Mixing
uniform Effect of K-means Clustering
In the following link is discussed the uniform Effect of K-means Clustering:
https://www.springer.com/cda/content/document/cda_downloaddocument/9783642298066-c2.pdf?SGWID=0-0-45-1338325-p174318763
He used the following equation:
$ frac{displaystyle2 d(C_1,C_2)}{displaystyle n_1n_2}=frac{displaystyle d(C_1,C_1)}{displaystyle n_1^2}+frac{displaystyle d(C_2,C_2)}{displaystyle n_2^2}+2||m_1−m_2||^2.$
where
$ d(C_1,C_2)=sum_{x_iin C_1}sum_{x_jin C_2}||x_i−x_j||^2$ with $ |C_1|=n_1$ and $|C_2 |=n_2$.
I tried to prove this, but I do not get this. He uses the fact that
$ d(C_1,C_1)=2(n_1-1)sum_{i=1}^{n_1} ||x_i||^2 -4 sum_{1leq i < j leq n_1}langle x_i,x_jrangle.$
$ d(C_2,C_2)=2(n_2-1)sum_{i=1}^{n_2} ||y_i||^2 -4 sum_{1leq i < j leq n_2}langle y_i,y_jrangle.$
$ d(C_1,C_2)=2n_2sum_{i=1}^{n_1} ||x_i||^2+2n_1sum_{i=1}^{n_2} ||y_i||^2 -4 sum_{1leq ileq n_1}sum_{1leq jleq n_2}langle x_i,y_jrangle.$
And furthermore with
$||m_1−m_2||^2=langle sum_{i=1}^{n_1}x_i/n_1-sum_{j=1}^{n_1}y_j/n_2,sum_{i=1}^{n_1}x_i/n_1-sum_{j=1}^{n_1}y_j/n_2 rangle = frac{1}{displaystyle n_1^2}sum_{i=1}^{n_1} ||x_i||^2 + frac{2}{displaystyle n_1^2}sum_{1leq i < jleq n_1} langle x_i, x_jrangle +frac{1}{displaystyle n_2^2}sum_{i=1}^{n_2} ||y_i||^2 + frac{2}{displaystyle n_2^2}sum_{1leq i < jleq n_2} langle y_i, y_jrangle - frac{2}{n_1n_2} sum_{1leq ileq n_1}sum_{1leq jleq n_2}langle x_i,y_jrangle, $
I obtain that
$ frac{displaystyle d(C_1,C_2)}{displaystyle n_1n_2}=frac{displaystyle d(C_1,C_1)}{displaystyle n_1^2}+frac{displaystyle d(C_2,C_2)}{displaystyle n_2^2}+2||m_1−m_2||^2.$
Where is my error?
analysis vector-analysis machine-learning clustering data-mining
asked Nov 20 '18 at 18:44
Patricio
17212
add a comment |
In the following link is discussed the uniform Effect of K-means Clustering:
https://www.springer.com/cda/content/document/cda_downloaddocument/9783642298066-c2.pdf?SGWID=0-0-45-1338325-p174318763
He used the following equation:
$ frac{displaystyle2 d(C_1,C_2)}{displaystyle n_1n_2}=frac{displaystyle d(C_1,C_1)}{displaystyle n_1^2}+frac{displaystyle d(C_2,C_2)}{displaystyle n_2^2}+2||m_1−m_2||^2.$
where
$ d(C_1,C_2)=sum_{x_iin C_1}sum_{x_jin C_2}||x_i−x_j||^2$ with $ |C_1|=n_1$ and $|C_2 |=n_2$.
