Word Game

For fun, and to prove to myself that I still have coding chops after my hiatus, I put together a generalization of the game Boggle few months ago.

It's an extension to Boggle because it supports alternative rectangular grids to the standard 4x4. As published it uses the bsd unix wordlist but other wordlists can be used; this allows easy portability to other languages, though it is still hard-wired to the 26 character alphabet used in English.
The task of finding all solutions, i.e. words of arbitrary length, in a standard-sized or larger boggle array is nontrivial. Pruning the search tree is necessary, as the number of all possible sequences is immense. One version of the game that I often play on my iPhone limits solutions to 7 letter lengths, which is exasperating when you find an 8 or 9 letter word! My version does not have this limitation.
Also I have a version, a little behind the main one,  played with the Python curses library, because I can be a bit old-fashioned.
All visible here: https://github.com/mtobis/mt_word_game

Work done at U Texas for ensemble management

Work done at U Texas for management of large ensembles of large scientific calculations, with robustness to job failure or resource allocation expiry, is visible here:

https://bitbucket.org/mtobis/tex-mecs

This is structured as a framework. The computation code and the analysis code are plugged into the framework by the end user. This is designed to be minimally intrusive. If the varied parameters of interest are already read in from a file and the code is runnable on the target machine, the user need only understand a fairly simple method for declaring the structure of the computation in ensemble parameter files.

Because the models are presumed coarse-grained (running for hours, not milliseconds) operating system calls are not costly. Consequently it is possible to wrap the executables into python objects, which makes the underlying code quite clean.

Re: "SEND ME YOUR DATA - PDF IS FINE," SAID NO ONE EVER

http://www.caitlinrivers.com/1/post/2013/04/send-me-your-data-pdf-is-fine-said-no-one-ever-how-to-share-your-data-effectively.html

There's a serious problem with the current state of shared data - it is almost completely unusable! Here are some ideas for sharing more effectively.

CAM3 fails on namelist read at TACC

Article moved here.

Environment variables in scripts

Nice tutorial here

Fixing Excel text exports in OS X

Excel exports .tsv and .csv with a broken linefeed character.

To fix: copy this file to "fixcel.py"

# fixcel.py

from sys import argv

infnam, outfnam = argv[1:]

inf = file(infnam)

indata = inf.read()

inf.close()

outdata = ""

for ch in indata:

if ord(ch) == 13:

  outdata += "\n"

else:

  outdata += ch

outf = file(outfnam,"w")

outf.write(outdata)

outf.close()

#

Then invoke with:

python fixcel.py oldfilename newfilename

Wordify

This code can count to a million in words. It could probably be refactored so it was shorter and could count to any number. If you were patient enough.

The "-b" flag puts "and" in the places a Brit would put it. It's optional but preferred.
I have random stopping points to increase the modest entertainment value of it running.

Can you recursively get wordify up to the quintillions? How should the "-b" flag be handled?

Can you do this in French?

from string import ascii_lowercase as alphabet
from sys import argv

units = [""] + "one two three four five six seven eight nine".split()
teens =  "ten eleven twelve thirteen fourteen fifteen sixteen seventeen eighteen nineteen".split()
tenties = 2 * [""] + "twenty thirty forty fifty sixty seventy eighty ninety".split() 


def name(number,terminate = True,continued = False,brit=False):
    assert type(number) == int
    assert number > 0 and number < 1001
    if number == 1000: return "one thousand"
    hundreds,residual = divmod(number,100)
    tens,ones = divmod(residual,10)
    result = ""
    if hundreds:
        result += units[hundreds] + " hundred"
    if residual and hundreds:
        result += " "
    if brit and residual and (hundreds or continued):
        result += "and "
    if (tens == 1):
        result += teens[ones] 
        if terminate:
            result += "."
    else:
        if tens:
            result += tenties[tens]
            if ones: result += "-"
        if ones:
            result += units[ones]
        if terminate:
            result += "."
    return result

