Simple example of invariants
$begingroup$
The combinatorics textbook I'm reading introduces invariants with the following example:
There are three piles with $n$ tokens each. In every step we are allowed
to choose two piles, take one token from each of those two piles and add a token to
the third pile. Using these moves, is it possible to end up having only one token?
Solution: To the tokens in the first pile we can assign the pair $(0, 1)$, to the tokens
in the second pile the pair $(1, 0)$ and to the tokens in the third pile the pair $(1, 1)$.
Notice that the sum modulo $2$ of any two of these pairs give us the third one. Thus,
in every step the sum modulo $2$ of all the assigned pairs is the same. However, the
sum of all the assigned pairs in the beginning is $(2n, 2n)$, which is equal to $(0, 0)$
modulo $2$. Since this pair was not assigned to any pile, it is not possible to end up
with only one token.
Although I already knew what invariants are, I do not understand how they're applied in this specific case, what is the meaning of the pairs that are assigned to each pile, and how is a pair affected by a step?
combinatorics invariance
asked Jan 19 at 14:10
Spasoje DurovicSpasoje Durovic
36510
$endgroup$
add a comment |
$begingroup$
The combinatorics textbook I'm reading introduces invariants with the following example:
There are three piles with $n$ tokens each. In every step we are allowed
to choose two piles, take one token from each of those two piles and add a token to
the third pile. Using these moves, is it possible to end up having only one token?
Solution: To the tokens in the first pile we can assign the pair $(0, 1)$, to the tokens
in the second pile the pair $(1, 0)$ and to the tokens in the third pile the pair $(1, 1)$.
Notice that the sum modulo $2$ of any two of these pairs give us the third one. Thus,
in every step the sum modulo $2$ of all the assigned pairs is the same. However, the
sum of all the assigned pairs in the beginning is $(2n, 2n)$, which is equal to $(0, 0)$
modulo $2$. Since this pair was not assigned to any pile, it is not possible to end up
with only one token.
Although I already knew what invariants are, I do not understand how they're applied in this specific case, what is the meaning of the pairs that are assigned to each pile, and how is a pair affected by a step?
combinatorics invariance
asked Jan 19 at 14:10
Spasoje DurovicSpasoje Durovic
36510
$endgroup$
add a comment |
$begingroup$
The combinatorics textbook I'm reading introduces invariants with the following example:
There are three piles with $n$ tokens each. In every step we are allowed
to choose two piles, take one token from each of those two piles and add a token to
the third pile. Using these moves, is it possible to end up having only one token?
Solution: To the tokens in the first pile we can assign the pair $(0, 1)$, to the tokens
in the second pile the pair $(1, 0)$ and to the tokens in the third pile the pair $(1, 1)$.
Notice that the sum modulo $2$ of any two of these pairs give us the third one. Thus,
in every step the sum modulo $2$ of all the assigned pairs is the same. However, the
sum of all the assigned pairs in the beginning is $(2n, 2n)$, which is equal to $(0, 0)$
modulo $2$. Since this pair was not assigned to any pile, it is not possible to end up
with only one token.
Although I already knew what invariants are, I do not understand how they're applied in this specific case, what is the meaning of the pairs that are assigned to each pile, and how is a pair affected by a step?
combinatorics invariance
asked Jan 19 at 14:10
Spasoje DurovicSpasoje Durovic
36510
$endgroup$
The combinatorics textbook I'm reading introduces invariants with the following example:
There are three piles with $n$ tokens each. In every step we are allowed
to choose two piles, take one token from each of those two piles and add a token to
the third pile. Using these moves, is it possible to end up having only one token?
Solution: To the tokens in the first pile we can assign the pair $(0, 1)$, to the tokens
in the second pile the pair $(1, 0)$ and to the tokens in the third pile the pair $(1, 1)$.
Notice that the sum modulo $2$ of any two of these pairs give us the third one. Thus,
in every step the sum modulo $2$ of all the assigned pairs is the same. However, the
sum of all the assigned pairs in the beginning is $(2n, 2n)$, which is equal to $(0, 0)$
modulo $2$. Since this pair was not assigned to any pile, it is not possible to end up
with only one token.
Although I already knew what invariants are, I do not understand how they're applied in this specific case, what is the meaning of the pairs that are assigned to each pile, and how is a pair affected by a step?
combinatorics invariance
combinatorics invariance
asked Jan 19 at 14:10
Spasoje DurovicSpasoje Durovic
36510
asked Jan 19 at 14:10
Spasoje DurovicSpasoje Durovic
36510
asked Jan 19 at 14:10
Spasoje DurovicSpasoje Durovic
36510
asked Jan 19 at 14:10
Spasoje DurovicSpasoje Durovic
36510
asked Jan 19 at 14:10
Spasoje DurovicSpasoje Durovic
36510
36510
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The solution can be simplified. The invariant is the condition that all three piles have the same parity of tokens. The initial state is all three piles have $,n,$ tokens which implies the invariant condition is true. In each step, the parity of each pile is changed from even to odd or odd to even, and therefore if the invariant condition is true before the step, then it is true after the step. By induction, the invariant condition always is true. Thus, if there is only one token left, the three parities are even, even, and odd and the invariant condition is false.
answered Jan 19 at 14:37
SomosSomos
13.9k11235
$endgroup$
add a comment |
$begingroup$
Let $a,b,c$ be the number of tokens in the first, second and third pile. Observe that in each step, the parity of $a+b$, $a+c$ are preserved after the operation. Let us make this observation formal. We have $2$ dimensional data $$(a+b,a+c)=a(1,1)+b(1,0)+c(0,1).$$ Since we are tracking its parity, we have this vector in $Bbb Z_2times Bbb Z_2$, i.e. in modulo $2$. This gives us the introduced invariant
$$
I=a(1,1)+b(1,0)+c(0,1)quad(text{mod }2).
