Talk:Mountain climbing problem

Some illustrations could be useful here. (Maybe I'll get to it myself some day if I live to be 200.....) Michael Hardy (talk) 17:21, 5 February 2009 (UTC)[reply]

This page is botched on my display. Five statements show up as FAILED TO PARSE.

Timbabwe (talk) 23:01, 20 February 2009 (UTC)timbabwe[reply]

If reloading doesn't help, try purging the server cache, per WP:purge:

At the end of the URL in the address bar of your browser, add the text ?action=purge

Michael Hardy (talk) 15:37, 21 February 2009 (UTC)[reply]

Why is the proof for the piecewise-linear case so convoluted

Here's a sketch of an elementary, constructive proof that generalises to any function with finitely many local extrema. We may divide the domains of f and g into maximal intervals on which the functions are monotonic. The base case that at least one function is strictly increasing is trivial. Choose a point p at which f attains the highest value h at any local extremum in (0,1), then choose a point q at which it attains the lowest value l at any local extremum in (p,1). These exist outside the base case. Let p' be the first point at which g attains h, and q' be the last point in [0,p'] at which g attains l. We have f([0,p])=g([0,p'])=[0,h], f([p,q])=g([q',p'])=[l,h] and f([q,1])=g([q',1])=[l,1]. By monotonically transforming these domains and ranges, we reduce to these three cases of the mountain climbing problem. But, on [0,p], [p,q] and [q,1], f must have fewer intervals than over the whole of [0,1]. So, by double induction, we are done. 101.112.156.92 (talk) 12:02, 28 November 2022 (UTC)[reply]

Okay I'm changing it 101.112.156.92 (talk) 08:02, 1 December 2022 (UTC)[reply]

Please don't change it without a published source. That would violate WP:OR. (The source there now also doesn't list a published source; that's no excuse.) —David Eppstein (talk) 08:04, 1 December 2022 (UTC)[reply]