Cycles on the torus
$begingroup$
Challenge
This challenge will have you write a program that takes in two integers n and m and outputs the number non-intersecting loops on the n by m torus made by only taking steps up and to the right. You can think of torus as the grid with wraparound both at the top and the bottom.
This is code-golf so fewest bytes wins.
Example
For example, if the input is n=m=5, one valid walk is
(0,0) -> (0,1) -> (0,2) -> (1,2) -> (2,2) -> (2,3) -> (2,4) ->
(2,0) -> (3,0) -> (4,0) -> (4,1) -> (4,2) -> (4,3) ->
(0,3) -> (1,3) -> (1,4) ->
(1,0) -> (1,1) -> (2,1) -> (3,1) -> (3,2) -> (3,3) -> (3,4) -> (4,4) -> (0,4) -> (0,0)
as shown in the graphic.
Some example input/outputs
f(1,1) = 2 (up or right)
f(1,2) = 2 (up or right-right)
f(2,2) = 4 (up-up, up-right-up-right, right-right, right-up-right-up)
f(2,3) = 7
f(3,3) = 22
f(2,4) = 13
f(3,4) = 66
f(4,4) = 258
code-golf combinatorics grid
asked 3 hours ago
Peter KageyPeter Kagey
878518
$endgroup$
add a comment |
$begingroup$
Challenge
This challenge will have you write a program that takes in two integers n and m and outputs the number non-intersecting loops on the n by m torus made by only taking steps up and to the right. You can think of torus as the grid with wraparound both at the top and the bottom.
This is code-golf so fewest bytes wins.
Example
For example, if the input is n=m=5, one valid walk is
(0,0) -> (0,1) -> (0,2) -> (1,2) -> (2,2) -> (2,3) -> (2,4) ->
(2,0) -> (3,0) -> (4,0) -> (4,1) -> (4,2) -> (4,3) ->
(0,3) -> (1,3) -> (1,4) ->
(1,0) -> (1,1) -> (2,1) -> (3,1) -> (3,2) -> (3,3) -> (3,4) -> (4,4) -> (0,4) -> (0,0)
as shown in the graphic.
Some example input/outputs
f(1,1) = 2 (up or right)
f(1,2) = 2 (up or right-right)
f(2,2) = 4 (up-up, up-right-up-right, right-right, right-up-right-up)
f(2,3) = 7
f(3,3) = 22
f(2,4) = 13
f(3,4) = 66
f(4,4) = 258
code-golf combinatorics grid
asked 3 hours ago
Peter KageyPeter Kagey
878518
$endgroup$
add a comment |
$begingroup$
Challenge
This challenge will have you write a program that takes in two integers n and m and outputs the number non-intersecting loops on the n by m torus made by only taking steps up and to the right. You can think of torus as the grid with wraparound both at the top and the bottom.
This is code-golf so fewest bytes wins.
Example
For example, if the input is n=m=5, one valid walk is
(0,0) -> (0,1) -> (0,2) -> (1,2) -> (2,2) -> (2,3) -> (2,4) ->
(2,0) -> (3,0) -> (4,0) -> (4,1) -> (4,2) -> (4,3) ->
(0,3) -> (1,3) -> (1,4) ->
(1,0) -> (1,1) -> (2,1) -> (3,1) -> (3,2) -> (3,3) -> (3,4) -> (4,4) -> (0,4) -> (0,0)
as shown in the graphic.
Some example input/outputs
f(1,1) = 2 (up or right)
f(1,2) = 2 (up or right-right)
f(2,2) = 4 (up-up, up-right-up-right, right-right, right-up-right-up)
f(2,3) = 7
f(3,3) = 22
f(2,4) = 13
f(3,4) = 66
f(4,4) = 258
code-golf combinatorics grid
asked 3 hours ago
Peter KageyPeter Kagey
878518
$endgroup$
Challenge
This challenge will have you write a program that takes in two integers n and m and outputs the number non-intersecting loops on the n by m torus made by only taking steps up and to the right. You can think of torus as the grid with wraparound both at the top and the bottom.
This is code-golf so fewest bytes wins.
