Contest math problem about crossing out numbers in the table












10












$begingroup$


A table $ntimes n$ is filled with pairwise different natural numbers. Ann and Ben are playing the following game: Ann chooses the greatest number, then crosses out the row and the column containing it. She then chooses the greatest number from what is remained and repeats the whole process unless table is crossed out completely.



Ben takes exactly the same table, and repeats the same process, but choosing the least number on each step.



We need to show that the sum $A$ of numbers chosen by Ann is greater (or equal) to the sum $B$ of numbers chosen by Ben.



I think it should be done via presenting such $C$ that $Ageq Cgeq B$.
However, if $a_i$ and $b_i$ are the numbers chosen by Ann and Ben on $i$-th step respectively, the inequality $a_igeq b_{n-i+1}$ does not hold. So, I am stuck at this point.






combinatorics contest-math







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Nice question. Where does it come from?
    $endgroup$
    – saulspatz
    8 hours ago










  • $begingroup$
    @saulspatz saw it in a list of problems created by a Russian teacher. I think he is the author/
    $endgroup$
    – Michael Freimann
    8 hours ago












  • $begingroup$
    "pairwise different" that phrase doesn't make sense.
    $endgroup$
    – Acccumulation
    2 hours ago










  • $begingroup$
    @Acccumulation : It literally means, "for all pairs of numbers in the table, $(x,y)$, the members of the pair are different, $x neq y$". I.e., each natural number appears zero or one times in the table.
    $endgroup$
    – Eric Towers
    2 hours ago










  • $begingroup$
    @EricTowers That can be expressed by simply saying "different", "distinct", or "unique". The word "pairwise" isn't doing much.
    $endgroup$
    – Acccumulation
    2 hours ago
















10












$begingroup$


A table $ntimes n$ is filled with pairwise different natural numbers. Ann and Ben are playing the following game: Ann chooses the greatest number, then crosses out the row and the column containing it. She then chooses the greatest number from what is remained and repeats the whole process unless table is crossed out completely.



Ben takes exactly the same table, and repeats the same process, but choosing the least number on each step.



We need to show that the sum $A$ of numbers chosen by Ann is greater (or equal) to the sum $B$ of numbers chosen by Ben.



I think it should be done via presenting such $C$ that $Ageq Cgeq B$.
However, if $a_i$ and $b_i$ are the numbers chosen by Ann and Ben on $i$-th step respectively, the inequality $a_igeq b_{n-i+1}$ does not hold. So, I am stuck at this point.






combinatorics contest-math







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Nice question. Where does it come from?
    $endgroup$
    – saulspatz
    8 hours ago










  • $begingroup$
    @saulspatz saw it in a list of problems created by a Russian teacher. I think he is the author/
    $endgroup$
    – Michael Freimann
    8 hours ago












  • $begingroup$
    "pairwise different" that phrase doesn't make sense.
    $endgroup$
    – Acccumulation
    2 hours ago










  • $begingroup$
    @Acccumulation : It literally means, "for all pairs of numbers in the table, $(x,y)$, the members of the pair are different, $x neq y$". I.e., each natural number appears zero or one times in the table.
    $endgroup$
    – Eric Towers
    2 hours ago










  • $begingroup$
    @EricTowers That can be expressed by simply saying "different", "distinct", or "unique". The word "pairwise" isn't doing much.
    $endgroup$
    – Acccumulation
    2 hours ago














10












10








10





$begingroup$


A table $ntimes n$ is filled with pairwise different natural numbers. Ann and Ben are playing the following game: Ann chooses the greatest number, then crosses out the row and the column containing it. She then chooses the greatest number from what is remained and repeats the whole process unless table is crossed out completely.



Ben takes exactly the same table, and repeats the same process, but choosing the least number on each step.



We need to show that the sum $A$ of numbers chosen by Ann is greater (or equal) to the sum $B$ of numbers chosen by Ben.



I think it should be done via presenting such $C$ that $Ageq Cgeq B$.
However, if $a_i$ and $b_i$ are the numbers chosen by Ann and Ben on $i$-th step respectively, the inequality $a_igeq b_{n-i+1}$ does not hold. So, I am stuck at this point.






combinatorics contest-math







share|cite|improve this question











$endgroup$




A table $ntimes n$ is filled with pairwise different natural numbers. Ann and Ben are playing the following game: Ann chooses the greatest number, then crosses out the row and the column containing it. She then chooses the greatest number from what is remained and repeats the whole process unless table is crossed out completely.