def wordify(number,brit=False):
    assert type(number) == int
    assert number > 0 and number < 1000000
    thousands,residual = divmod(number,1000)
    result = ""
    if thousands:
        result = name(thousands,False,brit=brit) + " thousand"
        if residual:
            result += " "
            result += name(residual,continued = True,brit=brit)
        else:
            result += "."
    else:
        result = name(number)
    return result

if __name__ == "__main__":

    from time import sleep
    from random import randint

    brit = "-b" in argv
    
    sleeper = 30
    for i in range(1,1000000):
        if i == sleeper:
            sleep(3)
            sleeper += randint(3000,90000)
        print wordify(i,brit=brit)

output:

one.
two.
three.
four.
five.
six.
seven.
eight.
nine.
ten.
eleven.
twelve.
...
nine hundred and ninety-nine thousand nine hundred and ninety-six.
nine hundred and ninety-nine thousand nine hundred and ninety-seven.
nine hundred and ninety-nine thousand nine hundred and ninety-eight.
nine hundred and ninety-nine thousand nine hundred and ninety-nine.

That's as far as it goes so far. Go fix it.

Up to 11

I managed to confuse myself for hours on #11 trying to be elegant. But I really like this solution:

grid = [[0]*23]*3 + [[int(x) for x in line.split()]+[0,0,0] for line in
'''08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48'''.split('\n')
] + [[0]*23]*3

import operator

print max([reduce(operator.mul, [grid[y+n*d[0]][x+n*d[1]] for n in (0,1,2,3)])
for x in xrange(0,20) for y in xrange(3,23)
for d in ((0,1),(1,0),(1,1),(-1,1))]) 

The halo of zeros is very clever and completely works around what made mine confusing, and despite everything, inelegant:

linelen = len(array[0])
for line in array:
    assert len(line) == linelen

numlines = len(array)
offset = 4

directions = {
              "row":(0,linelen-offset+1,numlines,(0,1)),
              "column":(0,linelen,numlines-offset+1,(1,0)),
              "slash":(offset-1,linelen-offset+1,numlines-offset+1,(-1,1)),
              "backslash":(0,linelen-offset+1,numlines-offset+1,(1,1))
             }

maxprod = None
valmax = 0

for direction in directions.keys():
    x0, xlen, yrg, delta = directions[direction]
    vec0 = [(x0 + i * delta[0] ,i * delta[1]) for i in range(offset)]
    for y in range(yrg):
        vec1 = [(item[0]+y,item[1]) for item in vec0]
        for x in range(xlen):
            vec = [(item[0],item[1]+x) for item in vec1]
            vals = [array[item[0]][item[1]] for item in vec]
            prod = reduce(mul,vals,1)
            if prod > maxprod:
                valmax = vals
                maxprod = prod

print valmax, maxprod

Puzzle 5

Profiting from my overwrought example 3

from bf7 import *
from sys import argv

if __name__ == "__main__":
    if len(argv) == 1:
        upto = 10
    else:
        upto = int(argv[1]) 
    allprimes = primesupto(upto)

    result = 1
    for prime in allprimes:
        factor = prime
        while factor < upto:
            result *= prime
            factor *= prime
    print result

For me it would have been less work to just write

print 2 * 2 * 2 * 2 * 3 * 3 * 5 * 7 * 11 * 13 * 17 * 19

Best solution, for those who can remember grade school:

def gcd(a, b):
 while(b != 0):
  a, b = b, a%b
 return a
 
def lcm(a,b):
 return a * b / gcd(a, b)
 
print reduce(lcm, range(2, 21))

Puzzle 4

What's the largest palindrome number that is the product of two three digit numbers.

