How Slow Can you Waterski?: and other puzzling questions by The Guardian, Simon Rogers

By The Guardian, Simon Rogers

While the powers that be lowered the rate restrict on Lake Windermere to ten knots, waterskiers complained that their activity used to be now thoroughly scuppered. So simply how sluggish are you able to waterski earlier than you begin to sink underneath the waves?

And, whereas we're approximately it, how lengthy are you able to continue to exist in a freezer? What are the possibilities of being struck via lightning in mattress? And why is it so esay to raed wrods eevn wehn the lteetrs are mdduled up?

Everyday existence can pose a few mind-boggling questions - yet the place do you discover the solutions? The Guardian's renowned 'This Week' column has been taking a look into the technological know-how in the back of the scoop for 3 years, and the way sluggish are you able to Waterski? attracts jointly a range of the main resourceful questions and the main superb solutions. If you've ever questioned what makes a planet a planet, why submarines continue bumping into issues or perhaps if it's suitable for eating dust, How gradual are you able to Waterski? will end up impossible to resist - and enlightening - reading.

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In other words, the compression ratio A r i t ( s ) / | s | is bounded by the entropy plus a small constant plus a term which vanishes as |s| ^ oo. In the following we denote with BWO the algorithm bw -I- mtf + Arit, that is, the algorithm which given s returns Arit(mtf (bw(s))). BWO is the basic algorithm •* Obviously, to completely determine the encoding we must specify the status of the mtf list at the beginning of the procedure. ^ In the following all logarithms are taken to the base 2, and we assume 0 log 0 = 0.

From (3) we see that to achieve the k-th order entropy "it suffices", for any w E. A'', to compress the string Wg up to its zeroth order entropy Ho{wa). One of the reasons for which this is not an easy task is that the symbols of Ws are scattered within the input string. But this problem is solved by the Burrows-Wheeler transform! In fact, from the discussion of the previous section we know that for any w the symbols of Ws are grouped together inside bw(s). More precisely, bw(s) contains as a substring a permutation of Ws- Permuting the symbols of a string does not change its zeroth order entropy.

That ci, C2 are both rational numbers. Let Jij := lij n (ci,C2) for i, j € N. Then Ai:= An (ci,C2) = flieNUjeN Jij is also a computable Gj-set. Fix a rational number b £ (a, C2) and define Vjj to be the longest rational open interval / which satisfies that b E I and I C Uj<^ Jjt. Suppose that Vij = {uij,Vij). Then {uij)ij^n and (%)t,j6N are all computable sequences of rational numbers which are non-increasing and nondecreasing, respectively. So, Ui := inf^gN Uij and Vi := sup gpj Vij are right and left computable, respectively.

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