Ben takes exactly the same table, and repeats the same process, but choosing the least number on each step.



We need to show that the sum $A$ of numbers chosen by Ann is greater (or equal) to the sum $B$ of numbers chosen by Ben.



I think it should be done via presenting such $C$ that $Ageq Cgeq B$.
However, if $a_i$ and $b_i$ are the numbers chosen by Ann and Ben on $i$-th step respectively, the inequality $a_igeq b_{n-i+1}$ does not hold. So, I am stuck at this point.






combinatorics contest-math




combinatorics contest-math






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 8 hours ago









darij grinberg

11k33167




11k33167










asked 8 hours ago









Michael FreimannMichael Freimann

295113




295113








  • 1




    $begingroup$
    Nice question. Where does it come from?
    $endgroup$
    – saulspatz
    8 hours ago










  • $begingroup$
    @saulspatz saw it in a list of problems created by a Russian teacher. I think he is the author/
    $endgroup$
    – Michael Freimann
    8 hours ago












  • $begingroup$
    "pairwise different" that phrase doesn't make sense.
    $endgroup$
    – Acccumulation
    2 hours ago










  • $begingroup$
    @Acccumulation : It literally means, "for all pairs of numbers in the table, $(x,y)$, the members of the pair are different, $x neq y$". I.e., each natural number appears zero or one times in the table.
    $endgroup$
    – Eric Towers
    2 hours ago










  • $begingroup$
    @EricTowers That can be expressed by simply saying "different", "distinct", or "unique". The word "pairwise" isn't doing much.
    $endgroup$
    – Acccumulation
    2 hours ago














  • 1




    $begingroup$
    Nice question. Where does it come from?
    $endgroup$
    – saulspatz
    8 hours ago










  • $begingroup$
    @saulspatz saw it in a list of problems created by a Russian teacher. I think he is the author/
    $endgroup$
    – Michael Freimann
    8 hours ago












  • $begingroup$
    "pairwise different" that phrase doesn't make sense.
    $endgroup$
    – Acccumulation
    2 hours ago










  • $begingroup$
    @Acccumulation : It literally means, "for all pairs of numbers in the table, $(x,y)$, the members of the pair are different, $x neq y$". I.e., each natural number appears zero or one times in the table.
    $endgroup$
    – Eric Towers
    2 hours ago










  • $begingroup$
    @EricTowers That can be expressed by simply saying "different", "distinct", or "unique". The word "pairwise" isn't doing much.
    $endgroup$
    – Acccumulation
    2 hours ago








1




1




$begingroup$
Nice question. Where does it come from?
$endgroup$
– saulspatz
8 hours ago




$begingroup$
Nice question. Where does it come from?
$endgroup$
– saulspatz
8 hours ago












$begingroup$
@saulspatz saw it in a list of problems created by a Russian teacher. I think he is the author/
$endgroup$
– Michael Freimann
8 hours ago






$begingroup$
@saulspatz saw it in a list of problems created by a Russian teacher. I think he is the author/
$endgroup$
– Michael Freimann
8 hours ago














$begingroup$
"pairwise different" that phrase doesn't make sense.
$endgroup$
– Acccumulation
2 hours ago




$begingroup$
"pairwise different" that phrase doesn't make sense.
$endgroup$
– Acccumulation
2 hours ago












$begingroup$
@Acccumulation : It literally means, "for all pairs of numbers in the table, $(x,y)$, the members of the pair are different, $x neq y$". I.e., each natural number appears zero or one times in the table.
$endgroup$
– Eric Towers
2 hours ago




$begingroup$
@Acccumulation : It literally means, "for all pairs of numbers in the table, $(x,y)$, the members of the pair are different, $x neq y$". I.e., each natural number appears zero or one times in the table.
$endgroup$
– Eric Towers
2 hours ago












$begingroup$
@EricTowers That can be expressed by simply saying "different", "distinct", or "unique". The word "pairwise" isn't doing much.
$endgroup$
– Acccumulation
2 hours ago




$begingroup$
@EricTowers That can be expressed by simply saying "different", "distinct", or "unique". The word "pairwise" isn't doing much.
$endgroup$
– Acccumulation
2 hours ago










2 Answers
2






active

oldest

votes


















12












$begingroup$

Let $a_1,a_2dots a_n$ be the numbers selected by Ann and $b_1,b_2,dots b_n$ be the numbers selected by Benjamin.



