Maximal zero sums in an order 4 projective plane, using -10 to 10 for the points












2












$begingroup$


q  abcd   idfo  t    
q efgh gnla t 1 2 3 4 9 4 6 15 The two 4x4 squares are part of
q ijkl pecj t 5 6 7 8 7 14 12 1 a resolvable block design. Any
q mnop bkmh t 9 10 11 12 16 5 3 10 two numbers are on one column,
13 14 15 16 2 11 13 8 row, or main diagonal.
s rrrr s uuuu s


The 2 4x4 squares give 16 variables with 20 lines of 4 in the rows, columns, diagonals. Adding 5 more variables gives an order-4 projective plane, with qrstu as the 21st line. Just to be clear, the 21 lines in the first representation are qabcd qefgh qijkl qmnop aeimr bfjnr cgkor dhlpr idfot gnlat pecjt bkmht igpbu dneku flcmu oajhu dgjms afkps olebs inchs qrstu. Another order-4 projective plane uses the lines mod((1,2,5,15,17)+n,21) for n=0-20.



If the 21 variables are labeled -10 to 10, what is the maximal number of 0 sums that are possible? I've gotten many arrangements with 11 zeros, but I've found no arrangements with 12 zeros.



order 4 prejective plane










share|cite|improve this question











$endgroup$












  • $begingroup$
    "21 vectors each with 5 variables" I don't see any vectors, nor any variables. What do you mean?
    $endgroup$
    – Gerry Myerson
    Jul 5 '11 at 5:41










  • $begingroup$
    The statement of the question is very unclear. Here's what I've gathered from trying to fathom it; if this is correct, I'd suggest you amend the question accordingly to make it less confusing: The number are redundant; the number squares are the same as the letter squares (if the letters are counted from 'a' $leftrightarrow$ 1). The letters stand for the points of a projective plane of order $4$. The lines of the plane are defined by the rows, columns and main diagonals of the two squares (where each of the two occurrences of a letter stands for the same point)...
    $endgroup$
    – joriki
    Jul 5 '11 at 7:29










  • $begingroup$
    ... One of the points corresponding to the letters 'q' to 'u' is added to each of these lines as indicated, but the rows and columns formed by these additional letters don't correspond to lines of the plane; instead, an additional line containing all of the five additional points labeled 'q' to 'u' is added to complete the plane. Then, when you look for "the maximal number of zero sums", you mean that you consider all possible assignments of the numbers $-10$ to $10$ to the points of the plane and then for each line you form the sum of the numbers assigned to the points on the line...
    $endgroup$
    – joriki
    Jul 5 '11 at 7:31












  • $begingroup$
    ... and you want to know the maximal number of lines with sum $0$ under any such assignment. Correct?
    $endgroup$
    – joriki
    Jul 5 '11 at 7:34












  • $begingroup$
    Thanks, joriki, that sounds like what Ed is trying to do, and I hope he'll edit accordingly. When you write "main diagonals," you want both upper-left to lower-right and upper-right to lower-left, don't you? That's the only way I can see 21 lines here.
    $endgroup$
    – Gerry Myerson
    Jul 5 '11 at 11:01
















2












$begingroup$


q  abcd   idfo  t    
q efgh gnla t 1 2 3 4 9 4 6 15 The two 4x4 squares are part of
q ijkl pecj t 5 6 7 8 7 14 12 1 a resolvable block design. Any
q mnop bkmh t 9 10 11 12 16 5 3 10 two numbers are on one column,
13 14 15 16 2 11 13 8 row, or main diagonal.
s rrrr s uuuu s


The 2 4x4 squares give 16 variables with 20 lines of 4 in the rows, columns, diagonals. Adding 5 more variables gives an order-4 projective plane, with qrstu as the 21st line. Just to be clear, the 21 lines in the first representation are qabcd qefgh qijkl qmnop aeimr bfjnr cgkor dhlpr idfot gnlat pecjt bkmht igpbu dneku flcmu oajhu dgjms afkps olebs inchs qrstu. Another order-4 projective plane uses the lines mod((1,2,5,15,17)+n,21) for n=0-20.



