Sum in a sudoku with sum total of 15 and 45
$begingroup$
I have created a sudoku puzzle with the following restrictions:
- Each row and column sum to $45$.
- Each row and column in the nine $3$ by $3$ sub-grids sum to $15$.
Is such a sudoku unique?
grid-deduction sudoku
edited 1 hour ago
Hugh
1,6961721
asked 2 hours ago
Aorica IresonAorica Ireson
111
New contributor
Aorica Ireson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I have created a sudoku puzzle with the following restrictions:
- Each row and column sum to $45$.
- Each row and column in the nine $3$ by $3$ sub-grids sum to $15$.
Is such a sudoku unique?
grid-deduction sudoku
edited 1 hour ago
Hugh
1,6961721
asked 2 hours ago
Aorica IresonAorica Ireson
111
New contributor
Aorica Ireson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
"...where each row and column adds to a total of 45..." is redundant, since it is in the nature of a sudoku puzzle that its rows and columns sum to $45$. This is because each row contains the numbers $1$ through $9$, and $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$.
$endgroup$
– Hugh
2 hours ago
$begingroup$
"where each (3*3) 9 individual squares of rows and columns total a sum of 15" >>> do you mean 45?
$endgroup$
– Kryesec
1 hour ago
$begingroup$
@Kryesec I've assumed that that means that each of the 3 by 3 sub-grids is a magic square with "magic constant" 15. In other words, each row/column in each 3 by 3 sub-grid sums to 15.
$endgroup$
– Hugh
1 hour ago
add a comment |
$begingroup$
I have created a sudoku puzzle with the following restrictions:
- Each row and column sum to $45$.
- Each row and column in the nine $3$ by $3$ sub-grids sum to $15$.
Is such a sudoku unique?
grid-deduction sudoku
edited 1 hour ago
Hugh
1,6961721
asked 2 hours ago
Aorica IresonAorica Ireson
111
New contributor
Aorica Ireson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I have created a sudoku puzzle with the following restrictions:
- Each row and column sum to $45$.
- Each row and column in the nine $3$ by $3$ sub-grids sum to $15$.
Is such a sudoku unique?
grid-deduction sudoku
grid-deduction sudoku
edited 1 hour ago
Hugh
1,6961721
asked 2 hours ago
Aorica IresonAorica Ireson
111
New contributor
Aorica Ireson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
Hugh
1,6961721
asked 2 hours ago
Aorica IresonAorica Ireson
111
New contributor
Aorica Ireson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
Hugh
1,6961721
edited 1 hour ago
Hugh
1,6961721
edited 1 hour ago
Hugh
1,6961721
1,6961721
asked 2 hours ago
Aorica IresonAorica Ireson
111
New contributor
Aorica Ireson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Aorica IresonAorica Ireson
111
asked 2 hours ago
Aorica IresonAorica Ireson
111
111
New contributor
Aorica Ireson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Aorica Ireson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Aorica Ireson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
"...where each row and column adds to a total of 45..." is redundant, since it is in the nature of a sudoku puzzle that its rows and columns sum to $45$. This is because each row contains the numbers $1$ through $9$, and $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$.
$endgroup$
– Hugh
2 hours ago
$begingroup$
"where each (3*3) 9 individual squares of rows and columns total a sum of 15" >>> do you mean 45?
$endgroup$
– Kryesec
1 hour ago
$begingroup$
@Kryesec I've assumed that that means that each of the 3 by 3 sub-grids is a magic square with "magic constant" 15. In other words, each row/column in each 3 by 3 sub-grid sums to 15.
$endgroup$
– Hugh
1 hour ago
add a comment |
$begingroup$
"...where each row and column adds to a total of 45..." is redundant, since it is in the nature of a sudoku puzzle that its rows and columns sum to $45$. This is because each row contains the numbers $1$ through $9$, and $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$.
$endgroup$
– Hugh
2 hours ago
$begingroup$
"where each (3*3) 9 individual squares of rows and columns total a sum of 15" >>> do you mean 45?
