Dichotomous search algorithm in numeric optimization
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 2 days ago
J. GarcíaJ. García
618
add a comment |
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 2 days ago
J. GarcíaJ. García
618
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?
– LinAlg
2 days ago
Correct. I must add continuity constraint.
– J. García
2 days ago
add a comment |
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 2 days ago
J. GarcíaJ. García
618
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 2 days ago
J. GarcíaJ. García
618
asked 2 days ago
J. GarcíaJ. García
618
edited 2 days ago
J. García
asked 2 days ago
J. GarcíaJ. García
618
asked 2 days ago
J. GarcíaJ. García
618
asked 2 days ago
J. GarcíaJ. García
618
618
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?
– LinAlg
2 days ago
Correct. I must add continuity constraint.
– J. García
2 days ago
add a comment |
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?
– LinAlg
2 days ago
Correct. I must add continuity constraint.
– J. García
2 days ago
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?
– LinAlg
2 days ago
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?
– LinAlg
2 days ago
Correct. I must add continuity constraint.
– J. García
2 days ago
Correct. I must add continuity constraint.
– J. García
2 days ago
add a comment |
1 Answer
1
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oldest
votes
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 2 days ago
LinAlgLinAlg
8,7761521
But, why I need to evaluate the function into $frac{a+b}{2} pm varepsilon$ and not in $frac{a+b}{2}$?
– J. García
2 days ago
@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.
– LinAlg
2 days ago
But, why I must pick two disctinct points? If I evaluate only in one point, $f(point)$, why can't do dichotomous search?
– J. García
2 days ago
@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.
– LinAlg
2 days ago
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 2 days ago
LinAlgLinAlg
8,7761521
But, why I need to evaluate the function into $frac{a+b}{2} pm varepsilon$ and not in $frac{a+b}{2}$?
– J. García
2 days ago
@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.
– LinAlg
2 days ago
But, why I must pick two disctinct points? If I evaluate only in one point, $f(point)$, why can't do dichotomous search?
– J. García
2 days ago
@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.
– LinAlg
2 days ago
add a comment |
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 2 days ago
LinAlgLinAlg
8,7761521
But, why I need to evaluate the function into $frac{a+b}{2} pm varepsilon$ and not in $frac{a+b}{2}$?
– J. García
2 days ago
@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.
– LinAlg
2 days ago
But, why I must pick two disctinct points? If I evaluate only in one point, $f(point)$, why can't do dichotomous search?
– J. García
2 days ago
@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.
– LinAlg
2 days ago
add a comment |
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 2 days ago
LinAlgLinAlg
8,7761521
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 2 days ago
LinAlgLinAlg
8,7761521
answered 2 days ago
LinAlgLinAlg
8,7761521
answered 2 days ago
LinAlgLinAlg
8,7761521
answered 2 days ago
LinAlgLinAlg
8,7761521
8,7761521
But, why I need to evaluate the function into $frac{a+b}{2} pm varepsilon$ and not in $frac{a+b}{2}$?
– J. García
2 days ago
@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.
– LinAlg
2 days ago
But, why I must pick two disctinct points? If I evaluate only in one point, $f(point)$, why can't do dichotomous search?
– J. García
2 days ago
@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.
– LinAlg
2 days ago
add a comment |
But, why I need to evaluate the function into $frac{a+b}{2} pm varepsilon$ and not in $frac{a+b}{2}$?
– J. García
2 days ago
@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.
– LinAlg
2 days ago
But, why I must pick two disctinct points? If I evaluate only in one point, $f(point)$, why can't do dichotomous search?
– J. García
2 days ago
@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.
– LinAlg
2 days ago
But, why I need to evaluate the function into $frac{a+b}{2} pm varepsilon$ and not in $frac{a+b}{2}$?
– J. García
2 days ago
But, why I need to evaluate the function into $frac{a+b}{2} pm varepsilon$ and not in $frac{a+b}{2}$?
– J. García
2 days ago
@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.
– LinAlg
2 days ago
@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.
– LinAlg
2 days ago
But, why I must pick two disctinct points? If I evaluate only in one point, $f(point)$, why can't do dichotomous search?
– J. García
2 days ago
But, why I must pick two disctinct points? If I evaluate only in one point, $f(point)$, why can't do dichotomous search?
– J. García
2 days ago
@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.
– LinAlg
2 days ago
@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.
– LinAlg
2 days ago
add a comment |
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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?
– LinAlg
2 days ago
Correct. I must add continuity constraint.
– J. García
2 days ago