I tried to prove this, but I do not get this. He uses the fact that
$ d(C_1,C_1)=2(n_1-1)sum_{i=1}^{n_1} ||x_i||^2 -4 sum_{1leq i < j leq n_1}langle x_i,x_jrangle.$
$ d(C_2,C_2)=2(n_2-1)sum_{i=1}^{n_2} ||y_i||^2 -4 sum_{1leq i < j leq n_2}langle y_i,y_jrangle.$
$ d(C_1,C_2)=2n_2sum_{i=1}^{n_1} ||x_i||^2+2n_1sum_{i=1}^{n_2} ||y_i||^2 -4 sum_{1leq ileq n_1}sum_{1leq jleq n_2}langle x_i,y_jrangle.$
And furthermore with
$||m_1−m_2||^2=langle sum_{i=1}^{n_1}x_i/n_1-sum_{j=1}^{n_1}y_j/n_2,sum_{i=1}^{n_1}x_i/n_1-sum_{j=1}^{n_1}y_j/n_2 rangle = frac{1}{displaystyle n_1^2}sum_{i=1}^{n_1} ||x_i||^2 + frac{2}{displaystyle n_1^2}sum_{1leq i < jleq n_1} langle x_i, x_jrangle +frac{1}{displaystyle n_2^2}sum_{i=1}^{n_2} ||y_i||^2 + frac{2}{displaystyle n_2^2}sum_{1leq i < jleq n_2} langle y_i, y_jrangle - frac{2}{n_1n_2} sum_{1leq ileq n_1}sum_{1leq jleq n_2}langle x_i,y_jrangle, $
I obtain that
$ frac{displaystyle d(C_1,C_2)}{displaystyle n_1n_2}=frac{displaystyle d(C_1,C_1)}{displaystyle n_1^2}+frac{displaystyle d(C_2,C_2)}{displaystyle n_2^2}+2||m_1−m_2||^2.$
Where is my error?
analysis vector-analysis machine-learning clustering data-mining
asked Nov 20 '18 at 18:44
Patricio
17212
I suggest working through the case $n_1=1, n_2=2$ to see where your derivation goes wrong (the paper is definitely correct). Maybe you've forgotten that $||x_i-x_j||^2$ appears twice in each $d(C_i,C_i)$?
– Matthew Towers
Nov 23 '18 at 17:59
add a comment |
In the following link is discussed the uniform Effect of K-means Clustering:
https://www.springer.com/cda/content/document/cda_downloaddocument/9783642298066-c2.pdf?SGWID=0-0-45-1338325-p174318763
He used the following equation:
$ frac{displaystyle2 d(C_1,C_2)}{displaystyle n_1n_2}=frac{displaystyle d(C_1,C_1)}{displaystyle n_1^2}+frac{displaystyle d(C_2,C_2)}{displaystyle n_2^2}+2||m_1−m_2||^2.$
where
$ d(C_1,C_2)=sum_{x_iin C_1}sum_{x_jin C_2}||x_i−x_j||^2$ with $ |C_1|=n_1$ and $|C_2 |=n_2$.
I tried to prove this, but I do not get this. He uses the fact that
$ d(C_1,C_1)=2(n_1-1)sum_{i=1}^{n_1} ||x_i||^2 -4 sum_{1leq i < j leq n_1}langle x_i,x_jrangle.$
$ d(C_2,C_2)=2(n_2-1)sum_{i=1}^{n_2} ||y_i||^2 -4 sum_{1leq i < j leq n_2}langle y_i,y_jrangle.$
$ d(C_1,C_2)=2n_2sum_{i=1}^{n_1} ||x_i||^2+2n_1sum_{i=1}^{n_2} ||y_i||^2 -4 sum_{1leq ileq n_1}sum_{1leq jleq n_2}langle x_i,y_jrangle.$
And furthermore with
$||m_1−m_2||^2=langle sum_{i=1}^{n_1}x_i/n_1-sum_{j=1}^{n_1}y_j/n_2,sum_{i=1}^{n_1}x_i/n_1-sum_{j=1}^{n_1}y_j/n_2 rangle = frac{1}{displaystyle n_1^2}sum_{i=1}^{n_1} ||x_i||^2 + frac{2}{displaystyle n_1^2}sum_{1leq i < jleq n_1} langle x_i, x_jrangle +frac{1}{displaystyle n_2^2}sum_{i=1}^{n_2} ||y_i||^2 + frac{2}{displaystyle n_2^2}sum_{1leq i < jleq n_2} langle y_i, y_jrangle - frac{2}{n_1n_2} sum_{1leq ileq n_1}sum_{1leq jleq n_2}langle x_i,y_jrangle, $
I obtain that
$ frac{displaystyle d(C_1,C_2)}{displaystyle n_1n_2}=frac{displaystyle d(C_1,C_1)}{displaystyle n_1^2}+frac{displaystyle d(C_2,C_2)}{displaystyle n_2^2}+2||m_1−m_2||^2.$
Where is my error?
analysis vector-analysis machine-learning clustering data-mining
asked Nov 20 '18 at 18:44
Patricio
17212
In the following link is discussed the uniform Effect of K-means Clustering:
https://www.springer.com/cda/content/document/cda_downloaddocument/9783642298066-c2.pdf?SGWID=0-0-45-1338325-p174318763
He used the following equation:
$ frac{displaystyle2 d(C_1,C_2)}{displaystyle n_1n_2}=frac{displaystyle d(C_1,C_1)}{displaystyle n_1^2}+frac{displaystyle d(C_2,C_2)}{displaystyle n_2^2}+2||m_1−m_2||^2.$
where
$ d(C_1,C_2)=sum_{x_iin C_1}sum_{x_jin C_2}||x_i−x_j||^2$ with $ |C_1|=n_1$ and $|C_2 |=n_2$.
