This one is not too hard, but a brain teaser nonetheless.
The key thing I learnt today is that strings in Julia are arrays of characters, so you can use set operators to check for things like "ab" ⊆ "cab", which should return true.
Part 1
Show challenge - day 8, part 1
You barely reach the safety of the cave when the whale smashes into the cave mouth, collapsing it. Sensors indicate another exit to this cave at a much greater depth, so you have no choice but to press on.
As your submarine slowly makes its way through the cave system, you notice that the four-digit seven-segment displays in your submarine are malfunctioning; they must have been damaged during the escape. You’ll be in a lot of trouble without them, so you’d better figure out what’s wrong.
Each digit of a seven-segment display is rendered by turning on or off any of seven segments named a through g:
0: 1: 2: 3: 4:
aaaa .... aaaa aaaa ....
b c . c . c . c b c
b c . c . c . c b c
.... .... dddd dddd dddd
e f . f e . . f . f
e f . f e . . f . f
gggg .... gggg gggg ....
5: 6: 7: 8: 9:
aaaa aaaa aaaa aaaa aaaa
b . b . . c b c b c
b . b . . c b c b c
dddd dddd .... dddd dddd
. f e f . f e f . f
. f e f . f e f . f
gggg gggg .... gggg gggg
So, to render a 1, only segments c and f would be turned on; the rest would be off.
To render a 7, only segments a, c, and f would be turned on.
The problem is that the signals which control the segments have been mixed up on each display.
The submarine is still trying to display numbers by producing output on signal wires a through g, but those wires are connected to segments randomly.
Worse, the wire/segment connections are mixed up separately for each four-digit display!
(All of the digits within a display use the same connections, though.)
So, you might know that only signal wires b and g are turned on, but that doesn’t mean segments b and g are turned on: the only digit that uses two segments is 1, so it must mean segments c and f are meant to be on.
With just that information, you still can’t tell which wire (b / g) goes to which segment (c / f).
For that, you’ll need to collect more information.
For each display, you watch the changing signals for a while, make a note of all ten unique signal patterns you see, and then write down a single four digit output value (your puzzle input). Using the signal patterns, you should be able to work out which pattern corresponds to which digit.
For example, here is what you might see in a single entry in your notes:
acedgfb cdfbe gcdfa fbcad dab cefabd cdfgeb eafb cagedb ab |
cdfeb fcadb cdfeb cdbaf
(The entry is wrapped here to two lines so it fits; in your notes, it will all be on a single line.)
Each entry consists of ten unique signal patterns, a | delimiter, and finally the four digit output value.
Within an entry, the same wire/segment connections are used (but you don’t know what the connections actually are).
The unique signal patterns correspond to the ten different ways the submarine tries to render a digit using the current wire/segment connections.
Because 7 is the only digit that uses three segments, dab in the above example means that to render a 7, signal lines d, a, and b are on.
Because 4 is the only digit that uses four segments, eafb means that to render a 4, signal lines e, a, f, and b are on.
Using this information, you should be able to work out which combination of signal wires corresponds to each of the ten digits.
Then, you can decode the four digit output value.
Unfortunately, in the above example, all of the digits in the output value (cdfeb fcadb cdfeb cdbaf) use five segments and are more difficult to deduce.
For now, focus on the easy digits. Consider this larger example:
be cfbegad cbdgef fgaecd cgeb fdcge agebfd fecdb fabcd edb |
fdgacbe cefdb cefbgd gcbe
edbfga begcd cbg gc gcadebf fbgde acbgfd abcde gfcbed gfec |
fcgedb cgb dgebacf gc
fgaebd cg bdaec gdafb agbcfd gdcbef bgcad gfac gcb cdgabef |
cg cg fdcagb cbg
fbegcd cbd adcefb dageb afcb bc aefdc ecdab fgdeca fcdbega |
efabcd cedba gadfec cb
aecbfdg fbg gf bafeg dbefa fcge gcbea fcaegb dgceab fcbdga |
gecf egdcabf bgf bfgea
fgeab ca afcebg bdacfeg cfaedg gcfdb baec bfadeg bafgc acf |
gebdcfa ecba ca fadegcb
dbcfg fgd bdegcaf fgec aegbdf ecdfab fbedc dacgb gdcebf gf |
cefg dcbef fcge gbcadfe
bdfegc cbegaf gecbf dfcage bdacg ed bedf ced adcbefg gebcd |
ed bcgafe cdgba cbgef
egadfb cdbfeg cegd fecab cgb gbdefca cg fgcdab egfdb bfceg |
gbdfcae bgc cg cgb
gcafb gcf dcaebfg ecagb gf abcdeg gaef cafbge fdbac fegbdc |
fgae cfgab fg bagce
Because the digits 1, 4, 7, and 8 each use a unique number of segments, you should be able to tell which combinations of signals correspond to those digits.
