Advent of Code

[tonycollinet]

Well this is rather fun. Not done any coding outside of simple VBA for years (about 20), and I'm trying this as a vehicle for learning python.

Day one:

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expense=[1956,1994,457,1654,2003,1902,1741,1494,1597,1129,1146,1589,1989,1093,1881,1288,1848,1371,1508,1035,1813,1335,1634,1102,1262,1637,1048,1807,1270,1528,1670,1803,1202,1294,1570,1640,1484,1872,1140,1207,1485,1781,1778,1772,1334,1267,1045,1194,1873,1441,1557,1414,1123,1980,1527,1591,1665,1916,1662,1139,1973,1258,1041,1134,1609,1554,1455,1124,1478,1938,1759,1281,1410,1511,930,1319,1302,1827,1216,1404,1460,2002,1590,1817,1341,1631,1608,1382,1158,1594,1049,1804,1555,1753,447,1021,1079,609,1766,1327,1851,1052,1737,1175,1043,1945,1573,1113,1724,1203,1856,1682,1623,1135,1015,1423,1412,1315,1375,1895,1351,1530,1758,1445,1518,1819,1567,1305,1919,1952,1432,1099,1476,1883,1871,1900,1442,1393,1214,1283,1538,1391,1008,1109,1621,1876,1998,1032,1324,1927,481,1732,1370,1683,1199,1465,1882,1293,1671,1456,1197,1506,1381,1469,1830,1957,1850,1184,1564,1170,1943,1131,1867,1208,1788,1337,1722,1760,1651,1069,1574,1959,1770,66,1190,1606,1899,1054,980,1693,1173,1479,1333,1579,1720,1782,1971,1438,1178,1306]
 
for index, num1 in enumerate(expense):
    for num2 in expense[index+1:]:
        if num1+num2 == 2020:
            print (num1, num2, num1+num2, num1*num2)

***************
And day 2: (for day two, I learned to read the input data from a text file into a list rather than declaring the data in the list)

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input_file = open("2aInput.txt", "r")
input_as_list = [(line.strip()).split() for line in input_file]
input_file.close()

valid=0

for password in input_as_list:
    occur = 0
    num1 = int(password[0].rsplit("-")[0])
    num2 = int(password[0].rsplit("-")[1])
    letter = password [1][0]
    if password[2][num1-1]==letter:occur+=1
    if password[2][num2-1]==letter:occur+=1
    if occur ==1: valid +=1

print (valid)

[tonycollinet]

Some ugly / semi-minimalist perl:-

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#!/usr/bin/perl

while( <> ) {
        @o = split( /\W+/ );
        %l=();
        map { $l{$_}++ } split( //, $o[3] );
        $r[0]++ if( $l{$o[2]} >= $o[0] && $l{$o[2]} <= $o[1] );
        $r[1]++ if( substr( $o[3], $o[0]-1, 1 ) eq $o[2] xor substr( $o[3], $o[1]-1, 1 ) eq $o[2] );
}
print "@r\n";

(I know it can be code golfed to a lot less than that, but that's about as minimalist as I'd go before it starts to become a puzzle rather than recognisable code.)

Call that recognisable? Let me give you a layman's 

[Greenbank]

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Today's lesson is arrays and the modulo operator.

Still waiting for the first nasty one.

[freeflow]

Yes.  The introduction is quite gentle this year.

[Pingu]

Day 3 done 

I've been putting my amateur JS code efforts here: http://www.pinniped.plus.com/AoC/index.htm

[Pingu]

Today's lesson is arrays and the modulo operator...

Could you put hints in a Spoiler, please?

[tonycollinet]

Day 3 done 

I've been putting my amateur JS code efforts here: http://www.pinniped.plus.com/AoC/index.htm

There is no source to view (firefox)

[Pingu]

Day 3 done 

I've been putting my amateur JS code efforts here: http://www.pinniped.plus.com/AoC/index.htm

There is no source to view (firefox)

Ctrl-u or right-click > View source

[tonycollinet]

Ahhhhh - of course 

[Greenbank]

Today's lesson is arrays and the modulo operator...

Could you put hints in a Spoiler, please?

Ah, ok, sorry, will do.

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And today we have regex parsing.

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"They aren't actually passport details, instead they are instructions for a special kind of computer ..."