Example
For example, if the input is n=m=5, one valid walk is
(0,0) -> (0,1) -> (0,2) -> (1,2) -> (2,2) -> (2,3) -> (2,4) ->
(2,0) -> (3,0) -> (4,0) -> (4,1) -> (4,2) -> (4,3) ->
(0,3) -> (1,3) -> (1,4) ->
(1,0) -> (1,1) -> (2,1) -> (3,1) -> (3,2) -> (3,3) -> (3,4) -> (4,4) -> (0,4) -> (0,0)
as shown in the graphic.
Some example input/outputs
f(1,1) = 2 (up or right)
f(1,2) = 2 (up or right-right)
f(2,2) = 4 (up-up, up-right-up-right, right-right, right-up-right-up)
f(2,3) = 7
f(3,3) = 22
f(2,4) = 13
f(3,4) = 66
f(4,4) = 258
code-golf combinatorics grid
code-golf combinatorics grid
asked 3 hours ago
Peter KageyPeter Kagey
878518
asked 3 hours ago
Peter KageyPeter Kagey
878518
asked 3 hours ago
Peter KageyPeter Kagey
878518
asked 3 hours ago
Peter KageyPeter Kagey
878518
asked 3 hours ago
Peter KageyPeter Kagey
878518
878518
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Python 2, 87 bytes
f=lambda m,n,z=0,l=:z==0if z in l else sum(f(m,n,(z+d)%m%(n*1j),l+[z])for d in(1,1j))
Try it online!
The interesting thing here is using a complex number z to store the coordinate of the current position. We can move up by adding 1 and move right by adding 1j. To my surprise, modulo works on complex numbers in a way that lets us handle the wrapping for each dimension separately: doing %m acts on the real part, and %(n*1j) acts on the imaginary part.
answered 2 hours ago
xnorxnor
91.8k18187444
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
Python 2, 87 bytes
f=lambda m,n,z=0,l=:z==0if z in l else sum(f(m,n,(z+d)%m%(n*1j),l+[z])for d in(1,1j))
Try it online!
The interesting thing here is using a complex number z to store the coordinate of the current position. We can move up by adding 1 and move right by adding 1j. To my surprise, modulo works on complex numbers in a way that lets us handle the wrapping for each dimension separately: doing %m acts on the real part, and %(n*1j) acts on the imaginary part.
answered 2 hours ago
xnorxnor
91.8k18187444
$endgroup$
add a comment |
$begingroup$
Python 2, 87 bytes
f=lambda m,n,z=0,l=:z==0if z in l else sum(f(m,n,(z+d)%m%(n*1j),l+[z])for d in(1,1j))
Try it online!
The interesting thing here is using a complex number z to store the coordinate of the current position. We can move up by adding 1 and move right by adding 1j. To my surprise, modulo works on complex numbers in a way that lets us handle the wrapping for each dimension separately: doing %m acts on the real part, and %(n*1j) acts on the imaginary part.
answered 2 hours ago
xnorxnor
91.8k18187444
$endgroup$
add a comment |
$begingroup$
Python 2, 87 bytes
f=lambda m,n,z=0,l=:z==0if z in l else sum(f(m,n,(z+d)%m%(n*1j),l+[z])for d in(1,1j))
Try it online!
The interesting thing here is using a complex number z to store the coordinate of the current position. We can move up by adding 1 and move right by adding 1j. To my surprise, modulo works on complex numbers in a way that lets us handle the wrapping for each dimension separately: doing %m acts on the real part, and %(n*1j) acts on the imaginary part.
answered 2 hours ago
xnorxnor
91.8k18187444
$endgroup$
Python 2, 87 bytes
f=lambda m,n,z=0,l=:z==0if z in l else sum(f(m,n,(z+d)%m%(n*1j),l+[z])for d in(1,1j))
Try it online!
The interesting thing here is using a complex number z to store the coordinate of the current position. We can move up by adding 1 and move right by adding 1j. To my surprise, modulo works on complex numbers in a way that lets us handle the wrapping for each dimension separately: doing %m acts on the real part, and %(n*1j) acts on the imaginary part.
answered 2 hours ago
xnorxnor
91.8k18187444
answered 2 hours ago
xnorxnor
91.8k18187444
answered 2 hours ago
xnorxnor
91.8k18187444
answered 2 hours ago
xnorxnor
91.8k18187444
91.8k18187444
add a comment |
add a comment |
If this is an answer to a challenge…
…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.
…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.
More generally…
…Please make sure to answer the question and provide sufficient detail.
…Avoid asking for help, clarification or responding to other answers (use comments instead).
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