Counting only digits in the output values (the part after | on each line), in the above example, there are 6 instances of digits that use a unique number of segments (highlighted above).
In the output values, how many times do digits 1, 4, 7, or 8 appear?
Here’s the provided input. It’s quite long so don’t try to scroll through:
Show input - day 8
ecbad fdeacg gaecbd gbae gfcdbea cadge fcagdb abc cfdbe ab | beag bac dacgbe aegb
gad agcfb afegcd afed gacdf gdfce ad cfdgbe cfgdeba bdcgea | gbdecf cdgeaf abcgde ad
ebadf ag efgcdab fgced edgbcf begcfa adgef gcaedf afg dgca | agcd agdc fagcbde gfa
bgaefd gfcbe fgeda dbf dafbgc dfbge bd bdgcaef dfecga ebad | dbf db edab dbf
ebfcgd fedbc adc da gafdce bdaec bdcegaf edbafc bdfa gbcea | afbcedg cedbf decfb eacbg
fdabec adefcg dcf gabcf fedgcba gdaebf dfcga gfead decg dc | beafgd fcd dfc dcf
cgfbd gfb cgfa dcgeb fg bdafge gcfbad acbfd fcdabe aefcgbd | abfdc cgfa gbf gf
dfae aeg cfeag ea fgedac fdcbega agebdc cdegf fgcedb abcfg | feda fcgba ae efcgda
eafdgb bfaceg fba bf bgdf agdecbf cdafe gabde dfbae agcdeb | bfeda ecdaf fb fab
agebf dgbcef dagc ebcgadf da dfcge egdaf dea dfecag fabecd | gefab feadg ead agdc
afbcg bdgcfe aecgf gb dcafgeb bgad agfcdb bfg bfdeca bafcd | gdba dabegfc decbagf acfdbe
ag acdgfb fecgbd edacbgf dgcfb gbad gaf ecbaf cagbf gefacd | ag cebaf gbdfce ga
ecbg gb acdbg dgceab acdegfb beadfc cfagd ecbda adfegb dbg | efdagb gebc gb bg
cdafgb gdebca edgbf becf fb fegcbd fgb egdfa cadefgb bgced | bf cbfaegd fecb fceb
bacgedf afdgb ebagf bfadec ebafcg ecabg fae fe ecfg cdabge | fe adcgebf gabef ef
cdgbaf ebadfc afcbdge gfeadc fgcde egdbf ce cagfd geac cef | dgfaec agec fgdaec ce
dcag dfeabc cba bfcdg ecdbfg egafb afgcb gcaefbd ca gfcabd | fcebdg dbegcf bcefadg gdfcb
cdgefa fdebg abfdg dba agcdf bcadfg cbadfe dabfegc acbg ab | fgadceb fbcgead bda facgd
fbg bf agbfde bgcdefa fcdb gcfebd efgcb ecgbd gefac gaedcb | cgbfe cfgbe fcaeg bgdfae
dafcg cefd fcg geafbd deagcf fcdgbea fbgcea abgdc cf gdaef | gcf fcg dfec fc
cbea gebdfa ecbgf cfgdbe afgcbe gae adgbefc eagfc ea gfdac | agbfec eacfbg ea begafc
ba dfaegb afbd gfbea cegbf caegdb fdaeg dcfgeba feagcd abg | bafge dfba feagd gceafdb
dea bgade geabf adcfbe adgebf adgf bdegacf cdebg da agcfbe | dfag dafecgb dcaegbf cedbfa
fedcag fdebgca dfeab gdcaeb be fdeag fbge fadegb aeb bcadf | eba bae ebcfgda egfda
dcfb cgf abgcf gdcab adcgfe cf dgcbfa fabeg beacgd fgdebac | cgfdab deagfc feabg cf
adbgc cabgef adec dfgbec gdbfa cbgdafe dgc begac cbaged dc | cade bgafec gdc edca
agef eac fbceg ecgbfa bcdgea fceadbg cbfda eabcf ebfdgc ea | gfae badgfce acgdeb ea
cfb cdegfa badf gcaeb dbacgf cagdf eacgfdb fcagb dfgebc fb | bf gfdebc fcgdba cafgdb
dbgfae bgfeac bc cgb gecfd facb fbega fdgbace gbefc adecbg | edafgb cbg dgeafb fabc
geafcdb ec gcedfb acgbfd ecd cedfg gaefd bcfdg ecgb dcafeb | cde dfcbge acdfgb fegcd
decab acegf bfgdec cabegf bfc aebfc cbdefga adgcfe bf bagf | aecbfgd cfb bf cfdgbea
egf gfbade dbecg gdfeb edaf cadgbf fe cfaebg bgedcfa fadbg | fe gdfabc gfdbe fdbcag
