Mixing
100 dot product riddle
$begingroup$
I was given this riddle by one of my professors, and was wondering if anyone could give some hints on this problem. Say I have 100 vectors, $x_1, x_2, ... x_{100}$. I compute every dot product pair, excluding self-pairs, so a vector is not dotted with itself, i.e $x_1 cdot x_2$, $x_1 cdot x_3$, ... $x_1 cdot x_{100}$, $x_2 cdot x_3$, $x_2 cdot x_4$, ... $x_2 cdot x_{100}$ ... $x_{99} cdot x_{100}$, and tabulate the result in a list $L_1$. I then create another list $L_2 = -L_1$, $L_2$ is just $L_1$ with the signs flipped. If I only gave you the two lists, could you tell which one is $L_1$ and which one is $L_2$, i.e which list is the true dot product pairs, and which list had it's sign flipped?
vectors
asked Jan 22 at 6:31
Shovik GuhaShovik Guha
433
$endgroup$
|
show 3 more comments
$begingroup$
I was given this riddle by one of my professors, and was wondering if anyone could give some hints on this problem. Say I have 100 vectors, $x_1, x_2, ... x_{100}$. I compute every dot product pair, excluding self-pairs, so a vector is not dotted with itself, i.e $x_1 cdot x_2$, $x_1 cdot x_3$, ... $x_1 cdot x_{100}$, $x_2 cdot x_3$, $x_2 cdot x_4$, ... $x_2 cdot x_{100}$ ... $x_{99} cdot x_{100}$, and tabulate the result in a list $L_1$. I then create another list $L_2 = -L_1$, $L_2$ is just $L_1$ with the signs flipped. If I only gave you the two lists, could you tell which one is $L_1$ and which one is $L_2$, i.e which list is the true dot product pairs, and which list had it's sign flipped?
vectors
asked Jan 22 at 6:31
Shovik GuhaShovik Guha
433
$endgroup$
$begingroup$
With two vectors this is clearly not possible. What about three or four? Try those, and see if you can't spot some pattern to use.
$endgroup$
– Arthur
Jan 22 at 6:50
2
$begingroup$
Is there any other information related to the vectors? Do you know anything about their values, dimension etc?
$endgroup$
– Matti P.
Jan 22 at 6:51
$begingroup$
No information is given about the vectors other than the fact they are all in the same dimension
$endgroup$
– Shovik Guha
Jan 22 at 6:57
2
$begingroup$
Obviously, if they are all orthogonal to each other, one couldn't tell.
$endgroup$
– Zeekless
Jan 22 at 7:09
1
$begingroup$
Zeekless: Of course, I was very vague in my comment. (Because I didn't think too much about the problem), but I only claimed "many" not "most". For example, in $mathbb{R}$, if you select $k geq 2$ (with $k$ even) vectors, there must be at least $frac{1}{2}( k^2/2 - k)$ positive products. (So you example is actually worst case scenario). There are $k^2/2 - k/2$ total distinct products, so, asymptotically, roughly half must be positive.
$endgroup$
– Jason DeVito
Jan 23 at 22:04
|
show 3 more comments
$begingroup$
I was given this riddle by one of my professors, and was wondering if anyone could give some hints on this problem. Say I have 100 vectors, $x_1, x_2, ... x_{100}$. I compute every dot product pair, excluding self-pairs, so a vector is not dotted with itself, i.e $x_1 cdot x_2$, $x_1 cdot x_3$, ... $x_1 cdot x_{100}$, $x_2 cdot x_3$, $x_2 cdot x_4$, ... $x_2 cdot x_{100}$ ... $x_{99} cdot x_{100}$, and tabulate the result in a list $L_1$. I then create another list $L_2 = -L_1$, $L_2$ is just $L_1$ with the signs flipped. If I only gave you the two lists, could you tell which one is $L_1$ and which one is $L_2$, i.e which list is the true dot product pairs, and which list had it's sign flipped?
