Uniqueness of spanning tree on a grid.
$begingroup$
When I was at the Graduate Student Combinatorics Conference earlier this month, someone introduced me to a puzzle game called Noodles!.
The game starts with a collection of "pipes" on a grid (centered on each vertex), clicking on a piece rotates it $90^circ$, and a piece can be rotated any number of times. The goal is to turn the final configuration of pipes into a spanning tree (of the grid graph), as shown in the screenshots below.
Example
Question
We left the conference with an unsolved question:
Are solutions to this puzzle always unique? Or is it possible to come up with a starting configuration (on any size grid) that has multiple trees as solutions?
(The prevailing guess is that solutions are unique, but nobody could manage to prove it.)
combinatorics graph-theory puzzle trees
$endgroup$
add a comment |
$begingroup$
When I was at the Graduate Student Combinatorics Conference earlier this month, someone introduced me to a puzzle game called Noodles!.
The game starts with a collection of "pipes" on a grid (centered on each vertex), clicking on a piece rotates it $90^circ$, and a piece can be rotated any number of times. The goal is to turn the final configuration of pipes into a spanning tree (of the grid graph), as shown in the screenshots below.
Example
Question
We left the conference with an unsolved question:
Are solutions to this puzzle always unique? Or is it possible to come up with a starting configuration (on any size grid) that has multiple trees as solutions?
(The prevailing guess is that solutions are unique, but nobody could manage to prove it.)
combinatorics graph-theory puzzle trees
$endgroup$
1
$begingroup$
This genre of logic puzzle is known originally as Netwalk.
$endgroup$
– Mike Earnest
4 hours ago
$begingroup$
@MikeEarnest, thanks for the reference. Based on this picture from this Reddit thread, it looks like (some versions of) Netwalk are on a torus instead of a grid—although that presents an interesting generalization of this question.
$endgroup$
– Peter Kagey
4 hours ago
add a comment |
$begingroup$
When I was at the Graduate Student Combinatorics Conference earlier this month, someone introduced me to a puzzle game called Noodles!.
The game starts with a collection of "pipes" on a grid (centered on each vertex), clicking on a piece rotates it $90^circ$, and a piece can be rotated any number of times. The goal is to turn the final configuration of pipes into a spanning tree (of the grid graph), as shown in the screenshots below.
Example
Question
We left the conference with an unsolved question:
Are solutions to this puzzle always unique? Or is it possible to come up with a starting configuration (on any size grid) that has multiple trees as solutions?
(The prevailing guess is that solutions are unique, but nobody could manage to prove it.)
combinatorics graph-theory puzzle trees
$endgroup$
When I was at the Graduate Student Combinatorics Conference earlier this month, someone introduced me to a puzzle game called Noodles!.
The game starts with a collection of "pipes" on a grid (centered on each vertex), clicking on a piece rotates it $90^circ$, and a piece can be rotated any number of times. The goal is to turn the final configuration of pipes into a spanning tree (of the grid graph), as shown in the screenshots below.
Example
Question
We left the conference with an unsolved question:
Are solutions to this puzzle always unique? Or is it possible to come up with a starting configuration (on any size grid) that has multiple trees as solutions?
(The prevailing guess is that solutions are unique, but nobody could manage to prove it.)
combinatorics graph-theory puzzle trees
combinatorics graph-theory puzzle trees
asked 4 hours ago
Peter KageyPeter Kagey
1,61272053
1,61272053
1
$begingroup$
This genre of logic puzzle is known originally as Netwalk.
$endgroup$
– Mike Earnest
4 hours ago
$begingroup$
@MikeEarnest, thanks for the reference. Based on this picture from this Reddit thread, it looks like (some versions of) Netwalk are on a torus instead of a grid—although that presents an interesting generalization of this question.
$endgroup$
– Peter Kagey
4 hours ago
add a comment |
1
$begingroup$
This genre of logic puzzle is known originally as Netwalk.
$endgroup$
– Mike Earnest
4 hours ago
$begingroup$
@MikeEarnest, thanks for the reference. Based on this picture from this Reddit thread, it looks like (some versions of) Netwalk are on a torus instead of a grid—although that presents an interesting generalization of this question.
$endgroup$
– Peter Kagey
4 hours ago
1
1
$begingroup$
This genre of logic puzzle is known originally as Netwalk.
$endgroup$
– Mike Earnest
4 hours ago
$begingroup$
This genre of logic puzzle is known originally as Netwalk.
$endgroup$
– Mike Earnest
4 hours ago
$begingroup$
@MikeEarnest, thanks for the reference. Based on this picture from this Reddit thread, it looks like (some versions of) Netwalk are on a torus instead of a grid—although that presents an interesting generalization of this question.
$endgroup$
– Peter Kagey
4 hours ago
$begingroup$
@MikeEarnest, thanks for the reference. Based on this picture from this Reddit thread, it looks like (some versions of) Netwalk are on a torus instead of a grid—although that presents an interesting generalization of this question.
