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In Pascal's triangle each number is the sum of the two numbers directly above it, treating empty spots as zero:

Source: https://en.wikipedia.org/wiki/File:Pascal_triangle_small.png

By rotating the triangle, we can cut out square matrices of varying sizes and rotations which I will call Pascal's matrices. Note that those matrices always need to contain the top \$1\$. Here are some examples:

1  1  1  1
1  2  3  4
1  3  6 10
1  4 10 20

6  3  1
3  2  1
1  1  1

1  5 15 35 70
1  4 10 20 35
1  3  6 10 15
1  2  3  4  5
1  1  1  1  1

1

1  1
2  1

The Task

Given a square matrix containing positive numbers in any reasonable format, decide if it is a Pascal's matrix.

Decide means to either return truthy or falsy values depending on whether the input is a Pascal's matrix, or to fix two constant values and return one for the true inputs and the other for false inputs.

This is , so try to use as few bytes as possible in the language of your choice. The shortest code in each language wins, thus I will not accept an answer.

Test cases

True

[[1, 1, 1, 1], [1, 2, 3, 4], [1, 3, 6, 10], [1, 4, 10, 20]]
[[6, 3, 1], [3, 2, 1], [1, 1, 1]]
[[1, 5, 15, 35, 70], [1, 4, 10, 20, 35], [1, 3, 6, 10, 15], [1, 2, 3, 4, 5], [1, 1, 1, 1, 1]]
[[1]]
[[1, 1], [2, 1]]

False

[[2]]
[[1, 2], [2, 1]]
[[1, 1], [3, 1]]
[[1, 1, 1, 1], [1, 2, 3, 4], [1, 4, 6, 10], [1, 4, 10, 20]]
[[6, 3, 1], [1, 1, 1], [3, 2, 1]]
[[2, 2, 2, 2], [2, 4, 6, 8], [2, 6, 12, 20], [2, 8, 20, 40]]
[[40, 20, 8, 2], [20, 12, 6, 2], [8, 6, 4, 2], [2, 2, 2, 2]] 
[[1, 5, 15, 34, 70], [1, 4, 10, 20, 34], [1, 3, 6, 10, 15], [1, 2, 3, 4, 5], [1, 1, 1, 1, 1]]
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  • \$\begingroup\$ Suggested test case: [[40, 20, 8, 2], [20, 12, 6, 2], [8, 6, 4, 2], [2, 2, 2, 2]]. My initial answer was incorrectly truthy for this one, but correct for all of the current test cases. \$\endgroup\$ – Kevin Cruijssen Mar 19 at 9:40
  • \$\begingroup\$ @KevinCruijssen Thanks, added. \$\endgroup\$ – Laikoni Mar 19 at 12:16

11 Answers 11

5
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Brachylog, 28 24 23 bytes

This feels quite long but here it is anyway

  • -4 bytes thanks to DLosc by compressing the optional flips
  • -1 bytes thanks to DLosc again by doing the partial sums in 1 go

{|↔}\↰₁{k{a₀ᶠ+ᵐ}ᵐ⊆?h=₁}

Explanation

{|↔}\↰₁{k{a₀ᶠ+ᵐ}ᵐ⊆?h=₁}       # Tests if this is a pascal matrix:
{|↔}\↰₁                       #     By trying to get a rows of 1's on top
{|↔}                          #       Through optionally mirroring vertically
     \                        #       Transposing
      ↰₁                      #       Through optionally mirroring vertically

       {k{a₀ᶠ+ᵐ}ᵐ⊆?h=₁}       #     and checking the following
                  ?h=₁        #        first row is a rows of 1's
        k{     }ᵐ             #        and for each row except the last
          a₀ᶠ+ᵐ               #          calculate the partial sum by
          a₀ᶠ                 #             take all prefixes of the input
             +ᵐ               #             and sum each
               ⊆?             #        => as a list is a subsequence of the rotated input

Try it online!

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4
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JavaScript (ES6), 114 bytes

m=>[m,m,m=m.map(r=>[...r].reverse()),m].some(m=>m.reverse(p=[1]).every(r=>p=!r.some((v,x)=>v-~~p[x]-~~r[x-1])&&r))

Try it online!

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4
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MATL, 17 bytes

4:"Gas2YLG@X!X=va

Try it online! Or verify all test cases.

Outputs 1 for Pascal matrices, 0 otherwise.

Explanation

4:      % Push [1 2 3 4]
"       % For each
  G     %   Push input: N×N
  a     %   1×N vector containing 1 for matrix columns that have at least a nonzero
        %   entry, and 0 otherwise. So it gives a vector containing 1 in all entries
  s     %   Sum. Gives N
  2YL   %   Pascal matrix of that size
  G     %   Push input
  @     %   Push current iteration index
  X!    %   Rotate the matrix that many times in steps of 90 degress
  X=    %   Are they equal?
  v     %   Concatenate with previous accumulated result
  a     %   Gives 1 if at least one entry of the vector is nonzero
        % End (implicit). Display (implicit)
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2
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R, 104 bytes

function(m,R=row(m)-1,y=nrow(m):1,Z=choose(R+t(R),R))any(sapply(list(Z,Z[,y],Z[y,y],Z[y,]),identical,m))

Try it online!

