## Archive for clouds

## another Nice shot [jatp]

Posted in Statistics with tags Alps, clouds, flight mode, French Alps, hiring committee, interview, jatp, Nice on May 22, 2019 by xi'an## shiny evening at the stadium [jatp]

Posted in Statistics with tags clouds, Coliseo, fori romani, Italia, Italy, jatp, Roma, Roman Forum, SPQR, sunset on May 5, 2019 by xi'an## end of [floating] town

Posted in Statistics with tags Angkor Wat, Cambodia, clouds, floating house, lake, locust village, Tonlé Sap lake on February 27, 2019 by xi'an## airport static

Posted in Statistics with tags airport, Birmingham, cellphone, clouds, noise on February 8, 2019 by xi'an*[An annoyingly loud businessman negotiating a deal for hours in Birmingham airport]*

…you are left with this money and you get the 60, we get 107 is the mathematics…if we discount 40% Werner still gets that, it’s not well-presented, you should get that minus that, do you mind if we do it from scratch, call me back in 20mn!

## unusual clouds [jatp]

Posted in pictures, Travel, Wines with tags ÌPA, beer, Brittany, Cité d'Ys, clouds, jatp, MCqMC 2018, Rennes, summer light, sunset, thunderstorm on July 19, 2018 by xi'an## a thread to bin them all [puzzle]

Posted in Books, Kids, R, Travel with tags Brittany, clouds, FiveThirtyEight, mathematical puzzle, meerkats, R, Rennes, sunset, The Riddler on July 9, 2018 by xi'an

**T**he most recent riddle on the Riddler consists in finding the shorter sequence of digits (in 0,1,..,9) such that all 10⁴ numbers between 0 (or 0000) and 9,999 can be found as a group of consecutive four digits. This sequence is obviously longer than 10⁴+3, but how long? On my trip to Brittany last weekend, I wrote an R code first constructing the sequence at random by picking with high preference the next digit among those producing a new four-digit number

tenz=10^(0:3) wn2dg=function(dz) 1+sum(dz*tenz) seqz=rep(0,10^4) snak=wndz=sample(0:9,4,rep=TRUE) seqz[wn2dg(wndz)]=1 while (min(seqz)==0){ wndz[1:3]=wndz[-1];wndz[4]=0 wndz[4]=sample(0:9,1,prob=.01+.99*(seqz[wn2dg(wndz)+0:9]==0)) snak=c(snak,wndz[4]) sek=wn2dg(wndz) seqz[sek]=seqz[sek]+1}

which usually returns a value above 75,000. I then looked through the sequence to eliminate useless replicas

for (i in sample(4:(length(snak)-5))){ if ((seqz[wn2dg(snak[(i-3):i])]>1) &(seqz[wn2dg(snak[(i-2):(i+1)])]>1) &(seqz[wn2dg(snak[(i-1):(i+2)])]>1) &(seqz[wn2dg(snak[i:(i+3)])]>1)){ seqz[wn2dg(snak[(i-3):i])]=seqz[wn2dg(snak[(i-3):i])]-1 seqz[wn2dg(snak[(i-2):(i+1)])]=seqz[wn2dg(snak[(i-2):(i+1)])]-1 seqz[wn2dg(snak[(i-1):(i+2)])]=seqz[wn2dg(snak[(i-1):(i+2)])]-1 seqz[wn2dg(snak[i:(i+3)])]=seqz[wn2dg(snak[i:(i+3)])]-1 snak=snak[-i] seqz[wn2dg(snak[(i-3):i])]=seqz[wn2dg(snak[(i-3):i])]+1 seqz[wn2dg(snak[(i-2):(i+1)])]=seqz[wn2dg(snak[(i-2):(i+1)])]+1 seqz[wn2dg(snak[(i-1):(i+2)])]=seqz[wn2dg(snak[(i-1):(i+2)])]+1}}

until none is found. A first attempt produced 12,911 terms in the sequence. A second one 12,913. A third one 12,871. Rather consistent figures but not concentrated enough to believe in achieving a true minimum. An overnight run produced 12,779 as the lowest value. Checking the answer the week after, it appears that 10⁴+3 *is* the correct answer!