$endgroup$
– Peter Kagey
4 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No, solutions are not unique. The four "T" shaped pieces in the grid below can be rotated into either of two configurations:
┏━━━┓
┗┓╻╻┃
╺┫┣┛┃
┏┫┣╸┃
╹╹┗━┛
┏━━━┓
┗┓╻╻┃
╺┻┻┛┃
┏┳┳╸┃
╹╹┗━┛
New contributor
$endgroup$
$begingroup$
Based on the picture in the question, all three ends of the T shapes have to connect to an adjacent piece; you can't have one end of the T run directly into the flat side of a | piece.
$endgroup$
– Misha Lavrov
3 hours ago
$begingroup$
It doesn't "run directly into the flat side of a | piece". There's actually a dead end "╸" piece in between, so this satisfies all the rules.
$endgroup$
– edderiofer
2 hours ago
$begingroup$
Oh, I see. I couldn't quite read it in your answer, but it showed up when I highlighted it and now everything makes sense.
$endgroup$
– Misha Lavrov
2 hours ago
1
$begingroup$
Here is an interactive illustration of the solution: chiark.greenend.org.uk/~sgtatham/puzzles/js/…
$endgroup$
– Mike Earnest
2 hours ago
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
var $window = $(window),
onScroll = function(e) {
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom)) {
StackExchange.using('gps', function() { StackExchange.gps.track('embedded_signup_form.view', { location: 'question_page' }); });
$window.unbind('scroll', onScroll);
}
};
$window.on('scroll', onScroll);
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3191645%2funiqueness-of-spanning-tree-on-a-grid%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No, solutions are not unique. The four "T" shaped pieces in the grid below can be rotated into either of two configurations:
┏━━━┓
┗┓╻╻┃
╺┫┣┛┃
┏┫┣╸┃
╹╹┗━┛
┏━━━┓
┗┓╻╻┃
╺┻┻┛┃
┏┳┳╸┃
╹╹┗━┛
New contributor
$endgroup$
$begingroup$
Based on the picture in the question, all three ends of the T shapes have to connect to an adjacent piece; you can't have one end of the T run directly into the flat side of a | piece.
$endgroup$
– Misha Lavrov
3 hours ago
$begingroup$
It doesn't "run directly into the flat side of a | piece". There's actually a dead end "╸" piece in between, so this satisfies all the rules.
$endgroup$
– edderiofer
2 hours ago
$begingroup$
Oh, I see. I couldn't quite read it in your answer, but it showed up when I highlighted it and now everything makes sense.
$endgroup$
– Misha Lavrov
2 hours ago
1
$begingroup$
Here is an interactive illustration of the solution: chiark.greenend.org.uk/~sgtatham/puzzles/js/…
$endgroup$
– Mike Earnest
2 hours ago
add a comment |
$begingroup$
No, solutions are not unique. The four "T" shaped pieces in the grid below can be rotated into either of two configurations:
┏━━━┓
┗┓╻╻┃
╺┫┣┛┃
┏┫┣╸┃
╹╹┗━┛
┏━━━┓
┗┓╻╻┃
╺┻┻┛┃
┏┳┳╸┃
╹╹┗━┛
New contributor
$endgroup$
$begingroup$
Based on the picture in the question, all three ends of the T shapes have to connect to an adjacent piece; you can't have one end of the T run directly into the flat side of a | piece.
$endgroup$
– Misha Lavrov
3 hours ago
$begingroup$
It doesn't "run directly into the flat side of a | piece". There's actually a dead end "╸" piece in between, so this satisfies all the rules.
$endgroup$
– edderiofer
2 hours ago
$begingroup$
Oh, I see. I couldn't quite read it in your answer, but it showed up when I highlighted it and now everything makes sense.
$endgroup$
– Misha Lavrov
2 hours ago
1
$begingroup$
Here is an interactive illustration of the solution: chiark.greenend.org.uk/~sgtatham/puzzles/js/…
$endgroup$
– Mike Earnest
2 hours ago
add a comment |
$begingroup$
No, solutions are not unique. The four "T" shaped pieces in the grid below can be rotated into either of two configurations:
┏━━━┓
┗┓╻╻┃
╺┫┣┛┃
┏┫┣╸┃
╹╹┗━┛
┏━━━┓
┗┓╻╻┃
╺┻┻┛┃
┏┳┳╸┃
╹╹┗━┛
New contributor
$endgroup$
No, solutions are not unique. The four "T" shaped pieces in the grid below can be rotated into either of two configurations:
┏━━━┓
┗┓╻╻┃
╺┫┣┛┃
┏┫┣╸┃
╹╹┗━┛
┏━━━┓
┗┓╻╻┃
╺┻┻┛┃
┏┳┳╸┃
╹╹┗━┛
New contributor
New contributor
answered 3 hours ago
edderioferedderiofer
1561
1561
New contributor
New contributor
$begingroup$
Based on the picture in the question, all three ends of the T shapes have to connect to an adjacent piece; you can't have one end of the T run directly into the flat side of a | piece.