Lemma: $a_igeq b_{n+1-i}$ for all $1leq i leq n$.



Proof: Consider the set $A$ of all numbers that were eligible when we selected $a_i$ and the set $B$ of all the numbers eligible when we selected $b_{n+1-i}$. Notice that these two sets intersect (because $n-i+1$ rows are eligible in $A$ and $i$ rows are eligible in $B$, and the same happens with the columns) so the claim is true, since $a_i=max(A)$ and $b_i=min B$.



The problem follows from the lemma:



$sumlimits_{i=1}^n a_i geq sumlimits_{i=1}^n b_{n+1-i} = sumlimits_{i=1}^n b_i$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What, exactly, does it mean for a row or column to be "eligible" and how does this translate into a proof that $Acap Bnot=emptyset$?
    $endgroup$
    – Barry Cipra
    8 hours ago










  • $begingroup$
    @BarryCipra excellent question ! A row is eligible if no previous number has been selected in that row and the same definition applies for column! Now we can notice that if an element is in an eligible row and column, then it is eligible. I hope this help. Best regards.
    $endgroup$
    – Jorge Fernández Hidalgo
    7 hours ago










  • $begingroup$
    @BarryCipra let $k$ be an index such that both (a) no entry in row $k$ was picked by Ann in her first $i-1$ picks and (b) no entry in row $k$ was picked by Ben in his first $n-i$ picks. There indeed is such a $k$ because $n-i + (i-1) = n-1 < n$. Let $A_k$ denote the $i$-th row. We now show that of the $n$ entries in $A$ there is at least one not picked by Ann or Ben. But now let $l$ be an index such that both (a) no entry in column $l$ was picked by Ann in her first $i-1$ picks and (b) no entry in column $l$ was picked by Ben in his first $n-i$ picks. ...
    $endgroup$
    – Mike
    7 hours ago










  • $begingroup$
    ... Then the entry $a_{kl}$ [the entry in the $k$-th row of $A$ and $l$-th column] is available to be picked by both Ann for her $i$-th pick and Ben's $n-i+1$-th pick. So what Ann picks then will be no larger than $a_{kl}$ and what Ben picks no larger.
    $endgroup$
    – Mike
    7 hours ago








  • 1




    $begingroup$
    I like how this proof generalizes, pretty much word-for-word, to the $k$-dimensional version of the problem. (+1)
    $endgroup$
    – Micah
    6 hours ago



















5












$begingroup$

And in fact to add to @Jorge's answer strict inequality is not always possible: For any integer $n$ there are $n times n$ tables $A$ that will force Player A and Player B to have the same sum: Let $A$ be any $n times n$ table where the entries get larger as you head down a column, and to the right on a row: e.g., $A =[a_{ij}]$ where the $ij$-th entry is $i(n+1) + j$. [Note that all entries of $A$ are distinct.] Then both Player A and Player B will end up picking the diagonal elements of $A$: Player A will end up picking the diagonals southeast to northwest when Player B northwest to southeast.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Good point! It makes it interesting as it raises the question as to why the numbers must be distinct. I guess it is so that the choosing number algorithm is easier to explain.
    $endgroup$
    – Jorge Fernández Hidalgo
    7 hours ago






  • 2




    $begingroup$
    @JorgeFernándezHidalgo - in fact it is clear from your very nice proof that even if the original matrix contains repeats, and even if a player can choose any of the multiple maxima (or minima), the claim is still true.
    $endgroup$
    – antkam
    5 hours ago











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%2f3131640%2fcontest-math-problem-about-crossing-out-numbers-in-the-table%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









12












$begingroup$

Let $a_1,a_2dots a_n$ be the numbers selected by Ann and $b_1,b_2,dots b_n$ be the numbers selected by Benjamin.



Lemma: $a_igeq b_{n+1-i}$ for all $1leq i leq n$.



Proof: Consider the set $A$ of all numbers that were eligible when we selected $a_i$ and the set $B$ of all the numbers eligible when we selected $b_{n+1-i}$. Notice that these two sets intersect (because $n-i+1$ rows are eligible in $A$ and $i$ rows are eligible in $B$, and the same happens with the columns) so the claim is true, since $a_i=max(A)$ and $b_i=min B$.



