added 69 characters in body

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edited Apr 10 at 17:23

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Uiua, 46 44 42 40 39 38 44 … 35 3534 bytes

⊏↧5/◇+\+{⇌.,¤,⊞=.⊙⟜¤⟜=⟜=}⇡+1×2.:$  |-\\|/O

Try it!Try it! I submitted this on the day of the 2024 solar eclipse, which is fitting for a challenge about the sun. (Edit - the eclipse was beautiful! I saw it at around 85%-90% coverage)

Explanation:

ExplanationNote: This explanation is very slightly outdated, but the strategy is the same, just a few things shifted around.

Uiua, 46 44 42 40 39 38 35 bytes

⊏↧5/◇+\+{⇌.⊞=.⊙⟜¤⟜=}⇡+1×2.:$  |-\/O

Try it! I submitted this on the day of the 2024 solar eclipse, which is fitting for a challenge about the sun. (Edit - the eclipse was beautiful! I saw it at around 85%-90% coverage)

Explanation:

Uiua, 46 44 … 35 34 bytes

⊏↧5/◇+\+{⇌,¤,⊞=.⟜=}⇡+1×2.:$  -\|/O

Try it! I submitted this on the day of the 2024 solar eclipse, which is fitting for a challenge about the sun. (Edit - the eclipse was beautiful! I saw it at around 85%-90% coverage)

Explanation:

Note: This explanation is very slightly outdated, but the strategy is the same, just a few things shifted around.

added 531 characters in body

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edited Apr 10 at 0:51

noodle person

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Uiua, 46 44 42 40 39 3838 35 bytes

⊏↧5/◇+×+1⇡4◇+\+{⇌.⊞=.⊙⟜¤⟜=}⇡+1×2.:$  /\|-|O\/O

Try it!Try it! I submitted this on the day of the 2024 solar eclipse, which is fitting for a challenge about the sun. (Edit - the eclipse was beautiful! I saw it at around 85%-90% coverage)

At one point this borrowed some logic from randomra’s very clever J solutionCertainly my most well-golfed Uiua submission to-date. I have had a lot of different ideas and methods to save bytes and frequently switched much of the approach around a lot, but finally I foundhave this at a shorter solution usingstate where I'd be happy if I don't find anything more.

Omnikar's very clever idea in the Uiua discord to use /+\+ instead of multiplying each term by a different method.factor helped me save 3 bytes! 38 → 35

Outdated explanationExplanation:

Consider the input 3. We double this and add 1 to get the side length 7, and take the range from 0 up to it.

[0 1 2 3 4 5 6]

FromUnderneath this we startpush a copy with a mask of where the identity matrixoriginal input appears, and a reversed"fixed" (wrapped in an array) copy: of that.

0[0 0 0 01 0 0 10]    1[[0 0 0 01 0 0 0
00]] 0 0 0 0[0 1 02 3 4 5 06]

Using the range at the top of the stack, push the identity matrix and a reversed copy.

 1 0 0 0 0 0
0 0 0 0 1 0 0    0 0 1 0 0 0 0
0 0 0 1 0 0 0    0 0 0 1 0 0 0
0 0 1[[1 0 0 0 0 0 0]  0 0[[0 0 0 10 0 0 1]
0                                         [0 1 0 0 0 0 00]    0[0 0 0 0 0 1 00]
1 0 0 0 0 0 0    0 0 0 0 0 0 1

Also make the array of this many zeros with a 1 at the position of the original input:

0 0 0 1 0 0 0

Create a square matrix of this, and a transposed copy:

0 0 0                 [0 0 1 0 0 0 0]    [0 0 0 0 1 0 0 00]
0[0 0 0 01 0 0 00]    0[[0 0 0 1 0 0 0
00]] 0 0   [0 0 0 1 0 0 0]   0 [0 0 0 1 0 0 00]
1 1 1 1 1 1 1    0 0 0 1 0 0 0
0 0 0                       [0 0 0 0 1 0 0]   0 0[0 0 1 0 0 0 0]
0 0 0                                       [0 0 0 0 0 1 0]  0 0 0[0 1 0 0 0
0 0 0]
                                         [0 0 0 0 0 0 1]]   0[1 0 0 1 0 0 0 0]]

Then we take these 4 matrices in an array and multiply them by 1, 2, 3, and 4 respectivelyTake their cumulative sums:

0[[0 0 0 1 0 0 0 1]  0 0[[1 0 0 0 0 0
0 01]   [[1 0 10 0 0 0 1]  0 0[[1 0 0 01 0 0 1]
 [0 0 0 0 0 1 00] 0 0  [0 01 0 0 0 01 00] 0
0 0 0 [0 1 0 0 0 1 0] 2 2 2 2[0 21 20 21 0 1 0]
 [0 0 0 0 1 0 00] 0   0[0 0 01 0 01 0 0
00] 0   [0 0 1 0 01 0 0]  0 0 0[0 0 01 01 1 0 0]
0 [0 0 0 1 0 0 00]   0 0[0 0 0 02 0 0

3 00] 0 0 0 0[1 01 1 3 01 01 01] 0 0 0 [1 1 1 4 1 1 1]
0 3[0 0 01 0 0 0 0]  0 0 0[0 0 1 0 41 0
0 00] 3 0  [0 0 1 0 1 0 0]  0 0 0[0 0 41 01 1 0 0]
0 0[0 01 30 0 0 0 0]  0 0 0[0 41 0 0 0
0 01 00] 0 3 0 0[0 1 0 0 0 41 00] 0   [0 1 0 1 0 1 0]
0 0[1 0 0 0 30 0 0]]  0 4[1 0 0 0 0 0
0 1]]   [1 0 0 0 0 0 31]]   4[1 0 0 01 0 0 01]]

Sum them all togetherNotice that adding the fixed version of the list adds to each column, while the non-fixed version adds to each row.

Now, we sum the cumulative sums:

3[[3 0 0  1 0 0 4  4]
0 [0 3 0  1 0 4 0  0]
0 [0 0 3  1 4 0 0  0]
2 [2 2 2 10 2 2 2 2]
0 [0 0 4  1 3 0 0  0]
0 [0 4 0  1 0 3 00]
4 0[4 0 0 1 0 0 33]]

3[[3 0 0 1 0 0 4  4]
0 [0 3 0 1 0 4 0  0]
0 [0 0 3 1 4 0 0  0]
2 [2 2 2 5 2 2 2 2]
0 [0 0 4 1 3 0 0  0]
0 [0 4 0 1 0 3 00]
4 [4 0 0 1 0 0 33]]

Uiua, 46 44 42 40 39 38 bytes

⊏↧5/◇+×+1⇡4{⇌.⊞=.⊙⟜¤⟜=}⇡+1×2.:$  /\-|O

Try it! I submitted this on the day of the 2024 solar eclipse, which is fitting for a challenge about the sun. (Edit - the eclipse was beautiful! I saw it at around 85%-90% coverage)

At one point this borrowed some logic from randomra’s very clever J solution, but I found a shorter solution using a different method.

Outdated explanation:

Consider the input 3. We double this and add 1 to get the side length 7.

From this we start with the identity matrix, and a reversed copy:

0 0 0 0 0 0 1    1 0 0 0 0 0 0
0 0 0 0 0 1 0    0 1 0 0 0 0 0
0 0 0 0 1 0 0    0 0 1 0 0 0 0
0 0 0 1 0 0 0    0 0 0 1 0 0 0
0 0 1 0 0 0 0    0 0 0 0 1 0 0
0 1 0 0 0 0 0    0 0 0 0 0 1 0
1 0 0 0 0 0 0    0 0 0 0 0 0 1

Also make the array of this many zeros with a 1 at the position of the original input:

0 0 0 1 0 0 0

Create a square matrix of this, and a transposed copy:

0 0 0 0 0 0 0    0 0 0 1 0 0 0
0 0 0 0 0 0 0    0 0 0 1 0 0 0
0 0 0 0 0 0 0    0 0 0 1 0 0 0
1 1 1 1 1 1 1    0 0 0 1 0 0 0
0 0 0 0 0 0 0    0 0 0 1 0 0 0
0 0 0 0 0 0 0    0 0 0 1 0 0 0
0 0 0 0 0 0 0    0 0 0 1 0 0 0

Then we take these 4 matrices in an array and multiply them by 1, 2, 3, and 4 respectively:

0 0 0 1 0 0 0   0 0 0 0 0 0 0
0 0 0 1 0 0 0   0 0 0 0 0 0 0
0 0 0 1 0 0 0   0 0 0 0 0 0 0
0 0 0 1 0 0 0   2 2 2 2 2 2 2
0 0 0 1 0 0 0   0 0 0 0 0 0 0
0 0 0 1 0 0 0   0 0 0 0 0 0 0
0 0 0 1 0 0 0   0 0 0 0 0 0 0

3 0 0 0 0 0 0   0 0 0 0 0 0 4
0 3 0 0 0 0 0   0 0 0 0 0 4 0
0 0 3 0 0 0 0   0 0 0 0 4 0 0
0 0 0 3 0 0 0   0 0 0 4 0 0 0
0 0 0 0 3 0 0   0 0 4 0 0 0 0
0 0 0 0 0 3 0   0 4 0 0 0 0 0
0 0 0 0 0 0 3   4 0 0 0 0 0 0

Sum them all together:

3 0 0  1 0 0 4  
0 3 0  1 0 4 0  
0 0 3  1 4 0 0  
2 2 2 10 2 2 2 
0 0 4  1 3 0 0  
0 4 0  1 0 3 0
4 0 0  1 0 0 3

3 0 0 1 0 0 4  
0 3 0 1 0 4 0  
0 0 3 1 4 0 0  
2 2 2 5 2 2 2 
0 0 4 1 3 0 0  
0 4 0 1 0 3 0
4 0 0 1 0 0 3

Uiua, 46 44 42 40 39 38 35 bytes

⊏↧5/◇+\+{⇌.⊞=.⊙⟜¤⟜=}⇡+1×2.:$  |-\/O

Try it! I submitted this on the day of the 2024 solar eclipse, which is fitting for a challenge about the sun. (Edit - the eclipse was beautiful! I saw it at around 85%-90% coverage)

Certainly my most well-golfed Uiua submission to-date. I have had a lot of different ideas and methods to save bytes and frequently switched much of the approach around a lot, but finally I have this at a state where I'd be happy if I don't find anything more.

Omnikar's very clever idea in the Uiua discord to use /+\+ instead of multiplying each term by a different factor helped me save 3 bytes! 38 → 35

Explanation:

Consider the input 3. We double this and add 1 to get the side length 7, and take the range from 0 up to it.

[0 1 2 3 4 5 6]

Underneath this we push a copy with a mask of where the original input appears, and a "fixed" (wrapped in an array) copy of that.

[0 0 0 1 0 0 0]    [[0 0 0 1 0 0 0]]    [0 1 2 3 4 5 6]

Using the range at the top of the stack, push the identity matrix and a reversed copy.

                                        [[1 0 0 0 0 0 0]   [[0 0 0 0 0 0 1]
                                         [0 1 0 0 0 0 0]    [0 0 0 0 0 1 0]
                                         [0 0 1 0 0 0 0]    [0 0 0 0 1 0 0]
[0 0 0 1 0 0 0]    [[0 0 0 1 0 0 0]]     [0 0 0 1 0 0 0]    [0 0 0 1 0 0 0]
                                         [0 0 0 0 1 0 0]    [0 0 1 0 0 0 0]
                                         [0 0 0 0 0 1 0]    [0 1 0 0 0 0 0]
                                         [0 0 0 0 0 0 1]]   [1 0 0 0 0 0 0]]

Take their cumulative sums:

[[0 0 0 0 0 0 1]   [[1 0 0 0 0 0 1]   [[1 0 0 0 0 0 1]   [[1 0 0 1 0 0 1]
 [0 0 0 0 0 1 0]    [0 1 0 0 0 1 0]    [0 1 0 0 0 1 0]    [0 1 0 1 0 1 0]
 [0 0 0 0 1 0 0]    [0 0 1 0 1 0 0]    [0 0 1 0 1 0 0]    [0 0 1 1 1 0 0]
 [0 0 0 1 0 0 0]    [0 0 0 2 0 0 0]    [1 1 1 3 1 1 1]    [1 1 1 4 1 1 1]
 [0 0 1 0 0 0 0]    [0 0 1 0 1 0 0]    [0 0 1 0 1 0 0]    [0 0 1 1 1 0 0]
 [0 1 0 0 0 0 0]    [0 1 0 0 0 1 0]    [0 1 0 0 0 1 0]    [0 1 0 1 0 1 0]
 [1 0 0 0 0 0 0]]   [1 0 0 0 0 0 1]]   [1 0 0 0 0 0 1]]   [1 0 0 1 0 0 1]]

Notice that adding the fixed version of the list adds to each column, while the non-fixed version adds to each row.

