Timeline for How to check if a box fits in a box?
Current License: CC BY-SA 3.0
11 events
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| Jan 9, 2020 at 12:34 | comment | added | bit | @მამუკაჯიბლაძე yes but in real-life you cannot just reorder the sizes of a box so that it fits into another box. The closest you can get to reordering sizes is to rotate the box and even then the inequalities do not hold for each of the dimensions. It turns out that the answer by Moritz Firsching does not work for the case of (2 3 4) and (1 3 5) where neither box should fit into one another despite the diagonal of one of the boxes being less than or equal to the diagonal of the other box. This question's quite old, is there an answer that works for all cases regardless of the boxes dimensions? | |
| Jan 9, 2020 at 6:46 | comment | added | მამუკა ჯიბლაძე | @MyWrathAcademia In your example there still is a reordering of sizes such that the inequalities hold for each of them, and this is already sufficient (but still not necessary). An example when no such reordering exists is given in a comment to the question by Denis T.: (1,1,12) fits in (9,9,9) | |
| Jan 8, 2020 at 18:58 | comment | added | bit | @მამუკაჯიბლაძე I just ran into this problem and the box fits into another box but condition i does not hold. Consider box A with width, depth and height measuring 30, 20 and 10 respectively and box B with width, depth and height measuring 10, 25 and 20 respectively. Box B fits into box A despite i not holding so as you said it is not true that if the box fits then i) holds. More tests show that for a lot of cases i needs to be satisfied in order to fit one box into another but for this particular case (and may be others) a box can fit another box when condition i is false. | |
| Sep 28, 2017 at 18:33 | comment | added | მამუკა ჯიბლაძე | @Philipp There can be no example satisfying i) since if $a_i\leqslant b_j$ for all $i,j$ then the box trivially fits. The condition i) is a sufficient condition which is not necessary. It is not true that if the box fits then i) holds. | |
| Sep 28, 2017 at 14:19 | comment | added | Philipp | I'm just saying that the conditions require that $a_i <= b_j \qquad j=1,2,3$ But I assumed that your example wanted to show a case where all conditions are met but still you couldn't fit to the box (thus showing that the conditions are insufficient). But for $a_2$ you find that it's bigger than $b_2$ and $b_3$ and thus in this case his conditions determine correctly that the box A doesn't fit into box B. | |
| Sep 28, 2017 at 14:05 | comment | added | მამუკა ჯიბლაძე | @Philipp Which condition 1? If you mean i) it is obviously (sufficient but) not necessary. I thought the last statement was meant to be unconditional. | |
| Sep 28, 2017 at 12:17 | comment | added | Philipp | Not a good example since here condition 1 is violated and thus shouldn't fit. | |
| Sep 27, 2017 at 23:05 | comment | added | მამუკა ჯიბლაძე | No you cannot fit A(1,1,1) in B(3,0,0) | |
| Sep 27, 2017 at 22:26 | review | Low quality posts | |||
| Sep 28, 2017 at 0:24 | |||||
| Sep 27, 2017 at 22:16 | review | First posts | |||
| Sep 27, 2017 at 22:25 | |||||
| Sep 27, 2017 at 22:02 | history | answered | MatheusLima | CC BY-SA 3.0 |