The problem follows from the lemma:



$sumlimits_{i=1}^n a_i geq sumlimits_{i=1}^n b_{n+1-i} = sumlimits_{i=1}^n b_i$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What, exactly, does it mean for a row or column to be "eligible" and how does this translate into a proof that $Acap Bnot=emptyset$?
    $endgroup$
    – Barry Cipra
    8 hours ago










  • $begingroup$
    @BarryCipra excellent question ! A row is eligible if no previous number has been selected in that row and the same definition applies for column! Now we can notice that if an element is in an eligible row and column, then it is eligible. I hope this help. Best regards.
    $endgroup$
    – Jorge Fernández Hidalgo
    7 hours ago










  • $begingroup$
    @BarryCipra let $k$ be an index such that both (a) no entry in row $k$ was picked by Ann in her first $i-1$ picks and (b) no entry in row $k$ was picked by Ben in his first $n-i$ picks. There indeed is such a $k$ because $n-i + (i-1) = n-1 < n$. Let $A_k$ denote the $i$-th row. We now show that of the $n$ entries in $A$ there is at least one not picked by Ann or Ben. But now let $l$ be an index such that both (a) no entry in column $l$ was picked by Ann in her first $i-1$ picks and (b) no entry in column $l$ was picked by Ben in his first $n-i$ picks. ...
    $endgroup$
    – Mike
    7 hours ago










  • $begingroup$
    ... Then the entry $a_{kl}$ [the entry in the $k$-th row of $A$ and $l$-th column] is available to be picked by both Ann for her $i$-th pick and Ben's $n-i+1$-th pick. So what Ann picks then will be no larger than $a_{kl}$ and what Ben picks no larger.
    $endgroup$
    – Mike
    7 hours ago








  • 1




    $begingroup$
    I like how this proof generalizes, pretty much word-for-word, to the $k$-dimensional version of the problem. (+1)
    $endgroup$
    – Micah
    6 hours ago
















12












$begingroup$

Let $a_1,a_2dots a_n$ be the numbers selected by Ann and $b_1,b_2,dots b_n$ be the numbers selected by Benjamin.



Lemma: $a_igeq b_{n+1-i}$ for all $1leq i leq n$.



Proof: Consider the set $A$ of all numbers that were eligible when we selected $a_i$ and the set $B$ of all the numbers eligible when we selected $b_{n+1-i}$. Notice that these two sets intersect (because $n-i+1$ rows are eligible in $A$ and $i$ rows are eligible in $B$, and the same happens with the columns) so the claim is true, since $a_i=max(A)$ and $b_i=min B$.



The problem follows from the lemma:



$sumlimits_{i=1}^n a_i geq sumlimits_{i=1}^n b_{n+1-i} = sumlimits_{i=1}^n b_i$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What, exactly, does it mean for a row or column to be "eligible" and how does this translate into a proof that $Acap Bnot=emptyset$?
    $endgroup$
    – Barry Cipra
    8 hours ago










  • $begingroup$
    @BarryCipra excellent question ! A row is eligible if no previous number has been selected in that row and the same definition applies for column! Now we can notice that if an element is in an eligible row and column, then it is eligible. I hope this help. Best regards.
    $endgroup$
    – Jorge Fernández Hidalgo
    7 hours ago










  • $begingroup$
    @BarryCipra let $k$ be an index such that both (a) no entry in row $k$ was picked by Ann in her first $i-1$ picks and (b) no entry in row $k$ was picked by Ben in his first $n-i$ picks. There indeed is such a $k$ because $n-i + (i-1) = n-1 < n$. Let $A_k$ denote the $i$-th row. We now show that of the $n$ entries in $A$ there is at least one not picked by Ann or Ben. But now let $l$ be an index such that both (a) no entry in column $l$ was picked by Ann in her first $i-1$ picks and (b) no entry in column $l$ was picked by Ben in his first $n-i$ picks. ...
    $endgroup$
    – Mike
    7 hours ago










  • $begingroup$
    ... Then the entry $a_{kl}$ [the entry in the $k$-th row of $A$ and $l$-th column] is available to be picked by both Ann for her $i$-th pick and Ben's $n-i+1$-th pick. So what Ann picks then will be no larger than $a_{kl}$ and what Ben picks no larger.
    $endgroup$
    – Mike
    7 hours ago








  • 1




    $begingroup$
    I like how this proof generalizes, pretty much word-for-word, to the $k$-dimensional version of the problem. (+1)
    $endgroup$
    – Micah
    6 hours ago














12












12








12





$begingroup$

Let $a_1,a_2dots a_n$ be the numbers selected by Ann and $b_1,b_2,dots b_n$ be the numbers selected by Benjamin.