$endgroup$
– Misha Lavrov
3 hours ago
$begingroup$
It doesn't "run directly into the flat side of a | piece". There's actually a dead end "╸" piece in between, so this satisfies all the rules.
$endgroup$
– edderiofer
2 hours ago
$begingroup$
Oh, I see. I couldn't quite read it in your answer, but it showed up when I highlighted it and now everything makes sense.
$endgroup$
– Misha Lavrov
2 hours ago
1
$begingroup$
Here is an interactive illustration of the solution: chiark.greenend.org.uk/~sgtatham/puzzles/js/…
$endgroup$
– Mike Earnest
2 hours ago
add a comment |
$begingroup$
Based on the picture in the question, all three ends of the T shapes have to connect to an adjacent piece; you can't have one end of the T run directly into the flat side of a | piece.
$endgroup$
– Misha Lavrov
3 hours ago
$begingroup$
It doesn't "run directly into the flat side of a | piece". There's actually a dead end "╸" piece in between, so this satisfies all the rules.
$endgroup$
– edderiofer
2 hours ago
$begingroup$
Oh, I see. I couldn't quite read it in your answer, but it showed up when I highlighted it and now everything makes sense.
$endgroup$
– Misha Lavrov
2 hours ago
1
$begingroup$
Here is an interactive illustration of the solution: chiark.greenend.org.uk/~sgtatham/puzzles/js/…
$endgroup$
– Mike Earnest
2 hours ago
$begingroup$
Based on the picture in the question, all three ends of the T shapes have to connect to an adjacent piece; you can't have one end of the T run directly into the flat side of a | piece.
$endgroup$
– Misha Lavrov
3 hours ago
$begingroup$
Based on the picture in the question, all three ends of the T shapes have to connect to an adjacent piece; you can't have one end of the T run directly into the flat side of a | piece.
$endgroup$
– Misha Lavrov
3 hours ago
$begingroup$
It doesn't "run directly into the flat side of a | piece". There's actually a dead end "╸" piece in between, so this satisfies all the rules.
$endgroup$
– edderiofer
2 hours ago
$begingroup$
It doesn't "run directly into the flat side of a | piece". There's actually a dead end "╸" piece in between, so this satisfies all the rules.
$endgroup$
– edderiofer
2 hours ago
$begingroup$
Oh, I see. I couldn't quite read it in your answer, but it showed up when I highlighted it and now everything makes sense.
$endgroup$
– Misha Lavrov
2 hours ago
$begingroup$
Oh, I see. I couldn't quite read it in your answer, but it showed up when I highlighted it and now everything makes sense.
$endgroup$
– Misha Lavrov
2 hours ago
1
1
$begingroup$
Here is an interactive illustration of the solution: chiark.greenend.org.uk/~sgtatham/puzzles/js/…
$endgroup$
– Mike Earnest
2 hours ago
$begingroup$
Here is an interactive illustration of the solution: chiark.greenend.org.uk/~sgtatham/puzzles/js/…
$endgroup$
– Mike Earnest
2 hours ago
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
var $window = $(window),
onScroll = function(e) {
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom)) {
StackExchange.using('gps', function() { StackExchange.gps.track('embedded_signup_form.view', { location: 'question_page' }); });
$window.unbind('scroll', onScroll);
}
};
$window.on('scroll', onScroll);
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3191645%2funiqueness-of-spanning-tree-on-a-grid%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
var $window = $(window),
onScroll = function(e) {
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom)) {
StackExchange.using('gps', function() { StackExchange.gps.track('embedded_signup_form.view', { location: 'question_page' }); });
$window.unbind('scroll', onScroll);
}
};
$window.on('scroll', onScroll);
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
var $window = $(window),
onScroll = function(e) {
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom)) {
StackExchange.using('gps', function() { StackExchange.gps.track('embedded_signup_form.view', { location: 'question_page' }); });
$window.unbind('scroll', onScroll);
}
};
$window.on('scroll', onScroll);
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
var $window = $(window),
onScroll = function(e) {
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom)) {
StackExchange.using('gps', function() { StackExchange.gps.track('embedded_signup_form.view', { location: 'question_page' }); });
$window.unbind('scroll', onScroll);
}
};
$window.on('scroll', onScroll);
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
This genre of logic puzzle is known originally as Netwalk.
$endgroup$
– Mike Earnest
4 hours ago
$begingroup$
@MikeEarnest, thanks for the reference. Based on this picture from this Reddit thread, it looks like (some versions of) Netwalk are on a torus instead of a grid—although that presents an interesting generalization of this question.
$endgroup$
– Peter Kagey
4 hours ago