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Explore Stack InternalEDIT: As pointed out by @GalenIvanov, these yield the wrong result because they were made from my original tradfn. I'll leave them here for the time being.
{⍵⍴(1⍵⍴1,∊{1∊1,⍨⍵⍴0 2 4 6 8}¨⍳⍵⍨¨j⍴¨(+\¯1+j←⍳⍵)⌽¨⊂0,2×⍳4} Try it online!Try it online!K⍴1⊢⍴1,∘∊{1∘∊1,⍨⍵⍴0 2 4 6 8}¨∘⍳K←⍨¨⍳⍴¨(⊂0,2×⍳4)⌽⍨¨(+\¯1+⍳) Try it online!Try it online!EDIT: As pointed out by @GalenIvanov, these yield the wrong result because they were made from my original tradfn. I'll leave them here for the time being.
{⍵⍴(1,∊{1,⍨⍵⍴0 2 4 6 8}¨⍳⍵)} Try it online!K⍴1,∘∊{1,⍨⍵⍴0 2 4 6 8}¨∘⍳K← Try it online!{⍵⍴1,∊1,⍨¨j⍴¨(+\¯1+j←⍳⍵)⌽¨⊂0,2×⍳4} Try it online!⊢⍴1,∘∊1,⍨¨⍳⍴¨(⊂0,2×⍳4)⌽⍨¨(+\¯1+⍳) Try it online!r←c k r n←1 0r←~n←0 :For j :In ⍳k n+←j-1 r,←1,⍨j⍴n⌽0,2×⍳4 :EndForEnd r←k⍴r Thanks to @GalenIvanov for pointing out an error in the function. Thanks to @Adám for 3 bytes.
r←c k ⍝ The function c takes the argument k and results in r r n←1 0 ⍝ Initializing the variables r and n, and setting them to 1 and 0, respectively. :For j :In ⍳k ⍝ For loop. ⍳k yields [1, 2, 3, ..., k], and j is the control variable. n+←j-1 ⍝ Accumulates j-1 into n, so it follows the progression (0, 1, 3, 6, 10, 15...) r,←1,⍨j⍴n⌽0,2×⍳4 ⍝ This line is explained below. :EndForEnd ⍝ Pretty self explanatory⍝ :pEnds the loop r←k⍴r ⍝ return the first k items of r. ⍝ ⍴ actually reshapes the vector r to the shape of k; ⍝ since k is a scalar, ⍴ reshapes r to a vector with k items. 2×⍳4 ⍝ APL is 1-indexed by default, so this yields the vector 2 4 6 8 0, ⍝ Prepend a 0 to it. We now have 0 2 4 6 8 n⌽ ⍝ Rotate the vector n times to the left. j⍴ ⍝ Reshape it to have j items, which cycles the vector. 1,⍨ ⍝ Append a 1, then r,← ⍝ Append everything to r. r←c k r n←1 0 :For j :In ⍳k n+←j-1 r,←1,⍨j⍴n⌽0,2×⍳4 :EndFor r←k⍴r Thanks to @GalenIvanov for pointing out an error in the function.
r←c k ⍝ The function c takes the argument k and results in r r n←1 0 ⍝ Initializing the variables r and n, and setting them to 1 and 0, respectively. :For j :In ⍳k ⍝ For loop. ⍳k yields [1, 2, 3, ..., k], and j is the control variable. n+←j-1 ⍝ Accumulates j-1 into n, so it follows the progression (0, 1, 3, 6, 10, 15...) r,←1,⍨j⍴n⌽0,2×⍳4 ⍝ This line is explained below. :EndFor ⍝ Pretty self explanatory :p r←k⍴r ⍝ return the first k items of r. ⍝ ⍴ actually reshapes the vector r to the shape of k; ⍝ since k is a scalar, ⍴ reshapes r to a vector with k items. 2×⍳4 ⍝ APL is 1-indexed by default, so this yields the vector 2 4 6 8 0, ⍝ Prepend a 0 to it. We now have 0 2 4 6 8 n⌽ ⍝ Rotate the vector n times to the left. j⍴ ⍝ Reshape it to have j items, which cycles the vector. 1,⍨ ⍝ Append a 1, then r,← ⍝ Append everything to r. r←c k r←~n←0 :For j :In ⍳k n+←j-1 r,←1,⍨j⍴n⌽0,2×⍳4 :End r←k⍴r Thanks to @GalenIvanov for pointing out an error in the function. Thanks to @Adám for 3 bytes.
r←c k ⍝ The function c takes the argument k and results in r r n←1 0 ⍝ Initializing the variables r and n, and setting them to 1 and 0, respectively. :For j :In ⍳k ⍝ For loop. ⍳k yields [1, 2, 3, ..., k], and j is the control variable. n+←j-1 ⍝ Accumulates j-1 into n, so it follows the progression (0, 1, 3, 6, 10, 15...) r,←1,⍨j⍴n⌽0,2×⍳4 ⍝ This line is explained below. :End ⍝ Ends the loop r←k⍴r ⍝ return the first k items of r. ⍝ ⍴ actually reshapes the vector r to the shape of k; ⍝ since k is a scalar, ⍴ reshapes r to a vector with k items. 2×⍳4 ⍝ APL is 1-indexed by default, so this yields the vector 2 4 6 8 0, ⍝ Prepend a 0 to it. We now have 0 2 4 6 8 n⌽ ⍝ Rotate the vector n times to the left. j⍴ ⍝ Reshape it to have j items, which cycles the vector. 1,⍨ ⍝ Append a 1, then r,← ⍝ Append everything to r. r←c k r←1r n←1 0 :For j :In ⍳k n+←j-1 r,←1,⍨j⍴0⍨j⍴n⌽0,2×⍳4 :EndFor r←k⍴r This is a tradfn (traditional function) taking one argument k and returning the first k items of the sequence.