Lemma: $a_igeq b_{n+1-i}$ for all $1leq i leq n$.



Proof: Consider the set $A$ of all numbers that were eligible when we selected $a_i$ and the set $B$ of all the numbers eligible when we selected $b_{n+1-i}$. Notice that these two sets intersect (because $n-i+1$ rows are eligible in $A$ and $i$ rows are eligible in $B$, and the same happens with the columns) so the claim is true, since $a_i=max(A)$ and $b_i=min B$.



The problem follows from the lemma:



$sumlimits_{i=1}^n a_i geq sumlimits_{i=1}^n b_{n+1-i} = sumlimits_{i=1}^n b_i$






share|cite|improve this answer











$endgroup$



Let $a_1,a_2dots a_n$ be the numbers selected by Ann and $b_1,b_2,dots b_n$ be the numbers selected by Benjamin.



Lemma: $a_igeq b_{n+1-i}$ for all $1leq i leq n$.



Proof: Consider the set $A$ of all numbers that were eligible when we selected $a_i$ and the set $B$ of all the numbers eligible when we selected $b_{n+1-i}$. Notice that these two sets intersect (because $n-i+1$ rows are eligible in $A$ and $i$ rows are eligible in $B$, and the same happens with the columns) so the claim is true, since $a_i=max(A)$ and $b_i=min B$.



The problem follows from the lemma:



$sumlimits_{i=1}^n a_i geq sumlimits_{i=1}^n b_{n+1-i} = sumlimits_{i=1}^n b_i$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 8 hours ago









darij grinberg

11k33167




11k33167










answered 8 hours ago









Jorge Fernández HidalgoJorge Fernández Hidalgo

75.6k1192193




75.6k1192193












  • $begingroup$
    What, exactly, does it mean for a row or column to be "eligible" and how does this translate into a proof that $Acap Bnot=emptyset$?
    $endgroup$
    – Barry Cipra
    8 hours ago










  • $begingroup$
    @BarryCipra excellent question ! A row is eligible if no previous number has been selected in that row and the same definition applies for column! Now we can notice that if an element is in an eligible row and column, then it is eligible. I hope this help. Best regards.
    $endgroup$
    – Jorge Fernández Hidalgo
    7 hours ago










  • $begingroup$
    @BarryCipra let $k$ be an index such that both (a) no entry in row $k$ was picked by Ann in her first $i-1$ picks and (b) no entry in row $k$ was picked by Ben in his first $n-i$ picks. There indeed is such a $k$ because $n-i + (i-1) = n-1 < n$. Let $A_k$ denote the $i$-th row. We now show that of the $n$ entries in $A$ there is at least one not picked by Ann or Ben. But now let $l$ be an index such that both (a) no entry in column $l$ was picked by Ann in her first $i-1$ picks and (b) no entry in column $l$ was picked by Ben in his first $n-i$ picks. ...
    $endgroup$
    – Mike
    7 hours ago










  • $begingroup$
    ... Then the entry $a_{kl}$ [the entry in the $k$-th row of $A$ and $l$-th column] is available to be picked by both Ann for her $i$-th pick and Ben's $n-i+1$-th pick. So what Ann picks then will be no larger than $a_{kl}$ and what Ben picks no larger.
    $endgroup$
    – Mike
    7 hours ago








  • 1




    $begingroup$
    I like how this proof generalizes, pretty much word-for-word, to the $k$-dimensional version of the problem. (+1)
    $endgroup$
    – Micah
    6 hours ago


















  • $begingroup$
    What, exactly, does it mean for a row or column to be "eligible" and how does this translate into a proof that $Acap Bnot=emptyset$?
    $endgroup$
    – Barry Cipra
    8 hours ago










  • $begingroup$
    @BarryCipra excellent question ! A row is eligible if no previous number has been selected in that row and the same definition applies for column! Now we can notice that if an element is in an eligible row and column, then it is eligible. I hope this help. Best regards.
    $endgroup$
    – Jorge Fernández Hidalgo
    7 hours ago