fdgecb geb gdcfeba edagf dcefga dgbac bfdgea be eadgb ebaf | fdegba caegdf deabfg dgbea
egfbad efgacb bgfea cea gaefdcb cdeagf aecfb cgeb dbfac ec | ecbaf bafedcg ce dfaebcg
gecad dbeag dfcgea bd deb eabgf eabdfgc agbdec bcad fdcgeb | edb adgfce eagbf db
gdbac cbgea cdb gadfb cd faegdb dgfc dbceagf fbcadg ecadfb | facbdg cafdbe dc gcfd
cagbe abdgcf ca fcbgea baedg gac gdbfce ecfbg aefc gcfadeb | cafe face abcge febgc
gaebd fcba debca adc dacfge cfbde afdbce dgbefc ac fbceagd | ac becfd fcdbgae cefagd
cfdgbae abcgdf afbcg aecbg eb gedac bfgcea cbef fdbage bea | eabcg gcfba be geacdbf
cbfga faeg bcfae gbcda efbdac bfgcde bdefgac gf gfc agfbec | dgaebfc bdcfaeg gfea fage
fb egacb bdacgfe afgbc cbf aebf adcgeb fbgace fedgbc cdafg | cfb cfbag ebaf fb
beafd cafbe gedfca dfebagc bceg ec fcbdag fbecga gacfb ace | bgcfad afdgbec ec efcab
abefdg bacgd ecfg fag ebfca fcgba gcefab beafdc fg agefcbd | fgec cfeba fga gf
degf adgcf edcfgba ead cbaefd bcega cafdeg ed dagce gbcdfa | ead aebdfcg agbdfc egdf
gfecdb dacfe dacgb fb facbd facedb bdf fbae bcdafeg aedfgc | bf cedaf fgacde dfb
gabdf cfbegad dbagfc decaf bgdecf aedgf eg egab gaedfb edg | eg gadfb abdgf gde
dgaf faecb bacfdg gbdac abdegc cbfgade fcbag efgcdb bfg gf | defacbg gcdfeb fgbadc bagcf
adfgcb feg aedbg fdgbc gafcebd bdgef fe fcdegb eafbcg decf | ef fdce eabgd faebcg
fb fgbde bdgefca cgbf fdb deacgb cefdgb bdafec aefdg degbc | bfcgde bdgec befcda dfb
efcgb fgcaed fcdbga fed acfdbge eadbfg edgfb de deab bfgda | fbgda ed fgdbae aebd
ecfabg dfegbac ecfbd cbdaf gceadb cgbdfa dagbc af fba gafd | fa bagfcd bafgdec af
bae dcbfag fadgebc acgbf fageb cbfega bdafce ea bfgde cega | eba efbgd eab bdecfa
dfgeacb fgcbed cfbdg gbcaed fcebg cd fced ecgbaf fdagb bdc | dc bdc bcgfde dcbfage
cgbeaf adefgb gbefcd cbfa bcfedag bfage cb ceb gcdae gcbea | ecb cb cegabf eacfdgb
gfbea adgfb dfbage fad fd agcbd defcga fdeb baecfg gcbdafe | df dgfab egbfa afegb
acedbg becgdfa dcfe gbfda gdceb bcf fc fgbdce cdgfb efcbga | fcgdb fbc bcf bcf
deb gfbdeca cead agbed dcefbg ecgdba bgacd gbfea de fdabgc | cegbfd feabg bed badfgce
acegdf cdaefbg acbgf df agced aedgbf gbedac adf cfed gfacd | dcegfa fbeagd edcf df
ecbfa dbegfc bcdga bgafedc bcadge gdcafb de bed bcade geda | caebf bdcafg eabgcd cbefa
edbcfg agedf bacgfd bdagf dfgcaeb gdabc bgf gdbaec fabc bf | bgdcef bf gbf adgcb
fg afegcb gbf aegf gbfceda fcbgad decbg fabec gcfbe bacdef | bgf dgbefca fg agfcebd
eb fgebdc dcafge bgdeacf acedf bdae fecbda ecb fcgba cbfae | efgcda facdeb eadb ebc
gbafecd bgaced cabed eg bega fgadc fadcbe dge cadge dcgbfe | edg cadbfe cfgad ageb
badc fbcgde bgfaed cb febca fbc eadcbf gceaf aebfd bdfcage | acdb fcb bcad bdac
ecgfa deafcg bcdgfea ed cdaeg acfgeb aedf ecd gcbad bgcfde | ecd befgac de cgbad
cbeda bc ecbfdag gcabdf egbdca bgce aebfdg bcd gbade dcfae | decaf gacfdb bgce cbeg
ecdgb dc dbc cfegb decf fgacbe dafcbg gbcefd bacdfge gbdea | fdabgc cd gecadbf cbfeg
afbcdg gcaef degfca ec cge aedc efadbcg fgaeb gcfad fcdbge | aefcg fcbgda ec acde
