,+/&i.) 5
1 2 4 7 10 3 5 8 11 13 6 9 12 14 15

,$ reshapes the list back into mxn table:

   3 (-@],\[:>:@/:@/:@,+/&i.) 5
1 2  4  7 10
3 5  8 11 13
6 9 12 14 15

Answer 6

APL+WIN, 38 or 22 bytes

Prompts for integer input column then row:

m[⍋+⌿1+(r,c)⊤m-1]←m←⍳(c←⎕)×r←⎕⋄(r,c)⍴m

or:

(r,c)⍴⍋⍋,(⍳r←⎕)∘.+⍳c←⎕

based on Dennis's double application of grade up. Missed that :(

Answer 7

Wolfram Language (Mathematica), 73 67 bytes

Count elements in rows above: Min[j+k,#2]~Sum~{k,i-1}

Count elements on the current row and below: Max[j-k+i-1,0]~Sum~{k,i,#}

Put into a table and add 1. Voila:

1+Table[Min[j+k,#2]~Sum~{k,i-1}+Max[j-k+i-1,0]~Sum~{k,i,#},{i,#},{j,#2}]&

Update: I realized there is a shorter way to count all the positions ahead of a normally specified position in the matrix with just one sum over two dimensions:

Table[1+Sum[Boole[s-i<j-t||s-i==j-t<0],{s,#},{t,#2}],{i,#},{j,#2}]&

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Answer 8

APL (Dyalog Unicode), 14 12 bytes

{⍵⍴⍋⍋∊+/↑⍳⍵}

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-2 thanks to ngn, due to his clever usage of ↑⍳.

Based off of Dennis's 5-byte Jelly solution.

Answer 9

05AB1E, 23 bytes

*L<DΣ¹LILâOsè}UΣXsè}Á>ô

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Answer 10

Python 3, 164 bytes

from numpy import*
r=range
def h(x,y):
 a,i,k,j=-array([i//y+i%y for i in r(x*y)]),1,2,0
 while j<x+y:a[a==-j],i,k,j=r(i,k),k,k+sum(a==~j),j+1
 a.shape=x,y;return a

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This is definitely not the shortest solution, but I thought it was a fun one.

Answer 11

Python 2, 93 bytes

def f(b,a):i=1;o=[];exec"if b:o+=[],;b-=1\nfor l in o:k=len(l)<a;l+=[i]*k;i+=k\n"*a*b;print o

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Semi-Ungolfed version:

def f(b,a):
    i=1
    o=[]
    for _ in range(a*b)
        if b:
            o+=[[]]
            b-=1

        for l in o:
            if len(l)<a:
                l+=[i]
                i+=1
    print o

Answer 12

Japt, 25 24 bytes

Hardly elegant, but gets the job done. Working with 2D data in Japt is tricky.

;N×Ç<U©Ap[] A®Ê<V©Zp°T
A

;                      // Set alternative default vars where A is an empty array.
 N×Ç                   // Multiply the inputs and map the range [0..U*V).
    <U                 // If the current item is less than the second input,
      ©Ap[]            // add a new empty subarray into A.
            A®         // Then, for each item in A,
              Ê<V      // if its length is less than the first input,
                 ©Zp°T // Add the next number in the sequence to it.
A                      // Output the results, stored in A.

I added the -Q flag in TIO for easier visualization of the results, it doesn't affect the solution.
Bit off one byte thanks to Oliver.

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Answer 13

JavaScript (Node.js), 103 bytes

(a,b,e=[...Array(b)].map(_=>[]))=>f=(x=0,y=i=0)=>(x<a&y<b&&(e[y][x]=++i),x?f(--x,++y):y>a+b?e:f(++y,0))

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Answer 14

TI-Basic, 76 bytes

Prompt A,B
{A,B🡒dim([A]
1🡒X
For(E,1,B+A
For(D,1,E
If D≤A and E-D<B
Then
X🡒[A](D,E-D+1
X+1🡒X
End
End
End
[A]

Prompts for user input and returns the matrix in Ans and prints it.

TI-Basic is a tokenized language; all tokens used here are one byte, other than [A] which is 2 bytes.