This time I have one of the better solutions:

def nextpal(low,high,reversed=False):
    assert low < high

    if reversed:
        candidate = high
        final = low
        increment = -1
    else:
        candidate = low
        final = high
        increment = 1
    while candidate != final:
        if str(candidate) == str(candidate)[-1::-1]:
            yield candidate
        candidate += increment

def getfactor(low,high,target):
    for test in range(low,high):
        if not target % test:
            residual = target/test
            if residual in range(low,high):
                return test, residual
    return None

if __name__ == "__main__":
    from time import time
    start = time()
    pal = nextpal(10000,1000000, reversed=True)
    while True:
        target = pal.next()
        result = getfactor(100,1000,target)
        if result:
            print target
            print result
            break
    end = time()
    print end - start

Most started from the factors to see if they built palindromes. Much better to go the other way; far fewer tests (the 94th palindrome succeeded, so less than 9400 calculations. The cast to string for the palindrome test is easy, but is it efficient?

The most interesting items on the discussion were the Pytho one-liner:

max(map(lambda s: bool(s) and max(s) or 0, [(filter(lambda n: n == int(str(n)[::-1]), map(lambda n: i*n, range(999, 99, -1)))) for i in range(999, 99, -1)]))

and the pen-and-paper solution

You can also do this with pen and paper. You have a number:

(100a + 10b + c)(100d + 10e + f)

Which is a palindrone. This factors out to:

100cd + 10ce + cf +
1000bd + 100be + 10bf +
10000ad + 1000ae + 100af

Assuming the first digit is 9, then cf must be equal to nine.
Also, both a and d must then be equal to nine. The only ways
to make the last digit of the product of two integers 9 would
be if those integers were either of:

1 and 9
3 and 3
7 and 7

So, both numbers must start with 9, end with either 1, 3, 7,
or 9, and one of them must be divisible by 11. The only
numbers divisible by 11 from 900 - 1000 are:

902
913
924
935
946
957
968
979
990

You can discard all of those that do not end in either 1, 3,
7, or 9, and you are left with:

913
957
979

So now the presumed answer is either:

(900 + 10 + 3)(900 + 10x + 3)
(900 + 50 + 7)(900 + 10x + 7)
(900 + 70 + 9)(900 + 10x + 1)

Factoring all those out, you get:

810000 + 9000x + 2700 + 9000 + 100x + 30 + 2700 + 30x + 9
824439 + 9130x

Now, for the first digit 824439 + 9130x to be 9, x must be 9
(if x were 8, then 824439 + 9130x = 897479, and the first
digit is 8). And so you have 913 * 993, which is the answer.
You can factor the others out to see if they produce a bigger
answer, which they don't.

Puzzle 3 without cheating

I am about 2800 times slower than the cheat I downloaded. But I get the right answer, and in a tolerable time (30 seconds). Pure brute forcing this seems like it would take much longer.

from sys import argv
from math import sqrt
from time import time

# PRIME GENERATOR

def nextprime():
    """ generate next prime """

    myprimes = [2]
    candidate = 3
    yield 2
    while True:
        factors = factorize(candidate,myprimes)
        if len(factors.keys()) == 1:
            myprimes.append(candidate)
            yield candidate
        candidate += 2

def primesupto(upto):
    """ collect primes from the generator"""

    result = []
    primes = nextprime()
    p = None
    while not p or p < upto:
        if p:
            result.append(p)
        p = primes.next()
    return result

def factorize(target,allprimes=None,verbose=False):
    """ get factors of an integer

    this works but is very slow if the primes aren't cached """

    factors = {}
    upto = int(sqrt(target))
    residual = target
    prime = nextprime()

    if allprimes:
        prime = allprimes.__iter__()

    while True:  # get factors and multiplicities up to sqrt(N)
        try:
            factor = prime.next()
        except StopIteration:
            break
        if factor > upto:
            break
        while not residual % factor:
            residual /= factor
            factors[factor] = factors.get(factor,0) + 1           

    if residual > 1: # if anything is left it is the sole prime factor > sqrt(N)
        factors[residual] = 1

    return factors
   
if __name__ == "__main__":
    target = int(argv[1]) # get target from command line
    upto = sqrt(target)
    starttime = time()
    allprimes = primesupto(upto)
    print max(factorize(int(argv[1]),allprimes).keys())
    endtime = time()
    print endtime - starttime

Update:

Best answer, wow!

from time import time
from sys import argv

n = int(argv[1])
d = 3

starttime = time()
while (d * d < n):
   if n % d == 0: n /= d
   else: d += 2
endtime = time()
print n
print endtime - startime

Runs in 0.7 msec. I feel like a complete idiot now. Oh well, it was interesting anyway.

