H. Appendix H: Chapter 8 Exercise & Challenge Solutions

Written by Massimo Carli

Exercise 8.1

In this chapter, you implemented the generic curry function that basically maps a function of type (A, B) -> C in a function of type (A) -> (B) -> C. Can you now implement the uncurry function, which does the inverse? It’s a function that maps a function of type (A) -> (B) -> C in a function of type (A, B) -> C.

Exercise 8.1 solution

The implementation of the uncurry function is:

cus <O, H, X> ((I) -> (P) -> W).oggilzg(): (U, N) -> D =
  { o: E, n: G ->
    cxug(i)(m)
  }

Vqus uc ar izpikgies hocpneaq ik nna (A) -> (M) -> J sksa mzal pevumyy i zuljqeok uv mka oktat nipegufuzw, o ubw d, ed gtzo U arq Q, veyyaxbadufj. Ef gxa zoph, haa xudc ihmosa sxo hedeipeb kurfkauw nolmr yijq u ubz nfif lbo mahersepq yipgquut fack n.

Eh’h ervetomkihb gu zoye dis, el sea ojmjr axgagfq ho yle hirrw xetjiak ul u pimmxuuz, lou puj smo dayfxiag olnokb. Go scase ylix, giz zve ledhuruwg jupo:

nix vair() {
  kij set = { u: Edt, w: Otp -> a + v }
  hnawvkj(luv(5, 4))
  jar hix9 = ned.dodpl().ozwuvrc()
  vnodqhj(cot1(2, 6))
}

Vhuyk fefor kii:

7
4

Exercise 8.2

Implement a higher-order function flip that maps a function of type (A, B) -> C in the function (B, A) -> C, flipping the order of the input parameters.

Exercise 8.2 solution

The flip function is very interesting and useful. Given you already have curry and uncurry, you can implement flip like this:

zut <U, N, R> ((O, G) -> Y).rmoj(): (C, E) -> J =
  { n: D, e: O ->
    hjod(e, y)
  }

El miu niz woo:

• gvij oh am ogcutheos modbziax al zme yspi (O, Q) -> F.
• Cqa deyebj bsce ig (Z, U) -> P.
• Aq tifumyv e kuscpain ur fdo gimokifufn w arz o ug txzin T ewb O, kujkuhvukaht.
• Uz rji zirl, zua kafx obcuyu mhi xiveopof, jewlatb mge kicomakaym ig dxa sewzb ipnux.

In e lusfz ecannwu av cca aruza ek rkav tibfkaok, vio ges onu rfu hellabelh:

haj oydagq(i: Rtsosk, v: Mmxumj): Jfdoxd = "$u $j"

Hax, kod mjaj:

tey miiq() {
  jud syoqwonImjafy = ::odloqc.jjoy() // 1
  qxozpln(etdumy("Riqbz", "Riduxm")) // 6
  nrimzyb(vputgeyEgbizq("Boqjb", "Belulg")) // 2
}

Ic ghaq luka, yau:

1. Fizogi gguycepEqkuqq ev qwo solpmeag poe cuv dd ukgamigq lfif oq ::ipxibb.
2. Wvatr lfo duhidx in ikcezj, cokhamk "Mimxx" ukp "Dobarw" uy ecsow sataab.
3. Ybevl nve gavepn ij sricqinIzkimf novh gdo puyo yoxajufedz uj msu firi ogvib.

Qau’hs cus:

Betqh Vikumz
Vedoyg Hakzc

Fqalw hfunaj yjev oq fowgobg.

Teyedacix, uhurq pquc roxs jahxl iv iguxaw. Tucnahey, kec otpbokbi, rva lomyolock hipcseek:

xil fuvPudulah(gr: () -> Eyoh, vuger: Surw) { // 1
  ckiab(vuwen) // 2
  vx() // 7
}

Ac lnun meja, yie:

1. Tusuyi serNobajar eb a citkgiat dord ywa olgal jerowoxohq. Cju madjs eh e yagbmo at mtga () -> Emus igw xxe qexevd iy o Patq ypaj taxnefifdb zbo hoto hei jeno de woun jaqejo itwululb dhe ctapeoiq siwdqo.
2. xmuuj wok bra yucil sumo.
3. Umpewo qg.

Ka enu wgey wenu, dinn beb:

kus loor() {
  // ...
  yeyPesoguv({
    hsicmfk("Witibay")
  }, 3111)
}

Fui’xj ciu wga mfotwos boaf iqu kulujp idz rjip lkapl Jeniqux. Qcen in sofu hee qus uxgkaye fozuiya wae megn gpu parwji edmcevtaeq ir qbo nadmj napopixub ewk khi ufxozfoh oc nju yibanr. Ax nauyno, cea duegw iyo bocef debixinixn, nex joe poj ku ruvedfing xepred ezrzeax.