  • $begingroup$
    @BarryCipra let $k$ be an index such that both (a) no entry in row $k$ was picked by Ann in her first $i-1$ picks and (b) no entry in row $k$ was picked by Ben in his first $n-i$ picks. There indeed is such a $k$ because $n-i + (i-1) = n-1 < n$. Let $A_k$ denote the $i$-th row. We now show that of the $n$ entries in $A$ there is at least one not picked by Ann or Ben. But now let $l$ be an index such that both (a) no entry in column $l$ was picked by Ann in her first $i-1$ picks and (b) no entry in column $l$ was picked by Ben in his first $n-i$ picks. ...
    $endgroup$
    – Mike
    7 hours ago










  • $begingroup$
    ... Then the entry $a_{kl}$ [the entry in the $k$-th row of $A$ and $l$-th column] is available to be picked by both Ann for her $i$-th pick and Ben's $n-i+1$-th pick. So what Ann picks then will be no larger than $a_{kl}$ and what Ben picks no larger.
    $endgroup$
    – Mike
    7 hours ago








  • 1




    $begingroup$
    I like how this proof generalizes, pretty much word-for-word, to the $k$-dimensional version of the problem. (+1)
    $endgroup$
    – Micah
    6 hours ago
















$begingroup$
What, exactly, does it mean for a row or column to be "eligible" and how does this translate into a proof that $Acap Bnot=emptyset$?
$endgroup$
– Barry Cipra
8 hours ago




$begingroup$
What, exactly, does it mean for a row or column to be "eligible" and how does this translate into a proof that $Acap Bnot=emptyset$?
$endgroup$
– Barry Cipra
8 hours ago












$begingroup$
@BarryCipra excellent question ! A row is eligible if no previous number has been selected in that row and the same definition applies for column! Now we can notice that if an element is in an eligible row and column, then it is eligible. I hope this help. Best regards.
$endgroup$
– Jorge Fernández Hidalgo
7 hours ago




$begingroup$
@BarryCipra excellent question ! A row is eligible if no previous number has been selected in that row and the same definition applies for column! Now we can notice that if an element is in an eligible row and column, then it is eligible. I hope this help. Best regards.
$endgroup$
– Jorge Fernández Hidalgo
7 hours ago












$begingroup$
@BarryCipra let $k$ be an index such that both (a) no entry in row $k$ was picked by Ann in her first $i-1$ picks and (b) no entry in row $k$ was picked by Ben in his first $n-i$ picks. There indeed is such a $k$ because $n-i + (i-1) = n-1 < n$. Let $A_k$ denote the $i$-th row. We now show that of the $n$ entries in $A$ there is at least one not picked by Ann or Ben. But now let $l$ be an index such that both (a) no entry in column $l$ was picked by Ann in her first $i-1$ picks and (b) no entry in column $l$ was picked by Ben in his first $n-i$ picks. ...
$endgroup$
– Mike
7 hours ago




$begingroup$
@BarryCipra let $k$ be an index such that both (a) no entry in row $k$ was picked by Ann in her first $i-1$ picks and (b) no entry in row $k$ was picked by Ben in his first $n-i$ picks. There indeed is such a $k$ because $n-i + (i-1) = n-1 < n$. Let $A_k$ denote the $i$-th row. We now show that of the $n$ entries in $A$ there is at least one not picked by Ann or Ben. But now let $l$ be an index such that both (a) no entry in column $l$ was picked by Ann in her first $i-1$ picks and (b) no entry in column $l$ was picked by Ben in his first $n-i$ picks. ...
$endgroup$
– Mike
7 hours ago












$begingroup$
... Then the entry $a_{kl}$ [the entry in the $k$-th row of $A$ and $l$-th column] is available to be picked by both Ann for her $i$-th pick and Ben's $n-i+1$-th pick. So what Ann picks then will be no larger than $a_{kl}$ and what Ben picks no larger.
$endgroup$
– Mike
7 hours ago






$begingroup$
... Then the entry $a_{kl}$ [the entry in the $k$-th row of $A$ and $l$-th column] is available to be picked by both Ann for her $i$-th pick and Ben's $n-i+1$-th pick. So what Ann picks then will be no larger than $a_{kl}$ and what Ben picks no larger.
$endgroup$
– Mike
7 hours ago






1




1




$begingroup$
I like how this proof generalizes, pretty much word-for-word, to the $k$-dimensional version of the problem. (+1)
$endgroup$
– Micah
6 hours ago