$endgroup$
– Kryesec
1 hour ago
$begingroup$
@Kryesec I've assumed that that means that each of the 3 by 3 sub-grids is a magic square with "magic constant" 15. In other words, each row/column in each 3 by 3 sub-grid sums to 15.
$endgroup$
– Hugh
1 hour ago
$begingroup$
"...where each row and column adds to a total of 45..." is redundant, since it is in the nature of a sudoku puzzle that its rows and columns sum to $45$. This is because each row contains the numbers $1$ through $9$, and $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$.
$endgroup$
– Hugh
2 hours ago
$begingroup$
"...where each row and column adds to a total of 45..." is redundant, since it is in the nature of a sudoku puzzle that its rows and columns sum to $45$. This is because each row contains the numbers $1$ through $9$, and $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$.
$endgroup$
– Hugh
2 hours ago
$begingroup$
"where each (3*3) 9 individual squares of rows and columns total a sum of 15" >>> do you mean 45?
$endgroup$
– Kryesec
1 hour ago
$begingroup$
"where each (3*3) 9 individual squares of rows and columns total a sum of 15" >>> do you mean 45?
$endgroup$
– Kryesec
1 hour ago
$begingroup$
@Kryesec I've assumed that that means that each of the 3 by 3 sub-grids is a magic square with "magic constant" 15. In other words, each row/column in each 3 by 3 sub-grid sums to 15.
$endgroup$
– Hugh
1 hour ago
$begingroup$
@Kryesec I've assumed that that means that each of the 3 by 3 sub-grids is a magic square with "magic constant" 15. In other words, each row/column in each 3 by 3 sub-grid sums to 15.
$endgroup$
– Hugh
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No.
Of course, there's rotation and reflection, but even beyond that you can swap any two rows within it's group of three rows (eg swap row 1 and 2 or swap 4 and 6), and you can swap a group of three with one of the other groups of three (eg swap 1,2,3 with 4,5,6), and the same applies to the columns.
answered 1 hour ago
Dr XorileDr Xorile
12.1k22567
$endgroup$
add a comment |
$begingroup$
I was beaten to a very similar answer, but you can create such a square based on the magic square below:
abc
def
ghi
Turn it into this:
abcdefghi
defghiabc
ghiabcdef
bcaefdhig
efdhigbca
higbcaefd
cabfdeigh
fdeighcab
ighcabfde
Since there are multiple possible magic squares, it's not unique. When you change the number at the very center (i), which you can, you get a different sudoku (not even a rotation or reflection).
answered 24 mins ago
NautilusNautilus
3,751523
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "559"
};
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: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
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
});
}
});
Aorica Ireson is a new contributor. Be nice, and check out our Code of Conduct.
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%2fpuzzling.stackexchange.com%2fquestions%2f79552%2fsum-in-a-sudoku-with-sum-total-of-15-and-45%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No.
Of course, there's rotation and reflection, but even beyond that you can swap any two rows within it's group of three rows (eg swap row 1 and 2 or swap 4 and 6), and you can swap a group of three with one of the other groups of three (eg swap 1,2,3 with 4,5,6), and the same applies to the columns.
answered 1 hour ago
Dr XorileDr Xorile
12.1k22567
$endgroup$
add a comment |
$begingroup$
No.
Of course, there's rotation and reflection, but even beyond that you can swap any two rows within it's group of three rows (eg swap row 1 and 2 or swap 4 and 6), and you can swap a group of three with one of the other groups of three (eg swap 1,2,3 with 4,5,6), and the same applies to the columns.
answered 1 hour ago
Dr XorileDr Xorile
12.1k22567
$endgroup$
add a comment |
$begingroup$
No.
Of course, there's rotation and reflection, but even beyond that you can swap any two rows within it's group of three rows (eg swap row 1 and 2 or swap 4 and 6), and you can swap a group of three with one of the other groups of three (eg swap 1,2,3 with 4,5,6), and the same applies to the columns.
answered 1 hour ago
Dr XorileDr Xorile
12.1k22567
$endgroup$
No.