vectors
asked Jan 22 at 6:31
Shovik GuhaShovik Guha
433
$endgroup$
I was given this riddle by one of my professors, and was wondering if anyone could give some hints on this problem. Say I have 100 vectors, $x_1, x_2, ... x_{100}$. I compute every dot product pair, excluding self-pairs, so a vector is not dotted with itself, i.e $x_1 cdot x_2$, $x_1 cdot x_3$, ... $x_1 cdot x_{100}$, $x_2 cdot x_3$, $x_2 cdot x_4$, ... $x_2 cdot x_{100}$ ... $x_{99} cdot x_{100}$, and tabulate the result in a list $L_1$. I then create another list $L_2 = -L_1$, $L_2$ is just $L_1$ with the signs flipped. If I only gave you the two lists, could you tell which one is $L_1$ and which one is $L_2$, i.e which list is the true dot product pairs, and which list had it's sign flipped?
vectors
vectors
asked Jan 22 at 6:31
Shovik GuhaShovik Guha
433
asked Jan 22 at 6:31
Shovik GuhaShovik Guha
433
asked Jan 22 at 6:31
Shovik GuhaShovik Guha
433
asked Jan 22 at 6:31
Shovik GuhaShovik Guha
433
asked Jan 22 at 6:31
Shovik GuhaShovik Guha
433
433
$begingroup$
With two vectors this is clearly not possible. What about three or four? Try those, and see if you can't spot some pattern to use.
$endgroup$
– Arthur
Jan 22 at 6:50
2
$begingroup$
Is there any other information related to the vectors? Do you know anything about their values, dimension etc?
$endgroup$
– Matti P.
Jan 22 at 6:51
$begingroup$
No information is given about the vectors other than the fact they are all in the same dimension
$endgroup$
– Shovik Guha
Jan 22 at 6:57
2
$begingroup$
Obviously, if they are all orthogonal to each other, one couldn't tell.
$endgroup$
– Zeekless
Jan 22 at 7:09
1
$begingroup$
Zeekless: Of course, I was very vague in my comment. (Because I didn't think too much about the problem), but I only claimed "many" not "most". For example, in $mathbb{R}$, if you select $k geq 2$ (with $k$ even) vectors, there must be at least $frac{1}{2}( k^2/2 - k)$ positive products. (So you example is actually worst case scenario). There are $k^2/2 - k/2$ total distinct products, so, asymptotically, roughly half must be positive.
$endgroup$
– Jason DeVito
Jan 23 at 22:04
|
show 3 more comments
$begingroup$
With two vectors this is clearly not possible. What about three or four? Try those, and see if you can't spot some pattern to use.
$endgroup$
– Arthur
Jan 22 at 6:50
2
$begingroup$
Is there any other information related to the vectors? Do you know anything about their values, dimension etc?
$endgroup$
– Matti P.
Jan 22 at 6:51
$begingroup$
No information is given about the vectors other than the fact they are all in the same dimension
$endgroup$
– Shovik Guha
Jan 22 at 6:57
2
$begingroup$
Obviously, if they are all orthogonal to each other, one couldn't tell.
$endgroup$
– Zeekless
Jan 22 at 7:09
1
$begingroup$
Zeekless: Of course, I was very vague in my comment. (Because I didn't think too much about the problem), but I only claimed "many" not "most". For example, in $mathbb{R}$, if you select $k geq 2$ (with $k$ even) vectors, there must be at least $frac{1}{2}( k^2/2 - k)$ positive products. (So you example is actually worst case scenario). There are $k^2/2 - k/2$ total distinct products, so, asymptotically, roughly half must be positive.
$endgroup$
– Jason DeVito
Jan 23 at 22:04
$begingroup$
With two vectors this is clearly not possible. What about three or four? Try those, and see if you can't spot some pattern to use.
$endgroup$
– Arthur
Jan 22 at 6:50
$begingroup$
With two vectors this is clearly not possible. What about three or four? Try those, and see if you can't spot some pattern to use.
$endgroup$
– Arthur
Jan 22 at 6:50
2
2
$begingroup$
Is there any other information related to the vectors? Do you know anything about their values, dimension etc?
$endgroup$
– Matti P.
Jan 22 at 6:51
$begingroup$
Is there any other information related to the vectors? Do you know anything about their values, dimension etc?
$endgroup$
– Matti P.