Cheating, sorta

What is the largest prime factor of the number 600851475143 ?

from os import system
from os.path import isfile
from sys import argv

try:
  import numbthy
except:
  if not isfile("numbthy.py"): # don't mess with it if it's already there and broken
     system("wget http://userpages.umbc.edu/~rcampbel/Computers/Python/lib/numb
thy.py")
  import numbthy

f = numbthy.factors(int(argv[1]))
if f: print f[-1]

Level 2

I'm mostly at level 2 in this chart, except for IDEs. I don't really use them. I like aquamacs for an editor but even there I'm not good at it. I'm also pretty lame at version control, never having had the opportunity to work closely with others; hence never having run into a merge problem in real life.

I read some of the sorts of books he calls level 3.

Number Two

Python shines on this one. Find the sum of all the even-valued terms in the Fibonacci sequence which do not exceed four million.

from itertools import ifilterfalse

def fib(n1=1,n2=1,fmax=4000000):
    "terminating fibonacci sequence"

    assert 0 < n1 <= n2 < fmax
    l = [n2,n1]
    while l:
        if l[0] < fmax/2:
            l.append(sum(l))
        yield l.pop(0)

if __name__ == "__main__":
    print sum(ifilterfalse(lambda x: x % 2,fib()))

Euler Project

prob 1

A one-liner.

>>> sum([i for i in range(10) if not (i%3 and i%5)])
23
>>> sum([i for i in range(1000) if not (i%3 and i%5)])
233168

So it's surprisingly interesting:

>>> for n in range(1,9):
...     print sum([i for i in range(10**n) if not (i%3 and i%5)])
... 
23
2318
233168
23331668
2333316668
233333166668
23333331666668
2333333316666668

Can't even come up with a succinct notation for the theorem, never mind any idea how to prove it.

Moving notes to self

Notes to self or those interested in my daily pain will be moved to http://aardvarknoplay.blogspot.com/ . Old notes being moved.

This blog will actually have content that others may find interesting.

Some Key Papers in Climate Modeling History

A first pass at a greatest hits list; updates will not necessarily be noted as such. I had a request for key papers in the history of modeling so the emphasis leans that way. Nominations welcome.

Charney, Fjortofft & von Neumann 1950: Numerical integration of the barotropic vorticity equation, Tellus 2:237-254

Lorenz 1963: Deterministic Nonperiodic Flow J. Atmos. Sci 20:130-141

Lorenz 1967: The nature and theory of the general circulation of the atmosphere. WMO. http://eapsweb.mit.edu/research/Lorenz/General_Circ_WMO_1967_Part1.pdf

Manabe & Bryan 1969: Climate calculations with a combined ocean-atmosphere model.

Bryan & Cox 1972: The circulation of the world ocean: a numerical study. Part I, a homogeneous model J Phys Oceanog
Arakawa & Lamb 1977: Computational Design of the basic processes of the UCLA general circulation model. Methods in Computational Physics 17:173-265

Charney et al 1979: Ad Hoc Study Group on Carbon Dioxide and Climate. http://www.atmos.ucla.edu/~brianpm/charneyreport.html

Manabe & Stouffer 1988: Two stable equilibria of a coupled ocean-atmosphere model

Ramanathan et al 1985: Trace gas trends and their potential role in climate change. J Geophys Res 90:5547-5566