If the 21 variables are labeled -10 to 10, what is the maximal number of 0 sums that are possible? I've gotten many arrangements with 11 zeros, but I've found no arrangements with 12 zeros.



order 4 prejective plane










share|cite|improve this question











$endgroup$












  • $begingroup$
    "21 vectors each with 5 variables" I don't see any vectors, nor any variables. What do you mean?
    $endgroup$
    – Gerry Myerson
    Jul 5 '11 at 5:41










  • $begingroup$
    The statement of the question is very unclear. Here's what I've gathered from trying to fathom it; if this is correct, I'd suggest you amend the question accordingly to make it less confusing: The number are redundant; the number squares are the same as the letter squares (if the letters are counted from 'a' $leftrightarrow$ 1). The letters stand for the points of a projective plane of order $4$. The lines of the plane are defined by the rows, columns and main diagonals of the two squares (where each of the two occurrences of a letter stands for the same point)...
    $endgroup$
    – joriki
    Jul 5 '11 at 7:29










  • $begingroup$
    ... One of the points corresponding to the letters 'q' to 'u' is added to each of these lines as indicated, but the rows and columns formed by these additional letters don't correspond to lines of the plane; instead, an additional line containing all of the five additional points labeled 'q' to 'u' is added to complete the plane. Then, when you look for "the maximal number of zero sums", you mean that you consider all possible assignments of the numbers $-10$ to $10$ to the points of the plane and then for each line you form the sum of the numbers assigned to the points on the line...
    $endgroup$
    – joriki
    Jul 5 '11 at 7:31












  • $begingroup$
    ... and you want to know the maximal number of lines with sum $0$ under any such assignment. Correct?
    $endgroup$
    – joriki
    Jul 5 '11 at 7:34












  • $begingroup$
    Thanks, joriki, that sounds like what Ed is trying to do, and I hope he'll edit accordingly. When you write "main diagonals," you want both upper-left to lower-right and upper-right to lower-left, don't you? That's the only way I can see 21 lines here.
    $endgroup$
    – Gerry Myerson
    Jul 5 '11 at 11:01














2












2








2


1



$begingroup$


q  abcd   idfo  t    
q efgh gnla t 1 2 3 4 9 4 6 15 The two 4x4 squares are part of
q ijkl pecj t 5 6 7 8 7 14 12 1 a resolvable block design. Any
q mnop bkmh t 9 10 11 12 16 5 3 10 two numbers are on one column,
13 14 15 16 2 11 13 8 row, or main diagonal.
s rrrr s uuuu s


The 2 4x4 squares give 16 variables with 20 lines of 4 in the rows, columns, diagonals. Adding 5 more variables gives an order-4 projective plane, with qrstu as the 21st line. Just to be clear, the 21 lines in the first representation are qabcd qefgh qijkl qmnop aeimr bfjnr cgkor dhlpr idfot gnlat pecjt bkmht igpbu dneku flcmu oajhu dgjms afkps olebs inchs qrstu. Another order-4 projective plane uses the lines mod((1,2,5,15,17)+n,21) for n=0-20.



If the 21 variables are labeled -10 to 10, what is the maximal number of 0 sums that are possible? I've gotten many arrangements with 11 zeros, but I've found no arrangements with 12 zeros.



order 4 prejective plane










share|cite|improve this question











$endgroup$




q  abcd   idfo  t    
q efgh gnla t 1 2 3 4 9 4 6 15 The two 4x4 squares are part of
q ijkl pecj t 5 6 7 8 7 14 12 1 a resolvable block design. Any
q mnop bkmh t 9 10 11 12 16 5 3 10 two numbers are on one column,
13 14 15 16 2 11 13 8 row, or main diagonal.
s rrrr s uuuu s


The 2 4x4 squares give 16 variables with 20 lines of 4 in the rows, columns, diagonals. Adding 5 more variables gives an order-4 projective plane, with qrstu as the 21st line. Just to be clear, the 21 lines in the first representation are qabcd qefgh qijkl qmnop aeimr bfjnr cgkor dhlpr idfot gnlat pecjt bkmht igpbu dneku flcmu oajhu dgjms afkps olebs inchs qrstu. Another order-4 projective plane uses the lines mod((1,2,5,15,17)+n,21) for n=0-20.