Now, we sum the cumulative sums:

[[3 0 0 1 0 0 4]
 [0 3 0 1 0 4 0]
 [0 0 3 1 4 0 0]
 [2 2 2 10 2 2 2]
 [0 0 4 1 3 0 0]
 [0 4 0 1 0 3 0]
 [4 0 0 1 0 0 3]]

[[3 0 0 1 0 0 4]
 [0 3 0 1 0 4 0]
 [0 0 3 1 4 0 0]
 [2 2 2 5 2 2 2]
 [0 0 4 1 3 0 0]
 [0 4 0 1 0 3 0]
 [4 0 0 1 0 0 3]]

deleted 726 characters in body

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edited Apr 9 at 15:48

noodle person

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Uiua, 46 44 42 40 3939 38 bytes

⊏↧5/+×+1⇡4⇌=0[⍉◇+×+1⇡4{⇌.⊞⟜⊃-+⊞=.]-⊙⟜¤⟜=}⇡+1×2.:$  /\-|O

Try it! I’m submittingTry it! I submitted this on the day of the 2024 solar eclipse, which is fitting for a challenge about the sun. (Edit - the eclipse was beautiful! I saw it at around 85%-90% coverage)

~~I also like that all of my improvements have been by 2 bytes each :)~~ This time I only saved a single byte. Still, 1 is better than 0!

As of my revision 42 → 40 bytes,At one point this borrowsborrowed some logic from randomra’s very clever J solution. This still isn’t a direct port as some parts didn’t translate so well to Uiua. (Keep in mind that although this submission is shorter than randomra's, that's not quite a fair comparison since that outputs to STDOUT and this returns a list of lines. J would save 6 bytes by outputting a character matrix.)

but I also havefound a 40 bytesshorter solution which doesn’t do any porting:

⊏↧5/+×+1⇡4[⍉,⇌,]↯⧻.:⊞=.⟜=⇡+1×2.:$  -/|\O

Try it!using a different method.

Explanation of the 40 bytesOutdated explanation:

Uiua, 46 44 42 40 39 bytes

⊏↧5/+×+1⇡4⇌=0[⍉.⊞⟜⊃-+.]-⇡+1×2.:$  /\-|O

Try it! I’m submitting this on the day of the 2024 solar eclipse, which is fitting for a challenge about the sun. (Edit - the eclipse was beautiful! I saw it at around 85%-90% coverage)

~~I also like that all of my improvements have been by 2 bytes each :)~~ This time I only saved a single byte. Still, 1 is better than 0!

As of my revision 42 → 40 bytes, this borrows some logic from randomra’s very clever J solution. This still isn’t a direct port as some parts didn’t translate so well to Uiua. (Keep in mind that although this submission is shorter than randomra's, that's not quite a fair comparison since that outputs to STDOUT and this returns a list of lines. J would save 6 bytes by outputting a character matrix.)

I also have a 40 bytes solution which doesn’t do any porting:

⊏↧5/+×+1⇡4[⍉,⇌,]↯⧻.:⊞=.⟜=⇡+1×2.:$  -/|\O

Try it!

Explanation of the 40 bytes:

Uiua, 46 44 42 40 39 38 bytes

⊏↧5/◇+×+1⇡4{⇌.⊞=.⊙⟜¤⟜=}⇡+1×2.:$  /\-|O

Try it! I submitted this on the day of the 2024 solar eclipse, which is fitting for a challenge about the sun. (Edit - the eclipse was beautiful! I saw it at around 85%-90% coverage)

At one point this borrowed some logic from randomra’s very clever J solution, but I found a shorter solution using a different method.

Outdated explanation:

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noodle person

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Uiua, 46 44 42 40 39 38 44 … 35 3534 bytes

Explanation:

Uiua, 46 44 42 40 39 38 35 bytes

Uiua, 46 44 … 35 34 bytes

Explanation:

Uiua, 46 44 42 40 39 3838 35 bytes

Uiua, 46 44 42 40 39 38 bytes

Uiua, 46 44 42 40 39 38 35 bytes

Uiua, 46 44 42 40 3939 38 bytes

Uiua, 46 44 42 40 39 bytes

Uiua, 46 44 42 40 39 38 bytes