Note: TI-Basic (at least on the TI-84 Plus CE) only supports matrices up to 99x99, and so does this program.

Explanation:

Prompt A,B        # 5 bytes, prompt for user input
{A,B🡒dim([A]      # 9 bytes, make the matrix the right size
1🡒X               # 4 bytes, counter variable starts at 1
For(E,1,B+A       # 9 bytes, Diagonal counter, 1 to A+B-1, but we can over-estimate since we have to check later anyway.
For(D,1,E         # 7 bytes, Row counter, 1 to diagonal count
If D≤A and E-D<B  # 10 bytes, Check if we are currently on a valid point in the matrix
Then              # 2 bytes, If so,
X🡒[A](D,E-D+1     # 13 bytes, Store the current number in the current point in the matrix
X+1🡒X             # 6 bytes, Increment counter
End               # 2 bytes, End dimension check if statement
End               # 2 bytes, End row for loop
End               # 2 bytes, End dimension for loop
[A]               # 2 bytes, Implicitly return the matrix in Ans and print it

Answer 15

Perl 6, 61 59 bytes

{($!={sort($_ Z=>1..*)>>.{*}})($!([X+] ^<<$_)).rotor(.[1])}

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Another port of Dennis' Jelly solution.

Answer 16

Java (JDK 10), 142 131 bytes

X->Y->{var A=new int[X][Y];int E=1;for(int y=0;y<Y+X-1;y++)for(int x=0;x<X;x++){if(y-x<0|y-x>Y-1)continue;A[x][y-x]=E++;}return A;}

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Explanation:

X->Y->{                            // Method with two integer parameters and integer-matrix return-type
    var A=new int[X][Y];           // The Matrix with the size of X and Y
    int E=1;                       // It's a counter
        for(int y=0;y<Y+X-1;y++)   // For each column plus the number of rows minus one so it will run as long as the bottom right corner will be reached
            for(int x=0;x<X;x++){  // For each row
                if(y-x<0|y-x>Y-1)  // If the cell does not exist becouse it's out of range
                    continue;      // Skip this loop cycle
                A[x][y-x]=E++;     // Set the cell to the counter plus 1
            }
    return A;                      // Return the filled Array
}

Big thank to Kevin Cruijssen because I didn't know how to run my code on tio.
Some code like the header and footer are stolen from him. -> His answer

Answer 17

Husk, 9 bytes

ṠCoηÖ¤×,ḣ

Try it online! Most of the heavy lifting is done by ×, which creates pairs in the desired order so that it can support infinite lists.

Explanation

CηÖ×,ḣ²ḣ⁰⁰  (Expanded; let N and K denote the arguments.)
 ηÖ          Take the list of indices of the sorted values of
   ×,         the list of all pairs taken from
     ḣ²        the range from 1 to N and
       ḣ⁰      the range from 1 to K, and
C        ⁰  cut it into lengths of K.

Answer 18

PHP, 115 bytes

a pretty lazy approach; probably not the shortest possible.

function($w,$h){for(;$i++<$h*$w;$r[+$y][+$x]=$i,$x--&&++$y<$h||$x=++$d+$y=0)while($x>=$w|$y<0)$y+=!!$x--;return$r;}

anonymous function, takes width and height as parameters, returns 2d matrix

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Answer 19

JavaScript (Node.js), 108 105 101 100 bytes

n=>(l=>{for(r=[];i<n*n;n*~-n/2+2>i?l++:l--*y++)for(T=y,t=l;t--;)r[T]=[...r[T++]||[],++i]})(y=i=0)||r

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Answer 20

Attache, 45 bytes

{Chop[Grade//2<|Flat!Table[`+,1:_2,1:_],_]+1}

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Anonymous lambda, where paramaters are switched. This can be fixed for +1 byte, by prepending ~ to the program. The test suite does this already.

Explanation

This approach is similar to the J answer and the Jelly answer.