Jan 22 at 6:51
$begingroup$
No information is given about the vectors other than the fact they are all in the same dimension
$endgroup$
– Shovik Guha
Jan 22 at 6:57
$begingroup$
No information is given about the vectors other than the fact they are all in the same dimension
$endgroup$
– Shovik Guha
Jan 22 at 6:57
2
2
$begingroup$
Obviously, if they are all orthogonal to each other, one couldn't tell.
$endgroup$
– Zeekless
Jan 22 at 7:09
$begingroup$
Obviously, if they are all orthogonal to each other, one couldn't tell.
$endgroup$
– Zeekless
Jan 22 at 7:09
1
1
$begingroup$
Zeekless: Of course, I was very vague in my comment. (Because I didn't think too much about the problem), but I only claimed "many" not "most". For example, in $mathbb{R}$, if you select $k geq 2$ (with $k$ even) vectors, there must be at least $frac{1}{2}( k^2/2 - k)$ positive products. (So you example is actually worst case scenario). There are $k^2/2 - k/2$ total distinct products, so, asymptotically, roughly half must be positive.
$endgroup$
– Jason DeVito
Jan 23 at 22:04
$begingroup$
Zeekless: Of course, I was very vague in my comment. (Because I didn't think too much about the problem), but I only claimed "many" not "most". For example, in $mathbb{R}$, if you select $k geq 2$ (with $k$ even) vectors, there must be at least $frac{1}{2}( k^2/2 - k)$ positive products. (So you example is actually worst case scenario). There are $k^2/2 - k/2$ total distinct products, so, asymptotically, roughly half must be positive.
$endgroup$
– Jason DeVito
Jan 23 at 22:04
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Consider a set of vectors $x_kin mathbb{R}^{100}:$
$$
Big{ x_1 = begin{bmatrix} 1 \ 0 \ 0 \ vdots \ 0 \0 end{bmatrix}, ; x_2 = begin{bmatrix} 1 \ 1 \ 0 \ vdots \ 0 \0end{bmatrix}; x_3 = begin{bmatrix} 1 \ 1 \ 1 \ vdots \ 0\0 end{bmatrix}, ; dots,; x_{99} = begin{bmatrix} 1 \ 1 \ 1 \ vdots\1 \ 0end{bmatrix},;x_{100} = begin{bmatrix} 1 \ 1 \ 1 \ vdots \ 1\1end{bmatrix}Big}
$$
Then $L_1=big{(1)_{times 99},(2)_{times 98},dots,(99)_{times 1}big}, ; L_2=big{(-1)_{times 99},(-2)_{times 98},dots,(-99)_{times 1}big}$.
Now consider a set of vectors $x_kin mathbb{R}^{100}:$
$$
Big{ x_1 = begin{bmatrix} -1 \ 0 \ 0 \ vdots \ 0 \0 end{bmatrix}, ; x_2 = begin{bmatrix} 1 \ -3 \ 0 \ vdots \ 0 \0end{bmatrix}; x_3 = begin{bmatrix} 1 \ 1 \ -5 \ vdots \ 0\0 end{bmatrix}, ; dots,; x_{99} = begin{bmatrix} 1 \ 1 \ 1 \ vdots\-197 \ 0end{bmatrix},;x_{100} = begin{bmatrix} 1 \ 1 \ 1 \ vdots \ 1\1end{bmatrix}Big}
$$
Then $L_1=big{(-1)_{times 99},(-2)_{times 98},dots,(-99)_{times 1}big},; L_2=big{(1)_{times 99},(2)_{times 98},dots,(99)_{times 1}big}$.
answered Jan 22 at 7:35
ZeeklessZeekless
577111
$endgroup$
$begingroup$
This is clever, but I'm not quite sure how this answers the question. In the question, we are only given the 2 lists, $L_1$ and $L_2$, and from there we have to derive which list is flipped and which is the true dot products. We don't get to choose the set of vectors.
$endgroup$
– Shovik Guha
Jan 22 at 16:26
$begingroup$
@Shovik Guha If you get two lists: $big{(-1)_{times 99},(-2)_{times 98},dots,(-99)_{times 1}big}$ and $big{(1)_{times 99},(2)_{times 98},dots,(99)_{times 1}big}$ you won't be able to decide which one is true and which one is inverted. Each one of these lists is the true one - either for the first or the second set of vectors. This works as a counterexample and thus answers the question: "could you tell which one is which?". The answer is: "No, for this pair of lists, I can't. Both could be $L_1$ and both could be $L_2$ depending on the initial set of vectors".
