This gets us back into the same types of questions as my last article on the Besicovich 1/2 Conjecture:That article is probably a good starting place, but I won’t explicitly use anything from it here.The name of the problem comes from the idea of a traveling salesperson wanting to minimize the driving distance between a bunch of cities they would be visiting to make sales.There are a few easy cases to think about. endobj If you aren’t familiar with purely unrectifiable sets, you can just think of those two properties as roughly the definition.Sets with these properties have to also be ruled out just as a matter of definition. %äüöß In other words, β is small and hence a and c are fairly close in size. 3 0 obj
These are just the squares you get by taking side lengths of 2ⁿ (where n can be a positive or negative integer) and you have (0,0) be a corner.The Analyst Traveling Salesman Theorem (for ℝ²) was proved by Peter Jones in 1990. @���|٦yЪu��'>�����X@��*�남��^��T�Ӫ�k�����ˢ.�Q���⃚8Xԉm@yϬ�9�C��3s�L(�.�S������p:��N_ �yĩQ��ⷿ���;ox\ʴ 534 ), and you can just brute force check all three possibilities to find the shortest.I won’t go into all the history and difficulty of the problem. �e���̤$�"3���E���U�!+Ox"�����ZM�>;��,���6iRk ���#�" +�`pz!�ի'�*���[|�[��Tu>��M�^F�6)�{��H��{��B��t�X8��y�d1��>��wGK���v�hU�
2 0 obj 4 0 obj The detour adds a length of There are more sophisticated ways to keep track of the error terms and make the estimations without so many assumptions, but you should now believe that these numbers have something to do with how far off we are from making a straight line to solve the Traveling Salesman Problem.P.S. Beta is 0 in the limit.I’ll just rephrase the above in as compact a way as possible to make talking about it in the future easier.The beta number of a set E on a square Q is just taking the line that approximates E the best and figuring out the worst “error” of this approximation (and dividing by the side-length of Q).The technical definition is more complicated because there may not be a clear best line that does the approximation or an exact worst error, so you take an infimum over the set of all such things.If all of this is a bit overwhelming, let’s take a moment to figure out Let’s consider a simple scenario where the set E consists of three points A, B, and C that are close to being flat:The “best line” is just the line between A and C and the error is Here’s the idea: we’d really like for our shortest path to just go A to C, but we have to go out of our way to hit B. I’ll call this a The Pythagorean Theorem says a²=c²+e². The clever way to do this is to use dyadic squares.
Note that this really is measuring the flatness somehow. Also, (Hausdorff) dimension is a slightly misleading concept here because there are 0-dimensional sets for which you can’t solve the problem!The truly interesting part is to find some sort of quantifiable or numerical condition that doesn’t require us to actually produce the path or already know if a set is rectifiable by some other means.Here’s the motivating idea: we want a number that tells us how far away from being “flat” the set is at various zoomed-in scales.As usual, just think about the nicest types of sets first. We consider all lines through the square Q. This gives us the first criterion: we’ll only want to consider sets with dimension at most 1.If you’ve seen the construction of the Cantor set, then this should feel familiar.Start with a square.
:��_���x�v�8b�LT���D?ƒy*�X ����ʮ'i����x���C�;8�(JB���HIԄ�lk����Ɲ ,l��%:�?��J��yOB�/��q�� ��sH���h5�BN�I�bu9�B�S������(�?��zj y�qY�V±�%}��bzf���+�C���]+w�fi����g�[��EOU5��0����4M�/t��K�G�aa��=��"��� ��Q;������@��J��`VO,����x.�2�X��y� H����]V϶��^��p�������ލ�VoQ��y��d{iR�u�/�+Ƽ VO 31.10.2005 Exakte Lösungsverfahren für NP-schwierige kombinatorische Optimierungsprobleme. Dazu gehört auch das vieldiskutierte Problem des Handlungsreisenden (Travelling-Salesman). Das Problem des Handlungsreisenden (auch Botenproblem, Rundreiseproblem, engl. I’ll jump to the punchline to keep your interest.If E is just a flat line, we can go from one side to the other with no detours.
We’ll think about When we actually pick the line itself as one of the possible lines, all points are a distance 0 away:This brings us to the beta number of E on the scale of Q (a square).
/Mask 6 0 R
The line segment itself touches every one of those points and it has length 1.
The Traveling Salesman Problem is one of the great classic problems in mathematics. It’s NP-complete, which means that no computer algorithm can solve it “quickly.” (It’s outside the scope of this article to define what is meant by that).Considering the classical problem has been an active area of research for 100 years, you might think there’d be no hope when we switch the problem from a finite number of points to any arbitrary (bounded) set.But it turns out the problem is essentially solved in this case! Second, we’re getting “corners” everywhere which makes the definition of a tangent impossible.
Documenta Mathematica. For example, just take a filled-in square.