Cess et al 1990: Intercomparison and interpretation of climate feedback processes in seventeen atmospheric general circulation models

Bryan 1991: Poleward heat transport in the ocean: a reviwe of a hierarchy of models of incerasing resolution. J Phys Oceanog 18:851-867

Cubasch et al 1992: Time-dependent greenhouse warming computations with a coupled ocean-atmosphere model

Hansen et al 1992: Potential climate impact of Mount Pinatubo eruption GEOPHYSICAL RESEARCH LETTERS, VOL. 19, NO. 2, PP. 215-218, 1992
doi:10.1029/91GL02788

Hasselmann 1997: Multi-pattern fingerprint method for detection and attribution of climate change. Climate Dynamics 13:601-612

Hansen et al 1998: Climate Forings in the Industrial Era. PNAS

Bengtsson 1999: From short-range barotropic modeling to extended-range global weather prediction. Tellus 51 A-B:13-23

Randall 2000: General Circulation Model Development: Past, Present, and Future (International Geophysics) [Hardcover]

Hansen et al 2005: Earth's Energy Imbalance: Confirmation and Implications. Science.

Hack et al 2006: Simulation of the Global Hydrological Cycle in the CCSM Community Atmosphere Model Version 3. J Clim Volume 19, Issue 11 (June 2006)

Schmidt et al 2006: Present-Day Atmospheric Simulations Using GISS ModelE: Comparison to In Situ, Satellite, and Reanalysis Data. J. Climate, 19, 153–192.

Rahmstorf et al 2007: Recent Climate Observations Compared to Predictions. Science 316:709 DOI: 10.1126/science.1136843

More Adventures in Climate Analysis Installs

So here I document getting Ferret to work on a Mac; getting it to run on the super is another problem. I assumed no privileges and more or less succeeded. However, the install instructions are (as is normal with scientist code) broken.

Here are the absurd installation instructions:

http://ferret.pmel.noaa.gov/static/Downloads/mac_intel_downloads.html

This is basically just three useful links to the actual binaries under "Environment and Executable files"; ignore the rest of it.

Then proceed to

http://ferret.pmel.noaa.gov/static/Downloads/ferret_installation_and_update_guide_v600.html

If you don;t knwo about environment variables in bash, go study up on them.

The first step 1 should be

tar xf $GET_LOCATION/fer_environment.tar

The second step 1 and step 2 [sic] work as noted. Step 3 should be

tar xf $GET_LOCATION/fer_dsets.tar

as the Mac ungzips behind your back. Your next step will have to be backwards, as the Finstall script expects the exectuables to be gzipped. So go back to your download directory and type

gzip fer_executables.tar

Also, you will have to find a place for an extra file in the middle of the next step. Be sensible, create a ~/.Ferret directory before doing the next step. Use that for your $SET_FER .

Now you can run the Finstall per instructions. But the script you ghet out will be a csh script not a bash script. The following edits worked for me:

export FER_DIR=/Users/tobis/Ferret
export FER_DSETS=/Users/tobis/FerSamp

#   if ( $PATH !~ *ferret* ) then
#       export PATH={$PATH}:$FER_DIR/bin
#   else
#        echo replacing ferret path
#      export PATH=`echo $PATH | awk -F: '{for (i=1; i<=NF; i++) {if ($i \!~ /ferret\/bin/) {printf "%s:",$i}}}'`
#      export PATH={$PATH}$FER_DIR/bin
#   endif
 
# =========== Initially make no modifications below this line ===========

#  Default Ferret document browser
#   export FER_WEB_BROWSER="netscape -ncols 60"
   export FER_EXTERNAL_FUNCTIONS="$FER_DIR/ext_func/libs"
   export FER_GO=". $FER_DIR/go $FER_DIR/examples $FER_DIR/contrib"
   export FER_DATA=". $FER_DSETS/data $FER_DIR/go $FER_DIR/examples $FER_DIR/contrib"
   export FER_DESCR=". $FER_DSETS/descr"
   export FER_GRIDS=". $FER_DSETS/grids"
   export TMAP="$FER_DIR/fmt"
   export PLOTFONTS="$FER_DIR/ppl/fonts"
   export SPECTRA="$FER_DIR/ppl"         # for old ferret versions
   export FER_PALETTE=". $FER_DIR/ppl"       # palette search list

So basically, change all the setenvs to exports, put in the equal sign, comment out the if/else block. If you are better with bash scripts, the if/else thing looks harmless.