adcfgb afe acegf cdefgab dafegc aedbfc dfcga fegd ef ceagb | dafceb cdaegf gdfe fe
gacdf bedgac bgfe bg gdb feabcdg fabdeg gdbaf bfade afdbce | begf bg egbcad gdb
beda aecfdg dgfaceb fegbc gbaec cba dbfacg gaedc ba dbcage | acdegf gcbfda aedcg edab
dbce ecgab cadge fcabg bfadegc ebdafg beg eb becdag fedacg | eb be efgdba be
cagebd df agdec fegcdba abfdce gfda cdgfe gdecfa cbgfe fde | egadfc fd cgdef dfga
da dgcafb acfbeg abgd dfbec fbacg fad fdagecb egafcd bfadc | feabcg fgaceb cbfeag bfgca
dcegaf dabgcf bedac febgadc fcdaeb eagcb cdfae dab db fedb | efdagbc bfdcga ebcagfd dab
bfc facgb deagbc fgeac badf bgdac bfdgac gcadfeb bf cdfbge | abgdecf afegc efgabcd fb
afbcde da bcfdge dbecg cad dgfebac dgbac edag cgbfa baedgc | gdeabc dega abdegcf dage
abfdc fa gabdcf gcbfd dfaebcg gbfeac bdeca acf cgbfde agdf | gdfa af caf fa
cfdg bedfca cfdbeg fdceb edcbga agbfe cg fcgbe bgc gaecdbf | bfaegcd bgc bcg cg
gcdafe fgec eafdcgb agdcf cfdea dgf fcaebd gf eadgbf adcbg | efcg cfge abgdef fg
cadgf dfacb cb ebdcafg cbf ebcdgf dabfe fbaged bcea fdacbe | gdcaf gbcefd dagcf cfabed
gcbdafe acd bacegd feca ac fgbdc facdg adegbf adegf dcgeaf | ebgacdf begdfca cgdaf cad
bfdc debfac acdeg baegcf afecd afd fd beacf bfedga acdgfeb | baedfc egcafb dfa afecd
geacfb efdacg eag cagdf cdea dgebf eagdf fgbacd dbgface ea | cdae acdgf acdgf efgdb
gafced eacgf cgbfe egdca eaf af eacbgd bacefdg adcf ebafgd | dbgaef agedbc fcda geacdb
eabcg gadb acgdeb db cegabf fecad dgefcab daceb dcb cbgdfe | db dgab gdba defgbc
eacf acgdfeb dgace cgfda gbcfad ebdgc efbdga aed ae cegdfa | cfgad cgdfa ae gadfec
bfdacg gdbfc cfbadge fcagb bcfdeg adfg dgebac ga agb bfaec | fgbeacd bdgeca ag ga
fabedc ed abegfcd bdfga feacb fecd ebadf agecfb aed adcbeg | ed fcde agcbdef fadeb
afgdce ecgabdf fc debaf cadbf gdabc cgadbf cebadg gfbc fca | fgbc eafdcg ecfdbag gdcabf
gad agedcf gebfadc gafbd fbcgd da abdc begfa bcgdaf dcefbg | efdcga gbfecad befga bagfe
feg afgb abdceg ecdgaf faedbgc afbcge cebdf efbcg cbage gf | gafbcde bgecaf feg egbfc
ba fdcagb dab adcefbg edacg dfecb gdabec cgfead gbea cedab | adb abd baecd cedbf
ged acgd dg cgabdfe cedgba fdcaeb baegd adecb bagef dgcbef | edg edg cdaeb ebfcad
bgdfe dfgae eabgfdc fdaebg egb afbg fcdbe bg agdebc degacf | egb egb bge bcfedga
adbfc ceabdgf aced facedb gadfcb dbe cdfbeg fdaeb efabg de | deabfc geadcfb ed fabdc
bc badgce gecb dabcg egfcad bdfaec cdgae abgdf beagdfc bca | dcaegbf cb dcbfage gdbca
debgafc becgda bdgfca dea bcdfe dcagb ae ecgadf egba ebdac | ebgdac gdaefc bgae eagb
gbdae adegbc daf gadbf fd dgef bfacg eafdcb cgefadb eadgbf | dfgab df daf adf
gbad fbdecg bcgfea bdafc defac dgbfc cgbfad ba bafecdg fba | bcgdf ba edcgbf efgdcb
ged gfecab cgfbdea aedc agefdb aebcgd ed dbceg abgce gdcbf | ged ged aegfbc dge
gbedcf be bfe afgdb aecb decafgb cgadef aebfd dacbfe faedc | be dcegbf cadbfeg cdfaeb
adfeb ecfadgb cbdefg ebdfag efag af daf acdbfg caedb fbedg | fdgbe daf daf af