$endgroup$
– Zeekless
Jan 22 at 17:18
1
$begingroup$
Ah, I see what you are saying now!
$endgroup$
– Shovik Guha
Jan 22 at 17:50
add a comment |
$begingroup$
First, let's consider all the possible answers this question could have:
- It it always possible t otell which list is which
- It is always impossible to tell which list is which
- It is sometimes possible and sometimes impossible to tell, depending on the list.
Since both the dimension and amount of vectors can be arbitrary, we can consider 3 1-dimensional vectors resulting in lists [1,1,1] and [-1,-1,-1]. It is clear that [1,1,1] is the correct one, since you can never have 3 real numbers so that the product of any 2 of them is negative.
On the other hand, consider 2 1-dimensional vectors with lists [1,1] and [-1,-1]. Both are possible from vectors (1), (1) and (1), (-1).
Therefore, the answer 3) is correct. Sometimes it is possible to tell, sometimes it isn't. The fun begins when you try to investigate when it is possible to tell...
answered Jan 22 at 12:34
kajacxkajacx
1233
$endgroup$
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%2f3082813%2f100-dot-product-riddle%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
$begingroup$
Consider a set of vectors $x_kin mathbb{R}^{100}:$
$$
Big{ x_1 = begin{bmatrix} 1 \ 0 \ 0 \ vdots \ 0 \0 end{bmatrix}, ; x_2 = begin{bmatrix} 1 \ 1 \ 0 \ vdots \ 0 \0end{bmatrix}; x_3 = begin{bmatrix} 1 \ 1 \ 1 \ vdots \ 0\0 end{bmatrix}, ; dots,; x_{99} = begin{bmatrix} 1 \ 1 \ 1 \ vdots\1 \ 0end{bmatrix},;x_{100} = begin{bmatrix} 1 \ 1 \ 1 \ vdots \ 1\1end{bmatrix}Big}
$$
Then $L_1=big{(1)_{times 99},(2)_{times 98},dots,(99)_{times 1}big}, ; L_2=big{(-1)_{times 99},(-2)_{times 98},dots,(-99)_{times 1}big}$.
Now consider a set of vectors $x_kin mathbb{R}^{100}:$
$$
Big{ x_1 = begin{bmatrix} -1 \ 0 \ 0 \ vdots \ 0 \0 end{bmatrix}, ; x_2 = begin{bmatrix} 1 \ -3 \ 0 \ vdots \ 0 \0end{bmatrix}; x_3 = begin{bmatrix} 1 \ 1 \ -5 \ vdots \ 0\0 end{bmatrix}, ; dots,; x_{99} = begin{bmatrix} 1 \ 1 \ 1 \ vdots\-197 \ 0end{bmatrix},;x_{100} = begin{bmatrix} 1 \ 1 \ 1 \ vdots \ 1\1end{bmatrix}Big}
$$
Then $L_1=big{(-1)_{times 99},(-2)_{times 98},dots,(-99)_{times 1}big},; L_2=big{(1)_{times 99},(2)_{times 98},dots,(99)_{times 1}big}$.
answered Jan 22 at 7:35
ZeeklessZeekless
577111
$endgroup$
$begingroup$
This is clever, but I'm not quite sure how this answers the question. In the question, we are only given the 2 lists, $L_1$ and $L_2$, and from there we have to derive which list is flipped and which is the true dot products. We don't get to choose the set of vectors.
$endgroup$
– Shovik Guha
Jan 22 at 16:26
$begingroup$
@Shovik Guha If you get two lists: $big{(-1)_{times 99},(-2)_{times 98},dots,(-99)_{times 1}big}$ and $big{(1)_{times 99},(2)_{times 98},dots,(99)_{times 1}big}$ you won't be able to decide which one is true and which one is inverted. Each one of these lists is the true one - either for the first or the second set of vectors. This works as a counterexample and thus answers the question: "could you tell which one is which?". The answer is: "No, for this pair of lists, I can't. Both could be $L_1$ and both could be $L_2$ depending on the initial set of vectors".
