Dichotomous search algorithm in numeric optimization
$begingroup$
If we assume that $phi$ is convex and continuous in $[a,b]$, it is obvius that, in this interval, $phi$ has a minimizer. The goal is know what point is it.
The main idea of Dichotomous search is reduce the size of the interval what is the minimizer evaluating $phi$ into two points, $overline{a}, overline{b} in (a,b) : overline{a} < overline{b} $.
For this, the algorithm uses the fact that if $phi(overline{a}) < phi(overline{b})$, the minium of $phi$ in $[a,b]$ is contained in the interval $[a,overline{b}]$. And $phi(overline{a}) geq phi(overline{b})$, the minium of $phi$ in $[a,b]$ is contained in the interval $[overline{a},b]$
Therefore, the Dichotomous algorithm computes the midpoint of the interval (step-by-step), $frac{a+b}{2}$ and, then moves slightly to either side of the midpoint to compute two test points: $frac{a+b}{2} pm varepsilon$, where $varepsilon$ is a very small number.
Why need to move slightly to either side of the midpoint?
optimization convex-optimization
asked Jan 5 at 16:56
J. GarcíaJ. García
618
$endgroup$
add a comment |
$begingroup$
If we assume that $phi$ is convex and continuous in $[a,b]$, it is obvius that, in this interval, $phi$ has a minimizer. The goal is know what point is it.
The main idea of Dichotomous search is reduce the size of the interval what is the minimizer evaluating $phi$ into two points, $overline{a}, overline{b} in (a,b) : overline{a} < overline{b} $.
For this, the algorithm uses the fact that if $phi(overline{a}) < phi(overline{b})$, the minium of $phi$ in $[a,b]$ is contained in the interval $[a,overline{b}]$. And $phi(overline{a}) geq phi(overline{b})$, the minium of $phi$ in $[a,b]$ is contained in the interval $[overline{a},b]$
Therefore, the Dichotomous algorithm computes the midpoint of the interval (step-by-step), $frac{a+b}{2}$ and, then moves slightly to either side of the midpoint to compute two test points: $frac{a+b}{2} pm varepsilon$, where $varepsilon$ is a very small number.
Why need to move slightly to either side of the midpoint?
optimization convex-optimization
asked Jan 5 at 16:56
J. GarcíaJ. García
618
$endgroup$
$begingroup$
It is not obvious at all that $phi$ has a minimizer; is $phi$ perhaps assumed to be continuous so we can apply the Weierstrass theorem?
$endgroup$
– LinAlg
Jan 5 at 18:15
$begingroup$
Correct. I must add continuity constraint.
$endgroup$
– J. García
Jan 5 at 19:41
add a comment |
$begingroup$
If we assume that $phi$ is convex and continuous in $[a,b]$, it is obvius that, in this interval, $phi$ has a minimizer. The goal is know what point is it.
The main idea of Dichotomous search is reduce the size of the interval what is the minimizer evaluating $phi$ into two points, $overline{a}, overline{b} in (a,b) : overline{a} < overline{b} $.
For this, the algorithm uses the fact that if $phi(overline{a}) < phi(overline{b})$, the minium of $phi$ in $[a,b]$ is contained in the interval $[a,overline{b}]$. And $phi(overline{a}) geq phi(overline{b})$, the minium of $phi$ in $[a,b]$ is contained in the interval $[overline{a},b]$
Therefore, the Dichotomous algorithm computes the midpoint of the interval (step-by-step), $frac{a+b}{2}$ and, then moves slightly to either side of the midpoint to compute two test points: $frac{a+b}{2} pm varepsilon$, where $varepsilon$ is a very small number.
Why need to move slightly to either side of the midpoint?
optimization convex-optimization
asked Jan 5 at 16:56
J. GarcíaJ. García
618
$endgroup$
If we assume that $phi$ is convex and continuous in $[a,b]$, it is obvius that, in this interval, $phi$ has a minimizer. The goal is know what point is it.
The main idea of Dichotomous search is reduce the size of the interval what is the minimizer evaluating $phi$ into two points, $overline{a}, overline{b} in (a,b) : overline{a} < overline{b} $.
