how to formally find decision boundary of linear separable problem with threshold

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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Hi!



How do I find threshold $theta=3/2$ here?



Should I do sum over inequalities:



P1: $theta > 0$



P2, P3: $w_1 * xi_1^3 + w_2*xi_2^3 - theta < 0 $



P4 : $w_1 * xi_1^4 + w_2*xi_2^4 - theta geq 0 $



then:



$theta > 1$



$theta leq 2$



summing to get mean:



$theta + theta = 2 + 1$



so



$theta = 3/2$



Is it correct?



And boundary line will be: $[-w_2, w_1] + theta$ ?



EDIT



Simply guest that its 3/2 its not enough on exam. Besides only solution what comes to my mind is to decide on $W$ vector, then draw orthogonal to it line passing by points (0,1) and (1,0). Vector W will cross line in point D(0.5,0.5). Then we move this line to line' passing by point C in the middle between D and (1,1). Point C should be in (0.75,0.75) so that distance from each side will be maximized (distance to point (0.5,0.5) and (1,1) are the same). Equation of line is $xi_1 + xi_2 = 1$ then equation of line' will be $xi_1 + xi_2 = 3/2$. If there is better solution I would be happy to see it :D




linear-algebra




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  • Is this your first question here? Make sure to not use images, but rather text, for your questions
    – Rushabh Mehta
    Aug 6 at 15:10






  • 2




    text below image, so whats the problem?
    – yourstruly
    Aug 6 at 15:19










  • The text in the image is not below the image
    – Rushabh Mehta
    Aug 6 at 15:21










  • @yourstruly: your comment is not nice. Please refrain from harassing new users.
    – amWhy
    Aug 6 at 15:43






  • 1




    ohh, actually i forgot to feed my cats too
    – yourstruly
    Aug 6 at 18:17














up vote
1
down vote

favorite












enter image description here



Hi!



How do I find threshold $theta=3/2$ here?



Should I do sum over inequalities:



P1: $theta > 0$



P2, P3: $w_1 * xi_1^3 + w_2*xi_2^3 - theta < 0 $



P4 : $w_1 * xi_1^4 + w_2*xi_2^4 - theta geq 0 $



then:



$theta > 1$



$theta leq 2$



summing to get mean:



$theta + theta = 2 + 1$



so



$theta = 3/2$



Is it correct?



And boundary line will be: $[-w_2, w_1] + theta$ ?



EDIT



Simply guest that its 3/2 its not enough on exam. Besides only solution what comes to my mind is to decide on $W$ vector, then draw orthogonal to it line passing by points (0,1) and (1,0). Vector W will cross line in point D(0.5,0.5). Then we move this line to line' passing by point C in the middle between D and (1,1). Point C should be in (0.75,0.75) so that distance from each side will be maximized (distance to point (0.5,0.5) and (1,1) are the same). Equation of line is $xi_1 + xi_2 = 1$ then equation of line' will be $xi_1 + xi_2 = 3/2$. If there is better solution I would be happy to see it :D




linear-algebra




share|cite|improve this question





















  • Is this your first question here? Make sure to not use images, but rather text, for your questions
    – Rushabh Mehta
    Aug 6 at 15:10






  • 2




    text below image, so whats the problem?
    – yourstruly
    Aug 6 at 15:19










  • The text in the image is not below the image
    – Rushabh Mehta
    Aug 6 at 15:21










  • @yourstruly: your comment is not nice. Please refrain from harassing new users.
    – amWhy
    Aug 6 at 15:43






  • 1




    ohh, actually i forgot to feed my cats too
    – yourstruly
    Aug 6 at 18:17












up vote
1
down vote

favorite









up vote
1
down vote

favorite











enter image description here



Hi!



How do I find threshold $theta=3/2$ here?



Should I do sum over inequalities:



P1: $theta > 0$



P2, P3: $w_1 * xi_1^3 + w_2*xi_2^3 - theta < 0 $



P4 : $w_1 * xi_1^4 + w_2*xi_2^4 - theta geq 0 $



then:



$theta > 1$



$theta leq 2$



summing to get mean:



$theta + theta = 2 + 1$



so



$theta = 3/2$



Is it correct?