$endgroup$
– Zeekless
Jan 22 at 17:18
1
$begingroup$
Ah, I see what you are saying now!
$endgroup$
– Shovik Guha
Jan 22 at 17:50
add a comment |
$begingroup$
Consider a set of vectors $x_kin mathbb{R}^{100}:$
$$
Big{ x_1 = begin{bmatrix} 1 \ 0 \ 0 \ vdots \ 0 \0 end{bmatrix}, ; x_2 = begin{bmatrix} 1 \ 1 \ 0 \ vdots \ 0 \0end{bmatrix}; x_3 = begin{bmatrix} 1 \ 1 \ 1 \ vdots \ 0\0 end{bmatrix}, ; dots,; x_{99} = begin{bmatrix} 1 \ 1 \ 1 \ vdots\1 \ 0end{bmatrix},;x_{100} = begin{bmatrix} 1 \ 1 \ 1 \ vdots \ 1\1end{bmatrix}Big}
$$
Then $L_1=big{(1)_{times 99},(2)_{times 98},dots,(99)_{times 1}big}, ; L_2=big{(-1)_{times 99},(-2)_{times 98},dots,(-99)_{times 1}big}$.
Now consider a set of vectors $x_kin mathbb{R}^{100}:$
$$
Big{ x_1 = begin{bmatrix} -1 \ 0 \ 0 \ vdots \ 0 \0 end{bmatrix}, ; x_2 = begin{bmatrix} 1 \ -3 \ 0 \ vdots \ 0 \0end{bmatrix}; x_3 = begin{bmatrix} 1 \ 1 \ -5 \ vdots \ 0\0 end{bmatrix}, ; dots,; x_{99} = begin{bmatrix} 1 \ 1 \ 1 \ vdots\-197 \ 0end{bmatrix},;x_{100} = begin{bmatrix} 1 \ 1 \ 1 \ vdots \ 1\1end{bmatrix}Big}
$$
Then $L_1=big{(-1)_{times 99},(-2)_{times 98},dots,(-99)_{times 1}big},; L_2=big{(1)_{times 99},(2)_{times 98},dots,(99)_{times 1}big}$.
answered Jan 22 at 7:35
ZeeklessZeekless
577111
$endgroup$
$begingroup$
This is clever, but I'm not quite sure how this answers the question. In the question, we are only given the 2 lists, $L_1$ and $L_2$, and from there we have to derive which list is flipped and which is the true dot products. We don't get to choose the set of vectors.
$endgroup$
– Shovik Guha
Jan 22 at 16:26
$begingroup$
@Shovik Guha If you get two lists: $big{(-1)_{times 99},(-2)_{times 98},dots,(-99)_{times 1}big}$ and $big{(1)_{times 99},(2)_{times 98},dots,(99)_{times 1}big}$ you won't be able to decide which one is true and which one is inverted. Each one of these lists is the true one - either for the first or the second set of vectors. This works as a counterexample and thus answers the question: "could you tell which one is which?". The answer is: "No, for this pair of lists, I can't. Both could be $L_1$ and both could be $L_2$ depending on the initial set of vectors".
$endgroup$
– Zeekless
Jan 22 at 17:18
1
$begingroup$
Ah, I see what you are saying now!
$endgroup$
– Shovik Guha
Jan 22 at 17:50
add a comment |
$begingroup$
Consider a set of vectors $x_kin mathbb{R}^{100}:$
$$
Big{ x_1 = begin{bmatrix} 1 \ 0 \ 0 \ vdots \ 0 \0 end{bmatrix}, ; x_2 = begin{bmatrix} 1 \ 1 \ 0 \ vdots \ 0 \0end{bmatrix}; x_3 = begin{bmatrix} 1 \ 1 \ 1 \ vdots \ 0\0 end{bmatrix}, ; dots,; x_{99} = begin{bmatrix} 1 \ 1 \ 1 \ vdots\1 \ 0end{bmatrix},;x_{100} = begin{bmatrix} 1 \ 1 \ 1 \ vdots \ 1\1end{bmatrix}Big}
$$
Then $L_1=big{(1)_{times 99},(2)_{times 98},dots,(99)_{times 1}big}, ; L_2=big{(-1)_{times 99},(-2)_{times 98},dots,(-99)_{times 1}big}$.