For this, the algorithm uses the fact that if $phi(overline{a}) < phi(overline{b})$, the minium of $phi$ in $[a,b]$ is contained in the interval $[a,overline{b}]$. And $phi(overline{a}) geq phi(overline{b})$, the minium of $phi$ in $[a,b]$ is contained in the interval $[overline{a},b]$
Therefore, the Dichotomous algorithm computes the midpoint of the interval (step-by-step), $frac{a+b}{2}$ and, then moves slightly to either side of the midpoint to compute two test points: $frac{a+b}{2} pm varepsilon$, where $varepsilon$ is a very small number.
Why need to move slightly to either side of the midpoint?
optimization convex-optimization
optimization convex-optimization
asked Jan 5 at 16:56
J. GarcíaJ. García
618
asked Jan 5 at 16:56
J. GarcíaJ. García
618
edited Jan 5 at 19:41
J. García
asked Jan 5 at 16:56
J. GarcíaJ. García
618
asked Jan 5 at 16:56
J. GarcíaJ. García
618
asked Jan 5 at 16:56
J. GarcíaJ. García
618
618
$begingroup$
It is not obvious at all that $phi$ has a minimizer; is $phi$ perhaps assumed to be continuous so we can apply the Weierstrass theorem?
$endgroup$
– LinAlg
Jan 5 at 18:15
$begingroup$
Correct. I must add continuity constraint.
$endgroup$
– J. García
Jan 5 at 19:41
add a comment |
$begingroup$
It is not obvious at all that $phi$ has a minimizer; is $phi$ perhaps assumed to be continuous so we can apply the Weierstrass theorem?
$endgroup$
– LinAlg
Jan 5 at 18:15
$begingroup$
Correct. I must add continuity constraint.
$endgroup$
– J. García
Jan 5 at 19:41
$begingroup$
It is not obvious at all that $phi$ has a minimizer; is $phi$ perhaps assumed to be continuous so we can apply the Weierstrass theorem?
$endgroup$
– LinAlg
Jan 5 at 18:15
$begingroup$
It is not obvious at all that $phi$ has a minimizer; is $phi$ perhaps assumed to be continuous so we can apply the Weierstrass theorem?
$endgroup$
– LinAlg
Jan 5 at 18:15
$begingroup$
Correct. I must add continuity constraint.
$endgroup$
– J. García
Jan 5 at 19:41
$begingroup$
Correct. I must add continuity constraint.
$endgroup$
– J. García
Jan 5 at 19:41
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The reason is that you need to evaluate the function at two points, $bar{a}$ and $bar{b}$. The closer those $bar{b}$ is to $a$, and the closer $bar{a}$ is to $b$, the further you can shrink the interval, so you want $bar{a}$ and $bar{b}$ to be close to each other.
answered Jan 5 at 18:19
LinAlgLinAlg
10.1k1521
$endgroup$
$begingroup$
But, why I need to evaluate the function into $frac{a+b}{2} pm varepsilon$ and not in $frac{a+b}{2}$?
$endgroup$
– J. García
Jan 5 at 19:43
$begingroup$
@J.García you have to pick two distinct points. You can pick $(a+b)/2$ and $(a+b)/2+varepsilon$ if you like. Or $(a+b)/2-varepsilon$ and $(a+b)/2$. Your teacher or book selected $(a+b)/2-varepsilon$ and $(a+b)/2+varepsilon$, but you do not have to.
$endgroup$
– LinAlg
Jan 5 at 22:12
$begingroup$
But, why I must pick two disctinct points? If I evaluate only in one point, $f(point)$, why can't do dichotomous search?
$endgroup$
– J. García
Jan 6 at 0:52
$begingroup$
@J.García because your algorithm compares $f(bar{a})$ with $f(bar{b})$. The current algorithm needs two points because if you only know the function value at the midpoint you do not know in which subinterval to continue the search. You could use a different algorithm based on the derivative at the midpoint that only needs one derivative evaluation.
$endgroup$
– LinAlg
Jan 6 at 2:29
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
The reason is that you need to evaluate the function at two points, $bar{a}$ and $bar{b}$. The closer those $bar{b}$ is to $a$, and the closer $bar{a}$ is to $b$, the further you can shrink the interval, so you want $bar{a}$ and $bar{b}$ to be close to each other.
answered Jan 5 at 18:19
LinAlgLinAlg
10.1k1521
$endgroup$
$begingroup$
But, why I need to evaluate the function into $frac{a+b}{2} pm varepsilon$ and not in $frac{a+b}{2}$?