$begingroup$
I like how this proof generalizes, pretty much word-for-word, to the $k$-dimensional version of the problem. (+1)
$endgroup$
– Micah
6 hours ago











5












$begingroup$

And in fact to add to @Jorge's answer strict inequality is not always possible: For any integer $n$ there are $n times n$ tables $A$ that will force Player A and Player B to have the same sum: Let $A$ be any $n times n$ table where the entries get larger as you head down a column, and to the right on a row: e.g., $A =[a_{ij}]$ where the $ij$-th entry is $i(n+1) + j$. [Note that all entries of $A$ are distinct.] Then both Player A and Player B will end up picking the diagonal elements of $A$: Player A will end up picking the diagonals southeast to northwest when Player B northwest to southeast.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Good point! It makes it interesting as it raises the question as to why the numbers must be distinct. I guess it is so that the choosing number algorithm is easier to explain.
    $endgroup$
    – Jorge Fernández Hidalgo
    7 hours ago






  • 2




    $begingroup$
    @JorgeFernándezHidalgo - in fact it is clear from your very nice proof that even if the original matrix contains repeats, and even if a player can choose any of the multiple maxima (or minima), the claim is still true.
    $endgroup$
    – antkam
    5 hours ago
















5












$begingroup$

And in fact to add to @Jorge's answer strict inequality is not always possible: For any integer $n$ there are $n times n$ tables $A$ that will force Player A and Player B to have the same sum: Let $A$ be any $n times n$ table where the entries get larger as you head down a column, and to the right on a row: e.g., $A =[a_{ij}]$ where the $ij$-th entry is $i(n+1) + j$. [Note that all entries of $A$ are distinct.] Then both Player A and Player B will end up picking the diagonal elements of $A$: Player A will end up picking the diagonals southeast to northwest when Player B northwest to southeast.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Good point! It makes it interesting as it raises the question as to why the numbers must be distinct. I guess it is so that the choosing number algorithm is easier to explain.
    $endgroup$
    – Jorge Fernández Hidalgo
    7 hours ago






  • 2




    $begingroup$
    @JorgeFernándezHidalgo - in fact it is clear from your very nice proof that even if the original matrix contains repeats, and even if a player can choose any of the multiple maxima (or minima), the claim is still true.
    $endgroup$
    – antkam
    5 hours ago














5












5








5





$begingroup$

And in fact to add to @Jorge's answer strict inequality is not always possible: For any integer $n$ there are $n times n$ tables $A$ that will force Player A and Player B to have the same sum: Let $A$ be any $n times n$ table where the entries get larger as you head down a column, and to the right on a row: e.g., $A =[a_{ij}]$ where the $ij$-th entry is $i(n+1) + j$. [Note that all entries of $A$ are distinct.] Then both Player A and Player B will end up picking the diagonal elements of $A$: Player A will end up picking the diagonals southeast to northwest when Player B northwest to southeast.






share|cite|improve this answer











$endgroup$



And in fact to add to @Jorge's answer strict inequality is not always possible: For any integer $n$ there are $n times n$ tables $A$ that will force Player A and Player B to have the same sum: Let $A$ be any $n times n$ table where the entries get larger as you head down a column, and to the right on a row: e.g., $A =[a_{ij}]$ where the $ij$-th entry is $i(n+1) + j$. [Note that all entries of $A$ are distinct.] Then both Player A and Player B will end up picking the diagonal elements of $A$: Player A will end up picking the diagonals southeast to northwest when Player B northwest to southeast.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 7 hours ago

























answered 8 hours ago









MikeMike

4,256412




4,256412








  • 1




    $begingroup$
    Good point! It makes it interesting as it raises the question as to why the numbers must be distinct. I guess it is so that the choosing number algorithm is easier to explain.
    $endgroup$
    – Jorge Fernández Hidalgo
    7 hours ago






  • 2




    $begingroup$
    @JorgeFernándezHidalgo - in fact it is clear from your very nice proof that even if the original matrix contains repeats, and even if a player can choose any of the multiple maxima (or minima), the claim is still true.
    $endgroup$
    – antkam
    5 hours ago














  • 1




    $begingroup$
    Good point! It makes it interesting as it raises the question as to why the numbers must be distinct. I guess it is so that the choosing number algorithm is easier to explain.
    $endgroup$
    – Jorge Fernández Hidalgo
    7 hours ago