$$
answered Jan 19 at 14:26
SongSong
13.6k633
$endgroup$
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: "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');
});
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%2f3079386%2fsimple-example-of-invariants%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
$begingroup$
The solution can be simplified. The invariant is the condition that all three piles have the same parity of tokens. The initial state is all three piles have $,n,$ tokens which implies the invariant condition is true. In each step, the parity of each pile is changed from even to odd or odd to even, and therefore if the invariant condition is true before the step, then it is true after the step. By induction, the invariant condition always is true. Thus, if there is only one token left, the three parities are even, even, and odd and the invariant condition is false.
answered Jan 19 at 14:37
SomosSomos
13.9k11235
$endgroup$
add a comment |
$begingroup$
The solution can be simplified. The invariant is the condition that all three piles have the same parity of tokens. The initial state is all three piles have $,n,$ tokens which implies the invariant condition is true. In each step, the parity of each pile is changed from even to odd or odd to even, and therefore if the invariant condition is true before the step, then it is true after the step. By induction, the invariant condition always is true. Thus, if there is only one token left, the three parities are even, even, and odd and the invariant condition is false.
answered Jan 19 at 14:37
SomosSomos
13.9k11235
$endgroup$
add a comment |
$begingroup$
The solution can be simplified. The invariant is the condition that all three piles have the same parity of tokens. The initial state is all three piles have $,n,$ tokens which implies the invariant condition is true. In each step, the parity of each pile is changed from even to odd or odd to even, and therefore if the invariant condition is true before the step, then it is true after the step. By induction, the invariant condition always is true. Thus, if there is only one token left, the three parities are even, even, and odd and the invariant condition is false.
answered Jan 19 at 14:37
SomosSomos
13.9k11235
$endgroup$
The solution can be simplified. The invariant is the condition that all three piles have the same parity of tokens. The initial state is all three piles have $,n,$ tokens which implies the invariant condition is true. In each step, the parity of each pile is changed from even to odd or odd to even, and therefore if the invariant condition is true before the step, then it is true after the step. By induction, the invariant condition always is true. Thus, if there is only one token left, the three parities are even, even, and odd and the invariant condition is false.
answered Jan 19 at 14:37
SomosSomos
13.9k11235
answered Jan 19 at 14:37
SomosSomos
13.9k11235
answered Jan 19 at 14:37
SomosSomos
13.9k11235
answered Jan 19 at 14:37
SomosSomos
13.9k11235
13.9k11235
add a comment |
add a comment |
$begingroup$
Let $a,b,c$ be the number of tokens in the first, second and third pile. Observe that in each step, the parity of $a+b$, $a+c$ are preserved after the operation. Let us make this observation formal. We have $2$ dimensional data $$(a+b,a+c)=a(1,1)+b(1,0)+c(0,1).$$ Since we are tracking its parity, we have this vector in $Bbb Z_2times Bbb Z_2$, i.e. in modulo $2$. This gives us the introduced invariant
$$
I=a(1,1)+b(1,0)+c(0,1)quad(text{mod }2).
$$
answered Jan 19 at 14:26
SongSong
13.6k633
$endgroup$
add a comment |
$begingroup$
Let $a,b,c$ be the number of tokens in the first, second and third pile. Observe that in each step, the parity of $a+b$, $a+c$ are preserved after the operation. Let us make this observation formal. We have $2$ dimensional data $$(a+b,a+c)=a(1,1)+b(1,0)+c(0,1).$$ Since we are tracking its parity, we have this vector in $Bbb Z_2times Bbb Z_2$, i.e. in modulo $2$. This gives us the introduced invariant
$$
I=a(1,1)+b(1,0)+c(0,1)quad(text{mod }2).
$$
answered Jan 19 at 14:26
SongSong
13.6k633
$endgroup$
add a comment |
$begingroup$
Let $a,b,c$ be the number of tokens in the first, second and third pile. Observe that in each step, the parity of $a+b$, $a+c$ are preserved after the operation. Let us make this observation formal. We have $2$ dimensional data $$(a+b,a+c)=a(1,1)+b(1,0)+c(0,1).$$ Since we are tracking its parity, we have this vector in $Bbb Z_2times Bbb Z_2$, i.e. in modulo $2$. This gives us the introduced invariant
$$
I=a(1,1)+b(1,0)+c(0,1)quad(text{mod }2).
$$
answered Jan 19 at 14:26
SongSong
13.6k633
$endgroup$
Let $a,b,c$ be the number of tokens in the first, second and third pile. Observe that in each step, the parity of $a+b$, $a+c$ are preserved after the operation. Let us make this observation formal. We have $2$ dimensional data $$(a+b,a+c)=a(1,1)+b(1,0)+c(0,1).$$ Since we are tracking its parity, we have this vector in $Bbb Z_2times Bbb Z_2$, i.e. in modulo $2$. This gives us the introduced invariant
$$
I=a(1,1)+b(1,0)+c(0,1)quad(text{mod }2).
$$
answered Jan 19 at 14:26
SongSong
13.6k633
answered Jan 19 at 14:26
SongSong
13.6k633
answered Jan 19 at 14:26
SongSong
13.6k633
answered Jan 19 at 14:26
SongSong
13.6k633
13.6k633
add a comment |
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');
});
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%2f3079386%2fsimple-example-of-invariants%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');
});
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');
});
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');
});
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