Now consider a set of vectors $x_kin mathbb{R}^{100}:$
$$
Big{ x_1 = begin{bmatrix} -1 \ 0 \ 0 \ vdots \ 0 \0 end{bmatrix}, ; x_2 = begin{bmatrix} 1 \ -3 \ 0 \ vdots \ 0 \0end{bmatrix}; x_3 = begin{bmatrix} 1 \ 1 \ -5 \ vdots \ 0\0 end{bmatrix}, ; dots,; x_{99} = begin{bmatrix} 1 \ 1 \ 1 \ vdots\-197 \ 0end{bmatrix},;x_{100} = begin{bmatrix} 1 \ 1 \ 1 \ vdots \ 1\1end{bmatrix}Big}
$$
Then $L_1=big{(-1)_{times 99},(-2)_{times 98},dots,(-99)_{times 1}big},; L_2=big{(1)_{times 99},(2)_{times 98},dots,(99)_{times 1}big}$.
answered Jan 22 at 7:35
ZeeklessZeekless
577111
$endgroup$
Consider a set of vectors $x_kin mathbb{R}^{100}:$
$$
Big{ x_1 = begin{bmatrix} 1 \ 0 \ 0 \ vdots \ 0 \0 end{bmatrix}, ; x_2 = begin{bmatrix} 1 \ 1 \ 0 \ vdots \ 0 \0end{bmatrix}; x_3 = begin{bmatrix} 1 \ 1 \ 1 \ vdots \ 0\0 end{bmatrix}, ; dots,; x_{99} = begin{bmatrix} 1 \ 1 \ 1 \ vdots\1 \ 0end{bmatrix},;x_{100} = begin{bmatrix} 1 \ 1 \ 1 \ vdots \ 1\1end{bmatrix}Big}
$$
Then $L_1=big{(1)_{times 99},(2)_{times 98},dots,(99)_{times 1}big}, ; L_2=big{(-1)_{times 99},(-2)_{times 98},dots,(-99)_{times 1}big}$.
Now consider a set of vectors $x_kin mathbb{R}^{100}:$
$$
Big{ x_1 = begin{bmatrix} -1 \ 0 \ 0 \ vdots \ 0 \0 end{bmatrix}, ; x_2 = begin{bmatrix} 1 \ -3 \ 0 \ vdots \ 0 \0end{bmatrix}; x_3 = begin{bmatrix} 1 \ 1 \ -5 \ vdots \ 0\0 end{bmatrix}, ; dots,; x_{99} = begin{bmatrix} 1 \ 1 \ 1 \ vdots\-197 \ 0end{bmatrix},;x_{100} = begin{bmatrix} 1 \ 1 \ 1 \ vdots \ 1\1end{bmatrix}Big}
$$
Then $L_1=big{(-1)_{times 99},(-2)_{times 98},dots,(-99)_{times 1}big},; L_2=big{(1)_{times 99},(2)_{times 98},dots,(99)_{times 1}big}$.
answered Jan 22 at 7:35
ZeeklessZeekless
577111
answered Jan 22 at 7:35
ZeeklessZeekless
577111
answered Jan 22 at 7:35
ZeeklessZeekless
577111
answered Jan 22 at 7:35
ZeeklessZeekless
577111
577111
$begingroup$
This is clever, but I'm not quite sure how this answers the question. In the question, we are only given the 2 lists, $L_1$ and $L_2$, and from there we have to derive which list is flipped and which is the true dot products. We don't get to choose the set of vectors.
$endgroup$
– Shovik Guha
Jan 22 at 16:26
$begingroup$
@Shovik Guha If you get two lists: $big{(-1)_{times 99},(-2)_{times 98},dots,(-99)_{times 1}big}$ and $big{(1)_{times 99},(2)_{times 98},dots,(99)_{times 1}big}$ you won't be able to decide which one is true and which one is inverted. Each one of these lists is the true one - either for the first or the second set of vectors. This works as a counterexample and thus answers the question: "could you tell which one is which?". The answer is: "No, for this pair of lists, I can't. Both could be $L_1$ and both could be $L_2$ depending on the initial set of vectors".