$endgroup$
– J. García
Jan 5 at 19:43
$begingroup$
@J.García you have to pick two distinct points. You can pick $(a+b)/2$ and $(a+b)/2+varepsilon$ if you like. Or $(a+b)/2-varepsilon$ and $(a+b)/2$. Your teacher or book selected $(a+b)/2-varepsilon$ and $(a+b)/2+varepsilon$, but you do not have to.
$endgroup$
– LinAlg
Jan 5 at 22:12
$begingroup$
But, why I must pick two disctinct points? If I evaluate only in one point, $f(point)$, why can't do dichotomous search?
$endgroup$
– J. García
Jan 6 at 0:52
$begingroup$
@J.García because your algorithm compares $f(bar{a})$ with $f(bar{b})$. The current algorithm needs two points because if you only know the function value at the midpoint you do not know in which subinterval to continue the search. You could use a different algorithm based on the derivative at the midpoint that only needs one derivative evaluation.
$endgroup$
– LinAlg
Jan 6 at 2:29
add a comment |
$begingroup$
The reason is that you need to evaluate the function at two points, $bar{a}$ and $bar{b}$. The closer those $bar{b}$ is to $a$, and the closer $bar{a}$ is to $b$, the further you can shrink the interval, so you want $bar{a}$ and $bar{b}$ to be close to each other.
answered Jan 5 at 18:19
LinAlgLinAlg
10.1k1521
$endgroup$
$begingroup$
But, why I need to evaluate the function into $frac{a+b}{2} pm varepsilon$ and not in $frac{a+b}{2}$?
$endgroup$
– J. García
Jan 5 at 19:43
$begingroup$
@J.García you have to pick two distinct points. You can pick $(a+b)/2$ and $(a+b)/2+varepsilon$ if you like. Or $(a+b)/2-varepsilon$ and $(a+b)/2$. Your teacher or book selected $(a+b)/2-varepsilon$ and $(a+b)/2+varepsilon$, but you do not have to.
$endgroup$
– LinAlg
Jan 5 at 22:12
$begingroup$
But, why I must pick two disctinct points? If I evaluate only in one point, $f(point)$, why can't do dichotomous search?
$endgroup$
– J. García
Jan 6 at 0:52
$begingroup$
@J.García because your algorithm compares $f(bar{a})$ with $f(bar{b})$. The current algorithm needs two points because if you only know the function value at the midpoint you do not know in which subinterval to continue the search. You could use a different algorithm based on the derivative at the midpoint that only needs one derivative evaluation.
$endgroup$
– LinAlg
Jan 6 at 2:29
add a comment |
$begingroup$
The reason is that you need to evaluate the function at two points, $bar{a}$ and $bar{b}$. The closer those $bar{b}$ is to $a$, and the closer $bar{a}$ is to $b$, the further you can shrink the interval, so you want $bar{a}$ and $bar{b}$ to be close to each other.
answered Jan 5 at 18:19
LinAlgLinAlg
10.1k1521
$endgroup$
The reason is that you need to evaluate the function at two points, $bar{a}$ and $bar{b}$. The closer those $bar{b}$ is to $a$, and the closer $bar{a}$ is to $b$, the further you can shrink the interval, so you want $bar{a}$ and $bar{b}$ to be close to each other.
answered Jan 5 at 18:19
LinAlgLinAlg
10.1k1521
answered Jan 5 at 18:19
LinAlgLinAlg
10.1k1521
answered Jan 5 at 18:19
LinAlgLinAlg
10.1k1521
answered Jan 5 at 18:19
LinAlgLinAlg
10.1k1521
10.1k1521
$begingroup$
But, why I need to evaluate the function into $frac{a+b}{2} pm varepsilon$ and not in $frac{a+b}{2}$?
$endgroup$
– J. García
Jan 5 at 19:43
$begingroup$
@J.García you have to pick two distinct points. You can pick $(a+b)/2$ and $(a+b)/2+varepsilon$ if you like. Or $(a+b)/2-varepsilon$ and $(a+b)/2$. Your teacher or book selected $(a+b)/2-varepsilon$ and $(a+b)/2+varepsilon$, but you do not have to.
$endgroup$
– LinAlg
Jan 5 at 22:12
$begingroup$
But, why I must pick two disctinct points? If I evaluate only in one point, $f(point)$, why can't do dichotomous search?