The first idea is to generate a table of values:

Table[`+,1:_2,1:_]

This generates an addition table using ranges of both input parameters. For input [5, 3], this gives:

A> Table[`+,1:3,1:5]
 2 3 4 5 6
 3 4 5 6 7
 4 5 6 7 8

Then, we flatten this with Flat!:

A> Flat!Table[`+,1:3,1:5]
[2, 3, 4, 5, 6, 3, 4, 5, 6, 7, 4, 5, 6, 7, 8]

Using the approach in the J answer, we can grade the array (that is, return indices of sorted values) twice, with Grade//2:

A> Grade//2<|Flat!Table[`+,1:3,1:5]
[0, 1, 3, 6, 9, 2, 4, 7, 10, 12, 5, 8, 11, 13, 14]

Then, we need to chop the values up correctly, as in the Jelly answer. We can cut every _ elements to do this:

A> Chop[Grade//2<|Flat!Table[`+,1:3,1:5],5]
 0 1  3  6  9
 2 4  7 10 12
 5 8 11 13 14

Then, we just need to compensate for the 0-indexing of Attache with +1:

A> Chop[Grade//2<|Flat!Table[`+,1:3,1:5],5]+1
 1 2  4  7 10
 3 5  8 11 13
 6 9 12 14 15

And thus we have the result.

Answer 21

Python 3, 259 bytes

So I did this a weird way. I noticed that there were two patterns in the way the array forms.

The first is how the top rows pattern has the difference between each term increasing from 1 -> h where h is the height and l is the length. So I construct the top row based on that pattern

For a matrix of dim(3,4) giving a max RoC = 3 We will see the top row of the form

1, (1+1), (2+2), (4+3) = 1, 2, 4, 7

Suppose instead that the dim(3,9) giving a max RoC = 3 we will instead see a top row of

`1, (1+1), (2+2), (4+3), (7+3), (10+3), (13+3), (16+3), (19+3) = 1, 2, 4, 7, 10, 13, 16, 19, 22

The second pattern is how the rows change from one another. If we consider the matrix:

1   2   4   7   11
3   5   8   12  16
6   9   13  17  20
10  14  18  21  23
15  19  22  24  25

and subtract each row from the row below (ignoring the extra row) we get

2 3 4 5 5
3 4 5 5 4
4 5 5 4 3
5 5 4 3 2

Upon seeing this matrix we can notice this matrix is the sequence 2 3 4 5 5 4 3 2 where by each row is 5 terms of this pattern shifted by 1 for each row. See below for visual.

         |2 3 4 5 5| 4 3 2
       2 |3 4 5 5 4| 3 2
     2 3 |4 5 5 4 3| 2
   2 3 4 |5 5 4 3 2|

So to get the final matrix we take our first row we created and output that row added with the 5 needed terms of this pattern.

This pattern will always have the characteristics of beginning 2-> max value and ending max value -> 2 where the max value = min(h+1, l) and the number of times that the max value will appear is appearances of max = h + l -2*c -2 where c = min(h+1, l) - 2

So in whole my method of creating new rows looks like

1  2  3  7  11 +      |2 3 4 5 5|4 3 2  = 3  5  8  12 16

3  5  8  12 16 +     2|3 4 5 5 4|3 4 2  = 6  9  13 17 20

6  9  13 17 20 +   2 3|4 5 5 4 3|4 2    = 10 14 18 21 23

10 14 18 21 23 + 2 3 4|5 5 4 3 2|       = 15 19 22 24 25

Relevant code below. It didn't end up being short but I still like the method.

o,r=len,range
def m(l,h):
 a,t=[1+sum(([0]+[x for x in r(1,h)]+[h]*(l-h))[:x+1]) for x in r(l)],min(l,h+1);s,c=[x for x in r(2,t)],[a[:]]
 for i in r(h-1):
  for j in r(o(a)):
   a[j]+=(s+[t]*(l+h-2*(t-2)-2)+s[::-1])[0+i:l+i][j]
  c+=[a[:]]
 for l in c:print(l)

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Answer 22

Japt, 20 bytes

õ ïVõ)ñx
£bYgUñ¹ÄÃòV

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Answer 23

Haskell, 92 bytes

r=[1..]
(#)=take
n!k=[[sum$lookup(i,j)$zip[(i,j)|s<-r,i<-n#r,j<-k#r,i+j==s]r|j<-k#r]|i<-n#r]

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