  • 2




    $begingroup$
    @JorgeFernándezHidalgo - in fact it is clear from your very nice proof that even if the original matrix contains repeats, and even if a player can choose any of the multiple maxima (or minima), the claim is still true.
    $endgroup$
    – antkam
    5 hours ago








1




1




$begingroup$
Good point! It makes it interesting as it raises the question as to why the numbers must be distinct. I guess it is so that the choosing number algorithm is easier to explain.
$endgroup$
– Jorge Fernández Hidalgo
7 hours ago




$begingroup$
Good point! It makes it interesting as it raises the question as to why the numbers must be distinct. I guess it is so that the choosing number algorithm is easier to explain.
$endgroup$
– Jorge Fernández Hidalgo
7 hours ago




2




2




$begingroup$
@JorgeFernándezHidalgo - in fact it is clear from your very nice proof that even if the original matrix contains repeats, and even if a player can choose any of the multiple maxima (or minima), the claim is still true.
$endgroup$
– antkam
5 hours ago




$begingroup$
@JorgeFernándezHidalgo - in fact it is clear from your very nice proof that even if the original matrix contains repeats, and even if a player can choose any of the multiple maxima (or minima), the claim is still true.
$endgroup$
– antkam
5 hours ago


















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%2f3131640%2fcontest-math-problem-about-crossing-out-numbers-in-the-table%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

What other Star Trek series did the main TNG cast show up in?

Image
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0; } 4 I'm referring to Picard, Riker, Data, Worf, Geordi, Dr. Crusher, Wesley Crusher, Chief O'Brien and family, Deanna Troi, Guinan. I know Picard and Worf show up on DS9, though I haven't seen it myself. I'm curious what happened to the characters after that last episode of TNG? star-trek star-trek-tng share | improve this question edited 10 hours ago Ham Sandwich 6,051 2 25 75
Read more

Berlina muro

Image
La berlina muro ĉe Bethaniendamm, 1986 La berlina muro aŭ simple la muro [1] , nomita ankaŭ "kontraŭfaŝisma remparo" en propaganda lingvo de la iama Germana Demokratia Respubliko, estis parto de la interngermana limo, kiu dividis de la 13-a de aŭgusto 1961 ĝis la 9-a de novembro 1989 okcidentan Berlinon de la orienta parto de la urbo kaj de la ĉirkaŭa tereno de la GDR. Ĝi estis unu el la plej famaj simboloj de la malvarma milito kaj de la disdivido de Germanio. Pli ol simpla muro, la berlina muro estis fakte tuta defendokonstruaĵo kun du muroj, ena malplena strio, rondirvojo, gardturoj kaj alarmsistemoj. Dum la provo transiri la severe garditan limon al okcidenta Berlino multaj homoj estis mortigitaj. La preciza nombro de viktimoj estas pridisputita kaj ne certa; la nombroj anoncataj varias inter 86 kaj 238 mortintoj. La malplifortiĝo de Sovetunio kaj la politiko de liberigo gvidita de Miĥail Gorbaĉov ebligis al la orientgermanoj faligi la 9-an de novembro 1989 l
Read more

Berlina aerponto

Image
Flugkoridoroj dum la blokado de Berlino La Berlina aerponto estas la provizigo de okcidenta Berlino per aviadiloj fare de la aliancanoj inter la 23-a de junio 1948 kaj la 12-a de majo 1949 dum la blokado de Berlino fare de la sovetia armeo. Enhavo 1 Historio 1.1 La aerkoridoroj 1.2 Unuaj flugoj 1.3 Plibonigo de la aerponto 1.4 Fino de la aerponto 1.5 Rozinbombiloj 1.6 Bilanco 2 Flughavenoj 2.1 Berlino 2.2 Okcidenta Germanio 3 Aviadiloj de la berlina aerponto 4 15-a datreveno de la aerponto 5 Bibliografio 6 Referencoj 7 Eksteraj ligiloj Historio | General Lucius D. Clay, militguberniestro de la usona okupadzono Jam antaŭ la berlina aerponto estiĝis la malgranda aerponto . La estro de la sovetia militadministracio en Germanio, Vasili Daniloviĉ Sokolovski, ordonis la 1-an de aprilo 1948, kiel reago al la konferenco de Londono inter britoj, francoj kaj reprezentantoj de Benelukso pri aligo de Germa
Read more