$endgroup$
– Zeekless
Jan 22 at 17:18
1
$begingroup$
Ah, I see what you are saying now!
$endgroup$
– Shovik Guha
Jan 22 at 17:50
add a comment |
$begingroup$
This is clever, but I'm not quite sure how this answers the question. In the question, we are only given the 2 lists, $L_1$ and $L_2$, and from there we have to derive which list is flipped and which is the true dot products. We don't get to choose the set of vectors.
$endgroup$
– Shovik Guha
Jan 22 at 16:26
$begingroup$
@Shovik Guha If you get two lists: $big{(-1)_{times 99},(-2)_{times 98},dots,(-99)_{times 1}big}$ and $big{(1)_{times 99},(2)_{times 98},dots,(99)_{times 1}big}$ you won't be able to decide which one is true and which one is inverted. Each one of these lists is the true one - either for the first or the second set of vectors. This works as a counterexample and thus answers the question: "could you tell which one is which?". The answer is: "No, for this pair of lists, I can't. Both could be $L_1$ and both could be $L_2$ depending on the initial set of vectors".
$endgroup$
– Zeekless
Jan 22 at 17:18
1
$begingroup$
Ah, I see what you are saying now!
$endgroup$
– Shovik Guha
Jan 22 at 17:50
$begingroup$
This is clever, but I'm not quite sure how this answers the question. In the question, we are only given the 2 lists, $L_1$ and $L_2$, and from there we have to derive which list is flipped and which is the true dot products. We don't get to choose the set of vectors.
$endgroup$
– Shovik Guha
Jan 22 at 16:26
$begingroup$
This is clever, but I'm not quite sure how this answers the question. In the question, we are only given the 2 lists, $L_1$ and $L_2$, and from there we have to derive which list is flipped and which is the true dot products. We don't get to choose the set of vectors.
$endgroup$
– Shovik Guha
Jan 22 at 16:26
$begingroup$
@Shovik Guha If you get two lists: $big{(-1)_{times 99},(-2)_{times 98},dots,(-99)_{times 1}big}$ and $big{(1)_{times 99},(2)_{times 98},dots,(99)_{times 1}big}$ you won't be able to decide which one is true and which one is inverted. Each one of these lists is the true one - either for the first or the second set of vectors. This works as a counterexample and thus answers the question: "could you tell which one is which?". The answer is: "No, for this pair of lists, I can't. Both could be $L_1$ and both could be $L_2$ depending on the initial set of vectors".
$endgroup$
– Zeekless
Jan 22 at 17:18
$begingroup$
@Shovik Guha If you get two lists: $big{(-1)_{times 99},(-2)_{times 98},dots,(-99)_{times 1}big}$ and $big{(1)_{times 99},(2)_{times 98},dots,(99)_{times 1}big}$ you won't be able to decide which one is true and which one is inverted. Each one of these lists is the true one - either for the first or the second set of vectors. This works as a counterexample and thus answers the question: "could you tell which one is which?". The answer is: "No, for this pair of lists, I can't. Both could be $L_1$ and both could be $L_2$ depending on the initial set of vectors".
$endgroup$
– Zeekless
Jan 22 at 17:18
1
1
$begingroup$
Ah, I see what you are saying now!
$endgroup$
– Shovik Guha
Jan 22 at 17:50
$begingroup$
Ah, I see what you are saying now!
$endgroup$
– Shovik Guha
Jan 22 at 17:50
add a comment |
$begingroup$
First, let's consider all the possible answers this question could have:
- It it always possible t otell which list is which
- It is always impossible to tell which list is which
- It is sometimes possible and sometimes impossible to tell, depending on the list.
Since both the dimension and amount of vectors can be arbitrary, we can consider 3 1-dimensional vectors resulting in lists [1,1,1] and [-1,-1,-1]. It is clear that [1,1,1] is the correct one, since you can never have 3 real numbers so that the product of any 2 of them is negative.
On the other hand, consider 2 1-dimensional vectors with lists [1,1] and [-1,-1]. Both are possible from vectors (1), (1) and (1), (-1).