gcbefda agfebc cedaf gface fecgad dc eadfb acd cdeg gfbcda | cgfdbea acd dcfea dc
gecafd aebfgd cde gcdf gdafe adcbge ecfba cgbfade cd decfa | egfadcb gbfdace cagfed fgcd
fgcdb gefcbd agb ag dcfegab bgcfda fbadge gacf bcgda dbace | degbafc agb dcgba gab
abfgde acdgfbe gabd gfecdb dfeca abgfe fdaeg gfd dg begcfa | bdag dgab gdafebc dfcae
eadcb dbcfe dgecbaf agdcfb cf fbc cbdfea abgedc fedbg ecaf | cf bfdace fbc fcb
ad efdgc cdafbg fecadg dbcegfa fgbae gdfae fbecdg adg cdea | decgbf acgbfde gabef da
gbf cdbgfea acfeb bedfgc adgbe gfeadb cegbad fg fdag afegb | gfb bgf cbefdga cafbe
adbcf gefb agbce acfbg gbfaecd fg fcg gceafd edagbc cfaebg | beagc fdeagc gf gfc
gdfbcea cdaeb egdbc fcgdbe gc cfdage gdbfe eadbfg fcbg cge | fbdecg fcegda bfdge bagdfe
adfbgc aegdb dc ecgdab dceb cedag fcgea dac ebagdfc dfbgae | efacdgb abcgfd egdab dac
dfg eacgbd ceadg gfac cefadg fg efagd gcfdbe begdacf eabdf | afgc acegd gf fg
fe eagfbd cgfab egf cadge begacd cedf caefg ecfagd dgbfaec | adegcb fe gdecbfa ecdf
da cedabg cfegad edacb bgcfdae dbcge gabd efbdgc adc ceabf | ceagfd bcefa cgdeb da
cebgdf bgcad dgface agbecf dfceg ecdbfag ae edaf ecdag eac | eca decgbf daef acefdg
aebgcd fdgcabe gecbfa fbgec bgfad gbecdf face cab ac acgbf | bfceg ca agdefbc efbgc
cgf caedf dfgabce gf ceagb bgaf cegdab fagebc dfcgeb fcega | eabcdfg fg gfc fgadbec
egfabdc gbfaed fcdea egadb fdgecb beacdg gc cge cedag gabc | bdgcef dceag gbac acedf
ecfdgab bd cadb bdg cbgdfa dbfcge bdafg efgda acbfg afbcge | bgd bd bgd dbg
aef acbefg ebcgf ecafdb degcbf fcadg afcge gaefdcb ae bage | daegbfc fgeabdc abge beag
egdc dacgf cafegd ebfgca cadfe gc gdbfa gfc dgfcaeb dacbfe | dfgeacb gc cg fgc
ebadgf acegfd ad debgfac efdba ebdcfg gbad gdfeb beacf aed | fagdbe da adgb gebcfda
gecfb ebadcf gbdcae ecg fabce gbdef fabceg beacdgf cg cagf | ceg fcegb gce dfbeg
dabcge gbdafe facbed fbacdge ecgfa dabeg bdgf fb bef fageb | bdeag bcdaef fbgd bdaeg
ac cfabe ecfbgd cbefg gbceda efcagdb edbfa bca cafg ecagfb | cgefdb cfgaebd ac edgcfba
cef abefcgd cabfe facgde cabgef agdecb fabdc fbge cgbae fe | agbec fcdabge gebf dfcega
agcbed aefgdbc ca gfabd dca abgcdf cbfa efbadg fcagd fcgde | ca cad gedfc dca
gadfcbe dc bcdag cdb dbcegf dbagce dabge gfdeab agcfb eacd | ecfgbd adce fgcdeab cead
cbafed bcefag ga aefbdcg abcef fgabcd aegc bedfg afg gaefb | gfa gdbfe acbef ecag
gdefa agedcfb gadbec fegbd gad da gdaefc cgabfe gfeca cdaf | dcgabef agd fcgea fcgea
fgcea fab cgfbae cefab ecbfd cgfdae dgefab gebdfac ba agcb | deacfbg beadgfc cgab cbga
dag dfebca gd bcdfa adbgc dfgeba cfgbad ebagc badcegf fgdc | gceab dag gd dag
gefbd aedgbf dgebcf ea aefd gea fbgae decgbaf gbafc agbcde | bgfed efad agbfe fbecgd
ebfc gfe ef ebgacfd cefdgb egcbd degfc dcabeg dfbgae dagfc | faegdb fge gefdc egf
dgefab fbeadgc gcadbe bdgcef bcgae edca cbdeg gae cgabf ae | age ae cdgbe cfdgeb
agdbf fe eabdc efd bfdaecg dbcage bfdae bdefcg cefa acbdef | fe ef efcgadb bedfacg
fcebga badecg adcef cfg cdebg cbdegf gfbd gefcd dfbaegc fg | edcbfg bfgd fceda fg