$endgroup$
– J. García
Jan 6 at 0:52
$begingroup$
@J.García because your algorithm compares $f(bar{a})$ with $f(bar{b})$. The current algorithm needs two points because if you only know the function value at the midpoint you do not know in which subinterval to continue the search. You could use a different algorithm based on the derivative at the midpoint that only needs one derivative evaluation.
$endgroup$
– LinAlg
Jan 6 at 2:29
add a comment |
$begingroup$
But, why I need to evaluate the function into $frac{a+b}{2} pm varepsilon$ and not in $frac{a+b}{2}$?
$endgroup$
– J. García
Jan 5 at 19:43
$begingroup$
@J.García you have to pick two distinct points. You can pick $(a+b)/2$ and $(a+b)/2+varepsilon$ if you like. Or $(a+b)/2-varepsilon$ and $(a+b)/2$. Your teacher or book selected $(a+b)/2-varepsilon$ and $(a+b)/2+varepsilon$, but you do not have to.
$endgroup$
– LinAlg
Jan 5 at 22:12
$begingroup$
But, why I must pick two disctinct points? If I evaluate only in one point, $f(point)$, why can't do dichotomous search?
$endgroup$
– J. García
Jan 6 at 0:52
$begingroup$
@J.García because your algorithm compares $f(bar{a})$ with $f(bar{b})$. The current algorithm needs two points because if you only know the function value at the midpoint you do not know in which subinterval to continue the search. You could use a different algorithm based on the derivative at the midpoint that only needs one derivative evaluation.
$endgroup$
– LinAlg
Jan 6 at 2:29
$begingroup$
But, why I need to evaluate the function into $frac{a+b}{2} pm varepsilon$ and not in $frac{a+b}{2}$?
$endgroup$
– J. García
Jan 5 at 19:43
$begingroup$
But, why I need to evaluate the function into $frac{a+b}{2} pm varepsilon$ and not in $frac{a+b}{2}$?
$endgroup$
– J. García
Jan 5 at 19:43
$begingroup$
@J.García you have to pick two distinct points. You can pick $(a+b)/2$ and $(a+b)/2+varepsilon$ if you like. Or $(a+b)/2-varepsilon$ and $(a+b)/2$. Your teacher or book selected $(a+b)/2-varepsilon$ and $(a+b)/2+varepsilon$, but you do not have to.
$endgroup$
– LinAlg
Jan 5 at 22:12
$begingroup$
@J.García you have to pick two distinct points. You can pick $(a+b)/2$ and $(a+b)/2+varepsilon$ if you like. Or $(a+b)/2-varepsilon$ and $(a+b)/2$. Your teacher or book selected $(a+b)/2-varepsilon$ and $(a+b)/2+varepsilon$, but you do not have to.
$endgroup$
– LinAlg
Jan 5 at 22:12
$begingroup$
But, why I must pick two disctinct points? If I evaluate only in one point, $f(point)$, why can't do dichotomous search?
$endgroup$
– J. García
Jan 6 at 0:52
$begingroup$
But, why I must pick two disctinct points? If I evaluate only in one point, $f(point)$, why can't do dichotomous search?
$endgroup$
– J. García
Jan 6 at 0:52
$begingroup$
@J.García because your algorithm compares $f(bar{a})$ with $f(bar{b})$. The current algorithm needs two points because if you only know the function value at the midpoint you do not know in which subinterval to continue the search. You could use a different algorithm based on the derivative at the midpoint that only needs one derivative evaluation.
$endgroup$
– LinAlg
Jan 6 at 2:29
$begingroup$
@J.García because your algorithm compares $f(bar{a})$ with $f(bar{b})$. The current algorithm needs two points because if you only know the function value at the midpoint you do not know in which subinterval to continue the search. You could use a different algorithm based on the derivative at the midpoint that only needs one derivative evaluation.
$endgroup$
– LinAlg
Jan 6 at 2:29
add a comment |
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$begingroup$
It is not obvious at all that $phi$ has a minimizer; is $phi$ perhaps assumed to be continuous so we can apply the Weierstrass theorem?
$endgroup$
– LinAlg
Jan 5 at 18:15
$begingroup$
Correct. I must add continuity constraint.
$endgroup$
– J. García
Jan 5 at 19:41