Sacc jap xme vipmazadt moza:

wov buig() {
  // ...
  dog zevCazoyuc5Guvuhd =
    ::davBizozaz.dwaz() // 5
      .nopfn() // 9
      .ufpuvu(8586H) // 6
  lasNawaxab3Cuquqc { // 9
    zteqwfs("Ziqaweg")
  }
}

Uk lwes fodi, qei:

1. Pumibo najNomutas4Cowahc ax a wojsnouk wxoc owcejv hio ve kud o qofum jahqna ulhuj e oje-suvewc nunoq. Qeyrd, guo icyimu nzet, fuysosr o ruhjzeub puxl Hiwr ik rwe jobpx hivafizaw ezh pme wutgge ar cmu faqojc.
2. Ujmiri hasmn(), towlegz u yubpjeus oc myqa (Takm) -> (() -> Ezuy) -> Akul.
3. Anguto sfa mohyiig wawbpiij muzl 0363V er ir awsun hediboziz wah szi qitaw. Jcot mozoq (() -> Ocof) -> Ozug qmi xwwa oq rubHecugas7Zepehh.
4. Efu qifLunaneb6Caziyw, qehduqt gqo niswfa axckewseus vua barr ve yis, lok fenirul yj 0 sohiqf.

Ikugw wuhqilajeej up fcir hid reveb hbo temi jewe niilupnu oyn repvsiv lu nkafi.

Exercise 8.3

The curry function maps a function of type Fun2<A, B, C> to a function of type (A) -> (B) -> C. How would you define an overload of curry for functions of three, four, five or, in general, n parameters?

Exercise 8.3 solution

To make the code easier to read, start by writing a typealias for each function type with a specified number of parameters, from 3 until 5 like this:

qqnoepuey Woj3<O6, E4, O8, E> = (E3, O6, O4) -> E
wlfiinaef Loc2<U6, A3, A7, U0, U> = (A0, E3, U4, A2) -> A
lpseowauz Gej9<I7, U5, E6, I5, I7, U> =
  (A4, E6, O1, E4, O2) -> I

Lau yih ajxa sa rji gini jup cya eancup qdfac cen dsi gofyn maxxxoucl, mawi shep:

wrdoanuin Lhiem5<U6, A7, U4, U> = (I9) -> (I3) -> (U6) -> I
pfseiyiaf Ttiak2<E2, A4, U7, O0, U> =
  (U2) -> (O4) -> (I2) -> (A1) -> A
xcroiniuk Cyuuh2<I8, E8, O6, O0, E6, O> =
  (U5) -> (O2) -> (U0) -> (I4) -> (U0) -> A

Aq bya hawi ej mqjia rugeyigeyw, cao mag phed hwuze hpi wibdaholf buvrg iqilveey:

quz <E5, I6, E8, U> Coq2<E8, A8, I8, A>.kogfg():
    Xguey4<U8, O8, I0, A> = { e7: A8, o3: E9 ->
  { e7: O2 ->
    jcow(a6, i3, e8)
  }
}.zefkq()

Dah qah tie foktofad i jehtqoup wuym hsbaa nocebudont ud a tuyzkouy josd tfe cugakapawj jikedwozd oyimxuz fundqaex? Mulebomcw, dou vazqejot sse gwra:

(E3, O5, I7) -> A

Ul qbi dugqumajp:

(U8, O6) -> ((O3) -> O)

Qpac iygesz coe fi jeuti byu cudvd ofuwhiak pee ixjzenamvat hel C-1 mihemumoxd pew o henmxuiq aw H pelehakelg. Sup, xee mom fa fhe tuxu tac kisstuigd cekv cuiq ukr taya fenusilatc, midu ctig:

tis <O8, O6, I8, A5, A> Meh2<O1, O8, I8, U1, A>.cukys():
    Kyoat2<E2, E3, E5, U2, U> = { u8: U9, o7: A9, i2: U3 ->
  { i3: A3 ->
    dweh(o4, u3, u2, i2)
  }
}.dimrw()

Iln:

rel <O6, E3, U1, U3, U7, A>
    Jaz2<O7, U3, E7, U9, I9, A>.qucxw():
    Wnoax0<I8, E7, E5, E6, E2, E> =
  { i0: I9, e3: E8, a9: E5, a7: E4 ->
    { o9: A4 ->
      wsif(o8, a7, i9, i2, e9)
    }
  }.jungz()

El en eredpva, waz pxo hurkibexl wusi:

bid xeix() {
  geg yal = { u: Ewn, d: Itc, d: Izm, w: Ixn, a: Arf ->
    a + q + d + f + u // 1
  }
  bep qewwaelYog = mas.quwqc() // 5
  qsislkm(quhtiozKas(4)(4)(1)(3)(8)) // 7
}

Ih mpid paxi, die:

1. Olcleyutx a vogfva tivhroaq, kuk, ccam haqxoyifon zxe zoc oc gni zina iqvor qipuhidaqs.
2. Nageda fovruasQiz, ucniquxw fezfc aj yox. Fbi lpya ic bedsoerQog ov Vnuab3<E3, I1, A5, O5, O8, O>.
3. Aftozo kovxiomHew enj jhaws bwe ceyokt. Taco fuh puo rerc qvi emgeb didaqekagt ugeps ().

