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Is this Pascal's Matrix?


Diamondize a MatrixCalculate the Kronecker sum of two matricesMatrix TrigonometryFold up a matrix!Eigenvalues of a MatrixGenerate all square sub-matrices of a given sizeHermitian matrix?Is this a Weyr matrix?Is the matrix centrosymmetric… and so is the code?Matrix Jigsaw Puzzles













7












$begingroup$


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 code-golf, 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]]
[[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]]









share|improve this question









$endgroup$
















    7












    $begingroup$


    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 code-golf, 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]]
    [[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]]









    share|improve this question









    $endgroup$














      7












      7








      7


      1



      $begingroup$


      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 code-golf, 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]]
      [[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]]









      share|improve this question









      $endgroup$




      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 code-golf, 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]]
      [[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]]






      code-golf decision-problem matrix






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      share|improve this question











      share|improve this question




      share|improve this question










      asked 5 hours ago









      LaikoniLaikoni

      20.2k438101




      20.2k438101




















          3 Answers
          3






          active

          oldest

          votes


















          1












          $begingroup$


          Brachylog, 28 bytes



          This feels quite long but here it is anyway



          -4 bytes thanks to DLosc



          ↔↰₁h=₁&s₂ᶠ⟨a₀ᶠ+ᵐ⟩ᵐ


          Explanation



          ↔↰₁h=₁&s₂ᶠ⟨a₀ᶠ+ᵐ⟩ᵐ # 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

          h=₁&s₂ᶠ⟨a₀ᶠ+ᵐ⟩ᵐ # and checking the following
          h=₁ # first row is a rows of 1's
          s₂ᶠ # and for each 2 rows which follow each other
          ⟨a₀ᶠ+ᵐ⟩ᵐ # the 2nd is the partial sums of the 1st
          a₀ᶠ # take all prefixes of the 1st
          +ᵐ # which if summed are the 2nd


          Try it online!






          share|improve this answer











          $endgroup$












          • $begingroup$
            First thought on golfing: you can save 4 bytes by using for "optionally mirror" and calling the same predicate the second time with ↰₁: Try it online!
            $endgroup$
            – DLosc
            3 hours ago


















          0












          $begingroup$


          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 minimum of its first row is greater than 1,



          ≔⮌θθ


          then flip the input array.



          F‹¹⌈Eθ§ι⁰


          If the minimum 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.






          share|improve this answer









          $endgroup$




















            0












            $begingroup$

            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!






            share|improve this answer









            $endgroup$












              Your Answer





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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1












              $begingroup$


              Brachylog, 28 bytes



              This feels quite long but here it is anyway



              -4 bytes thanks to DLosc



              ↔↰₁h=₁&s₂ᶠ⟨a₀ᶠ+ᵐ⟩ᵐ


              Explanation



              ↔↰₁h=₁&s₂ᶠ⟨a₀ᶠ+ᵐ⟩ᵐ # 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

              h=₁&s₂ᶠ⟨a₀ᶠ+ᵐ⟩ᵐ # and checking the following
              h=₁ # first row is a rows of 1's
              s₂ᶠ # and for each 2 rows which follow each other
              ⟨a₀ᶠ+ᵐ⟩ᵐ # the 2nd is the partial sums of the 1st
              a₀ᶠ # take all prefixes of the 1st
              +ᵐ # which if summed are the 2nd


              Try it online!






              share|improve this answer











              $endgroup$












              • $begingroup$
                First thought on golfing: you can save 4 bytes by using for "optionally mirror" and calling the same predicate the second time with ↰₁: Try it online!
                $endgroup$
                – DLosc
                3 hours ago















              1












              $begingroup$


              Brachylog, 28 bytes



              This feels quite long but here it is anyway



              -4 bytes thanks to DLosc



              ↔↰₁h=₁&s₂ᶠ⟨a₀ᶠ+ᵐ⟩ᵐ


              Explanation



              ↔↰₁h=₁&s₂ᶠ⟨a₀ᶠ+ᵐ⟩ᵐ # 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

              h=₁&s₂ᶠ⟨a₀ᶠ+ᵐ⟩ᵐ # and checking the following
              h=₁ # first row is a rows of 1's
              s₂ᶠ # and for each 2 rows which follow each other
              ⟨a₀ᶠ+ᵐ⟩ᵐ # the 2nd is the partial sums of the 1st
              a₀ᶠ # take all prefixes of the 1st
              +ᵐ # which if summed are the 2nd


