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Showing posts from January 18, 2019

18.10 doesn't recognize my display while 18.04 did

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0 I have an Ubuntu Desktop 18.10 with a 4.18.0-13-generic kernel. I also have a VGA compatible controller RV770 [Radeon HD 4850] whose display is UNCLAIMED. The display is the unrecognized and the only resolution is 800x600 to 800x600 I tried to add via xrandr another mode 1280x1024 at 60MHz, but with no success. The screen is a DELL 1703FP, capable of 1280x1024 at 60Hz The old 16.04 and the 18.04 recognized the screen as a Dell 1703FP and provide the correct Width x Height combinations. But not the 18.1o. I'm now stuck with a display with gigantic fonts, where all the windows are greater than the screen, with no way to adjust the situation. Any help or suggestions? display 18.10 share | im...

Points distinguishable by set of functions

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0 1 $begingroup$ Let $F$ be the set of monotone functions from $[0, 1]^d$ to $mathbb{R}^+$ , with $d > 1$ . There is a well-known result [1] according to which no total order on $[0, 1]^d$ , $d > 1$ , can be represented by a function in $F$ . This means that, no matter what function $f in F$ I choose, there will always be two points $P$ and $Q$ in $[0, 1]^d$ that are indistinguishable by $f$ , i.e., $f(P)=f(Q)$ . Example: $f(x,y)=x+y$ , $P=(1,0)$ and $Q=(0,1)$ . My question: what if I consider a set of functions $F'subseteq F$ . Are there known conditions under which I reach distinguishability of all points in $[0, 1]^d$ (i.e., any two points in $[0, 1]^d$ are distinguishable by at least one function in $F'$ )? I guess, e.g., that having less than $d$ functions implies indistingui...