## Archive for sunset

## impressions soleil couchant [jatp]

Posted in Mountains, pictures, Running, Travel with tags Balmoral Beach, British Columbia, Canada, Claude Monet, Comox, Comox Glacier, Forbidden Plateau, forest fire, Georgia Strait, impressions soleil levant, jatp, mainland, Pacific North West, Pacific Ocean, smog, sunset, Van Isle, Vancouver Island on August 25, 2018 by xi'an## 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!