Of course, there's rotation and reflection, but even beyond that you can swap any two rows within it's group of three rows (eg swap row 1 and 2 or swap 4 and 6), and you can swap a group of three with one of the other groups of three (eg swap 1,2,3 with 4,5,6), and the same applies to the columns.
answered 1 hour ago
Dr XorileDr Xorile
12.1k22567
answered 1 hour ago
Dr XorileDr Xorile
12.1k22567
answered 1 hour ago
Dr XorileDr Xorile
12.1k22567
answered 1 hour ago
Dr XorileDr Xorile
12.1k22567
12.1k22567
add a comment |
add a comment |
$begingroup$
I was beaten to a very similar answer, but you can create such a square based on the magic square below:
abc
def
ghi
Turn it into this:
abcdefghi
defghiabc
ghiabcdef
bcaefdhig
efdhigbca
higbcaefd
cabfdeigh
fdeighcab
ighcabfde
Since there are multiple possible magic squares, it's not unique. When you change the number at the very center (i), which you can, you get a different sudoku (not even a rotation or reflection).
answered 24 mins ago
NautilusNautilus
3,751523
$endgroup$
add a comment |
$begingroup$
I was beaten to a very similar answer, but you can create such a square based on the magic square below:
abc
def
ghi
Turn it into this:
abcdefghi
defghiabc
ghiabcdef
bcaefdhig
efdhigbca
higbcaefd
cabfdeigh
fdeighcab
ighcabfde
Since there are multiple possible magic squares, it's not unique. When you change the number at the very center (i), which you can, you get a different sudoku (not even a rotation or reflection).
answered 24 mins ago
NautilusNautilus
3,751523
$endgroup$
add a comment |
$begingroup$
I was beaten to a very similar answer, but you can create such a square based on the magic square below:
abc
def
ghi
Turn it into this:
abcdefghi
defghiabc
ghiabcdef
bcaefdhig
efdhigbca
higbcaefd
cabfdeigh
fdeighcab
ighcabfde
Since there are multiple possible magic squares, it's not unique. When you change the number at the very center (i), which you can, you get a different sudoku (not even a rotation or reflection).
answered 24 mins ago
NautilusNautilus
3,751523
$endgroup$
I was beaten to a very similar answer, but you can create such a square based on the magic square below:
abc
def
ghi
Turn it into this:
abcdefghi
defghiabc
ghiabcdef
bcaefdhig
efdhigbca
higbcaefd
cabfdeigh
fdeighcab
ighcabfde
Since there are multiple possible magic squares, it's not unique. When you change the number at the very center (i), which you can, you get a different sudoku (not even a rotation or reflection).
answered 24 mins ago
NautilusNautilus
3,751523
answered 24 mins ago
NautilusNautilus
3,751523
answered 24 mins ago
NautilusNautilus
3,751523
answered 24 mins ago
NautilusNautilus
3,751523
3,751523
add a comment |
add a comment |
Aorica Ireson is a new contributor. Be nice, and check out our Code of Conduct.
Aorica Ireson is a new contributor. Be nice, and check out our Code of Conduct.
Aorica Ireson is a new contributor. Be nice, and check out our Code of Conduct.
Aorica Ireson is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Puzzling 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%2fpuzzling.stackexchange.com%2fquestions%2f79552%2fsum-in-a-sudoku-with-sum-total-of-15-and-45%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
$begingroup$
"...where each row and column adds to a total of 45..." is redundant, since it is in the nature of a sudoku puzzle that its rows and columns sum to $45$. This is because each row contains the numbers $1$ through $9$, and $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$.
$endgroup$
– Hugh
2 hours ago
$begingroup$
"where each (3*3) 9 individual squares of rows and columns total a sum of 15" >>> do you mean 45?
$endgroup$
– Kryesec
1 hour ago
$begingroup$
@Kryesec I've assumed that that means that each of the 3 by 3 sub-grids is a magic square with "magic constant" 15. In other words, each row/column in each 3 by 3 sub-grid sums to 15.
$endgroup$
– Hugh
1 hour ago