ImzweitenTeilwirdeineWebapplikationpräsentiert,dieimZugedieserArbeiter …
If there existed a rectifiable curve that passed through all the points, then it would be a rectifiable set.These are some necessary conditions on E, but it doesn’t come close to solving the problem. If B is placed roughly in the middle, then 2c≈2d≈(c+d).This shows that the detour adds roughly e²/(c+d). Divide it up into a 4x4 grid and remove all but the four corner squares.Keep iterating that process, and in the limit, you’ll end up with a set.
These are just the squares you get by taking side lengths of 2ⁿ (where n can be a positive or negative integer) and you have (0,0) be a corner.The Analyst Traveling Salesman Theorem (for ℝ²) was proved by Peter Jones in 1990. @���|٦yЪu��'>�����X@��*�남��^��T�Ӫ�k�����ˢ.�Q���⃚8Xԉm@yϬ�9�C��3s�L(�.�S������p:��N_ �yĩQ��ⷿ���;ox\ʴ 534 ), and you can just brute force check all three possibilities to find the shortest.I won’t go into all the history and difficulty of the problem. �e���̤$�"3���E���U�!+Ox"�����ZM�>;��,���6iRk ���#�" +�`pz!�ի'�*���[|�[��Tu>��M�^F�6)�{��H��{��B��t�X8��y�d1��>��wGK���v�hU�
2 0 obj 4 0 obj The detour adds a length of There are more sophisticated ways to keep track of the error terms and make the estimations without so many assumptions, but you should now believe that these numbers have something to do with how far off we are from making a straight line to solve the Traveling Salesman Problem.P.S. Beta is 0 in the limit.I’ll just rephrase the above in as compact a way as possible to make talking about it in the future easier.The beta number of a set E on a square Q is just taking the line that approximates E the best and figuring out the worst “error” of this approximation (and dividing by the side-length of Q).The technical definition is more complicated because there may not be a clear best line that does the approximation or an exact worst error, so you take an infimum over the set of all such things.If all of this is a bit overwhelming, let’s take a moment to figure out Let’s consider a simple scenario where the set E consists of three points A, B, and C that are close to being flat:The “best line” is just the line between A and C and the error is Here’s the idea: we’d really like for our shortest path to just go A to C, but we have to go out of our way to hit B. I’ll call this a The Pythagorean Theorem says a²=c²+e². The clever way to do this is to use dyadic squares.
Note that this really is measuring the flatness somehow. Also, (Hausdorff) dimension is a slightly misleading concept here because there are 0-dimensional sets for which you can’t solve the problem!The truly interesting part is to find some sort of quantifiable or numerical condition that doesn’t require us to actually produce the path or already know if a set is rectifiable by some other means.Here’s the motivating idea: we want a number that tells us how far away from being “flat” the set is at various zoomed-in scales.As usual, just think about the nicest types of sets first. We consider all lines through the square Q. This gives us the first criterion: we’ll only want to consider sets with dimension at most 1.If you’ve seen the construction of the Cantor set, then this should feel familiar.Start with a square.
:��_���x�v�8b�LT���D?ƒy*�X ����ʮ'i����x���C�;8�(JB���HIԄ�lk����Ɲ ,l��%:�?��J��yOB�/��q�� ��sH���h5�BN�I�bu9�B�S������(�?��zj y�qY�V±�%}��bzf���+�C���]+w�fi����g�[��EOU5��0����4M�/t��K�G�aa��=��"��� ��Q;������@��J��`VO,����x.�2�X��y� H����]V϶��^��p�������ލ�VoQ��y��d{iR�u�/�+Ƽ VO 31.10.2005 Exakte Lösungsverfahren für NP-schwierige kombinatorische Optimierungsprobleme. Dazu gehört auch das vieldiskutierte Problem des Handlungsreisenden (Travelling-Salesman). Das Problem des Handlungsreisenden (auch Botenproblem, Rundreiseproblem, engl. I’ll jump to the punchline to keep your interest.If E is just a flat line, we can go from one side to the other with no detours.
We’ll think about When we actually pick the line itself as one of the possible lines, all points are a distance 0 away:This brings us to the beta number of E on the scale of Q (a square).
/Mask 6 0 R
The line segment itself touches every one of those points and it has length 1.
The Traveling Salesman Problem is one of the great classic problems in mathematics. It’s NP-complete, which means that no computer algorithm can solve it “quickly.” (It’s outside the scope of this article to define what is meant by that).Considering the classical problem has been an active area of research for 100 years, you might think there’d be no hope when we switch the problem from a finite number of points to any arbitrary (bounded) set.But it turns out the problem is essentially solved in this case! Second, we’re getting “corners” everywhere which makes the definition of a tangent impossible.
Documenta Mathematica. For example, just take a filled-in square.
ImzweitenTeilwirdeineWebapplikationpräsentiert,dieimZugedieserArbeiter …
If there existed a rectifiable curve that passed through all the points, then it would be a rectifiable set.These are some necessary conditions on E, but it doesn’t come close to solving the problem. If B is placed roughly in the middle, then 2c≈2d≈(c+d).This shows that the detour adds roughly e²/(c+d). Divide it up into a 4x4 grid and remove all but the four corner squares.Keep iterating that process, and in the limit, you’ll end up with a set.