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#MATLAB, (削除) 119 (削除ここまで) 115

MATLAB, (削除) 119 (削除ここまで) 115

In an unusual turn of events, I found that this program actually worked better as I golfed it. First, it became much faster due to vectorization. Now, it displays a helpful prompt ~n:~ reminding the user of which quantity to enter!

Newlines are not part of the program.

x=exp(i*pi/3);
o='~n:~';
P=x.^o;
for l=2:input(o);
P=[kron(P(1:end-1),~~o)+kron(diff(P)/3,[0 1 1+1/x 2]) 1];
end;
plot(P)

n = 9: n=9

o is an arbitrary string which is equal to [0 2 4 0] modulo 6. eiπ/3 raised to these powers gives the vertices of an equilateral triangle in the complex plane. The first kron is used to make a copy of the list of points with each one duplicated 4 times. ~~o is the convenient way to get a vector of 4 ones. Secondly diff(P) finds the vector between each pair of consecutive points. Multiples of this vector (0, 1/3, (1 + e-iπ/3)/3, and 2/3) are added to each of the old points.

#MATLAB, (削除) 119 (削除ここまで) 115

In an unusual turn of events, I found that this program actually worked better as I golfed it. First, it became much faster due to vectorization. Now, it displays a helpful prompt ~n:~ reminding the user of which quantity to enter!

Newlines are not part of the program.

x=exp(i*pi/3);
o='~n:~';
P=x.^o;
for l=2:input(o);
P=[kron(P(1:end-1),~~o)+kron(diff(P)/3,[0 1 1+1/x 2]) 1];
end;
plot(P)

n = 9: n=9

o is an arbitrary string which is equal to [0 2 4 0] modulo 6. eiπ/3 raised to these powers gives the vertices of an equilateral triangle in the complex plane. The first kron is used to make a copy of the list of points with each one duplicated 4 times. ~~o is the convenient way to get a vector of 4 ones. Secondly diff(P) finds the vector between each pair of consecutive points. Multiples of this vector (0, 1/3, (1 + e-iπ/3)/3, and 2/3) are added to each of the old points.

MATLAB, (削除) 119 (削除ここまで) 115

In an unusual turn of events, I found that this program actually worked better as I golfed it. First, it became much faster due to vectorization. Now, it displays a helpful prompt ~n:~ reminding the user of which quantity to enter!

Newlines are not part of the program.

x=exp(i*pi/3);
o='~n:~';
P=x.^o;
for l=2:input(o);
P=[kron(P(1:end-1),~~o)+kron(diff(P)/3,[0 1 1+1/x 2]) 1];
end;
plot(P)

n = 9: n=9

o is an arbitrary string which is equal to [0 2 4 0] modulo 6. eiπ/3 raised to these powers gives the vertices of an equilateral triangle in the complex plane. The first kron is used to make a copy of the list of points with each one duplicated 4 times. ~~o is the convenient way to get a vector of 4 ones. Secondly diff(P) finds the vector between each pair of consecutive points. Multiples of this vector (0, 1/3, (1 + e-iπ/3)/3, and 2/3) are added to each of the old points.

added 527 characters in body
Source Link
feersum
  • 31.8k
  • 10
  • 66
  • 125

#MATLAB, 119(削除) 119 (削除ここまで) 115

In an unusual turn of events, I found that this program actually worked better as I golfed it. First, it became much faster due to vectorization. Now, it displays a helpful prompt ~n:~ reminding the user of which quantity to enter!

Newlines are not part of the program.

x=exp(i*pi/3);
o='~n:~';
P=x.^o;
for l=2:input(o);
P=[kron(P(1:end-1),~~o)+kron(diff(P)/3,[0 1 1+1/x 2]) 1];
end;
plot(P,P/i)

n = 9: n=9

o is an arbitrary string which is equal to [0 2 4 0] modulo 6. eiπ/3 raised to these powers gives the vertices of an equilateral triangle in the complex plane. The first kron is used to make a copy of the list of points with each one duplicated 4 times. ~~o is the convenient way to get a vector of 4 ones. Secondly diff(P) finds the vector between each pair of consecutive points. Multiples of this vector (0, 1/3, (1 + e-iπ/3)/3, and 2/3) are added to each of the old points.

#MATLAB, 119

In an unusual turn of events, I found that this program actually worked better as I golfed it. First, it became much faster due to vectorization. Now, it displays a helpful prompt ~n:~ reminding the user of which quantity to enter!

Newlines are not part of the program.

x=exp(i*pi/3);
o='~n:~';
P=x.^o;
for l=2:input(o);
P=[kron(P(1:end-1),~~o)+kron(diff(P)/3,[0 1 1+1/x 2]) 1];
end;
plot(P,P/i)

n = 9: n=9

#MATLAB, (削除) 119 (削除ここまで) 115

In an unusual turn of events, I found that this program actually worked better as I golfed it. First, it became much faster due to vectorization. Now, it displays a helpful prompt ~n:~ reminding the user of which quantity to enter!

Newlines are not part of the program.

x=exp(i*pi/3);
o='~n:~';
P=x.^o;
for l=2:input(o);
P=[kron(P(1:end-1),~~o)+kron(diff(P)/3,[0 1 1+1/x 2]) 1];
end;
plot(P)

n = 9: n=9

o is an arbitrary string which is equal to [0 2 4 0] modulo 6. eiπ/3 raised to these powers gives the vertices of an equilateral triangle in the complex plane. The first kron is used to make a copy of the list of points with each one duplicated 4 times. ~~o is the convenient way to get a vector of 4 ones. Secondly diff(P) finds the vector between each pair of consecutive points. Multiples of this vector (0, 1/3, (1 + e-iπ/3)/3, and 2/3) are added to each of the old points.

added 237 characters in body
Source Link
feersum
  • 31.8k
  • 10
  • 66
  • 125

#MATLAB, 119

In an unusual turn of events, I found that this program actually worked better as I golfed it. First, it became much faster due to vectorization. Now, it displays a helpful prompt ~n:~ reminding the user of which quantity to enter!

Newlines are not part of the program.

x=exp(i*pi/3);
o='fhjl';o='~n:~';
P=x.^o;
for l=2:input(o);
P=[kron(P(1:end-1),~~o)+kron(diff(P)/3,[0 1 1+1/x 2]) 1];
end;
plot(P,P/i)

n = 9: n=9

#MATLAB, 119

Newlines are not part of the program.

x=exp(i*pi/3);
o='fhjl';
P=x.^o;
for l=2:input(o);
P=[kron(P(1:end-1),~~o)+kron(diff(P)/3,[0 1 1+1/x 2]) 1];
end;
plot(P,P/i)

n = 9: n=9

#MATLAB, 119

In an unusual turn of events, I found that this program actually worked better as I golfed it. First, it became much faster due to vectorization. Now, it displays a helpful prompt ~n:~ reminding the user of which quantity to enter!

Newlines are not part of the program.

x=exp(i*pi/3);
o='~n:~';
P=x.^o;
for l=2:input(o);
P=[kron(P(1:end-1),~~o)+kron(diff(P)/3,[0 1 1+1/x 2]) 1];
end;
plot(P,P/i)

n = 9: n=9

deleted 1 character in body
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feersum
  • 31.8k
  • 10
  • 66
  • 125
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deleted 55 characters in body
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feersum
  • 31.8k
  • 10
  • 66
  • 125
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added 10 characters in body
Source Link
feersum
  • 31.8k
  • 10
  • 66
  • 125
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Source Link
feersum
  • 31.8k
  • 10
  • 66
  • 125
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