Thanks to @GalenIvanov for pointing out an error in the function.
r←c k ⍝ The function c takes the argument k and results in r r←1r n←1 0 ⍝ Initializing the variablevariables r and n, and setting itthem to 1 and 0, respectively. :For j :In ⍳k ⍝ For loop. ⍳k yields [1, 2, 3, ..., k], and j is the control variable. n+←j-1 ⍝ Accumulates j-1 into n, so it follows the progression (0, 1, 3, 6, 10, 15...) r,←1,⍨j⍴0⍨j⍴n⌽0,2×⍳4 ⍝ This line is explained below. :EndFor ⍝ Pretty self explanatory :p r←k⍴r ⍝ return the first k items of r. ⍝ ⍴ actually reshapes the vector r to the shape of k; ⍝ since k is a scalar, ⍴ reshapes r to a vector with k items. 2×⍳4 ⍝ APL is 1-indexed by default, so this yields the vector 2 4 6 8 0, ⍝ Prepend a 0 to it. We now have 0 2 4 6 8 n⌽ ⍝ Rotate the vector n times to the left. j⍴ ⍝ Reshape it to have j items, which cycles the vector. 1,⍨ ⍝ Append a 1, then r,← ⍝ Append everything to r. Below are a Dfn(direct function) and a tacit function that also solve the challenge, both kindly provided by @Adám.
EDIT: As pointed out by @GalenIvanov, these yield the wrong result because they were made from my original tradfn. I'll leave them here for the time being.
r←c k r←1 :For j :In ⍳k r,←1,⍨j⍴0,2×⍳4 :EndFor r←k⍴r This is a tradfn (traditional function) taking one argument k and returning the first k items of the sequence.
r←c k ⍝ The function c takes the argument k and results in r r←1 ⍝ Initializing the variable r and setting it to 1 :For j :In ⍳k ⍝ For loop. ⍳k yields [1, 2, 3, ..., k], and j is the control variable. r,←1,⍨j⍴0,2×⍳4 ⍝ This line is explained below. :EndFor ⍝ Pretty self explanatory :p r←k⍴r ⍝ return the first k items of r. ⍝ ⍴ actually reshapes the vector r to the shape of k; ⍝ since k is a scalar, ⍴ reshapes r to a vector with k items. 2×⍳4 ⍝ APL is 1-indexed by default, so this yields the vector 2 4 6 8 0, ⍝ Prepend a 0 to it. We now have 0 2 4 6 8 j⍴ ⍝ Reshape it to have j items, which cycles the vector. 1,⍨ ⍝ Append a 1, then r,← ⍝ Append everything to r. Below are a Dfn(direct function) and a tacit function that also solve the challenge, both kindly provided by @Adám.
r←c k r n←1 0 :For j :In ⍳k n+←j-1 r,←1,⍨j⍴n⌽0,2×⍳4 :EndFor r←k⍴r This is a tradfn (traditional function) taking one argument k and returning the first k items of the sequence.
Thanks to @GalenIvanov for pointing out an error in the function.
r←c k ⍝ The function c takes the argument k and results in r r n←1 0 ⍝ Initializing the variables r and n, and setting them to 1 and 0, respectively. :For j :In ⍳k ⍝ For loop. ⍳k yields [1, 2, 3, ..., k], and j is the control variable. n+←j-1 ⍝ Accumulates j-1 into n, so it follows the progression (0, 1, 3, 6, 10, 15...) r,←1,⍨j⍴n⌽0,2×⍳4 ⍝ This line is explained below. :EndFor ⍝ Pretty self explanatory :p r←k⍴r ⍝ return the first k items of r. ⍝ ⍴ actually reshapes the vector r to the shape of k; ⍝ since k is a scalar, ⍴ reshapes r to a vector with k items. 2×⍳4 ⍝ APL is 1-indexed by default, so this yields the vector 2 4 6 8 0, ⍝ Prepend a 0 to it. We now have 0 2 4 6 8 n⌽ ⍝ Rotate the vector n times to the left. j⍴ ⍝ Reshape it to have j items, which cycles the vector. 1,⍨ ⍝ Append a 1, then r,← ⍝ Append everything to r. Below are a Dfn(direct function) and a tacit function that also solve the challenge, both kindly provided by @Adám.
EDIT: As pointed out by @GalenIvanov, these yield the wrong result because they were made from my original tradfn. I'll leave them here for the time being.