df dfgbca gfed feacb bdf bdfegc dagbcfe cgbeda bdecf ebdgc | acdbfg edfcb debcf debcf
dbc beac edfgb afdce facdbe debcf cb abdcfg gdecfa aedcbfg | eacb becdfa deafcbg cbd
gcdbfa afdceg abdfge befag gba efbgc ba cfeabgd dfaeg baed | feagb agb ba acdbgfe
gfdace ecdgafb fdegb ebdagc gb fcegd dgfbce bge debaf fcbg | bcdage egfbd gdebcf gb
bfa gface fb fabgdc dacfeb becgad bcead fbed dgbcfea eacbf | bedf beacgd gceaf caegbd
adgcb de edafgc egfba eda abecgf geabd abfged egbcadf bdef | ebfdcga ebdf ed dae
cgbed ebdgfc aebdcg bcdga efcbadg cfgab cfaged aedb da dag | bagcfed cdefgab cgdeba aegcbd
efabdg aedc cd fdcge cbdafg dcbeafg ecbgf fdgae dgcfea gdc | gdc dc aegdf dc
edgfca febgac egcfb efcgbd begcfda acg gdbea aegbc bcfa ac | bcaf fcegb bcaefdg bedfcga
ecfba bg cgdafbe cbfag cgbe aegfbc dafecb efbgad agb dgafc | bag fcbea aefcbg cegb
ebagdc fgaecd gcaefb dgceb ebfgd gcb dgfbeca bdac cb decag | cafebgd fcgdea adcbgfe cb
caedfb bdcegf dgac fad cbfdg gfabd gacdfb feabcdg da afgbe | deagcfb adf dfcgb aefgb
adgfbe ed eda dbaec bcagde egcd gacbe bfdac bagecf edfcbag | cedab bgfeda bdfac ed
abcgf cd edbag gbacd fcgd gfebca dfbgca adc bcgdafe facebd | dac fcdg gacdb dc
fagde badfe acdeg gbefacd gf bdgecf ecdgba fagc fecdga fdg | bfdgec dfg gf egcadbf
aedcfb cadeb agbecd cfda aefbg ecabgfd fbecdg bfaec fc efc | dcfa acfd dcaeb fagbe
ecbdga bacefg ea dcefb bcdae adeg dgbac fcgdba aeb cgafbde | dcbea ae cbefd aecdbg
df ecabgf fbde efgba fad dbfgca fdebga fegad cdgbfae acdge | daf agced egfba fcdbega
bfaecg gfdaebc dgec dcb baegc agdfb cd befdac bcdga cdgeba | decfab dcb cedg agdbc
gbadecf gbef gadec ebdcg bg dgb gcbfad dcfbea bfdecg fcbed | dbg dgb gb bfeg
bdfca cbdea dae ebgdfac cabedg efcbgd egdcb ae agce badfeg | dabcf dcabf dbecag cabedfg
edcafb ed aedgbc cedbf dgfbc dafe abefc deb cdfbeag bagfec | fdea ecabgd fcbgd ecfab
afbdce agc dcgf bcdag cedgfba dbgcfa bacfge gaedb bdfca cg | cbafd dcgf cg edbacf
eabfgc cfaeg cbfde adgbfc bface eagb acdbefg ab acb edacgf | ba cfdgab ab ab
egdfb dfbgaec dfaeg abgedf defcga egba feb dgbfc fabdec eb | efb eabg fbgcead aegdcfb
df fbdace gdebcf efabdcg dbf cagbfe efgd gbcda cgdfb cebfg | bgfecd aedfcgb dfbcae abcgd
ebgacf adcebf gcbaf df adbfcg bdcfega cfgbd dbf gecbd fadg | fcbga agbefdc fdb fadg
bagfe efcab cbf cbfgad abgfed faecbg begc cb daebfcg feadc | bcafe ecgb fcb cb
afdecg adbfeg cdgebf fecbgad eafc eacgd ec ceg cagbd afedg | gec ce ecfa fegdca
fgcade fbadeg abcedf ebcgf adcgfbe eca dafge ac cagd agecf | agdc cdgfae begfc geafcdb
cadbeg beagc ba afgbed cefdga bacd fecgb caedg gfebcda bag | gebacdf fcegb gdebaf ba
fbgad fcad egdab fdb cbedfg cbafedg gdcfab fd agbfc abcegf | df dfagecb gbefcd aebgcf
gbacfd ebacfd fcadb eagfbc ag adbgf cedfgab bag dfbge acgd | ga ag ag acbfgd
gebdfc fbedac efbag cdebf dg decg fbgadce egfdb dfcgba bdg | fegdb dfbec fbdacge dg
dbgfc fabcd fcegb bacdfe bgd eabdgc gd bdagfc fadg afdebgc | bcgdf dg dgb gd
aefcgdb faegcb fead edc aecbd bgcad de bfaec bdegfc afdbce | badfec cfgeab gbacef baecdf
