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2017-09-13

-- Sep 13 In-Class Exercise
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-- Sep 13 In-Class Exercise
  • 2*m/epsilon
  • where
      epsilon = 1/n+1
      m = max(144*ln(1/1000)/2,m1)
      m1 = (144*(n+1)^2)/(25*(k*epsilon)^2)
      k = 1/6
    
(Edited: 2017-09-13)
*2*m/epsilon * where epsilon = 1/n+1 m = max(144*ln(1/1000)/2,m1) m1 = (144*(n+1)^2)/(25*(k*epsilon)^2) k = 1/6

-- Sep 13 In-Class Exercise
`m = max((144ln \delta)/2, m_1)`
`m_1 = r + s = (144(n+1)^2)/(25(k epsilon)^2)`
`epsilon = 1/(n+1)`
`k = 1/(2c)`. Since c = 3. K = 1/6
`delta = 1/1000` and `ln(1/1000)` will be negative. Since m1 will be positive that will be max. So substituting m in below formula will give the number of examples needed.
Number of examples needed `2m/epsilon`
(Edited: 2017-09-13)
@BT@m = max((144ln \delta)/2, m_1)@BT@ @BT@m_1 = r + s = (144(n+1)^2)/(25(k epsilon)^2)@BT@ @BT@epsilon = 1/(n+1)@BT@ @BT@k = 1/(2c)@BT@. Since c = 3. K = 1/6 @BT@delta = 1/1000@BT@ and @BT@ln(1/1000)@BT@ will be negative. Since m1 will be positive that will be max. So substituting m in below formula will give the number of examples needed. Number of examples needed @BT@2m/epsilon@BT@

-- Sep 13 In-Class Exercise
delta=1/1000 Epsilon=1/1+n
2*m/Epsilon
m=max(144 * ln Epsilon / 2, m1 )
delta=1/1000 Epsilon=1/1+n 2*m/Epsilon m=max(144 * ln Epsilon / 2, m1 )

-- Sep 13 In-Class Exercise
No.of examples = 2m /epsilon m = max(144lnδ/2,mi) m1 = 144(n+1)4 / 25 k2 Therefore, number of examples = 288 (n+1)5 / 25(1/6)2
No.of examples = 2m /epsilon m = max(144lnδ/2,mi) m1 = 144(n+1)4 / 25 k2 Therefore, number of examples = 288 (n+1)5 / 25(1/6)2

-- Sep 13 In-Class Exercise
m=max(144lnδ2,m1)m=max(144lnδ2,m1) m1=r+s=144(n+1)225(kε)2m1=r+s=144(n+1)225(kε)2 ε=1n+1, Number of examples = 88 (n+1)5 / 25(1/6)2
m=max(144lnδ2,m1)m=max(144lnδ2,m1) m1=r+s=144(n+1)225(kε)2m1=r+s=144(n+1)225(kε)2 ε=1n+1, Number of examples = 88 (n+1)5 / 25(1/6)2

-- Sep 13 In-Class Exercise
with probability 1-delta/2, the perceptron will have found an eps accurate hypothesis.
using 2* m/eps will ensure m updates occur.
we have an n bit NBF delta = 1/1000 eps = 1/(n+1)
m = max(144 ln(delta)/2 , m1) m1 = 144(n+1)^2 / 25(ke)^2
so we need 2*(m*(n+1)) examples
with probability 1-delta/2, the perceptron will have found an eps accurate hypothesis. using 2* m/eps will ensure m updates occur. we have an n bit NBF delta = 1/1000 eps = 1/(n+1) m = max(144 ln(delta)/2 , m1) m1 = 144(n+1)^2 / 25(ke)^2 so we need 2*(m*(n+1)) examples

-- Sep 13 In-Class Exercise
m = max( (144 ln delta )/2 , m1)
m1 = r+s = 144(n+1)^2 / 25(ke)^2
 e  = 1/n+1
m = max( (144 ln delta )/2 , m1) m1 = r+s = 144(n+1)^2 / 25(ke)^2 e = 1/n+1

-- Sep 13 In-Class Exercise
'/delta = 1/1000 m = max(144 in delta)/2 m = max(144*ln(1/1000)/2,m1) m1 = (144*(n+1)^2)/(25*(k*epsilon)^2)
(Edited: 2017-09-17)
'/delta = 1/1000 m = max(144 in delta)/2 m = max(144*ln(1/1000)/2,m1) m1 = (144*(n+1)^2)/(25*(k*epsilon)^2)

-- Sep 13 In-Class Exercise
`m=max(\frac{144ln\delta}{2},m_i)`
`m_1= \frac{144(n+1)^2} {25(k\epsilon)^2}`
 
provided, `\epsilon = \frac{1} {n+1}`
and,
`\delta = \frac{1} {1000}`
since, inserting delta in m, 1st component would be negative and m1 would result in maximum
m = `m_1`
`k = frac{1}{2c}`
since,
c=3
`k = frac{1}{6}`
substituting value of k and `epsilon` in equation gives
`m_1 = \frac{144((6)^2)(n+1)^4}{25}`
(Edited: 2017-09-13)
@BT@m=max(\frac{144ln\delta}{2},m_i)@BT@ @BT@m_1= \frac{144(n+1)^2} {25(k\epsilon)^2}@BT@ provided, @BT@\epsilon = \frac{1} {n+1}@BT@ and, @BT@\delta = \frac{1} {1000}@BT@ since, inserting delta in m, 1st component would be negative and m1 would result in maximum<br/> m = @BT@m_1@BT@<br/> @BT@k = frac{1}{2c}@BT@<br/> since, <br/> c=3<br/> @BT@k = frac{1}{6}@BT@<br/> substituting value of k and @BT@epsilon@BT@ in equation gives<br/> @BT@m_1 = \frac{144((6)^2)(n+1)^4}{25}@BT@
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