I did not uncomment the reference to netscape. I don;t know where to get a netscape for os X 10.6 anyway. Sigh.

This actually works!

Meanwhile Chas has bunches of hoops for me to jump through. Will keep you posted, if you exist.

PyNGL Even worse than CDAT?

Undecided as yet. More to follow. But plenty of the same signs of cluelessness.

I am still hopeful that the interpolation routine I need will work. And the install is much more straightforward, there being suitable binaries for the machines I care about.

But the test routine starts with an absolute path to a python installation at NCAR!

And this is unpythonic, though perhaps this is a Fortran 77 constraint rearing its ugly head

import Ngl
import Nio
import numpy

# create an array and make a netcdf file

ff = Nio.open_file("foo.nc","c")
ff.create_dimension("x",10)
ff.create_dimension("y",10)

ff.create_variable("u","f",("x","y"))

z = numpy.zeros((10,10),"f")

# ff.variables["u"] = z doesn't work

ff.variables["u"].assign_value(z)

# assignment is by value, as the name indicates

z[3,3] = 3.3

ff.variables["u"][4,4] = 4.4

# but after assignment you can pick up a reference

u = ff.variables["u"]

u[2,2] = 2.2

"""
>>> ff.variables["u"][:]
array([[ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.],
      [ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.],
      [ 0.,  0.,  2.2,  0.,  0.,  0.,  0.,  0.,  0.,  0.],
      [ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.],
      [ 0.,  0.,  0.,  0.,  4.4,  0.,  0.,  0.,  0.,  0.],
      [ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.],
      [ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.],
      [ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.],
      [ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.],
      [ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.]], dtype=float32)
"""

ff.close()

Anthony Doerr

Read the whole thing. It starts

During my sophomore year, 1992, 1,500 scientists, including more than half the living Nobel laureates, admonished in their Warning to Humanity: “A great change in our stewardship of the earth and the life on it is required if vast human misery is to be avoided and our global home on this planet is not to be irretrievably mutilated.”

So what have we done? Not much. From 1992 to 2007, global CO2 emissions from burning fossil fuels rose 38 percent. Emissions in 2008 rose a full 2 percent despite a global economic slump. Honeybees are dying by the billions1, amphibians by the millions, and shallow Caribbean reefs are mostly dead already.2 Our soil is disappearing faster than ever before, half of all mammals are in decline, and a recent climate change model predicts that the Arctic could have ice-free summers by 2013. Unchecked, carbon emissions from China alone will probably match the current global level by 2030.

“The god thou servest,” Marlowe wrote in Dr. Faustus, almost four hundred years before the invention of internet shopping, “is thine own appetite.” Was he wrong? How significantly have you reduced your own emissions since you first heard the phrase “climate change?” By a tenth? A quarter? A half? That’s better than I’m doing. The shirt I’m wearing was shipped here from Thailand. The Twinkie I just ate had 37 ingredients in it. I biked to work through 91-degree heat this morning but back at my house the air conditioner is grinding away, keeping all three bedrooms a pleasant 74 degrees.

My computer is on; my desk lamp is glowing. The vent on the wall is blowing a steady, soothing stream of cool air onto my shoes.

h/t Andrew Sullivan. Anthony Doerr, whom I had not heard of until today, lives in Idaho, writes "on Science" column for the Boston Globe, and is a 2010 Fellow of the John Simon Guggenheim Memorial Foundation.