fecdga cdgaf gba dbcafg dbfg cbfae agedcb dafgecb bgcaf bg | fdbg bgcfa dbagcf gefbdac
fea gfdbec abdf abefg edbgf bacge dcebgfa af egdabf cefadg | abfgdce dacfegb efa efa
bdgafc ebcgf bfea dgcbe abfcge egbcdaf efc gabcf ef cgdaef | fce fe fe fgbac
gadefbc dg egdcaf fdcbe caegb gcfedb dfbg gde edcbg fedbca | aebgc bfeacd cgdafbe gebdc
ec dcgfe aebgfd cagfbde fec dcgfa ebdc gebdf cbfdge ecbafg | egdbf cbegfa ebdc egcfba
dcbgf dcgeb cf dfc fgdbeac eacdfg gcdabf gafbd adfgbe bafc | cgdafb bfac dgeafc fdabg
aefbcd cadgf eacdg bgade facgbd eac ec dacegbf gfec acfged | gecf bedagfc geadc eac
egcab eagdfc egcdfb aegdf dec afcd cgead eadcfgb agdfbe cd | eafcdg cgdae aedgc dbfacge
ag bfdcge gca bgcef fgcae bagdfc egab bcdeafg afdce abcgfe | fdagcb ag egab faegc
abdfgec bgdca edcg abedgf bcd cd cgeadb fgcab dbgea dacefb | cbd dgbca gabdce dcfbaeg
eacbg gbe egacd dbecfg agedcb bg dgab edfacg cafeb fbgcaed | dgba dgab aecfb gb
fcgabe fecab fabedcg edbca bdgfce afeg efbcg dabfcg fab af | bgedfc afb baf ebgcafd
badcgf bef cgedbf fbagc ceafbdg begfa afgecb eabc egfad eb | abefcg fbgae gebafc gdcfeb
dafecg defbc fbgcae dbcg abfed gecfd ebc cgefbd cb dfagcbe | cb ceb bdcg bgcd
acgefb cae dgecbfa fcead abfdc gdcfe dbae bdcgaf ea dbface | cefdg cea eca cefdabg
fab egfda fb egbcda bgfc gafdb afdbgc dfebac gbadcef cdabg | dbcfag fb bgfad ecafdb
bgace badceg fdeacbg bedc gdebfa bge eb cdaeg dgefac fcgba | be fcabgde gdcafeb be
cg dcg gdabcfe dcgfae gcdea abgdef fgec abedc bgcafd gadef | gfdea cbead ecfg dcg
aecbfg edagc edc edcgfa gcbad gfcae gefd adecbgf bdafec de | cgaefd efcga cagef bgcaef
fedgc dafecg bcedfag agdbc dgbfec bge dbceg bcgfea be bdfe | cdgefab gdabc gbe eadgfc
eagf fcgbad edgabf cfegbd efb dbeaf decba fbgad bfedcag fe | dfbaeg ef ebf fbe
cebfag dbefgc cdgba dcgaeb dfbagce dg bgeca aged adcbf gdc | cbdegaf gefcab abcgd daeg
gafdb bdfea afbgcd fgd dgcfaeb gdbc facbg gd acdfeg efbcag | gdcb adfeb gcafb bagcef
fdaec adbcf acegfbd efbdag afcedg ea gbefdc geac cfdge ead | aegc dgcef fdcae gcae
afcbed dce fgadc gcea abfcgde dgefb ec bfagcd fadgce fcegd | efbdca bfdecag cde cefgd
Here’s the solution to part 1
using Pipe: @pipe
@pipe split(strip(input), "\n") .|> split(_, "|") .|>
strip(_[2]) .|> split(_, " ") .|>
map(length, _) .|>
count(∈([2, 3, 4, 7]), _) |> sum |>
print
Part 2
Show challenge - day 8, part 2
Through a little deduction, you should now be able to determine the remaining digits. Consider again the first example above:
acedgfb cdfbe gcdfa fbcad dab cefabd cdfgeb eafb cagedb ab |
cdfeb fcadb cdfeb cdbaf
After some careful analysis, the mapping between signal wires and segments only make sense in the following configuration:
dddd
e a
e a
ffff
g b
g b
cccc
So, the unique signal patterns would correspond to the following digits:
acedgfb:8cdfbe:5gcdfa:2fbcad:3dab:7cefabd:9cdfgeb:6eafb:4cagedb:0ab:1
Then, the four digits of the output value can be decoded:
cdfeb:5fcadb:3cdfeb:5cdbaf:3
Therefore, the output value for this entry is 5353.