And boundary line will be: $[-w_2, w_1] + theta$ ?



EDIT



Simply guest that its 3/2 its not enough on exam. Besides only solution what comes to my mind is to decide on $W$ vector, then draw orthogonal to it line passing by points (0,1) and (1,0). Vector W will cross line in point D(0.5,0.5). Then we move this line to line' passing by point C in the middle between D and (1,1). Point C should be in (0.75,0.75) so that distance from each side will be maximized (distance to point (0.5,0.5) and (1,1) are the same). Equation of line is $xi_1 + xi_2 = 1$ then equation of line' will be $xi_1 + xi_2 = 3/2$. If there is better solution I would be happy to see it :D




linear-algebra




share|cite|improve this question













enter image description here



Hi!



How do I find threshold $theta=3/2$ here?



Should I do sum over inequalities:



P1: $theta > 0$



P2, P3: $w_1 * xi_1^3 + w_2*xi_2^3 - theta < 0 $



P4 : $w_1 * xi_1^4 + w_2*xi_2^4 - theta geq 0 $



then:



$theta > 1$



$theta leq 2$



summing to get mean:



$theta + theta = 2 + 1$



so



$theta = 3/2$



Is it correct?



And boundary line will be: $[-w_2, w_1] + theta$ ?



EDIT



Simply guest that its 3/2 its not enough on exam. Besides only solution what comes to my mind is to decide on $W$ vector, then draw orthogonal to it line passing by points (0,1) and (1,0). Vector W will cross line in point D(0.5,0.5). Then we move this line to line' passing by point C in the middle between D and (1,1). Point C should be in (0.75,0.75) so that distance from each side will be maximized (distance to point (0.5,0.5) and (1,1) are the same). Equation of line is $xi_1 + xi_2 = 1$ then equation of line' will be $xi_1 + xi_2 = 3/2$. If there is better solution I would be happy to see it :D





linear-algebra





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edited Aug 6 at 19:23
























asked Aug 6 at 15:01









yourstruly

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  • Is this your first question here? Make sure to not use images, but rather text, for your questions
    – Rushabh Mehta
    Aug 6 at 15:10






  • 2




    text below image, so whats the problem?
    – yourstruly
    Aug 6 at 15:19










  • The text in the image is not below the image
    – Rushabh Mehta
    Aug 6 at 15:21










  • @yourstruly: your comment is not nice. Please refrain from harassing new users.
    – amWhy
    Aug 6 at 15:43






  • 1




    ohh, actually i forgot to feed my cats too
    – yourstruly
    Aug 6 at 18:17
















  • Is this your first question here? Make sure to not use images, but rather text, for your questions
    – Rushabh Mehta
    Aug 6 at 15:10






  • 2




    text below image, so whats the problem?
    – yourstruly
    Aug 6 at 15:19










  • The text in the image is not below the image
    – Rushabh Mehta
    Aug 6 at 15:21










  • @yourstruly: your comment is not nice. Please refrain from harassing new users.
    – amWhy
    Aug 6 at 15:43






  • 1




    ohh, actually i forgot to feed my cats too
    – yourstruly
    Aug 6 at 18:17















Is this your first question here? Make sure to not use images, but rather text, for your questions
– Rushabh Mehta
Aug 6 at 15:10




Is this your first question here? Make sure to not use images, but rather text, for your questions
– Rushabh Mehta
Aug 6 at 15:10




2




2




text below image, so whats the problem?
– yourstruly
Aug 6 at 15:19




text below image, so whats the problem?
– yourstruly
Aug 6 at 15:19












The text in the image is not below the image
– Rushabh Mehta
Aug 6 at 15:21




The text in the image is not below the image
– Rushabh Mehta
Aug 6 at 15:21












@yourstruly: your comment is not nice. Please refrain from harassing new users.
– amWhy
Aug 6 at 15:43




@yourstruly: your comment is not nice. Please refrain from harassing new users.
– amWhy
Aug 6 at 15:43




1




1




ohh, actually i forgot to feed my cats too
– yourstruly
Aug 6 at 18:17




ohh, actually i forgot to feed my cats too
– yourstruly
Aug 6 at 18:17















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