Nasty...

Creates a canonical Pascal's matrix Z with dimensions equal to that of m, then tests if the input matrix m is identical to any of the rotations of Z.

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2
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Charcoal, 41 bytes

F‹¹⌈§θ⁰≔⮌θθF‹¹⌈Eθ§ι⁰≦⮌θ⌊⭆θ⭆ι⁼λ∨¬κΣ…§θ⊖κ⊕μ

Try it online! Link is to verbose version of code. Explanation:

F‹¹⌈§θ⁰

If the maximum of its first row is greater than 1,

≔⮌θθ

then flip the input array.

F‹¹⌈Eθ§ι⁰

If the maximum of its first column is greater than 1,

≦⮌θ

then mirror the input array.

⌊⭆θ⭆ι

Loop over the elements of the input array and print the minimum result (i.e. the logical And of all of the results),

⁼λ∨¬κΣ…§θ⊖κ⊕μ

comparing each value to 1 if it is on the first row otherwise the sum of the row above up to and including the cell above.

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1
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Python 2, 129 bytes

f=lambda M,i=4:i and(set(M[0])=={1}and all(a+b==c for x,y in zip(M,M[1:])for a,b,c in zip(x[1:],y,y[1:]))or f(zip(*M[::-1]),i-1))

Try it online!

Returns True if M is a Pascal's Matrix, else 0.

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1
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05AB1E, 29 bytes

¬P≠iR}DøнP≠ií}¬PΘsü)ε`sηOQ}P*

Try it online or verify all test cases.

Explanation:

¬P≠i }        # If the product of the first row of the (implicit) input-matrix is NOT 1:
    R         #  Reverse the order of the rows
D             # Duplicate the resulting matrix
 øнP≠i }      # If the product of the first column is NOT 1:
      í       #  Reverse each row individually
¬PΘ           # Check if the product of the first row is exactly 1
           *  # AND
          P   # And check if everything after the following map is truthy:
sü)ε     }    #  Map over each pair of rows:
    `sη       #   Get the prefixes of the first row
       O      #   Sum each prefix
        Q     #   And check if it's equal to the second row
              # (and output the result implicitly)
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1
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Kotlin, 269 bytes

{m:List<List<Int>>->val n=m.size
var r=0
var c=0
fun f()=if(m[0][0]!=1)m[n-r-1][n-c-1]
else if(m[n-1][0]!=1)m[r][n-c-1]
else if(m[0][n-1]!=1)m[n-r-1][c]
else m[r][c]
var g=0<1
for(l in 0..n*2-2){r=l
c=0
var v=1
do{if(r<n&&c<n)g=f()==v&&g
v=v*(l-c)/++c}while(--r>=0)}
g}

Try it online!

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1
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Julia 0.7, 78 bytes

m->any(i->(n=rotr90(m,i))[1]<2&&all(cumsum(n)'[1:end-1,:]-n[2:end,:].==0),0:3)

Try it online!

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1
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Java (JDK), 234 bytes

m->{int l=m.length,L=l-1,p=1,s=0,S=0,e=l,E=l,d=1,D=1,i,j;if(m[0][0]>1|m[0][L]>1){s=L;e=d=-1;}if(m[0][0]>1|m[L][0]>1){S=L;E=D=-1;}for(i=s;i!=e;i+=d)for(j=S;j!=E;j+=D)p=(i==s|j==S?m[i][j]<2:m[i][j]==m[i-d][j]+m[i][j-D])?p:0;return p>0;}

Try it online!

Credits

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  • 1
    \$\begingroup\$ Nice answer, but dang, loads of variables. ;) Oh, and -1: i==s||j==S to i==s|j==S. \$\endgroup\$ – Kevin Cruijssen Mar 20 at 14:26
  • \$\begingroup\$ @KevinCruijssen if you know a better algorithm I take it! But the rotation is the cause for all the variables. Some languages can handle them in 1-2 bytes, in Java, you have to think the code around them. The core algorithm is actually pretty short: m->{int l=m.length,i=0,j;for(;i<l;i++)for(j=0;j<l;j++)p=(i<1|j<1?m[i][j]<2:m[i][j]==m[i-1][j]+m[i][j-1])?p:0;return p>0;} (122 bytes) \$\endgroup\$ – Olivier Grégoire Mar 20 at 15:41
0
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Jelly, 22 bytes

Ż€Iṫ2⁼ṖaFḢ=1Ʋ
,Ṛ;U$Ç€Ẹ

Try it online!

Explanation

Helper link, checks whether this rotation of matrix valid

Ż€            | prepend each row with zero
  I           | find differences within rows
   ṫ2         | drop the first row
     ⁼Ṗ       | compare to the original matrix
              |   with the last row removed
       a      | logical and
        FḢ=1Ʋ | top left cell is 1

Main link

,Ṛ            | copy the matrix and reverse the rows
  ;U$         | append a copy of both of these
              |   with the columns reversed
     ǀ       | run each version of the matrix
              |   through the helper link
       Ẹ      | check if any are valid
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