Following this same process for each entry in the second, larger example above, the output value of each entry can be determined:
fdgacbe cefdb cefbgd gcbe:8394fcgedb cgb dgebacf gc:9781cg cg fdcagb cbg:1197efabcd cedba gadfec cb:9361gecf egdcabf bgf bfgea:4873gebdcfa ecba ca fadegcb:8418cefg dcbef fcge gbcadfe:4548ed bcgafe cdgba cbgef:1625gbdfcae bgc cg cgb:8717fgae cfgab fg bagce:4315
Adding all of the output values in this larger example produces 61229.
For each entry, determine all of the wire/segment connections and decode the four-digit output values. What do you get if you add up all of the output values?
Here’s the solution to part 2.
using Pipe: @pipe
lines = @pipe split(strip(input), "\n") .|> split(_, "|") .|>
map(strip, _) .|> map(split, _)
function decode(line)
clue, out = line[1], line[2]
clue = clue .|> collect .|> sort .|> join
out = out .|> collect .|> sort .|> join
len = length.(clue)
dict = Dict()
dict[1] = clue[len .== 2][1]
dict[7] = clue[len .== 3][1]
dict[4] = clue[len .== 4][1]
dict[8] = clue[len .== 7][1]
dict[6] = clue[(len .== 6) .& (dict[7] .⊈ clue)][1]
dict[9] = clue[(len .== 6) .& (dict[4] .⊆ clue)][1]
dict[0] = setdiff(clue[(len .== 6)], [dict[6], dict[9]])[1]
dict[3] = clue[(len .== 5) .& (dict[1] .⊆ clue)][1]
dict[5] = clue[(len .== 5) .& (dict[6] .⊇ clue)][1]
dict[2] = setdiff(clue[(len .== 5)], [dict[3], dict[5]])[1]
dict = Dict(value => key for (key, value) in dict)
return @pipe [dict[o] for o in out] .|> string |> join |> parse(Int, _)
end
@time decode.(lines) |> sum |> print
Benchmark
Some folks in Discord pointed out that using Dict is not good for perfomance.
So let’s create two version of the code and see if it’s any different.
using Pipe: @pipe
using BenchmarkTools
function decode_dict(line)
clue, out = line[1], line[2]
clue = clue .|> collect .|> sort .|> join
out = out .|> collect .|> sort .|> join
len = length.(clue)
dict = Dict()
dict[1] = clue[len .== 2][1]
dict[7] = clue[len .== 3][1]
dict[4] = clue[len .== 4][1]
dict[8] = clue[len .== 7][1]
dict[6] = clue[(len .== 6) .& (dict[7] .⊈ clue)][1]
dict[9] = clue[(len .== 6) .& (dict[4] .⊆ clue)][1]
dict[0] = setdiff(clue[(len .== 6)], [dict[6], dict[9]])[1]
dict[3] = clue[(len .== 5) .& (dict[1] .⊆ clue)][1]
dict[5] = clue[(len .== 5) .& (dict[6] .⊇ clue)][1]
dict[2] = setdiff(clue[(len .== 5)], [dict[3], dict[5]])[1]
dict = Dict(value => key for (key, value) in dict)
return @pipe [dict[o] for o in out] .|> string |> join |> parse(Int, _)
end
function decode_vector(line)
clue, out = line[1], line[2]
clue = clue .|> collect .|> sort .|> join
out = out .|> collect .|> sort .|> join
len = length.(clue)
l = repeat([""], 10)
l[1 + 1] = clue[len .== 2][1]
l[1 + 7] = clue[len .== 3][1]
l[1 + 4] = clue[len .== 4][1]
l[1 + 8] = clue[len .== 7][1]
l[1 + 6] = clue[(len .== 6) .& (l[1 + 7] .⊈ clue)][1]
l[1 + 9] = clue[(len .== 6) .& (l[1 + 4] .⊆ clue)][1]
l[1 + 0] = setdiff(clue[(len .== 6)], l[1 .+ [6,9]])[1]
l[1 + 3] = clue[(len .== 5) .& (l[1 + 1] .⊆ clue)][1]
l[1 + 5] = clue[(len .== 5) .& (l[1 + 6] .⊇ clue)][1]
l[1 + 2] = setdiff(clue[(len .== 5)], l[1 .+ [3,5]])[1]
return (@pipe out .|> findfirst(==(_), l) .- 1) .* (10 .^ (3👎0)) |> sum
end
lines = @pipe split(strip(input), "\n") .|> split(_, "|") .|>
map(strip, _) .|> map(split, _)
@benchmark decode_dict.(lines)
@benchmark decode_vector.(lines)
Here’s the output:
It seems that the implementation using Dict is almost twice slower than using simple Array.