If the 21 variables are labeled -10 to 10, what is the maximal number of 0 sums that are possible? I've gotten many arrangements with 11 zeros, but I've found no arrangements with 12 zeros.



order 4 prejective plane







puzzle recreational-mathematics projective-geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 26 '18 at 9:30









Glorfindel

3,42981830




3,42981830










asked Jul 5 '11 at 4:44









Ed PeggEd Pegg

10k32593




10k32593












  • $begingroup$
    "21 vectors each with 5 variables" I don't see any vectors, nor any variables. What do you mean?
    $endgroup$
    – Gerry Myerson
    Jul 5 '11 at 5:41










  • $begingroup$
    The statement of the question is very unclear. Here's what I've gathered from trying to fathom it; if this is correct, I'd suggest you amend the question accordingly to make it less confusing: The number are redundant; the number squares are the same as the letter squares (if the letters are counted from 'a' $leftrightarrow$ 1). The letters stand for the points of a projective plane of order $4$. The lines of the plane are defined by the rows, columns and main diagonals of the two squares (where each of the two occurrences of a letter stands for the same point)...
    $endgroup$
    – joriki
    Jul 5 '11 at 7:29










  • $begingroup$
    ... One of the points corresponding to the letters 'q' to 'u' is added to each of these lines as indicated, but the rows and columns formed by these additional letters don't correspond to lines of the plane; instead, an additional line containing all of the five additional points labeled 'q' to 'u' is added to complete the plane. Then, when you look for "the maximal number of zero sums", you mean that you consider all possible assignments of the numbers $-10$ to $10$ to the points of the plane and then for each line you form the sum of the numbers assigned to the points on the line...
    $endgroup$
    – joriki
    Jul 5 '11 at 7:31












  • $begingroup$
    ... and you want to know the maximal number of lines with sum $0$ under any such assignment. Correct?
    $endgroup$
    – joriki
    Jul 5 '11 at 7:34












  • $begingroup$
    Thanks, joriki, that sounds like what Ed is trying to do, and I hope he'll edit accordingly. When you write "main diagonals," you want both upper-left to lower-right and upper-right to lower-left, don't you? That's the only way I can see 21 lines here.
    $endgroup$
    – Gerry Myerson
    Jul 5 '11 at 11:01


















  • $begingroup$
    "21 vectors each with 5 variables" I don't see any vectors, nor any variables. What do you mean?
    $endgroup$
    – Gerry Myerson
    Jul 5 '11 at 5:41










  • $begingroup$
    The statement of the question is very unclear. Here's what I've gathered from trying to fathom it; if this is correct, I'd suggest you amend the question accordingly to make it less confusing: The number are redundant; the number squares are the same as the letter squares (if the letters are counted from 'a' $leftrightarrow$ 1). The letters stand for the points of a projective plane of order $4$. The lines of the plane are defined by the rows, columns and main diagonals of the two squares (where each of the two occurrences of a letter stands for the same point)...
    $endgroup$
    – joriki
    Jul 5 '11 at 7:29










  • $begingroup$
    ... One of the points corresponding to the letters 'q' to 'u' is added to each of these lines as indicated, but the rows and columns formed by these additional letters don't correspond to lines of the plane; instead, an additional line containing all of the five additional points labeled 'q' to 'u' is added to complete the plane. Then, when you look for "the maximal number of zero sums", you mean that you consider all possible assignments of the numbers $-10$ to $10$ to the points of the plane and then for each line you form the sum of the numbers assigned to the points on the line...
    $endgroup$
    – joriki
    Jul 5 '11 at 7:31












  • $begingroup$
    ... and you want to know the maximal number of lines with sum $0$ under any such assignment. Correct?
    $endgroup$
    – joriki
    Jul 5 '11 at 7:34












  • $begingroup$
    Thanks, joriki, that sounds like what Ed is trying to do, and I hope he'll edit accordingly. When you write "main diagonals," you want both upper-left to lower-right and upper-right to lower-left, don't you? That's the only way I can see 21 lines here.
    $endgroup$
    – Gerry Myerson
    Jul 5 '11 at 11:01
