Therefore, the answer 3) is correct. Sometimes it is possible to tell, sometimes it isn't. The fun begins when you try to investigate when it is possible to tell...
answered Jan 22 at 12:34
kajacxkajacx
1233
$endgroup$
add a comment |
$begingroup$
First, let's consider all the possible answers this question could have:
- It it always possible t otell which list is which
- It is always impossible to tell which list is which
- It is sometimes possible and sometimes impossible to tell, depending on the list.
Since both the dimension and amount of vectors can be arbitrary, we can consider 3 1-dimensional vectors resulting in lists [1,1,1] and [-1,-1,-1]. It is clear that [1,1,1] is the correct one, since you can never have 3 real numbers so that the product of any 2 of them is negative.
On the other hand, consider 2 1-dimensional vectors with lists [1,1] and [-1,-1]. Both are possible from vectors (1), (1) and (1), (-1).
Therefore, the answer 3) is correct. Sometimes it is possible to tell, sometimes it isn't. The fun begins when you try to investigate when it is possible to tell...
answered Jan 22 at 12:34
kajacxkajacx
1233
$endgroup$
add a comment |
$begingroup$
First, let's consider all the possible answers this question could have:
- It it always possible t otell which list is which
- It is always impossible to tell which list is which
- It is sometimes possible and sometimes impossible to tell, depending on the list.
Since both the dimension and amount of vectors can be arbitrary, we can consider 3 1-dimensional vectors resulting in lists [1,1,1] and [-1,-1,-1]. It is clear that [1,1,1] is the correct one, since you can never have 3 real numbers so that the product of any 2 of them is negative.
On the other hand, consider 2 1-dimensional vectors with lists [1,1] and [-1,-1]. Both are possible from vectors (1), (1) and (1), (-1).
Therefore, the answer 3) is correct. Sometimes it is possible to tell, sometimes it isn't. The fun begins when you try to investigate when it is possible to tell...
answered Jan 22 at 12:34
kajacxkajacx
1233
$endgroup$
First, let's consider all the possible answers this question could have:
- It it always possible t otell which list is which
- It is always impossible to tell which list is which
- It is sometimes possible and sometimes impossible to tell, depending on the list.
Since both the dimension and amount of vectors can be arbitrary, we can consider 3 1-dimensional vectors resulting in lists [1,1,1] and [-1,-1,-1]. It is clear that [1,1,1] is the correct one, since you can never have 3 real numbers so that the product of any 2 of them is negative.
On the other hand, consider 2 1-dimensional vectors with lists [1,1] and [-1,-1]. Both are possible from vectors (1), (1) and (1), (-1).
Therefore, the answer 3) is correct. Sometimes it is possible to tell, sometimes it isn't. The fun begins when you try to investigate when it is possible to tell...
answered Jan 22 at 12:34
kajacxkajacx
1233
answered Jan 22 at 12:34
kajacxkajacx
1233
answered Jan 22 at 12:34
kajacxkajacx
1233
answered Jan 22 at 12:34
kajacxkajacx
1233
1233
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%2f3082813%2f100-dot-product-riddle%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
$begingroup$
With two vectors this is clearly not possible. What about three or four? Try those, and see if you can't spot some pattern to use.
$endgroup$
– Arthur
Jan 22 at 6:50
2
$begingroup$
Is there any other information related to the vectors? Do you know anything about their values, dimension etc?
$endgroup$
– Matti P.
Jan 22 at 6:51
$begingroup$
No information is given about the vectors other than the fact they are all in the same dimension
$endgroup$
– Shovik Guha
Jan 22 at 6:57
2
$begingroup$
Obviously, if they are all orthogonal to each other, one couldn't tell.
$endgroup$
– Zeekless
Jan 22 at 7:09
1
$begingroup$
Zeekless: Of course, I was very vague in my comment. (Because I didn't think too much about the problem), but I only claimed "many" not "most". For example, in $mathbb{R}$, if you select $k geq 2$ (with $k$ even) vectors, there must be at least $frac{1}{2}( k^2/2 - k)$ positive products. (So you example is actually worst case scenario). There are $k^2/2 - k/2$ total distinct products, so, asymptotically, roughly half must be positive.
$endgroup$
– Jason DeVito
Jan 23 at 22:04