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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
$begingroup$
In Pascal's triangle each number is the sum of the two numbers directly above it, treating empty spots as zero:
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
$endgroup$
add a comment |
$begingroup$
In Pascal's triangle each number is the sum of the two numbers directly above it, treating empty spots as zero:
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
$endgroup$
add a comment |
$begingroup$
In Pascal's triangle each number is the sum of the two numbers directly above it, treating empty spots as zero:
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
$endgroup$
In Pascal's triangle each number is the sum of the two numbers directly above it, treating empty spots as zero:
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
code-golf decision-problem matrix
asked 4 hours ago
LaikoniLaikoni
20.2k438101
20.2k438101
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Brachylog, 28 bytes
This feels quite long but here it is anyway
⟨≡∋↔⟩⟨≡∋↔⟩{h=₁&s₂ᶠ⟨a₀ᶠ+ᵐ⟩ᵐ}
Explanation
⟨≡∋↔⟩⟨≡∋↔⟩{h=₁&s₂ᶠ⟨a₀ᶠ+ᵐ⟩ᵐ} # Tests if this is a pascal matrix:
⟨≡∋↔⟩⟨≡∋↔⟩ # By trying to get a rows of 1's on top
⟨≡∋↔⟩ # Through optionally mirroring vertically
# Transposing
⟨≡∋↔⟩ # 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!
$endgroup$
$begingroup$
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
2 hours ago
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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!
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Brachylog, 28 bytes
This feels quite long but here it is anyway
⟨≡∋↔⟩⟨≡∋↔⟩{h=₁&s₂ᶠ⟨a₀ᶠ+ᵐ⟩ᵐ}
Explanation
⟨≡∋↔⟩⟨≡∋↔⟩{h=₁&s₂ᶠ⟨a₀ᶠ+ᵐ⟩ᵐ} # Tests if this is a pascal matrix:
⟨≡∋↔⟩⟨≡∋↔⟩ # By trying to get a rows of 1's on top
⟨≡∋↔⟩ # Through optionally mirroring vertically
# Transposing
⟨≡∋↔⟩ # 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!
$endgroup$
$begingroup$
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
2 hours ago
add a comment |
$begingroup$
Brachylog, 28 bytes
This feels quite long but here it is anyway
⟨≡∋↔⟩⟨≡∋↔⟩{h=₁&s₂ᶠ⟨a₀ᶠ+ᵐ⟩ᵐ}
Explanation
⟨≡∋↔⟩⟨≡∋↔⟩{h=₁&s₂ᶠ⟨a₀ᶠ+ᵐ⟩ᵐ} # Tests if this is a pascal matrix:
⟨≡∋↔⟩⟨≡∋↔⟩ # By trying to get a rows of 1's on top
⟨≡∋↔⟩ # Through optionally mirroring vertically
# Transposing
⟨≡∋↔⟩ # 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!
$endgroup$
$begingroup$
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
2 hours ago
add a comment |
$begingroup$
Brachylog, 28 bytes
This feels quite long but here it is anyway
⟨≡∋↔⟩⟨≡∋↔⟩{h=₁&s₂ᶠ⟨a₀ᶠ+ᵐ⟩ᵐ}
Explanation
⟨≡∋↔⟩⟨≡∋↔⟩{h=₁&s₂ᶠ⟨a₀ᶠ+ᵐ⟩ᵐ} # Tests if this is a pascal matrix:
⟨≡∋↔⟩⟨≡∋↔⟩ # By trying to get a rows of 1's on top
⟨≡∋↔⟩ # Through optionally mirroring vertically
# Transposing
⟨≡∋↔⟩ # 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!
$endgroup$
Brachylog, 28 bytes
This feels quite long but here it is anyway
⟨≡∋↔⟩⟨≡∋↔⟩{h=₁&s₂ᶠ⟨a₀ᶠ+ᵐ⟩ᵐ}
Explanation
⟨≡∋↔⟩⟨≡∋↔⟩{h=₁&s₂ᶠ⟨a₀ᶠ+ᵐ⟩ᵐ} # Tests if this is a pascal matrix:
⟨≡∋↔⟩⟨≡∋↔⟩ # By trying to get a rows of 1's on top
⟨≡∋↔⟩ # Through optionally mirroring vertically
# Transposing
⟨≡∋↔⟩ # 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!
edited 3 hours ago
answered 3 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
2 hours ago
add a comment |
$begingroup$
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
2 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
2 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
2 hours ago
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
answered 1 hour ago
NeilNeil
81.7k745178
81.7k745178
add a comment |
add a comment |
$begingroup$
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!
$endgroup$
add a comment |
$begingroup$
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!
$endgroup$
add a comment |
$begingroup$
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!
$endgroup$
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!
answered 1 hour ago
ArnauldArnauld
79k795328
79k795328
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