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May 10[edit]

Finding a bijection between a permutation of n digits and the range [1,n!][edit]

Hello,

I am given a permutation of the digits 1-9, representing an ID number. I try to find a bijection that would link each number to an integer in the range 1-9!, according to its relative order. For example:

f(123456789) = 1, f(123456798) = 2, ....,f(987654312) = 9!-1 f(987654321) = 9!

Any hints or suggestions regarding the finding of such a function?

Thanks! — Preceding unsigned comment added by 212.179.21.194 (talk) 07:44, 10 May 2016 (UTC)[reply]

I think it is slightly easier to do this for g(N)=f(N)-1. You can write

g(N)=a_{1}8!+a_{2}7!+a_{3}6!+\dots +a_{9}

,

where

a_{1}

is one less than the first digit of N (and it may be easier to do this with digits 0..8 instead of 1..9). To find

a_{2}

, you map the remaining digits to 1..8 and continue recursively. But I have no idea whether this is the most efficient way of going about this. —Kusma (t·c) 09:54, 10 May 2016 (UTC)[reply]

Its not particularly pretty, but you can construct the answer recursively by noting that the first 8! numbers start with 1, the next 8! with 2 etc... and applying the same logic to the remaining digits. — Preceding unsigned comment added by 128.40.61.82 (talk) 10:08, 10 May 2016 (UTC)[reply]

Thank you both for your answers, why did you choose to represents the number as a linear combination of factorial terms? 212.179.21.194 (talk) 11:11, 10 May 2016 (UTC)[reply]

It seems natural, as the first digit is 1 for 8! times, the second digit is 2 for 7! times, the third digit is 3 for 6! times, ... (and similar things with other digits). Really, as 128.X has said, you just look at how many numbers start with what digits. —Kusma (t·c) 14:48, 10 May 2016 (UTC)[reply]

It doesn't add much to what's already here but you might be interested to read the section of the article Permutation#Permutations in computing which discusses this problem. Dmcq (talk) 15:05, 10 May 2016 (UTC)[reply]

Relation vocabulary[edit]

WP:RDL#Fill in the blanks includes a question on mathematical vocabulary. (Answers would be best placed there, on the Language Desk.) -- ToE 14:14, 10 May 2016 (UTC)[reply]

Railroad tracks[edit]

I stand on a bridge over railroad tracks, directly above the symmetry axis between the two rails, and take a picture of the (totally straight) rails all the way to the horizon, while pointing the camera along the symmetry axis. The images of the rails get closer and closer the higher you look on the photo. What is the equation for the curve of the image of one of the rails? And how about in the special case where the camera is no higher than the rails? Loraof (talk) 17:03, 10 May 2016 (UTC)[reply]

It is a straight line. The articles Graphical projection and vanishing point have a few details. Sławomir Biały (talk) 17:09, 10 May 2016 (UTC)[reply]

Sławomir Biały's answer is correct for a rectilinear projection, which is what you get with a typical camera and lens (modulo some distortion). In the case of a vertical cylindrical panorama, like this one, the image is a sinusoid. (If the camera is in the plane of the rails, then it's a straight line, which in this case is a sinusoid of amplitude 0.) -- BenRG (talk) 19:02, 10 May 2016 (UTC)[reply]

Thanks. Now if I'm on a small asteroid with a very near horizon, presumably both rails seem to just end at the height of the horizon, without merging. But what if I'm on a flat planet of infinite extent? Do the two straight lines intersect at a place on the photo representing the same elevation as the camera, with neither continuing beyond there? Loraof (talk) 19:31, 10 May 2016 (UTC)[reply]

Assuming a flat planet and an ideal pinhole camera, the horizon is the intersection of the film with the plane parallel to the ground and containing the aperture (so any object at the same height as the camera will also project to the horizon). The image of a track is the intersection of the film with the plane containing the track and the aperture. The track images intersect at a point on the horizon, namely the intersection of the film with the line parallel to the tracks and containing the aperture.

On a curved asteroid there are no straight lines (unless you mean great circles on a sphere, in which case there are no parallel lines), so you'd have to be more precise about how the tracks are laid out, but most likely they will not intersect anywhere in the image. -- BenRG (talk) 20:24, 10 May 2016 (UTC)[reply]

The same effect can be seen close taken on photos from buildings. See original[1] and modified[2] picture. The modification is an stretch of the original photo from |__| to \__/. The original picture appears like /__\. The reason is the photographer on the flor and the relative distance form the camera lens to each corner of the building. Such fixing can make a fat tower appear more fat. In Jim Button and Luke the Engine Driver, Michael Ende created the fictional Mr. Tur Tur, who against physical law, appered even more tall from the distance he was seen, described as a Scheinriese in fiction. During the Cold War, the borders of nations were shown on a globe or map to make the enenmy apper smaller. Statistic sometimes is being displayed linear or scaled in a logarithm to make the smaller values appear bigger and in a smaller difference. --Hans Haase (有问题吗) 12:06, 12 May 2016 (UTC)[reply]