I tried to prove this, but I do not get this. He uses the fact that
$ d(C_1,C_1)=2(n_1-1)sum_{i=1}^{n_1} ||x_i||^2 -4 sum_{1leq i < j leq n_1}langle x_i,x_jrangle.$
$ d(C_2,C_2)=2(n_2-1)sum_{i=1}^{n_2} ||y_i||^2 -4 sum_{1leq i < j leq n_2}langle y_i,y_jrangle.$
$ d(C_1,C_2)=2n_2sum_{i=1}^{n_1} ||x_i||^2+2n_1sum_{i=1}^{n_2} ||y_i||^2 -4 sum_{1leq ileq n_1}sum_{1leq jleq n_2}langle x_i,y_jrangle.$
And furthermore with
$||m_1−m_2||^2=langle sum_{i=1}^{n_1}x_i/n_1-sum_{j=1}^{n_1}y_j/n_2,sum_{i=1}^{n_1}x_i/n_1-sum_{j=1}^{n_1}y_j/n_2 rangle = frac{1}{displaystyle n_1^2}sum_{i=1}^{n_1} ||x_i||^2 + frac{2}{displaystyle n_1^2}sum_{1leq i < jleq n_1} langle x_i, x_jrangle +frac{1}{displaystyle n_2^2}sum_{i=1}^{n_2} ||y_i||^2 + frac{2}{displaystyle n_2^2}sum_{1leq i < jleq n_2} langle y_i, y_jrangle - frac{2}{n_1n_2} sum_{1leq ileq n_1}sum_{1leq jleq n_2}langle x_i,y_jrangle, $
I obtain that
$ frac{displaystyle d(C_1,C_2)}{displaystyle n_1n_2}=frac{displaystyle d(C_1,C_1)}{displaystyle n_1^2}+frac{displaystyle d(C_2,C_2)}{displaystyle n_2^2}+2||m_1−m_2||^2.$
Where is my error?
analysis vector-analysis machine-learning clustering data-mining
analysis vector-analysis machine-learning clustering data-mining
asked Nov 20 '18 at 18:44
Patricio
17212
asked Nov 20 '18 at 18:44
Patricio
17212
edited Nov 21 '18 at 11:19
asked Nov 20 '18 at 18:44
Patricio
17212
asked Nov 20 '18 at 18:44
Patricio
17212
asked Nov 20 '18 at 18:44
Patricio
17212
17212
I suggest working through the case $n_1=1, n_2=2$ to see where your derivation goes wrong (the paper is definitely correct). Maybe you've forgotten that $||x_i-x_j||^2$ appears twice in each $d(C_i,C_i)$?
– Matthew Towers
Nov 23 '18 at 17:59
add a comment |
I suggest working through the case $n_1=1, n_2=2$ to see where your derivation goes wrong (the paper is definitely correct). Maybe you've forgotten that $||x_i-x_j||^2$ appears twice in each $d(C_i,C_i)$?
– Matthew Towers
Nov 23 '18 at 17:59
I suggest working through the case $n_1=1, n_2=2$ to see where your derivation goes wrong (the paper is definitely correct). Maybe you've forgotten that $||x_i-x_j||^2$ appears twice in each $d(C_i,C_i)$?
– Matthew Towers
Nov 23 '18 at 17:59
I suggest working through the case $n_1=1, n_2=2$ to see where your derivation goes wrong (the paper is definitely correct). Maybe you've forgotten that $||x_i-x_j||^2$ appears twice in each $d(C_i,C_i)$?
– Matthew Towers
Nov 23 '18 at 17:59
add a comment |
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I suggest working through the case $n_1=1, n_2=2$ to see where your derivation goes wrong (the paper is definitely correct). Maybe you've forgotten that $||x_i-x_j||^2$ appears twice in each $d(C_i,C_i)$?
– Matthew Towers
Nov 23 '18 at 17:59