              Try it online!






              share|improve this answer











              $endgroup$












              • $begingroup$
                First thought on golfing: you can save 4 bytes by using for "optionally mirror" and calling the same predicate the second time with ↰₁: Try it online!
                $endgroup$
                – DLosc
                3 hours ago













              1












              1








              1





              $begingroup$


              Brachylog, 28 bytes



              This feels quite long but here it is anyway



              -4 bytes thanks to DLosc



              ↔↰₁h=₁&s₂ᶠ⟨a₀ᶠ+ᵐ⟩ᵐ


              Explanation



              ↔↰₁h=₁&s₂ᶠ⟨a₀ᶠ+ᵐ⟩ᵐ # 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

              h=₁&s₂ᶠ⟨a₀ᶠ+ᵐ⟩ᵐ # and checking the following
              h=₁ # first row is a rows of 1's
              s₂ᶠ # and for each 2 rows which follow each other
              ⟨a₀ᶠ+ᵐ⟩ᵐ # the 2nd is the partial sums of the 1st
              a₀ᶠ # take all prefixes of the 1st
              +ᵐ # which if summed are the 2nd


              Try it online!






              share|improve this answer











              $endgroup$




              Brachylog, 28 bytes



              This feels quite long but here it is anyway



              -4 bytes thanks to DLosc



              ↔↰₁h=₁&s₂ᶠ⟨a₀ᶠ+ᵐ⟩ᵐ


              Explanation



              ↔↰₁h=₁&s₂ᶠ⟨a₀ᶠ+ᵐ⟩ᵐ # 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

              h=₁&s₂ᶠ⟨a₀ᶠ+ᵐ⟩ᵐ # and checking the following
              h=₁ # first row is a rows of 1's
              s₂ᶠ # and for each 2 rows which follow each other
              ⟨a₀ᶠ+ᵐ⟩ᵐ # the 2nd is the partial sums of the 1st
              a₀ᶠ # take all prefixes of the 1st
              +ᵐ # which if summed are the 2nd


              Try it online!







              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited 35 mins ago

























              answered 4 hours ago









              KroppebKroppeb

              1,326210




              1,326210











              • $begingroup$
                First thought on golfing: you can save 4 bytes by using for "optionally mirror" and calling the same predicate the second time with ↰₁: Try it online!
                $endgroup$
                – DLosc
                3 hours ago
















              • $begingroup$
                First thought on golfing: you can save 4 bytes by using for "optionally mirror" and calling the same predicate the second time with ↰₁: Try it online!
                $endgroup$
                – DLosc
                3 hours ago















              $begingroup$
              First thought on golfing: you can save 4 bytes by using for "optionally mirror" and calling the same predicate the second time with ↰₁: Try it online!
              $endgroup$
              – DLosc
              3 hours ago




              $begingroup$
              First thought on golfing: you can save 4 bytes by using for "optionally mirror" and calling the same predicate the second time with ↰₁: Try it online!
              $endgroup$
              – DLosc
              3 hours ago











              0












              $begingroup$


              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 minimum of its first row is greater than 1,



              ≔⮌θθ


              then flip the input array.



              F‹¹⌈Eθ§ι⁰


              If the minimum 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.






              share|improve this answer









              $endgroup$

















                0












                $begingroup$


                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 minimum of its first row is greater than 1,



                ≔⮌θθ


                then flip the input array.



                F‹¹⌈Eθ§ι⁰


                If the minimum 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.






                share|improve this answer









                $endgroup$















                  0












                  0








                  0





                  $begingroup$


                  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 minimum of its first row is greater than 1,



                  ≔⮌θθ


                  then flip the input array.



                  F‹¹⌈Eθ§ι⁰


                  If the minimum 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.






                  share|improve this answer









                  $endgroup$




                  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 minimum of its first row is greater than 1,



                  ≔⮌θθ


                  then flip the input array.



                  F‹¹⌈Eθ§ι⁰


                  If the minimum 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.







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 2 hours ago









                  NeilNeil

                  81.7k745178




                  81.7k745178





















                      0












                      $begingroup$

                      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!






                      share|improve this answer









                      $endgroup$

















                        0












                        $begingroup$

                        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!






                        share|improve this answer









                        $endgroup$















                          0












                          0








                          0





                          $begingroup$

                          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!






                          share|improve this answer









                          $endgroup$



                          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!







                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered 2 hours ago









                          ArnauldArnauld

                          79k795328




                          79k795328



























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