[Ben T]

Found day 4 a bit boring to be honest.
Reading arbitrary rules and writing validation code is a bit too much like actual work...

[Ben T]

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"They aren't actually passport details, instead they are instructions for a special kind of computer ..."

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the presence of colours makes me think it might make a picture. Years in fact being pixel positions, perhaps

[grams]

Still sticking with Google Sheets until they force me to stop.

Day 3:
https://docs.google.com/spreadsheets/d/1­7_5yPJ8oNENFlRtqBnbCOLqE50RDK5QlLC_3p1v8­Csg/edit

Day 4:
https://docs.google.com/spreadsheets/d/1Gj7tgbzQUkiwS5_wJ62lvNfRNmbqBpAjdYLCAD1ulQw/edit

[freeflow]

Day04 has me scratching my head.  My VBA (with interfaces and code injection!!)  works OK with the test  data but fails on the real stuff < mumble mumble mumble go read the question again again again again>

[Pingu]

Day 4 done.

http://www.pinniped.plus.com/AoC/index.htm

[Pingu]

Still sticking with Google Sheets until they force me to stop.

Do you have a link to your Day 1 solution?

[Greenbank]

Still sticking with Google Sheets until they force me to stop.

Do you have a link to your Day 1 solution?

He posted it elsewhere: https://docs.google.com/spreadsheets/d/1joA1oMsoGaDCwX0u8CAK7IUuMeFqyfLEYzzIWzGB1d0/edit#gid=0

[Pingu]

Still sticking with Google Sheets until they force me to stop.

Do you have a link to your Day 1 solution?

He posted it elsewhere: https://docs.google.com/spreadsheets/d/1joA1oMsoGaDCwX0u8CAK7IUuMeFqyfLEYzzIWzGB1d0/edit#gid=0

Ta 

[vorsprung]

I spent all day writing 4 unit tests for a react state filter

So excuse me if ICBA

[tonycollinet]

Found day 4 a bit boring to be honest.
Reading arbitrary rules and writing validation code is a bit too much like actual work...

This

Though I am using it as an excuse to learn

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regular expressions

[Pingu]

Found day 4 a bit boring to be honest.
Reading arbitrary rules and writing validation code is a bit too much like actual work...

I learnt more about how JSON works 

[Ben T]

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sed -E 's/([R|L]{3})/ \1/g' | sed -E 's/[F|L]/0/g; s/[B|R]/1/g' | perl -lane 'print oct("0b"."@F[0,2]")." ".oct("0b"."@F[1,2]")' | awk '{print $1*8+$2}' | awk 'BEGIN{max=0} {if ($1>0+max) max=$1} END {print max}'

./day5.sh < day5.txt

part 2 just replace the last awk with 'sort -n' and look through the list, with the help of 'select * from Day5 d where not exists (select * from Day5 d2 where d2.num = d.num + 1 );' (answer is the lowest row returned + 1)

[Greenbank]

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sed -E 's/([R|L]{3})/ \1/g' | sed -E 's/[F|L]/0/g; s/[B|R]/1/g' | perl -lane 'print oct("0b"."@F[0,2]")." ".oct("0b"."@F[1,2]")' | awk '{print $1*8+$2}' | awk 'BEGIN{max=0} {if ($1>0+max) max=$1} END {print max}'

No need to split things up, just treat the whole input as a single binary number. Also use tr instead of sed. Use sort -rn and head -1 to find the max.

cat 5.inp | tr FLBR 0011 | perl -lane 'print oct("0b".$_)' | sort -rn | head -1

[Pingu]

Day 5 done.

http://www.pinniped.plus.com/AoC/index.htm

[Ben T]

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sed -E 's/([R|L]{3})/ \1/g' | sed -E 's/[F|L]/0/g; s/[B|R]/1/g' | perl -lane 'print oct("0b"."@F[0,2]")." ".oct("0b"."@F[1,2]")' | awk '{print $1*8+$2}' | awk 'BEGIN{max=0} {if ($1>0+max) max=$1} END {print max}'

No need to split things up, just treat the whole input as a single binary number. Also use tr instead of sed. Use sort -rn and head -1 to find the max.

cat 5.inp | tr FLBR 0011 | perl -lane 'print oct("0b".$_)' | sort -rn | head -1

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Interesting. Would however argue that using awk to get the max is more performant than sort because it only requires one pass, at the expense of less simplicity.