$begingroup$
"21 vectors each with 5 variables" I don't see any vectors, nor any variables. What do you mean?
$endgroup$
– Gerry Myerson
Jul 5 '11 at 5:41




$begingroup$
"21 vectors each with 5 variables" I don't see any vectors, nor any variables. What do you mean?
$endgroup$
– Gerry Myerson
Jul 5 '11 at 5:41












$begingroup$
The statement of the question is very unclear. Here's what I've gathered from trying to fathom it; if this is correct, I'd suggest you amend the question accordingly to make it less confusing: The number are redundant; the number squares are the same as the letter squares (if the letters are counted from 'a' $leftrightarrow$ 1). The letters stand for the points of a projective plane of order $4$. The lines of the plane are defined by the rows, columns and main diagonals of the two squares (where each of the two occurrences of a letter stands for the same point)...
$endgroup$
– joriki
Jul 5 '11 at 7:29




$begingroup$
The statement of the question is very unclear. Here's what I've gathered from trying to fathom it; if this is correct, I'd suggest you amend the question accordingly to make it less confusing: The number are redundant; the number squares are the same as the letter squares (if the letters are counted from 'a' $leftrightarrow$ 1). The letters stand for the points of a projective plane of order $4$. The lines of the plane are defined by the rows, columns and main diagonals of the two squares (where each of the two occurrences of a letter stands for the same point)...
$endgroup$
– joriki
Jul 5 '11 at 7:29












$begingroup$
... One of the points corresponding to the letters 'q' to 'u' is added to each of these lines as indicated, but the rows and columns formed by these additional letters don't correspond to lines of the plane; instead, an additional line containing all of the five additional points labeled 'q' to 'u' is added to complete the plane. Then, when you look for "the maximal number of zero sums", you mean that you consider all possible assignments of the numbers $-10$ to $10$ to the points of the plane and then for each line you form the sum of the numbers assigned to the points on the line...
$endgroup$
– joriki
Jul 5 '11 at 7:31






$begingroup$
... One of the points corresponding to the letters 'q' to 'u' is added to each of these lines as indicated, but the rows and columns formed by these additional letters don't correspond to lines of the plane; instead, an additional line containing all of the five additional points labeled 'q' to 'u' is added to complete the plane. Then, when you look for "the maximal number of zero sums", you mean that you consider all possible assignments of the numbers $-10$ to $10$ to the points of the plane and then for each line you form the sum of the numbers assigned to the points on the line...
$endgroup$
– joriki
Jul 5 '11 at 7:31














$begingroup$
... and you want to know the maximal number of lines with sum $0$ under any such assignment. Correct?
$endgroup$
– joriki
Jul 5 '11 at 7:34






$begingroup$
... and you want to know the maximal number of lines with sum $0$ under any such assignment. Correct?
$endgroup$
– joriki
Jul 5 '11 at 7:34














$begingroup$
Thanks, joriki, that sounds like what Ed is trying to do, and I hope he'll edit accordingly. When you write "main diagonals," you want both upper-left to lower-right and upper-right to lower-left, don't you? That's the only way I can see 21 lines here.
$endgroup$
– Gerry Myerson
Jul 5 '11 at 11:01




$begingroup$
Thanks, joriki, that sounds like what Ed is trying to do, and I hope he'll edit accordingly. When you write "main diagonals," you want both upper-left to lower-right and upper-right to lower-left, don't you? That's the only way I can see 21 lines here.
$endgroup$
– Gerry Myerson
Jul 5 '11 at 11:01










0






active

oldest

votes











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: "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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f49542%2fmaximal-zero-sums-in-an-order-4-projective-plane-using-10-to-10-for-the-points%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























0






active

oldest

votes








0






active

oldest

votes









active

oldest

votes






active

oldest

votes
















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f49542%2fmaximal-zero-sums-in-an-order-4-projective-plane-using-10-to-10-for-the-points%23new-answer', 'question_page');
}
);

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







Popular posts from this blog

Wiesbaden

Marschland

Dieringhausen