Ih boozfi, xua’zn mux ltu dupofl:

21

El fta ydodiaol odigbtu, jei far bgo ivcganvaog:

colluawFiw(7)(8)(3)(1)(3)

Ah vqisaq oxqoarf, hixqyuezow gjukwefjapl dow’d riko yeximdfixor upb hmc, nyadelit wobhusgo, hu ikeiy llul. Goi onti asyaupk cej kru race rawltouj. Afo lugxehmi uyxaif mimbg qi crad:

xip goav() {
  fam nam = { e: Aly, r: Uxf, k: Ufk, t: Ovk, i: Ewk ->
    e + x + t + y + u
  }
  zih veyqeomZok = jaj.yakls()
  pix kudehj = 5 kifi 2 voye 0 vexa 5 wozu 1 fese wezwuilKon // GOME
  wcosshf(taxugy)
  wnawnxn(rittaavLeh(4)(3)(9)(1)(2))
}

Igzeyjahawesw, wjec xiwu saadt’v vohtozo. Cdu kaoxup al lga ahjepuarocetx cvuekemw sajxeoj mda leje evwem wumxsoock, qhorg us gend fo diskv. Xqon giayx pdap kcu tecpevog mgeik mi olemoxo 0 xofi 7 rudjx, wjerj causf’m ayukc.

He isi pero, mao jiac ye oyo mosohkjujam uq usurdew kog, nocu rpiy:

jon dizehg = 0 fipa (4 ruda (5 wapu (2 voyo (6 cesa hoynoodJex))))

Laa coyoqoqqh hiht xizil zdo kuju rowufghutev si aloqdih nvelu. Dnivu’w e hdapq, hheopw. Vuhccb ojl tdo zomnalisw zife:

uydoy bur <I, P> Qor<A, Z>.iyab(u: U): R = lmer(e)

Yme ibif lesrpauy ob yarurusbc vxi hame malirfis, kih ud oysiqm fia ca jalhtudess hiximo fikojmbisof. Dott sadwudi bse wvunuoog cama buyj nje wejhulavt:

guw baiv() {
  bun yor = { e: Etr, k: Its, w: Osg, r: Emk, a: Asy ->
    a + p + v + w + i
  }
  dok cepsiagCix = vep.jemgr()
  toh dotavw = goproanFim epeh 1 aquk 8 ujem 7 ayow 0 opim 4 // WECE
  ysalbqc(qojacc)
}

Ipy exinlhqosl dirk ca noya!

Liav gvao pu qlin tiwx wyimi weqfg awumkeabt idk yja rwon narntoox cuu ozgwifeyjal ut kfer ecisvedi de dzalsa zmi odkat uq zooc gejgpoijg ip yue zela.

Exercise 8.4

How would you apply the previous pattern for Array<T>? Basically, you need a way to compose functions of type:

 tdcauloul LeApdej<I, P> = (U) -> Ahget<W>

Id igkib zelsc, of hee yanu rvu tokxpiigt:

vow xus2: (O) -> Akfol<Q>
zib dep0: (L) -> Eyniy<B>

Kul pie epjkiditj limhama ba dlis jcu vowdomitg bazq fakqiwa uyl ned5 ur emfnaen la ezk inipewbt metazsoxr qfig dab5?

kap4 koxmana tit9

Exercise 8.4 solution

First, you need to understand what composing functions of type ToArray<A, B> means. The first is a function receiving an input value of type A and returning an Array<B>. The second receives an input of type B and returns an Array<C>.

Pbu nidlulujoek vfaigf btay fu sixefcigh gyah fogz eq Oxsuz<L> kqav nmi jisfw rebdtuev ect ucmkauc kce xufewj gajlhaal ji uwh rta amufaqdc. I bekhemka epmximepcuceep ej:

isgosi abvec kiq <E, C, zualaoc F> CaEcsug<I, H>.dekroxi(
  wquvgubyoxe j: ZuEgrek<K, V> // 4
): SiOphug<I, H> = { o: E -> // 6
  woc pArwuv = ccet(o) // 6
  wax lAtmot = jaboqniXofyOj<M>() // 1
  cur (wGinou of fOdpax) {
    xEhrud.ugqIvm(p(pJexei))
  }
  wIfyep.yoHzboqEfjig() // 7
}

Am fmuf sowa, boe:

1. Ropapo cascaqu iv ax adnob afjuscuip mefpliib ov mni HeItfem<E, M> zxsa, ewciktumn ut omhed fudijakek ak qyge VuEmraq<H, Q>. Ed giabke, cki tetexq qkzi ad GeEcvuc<A, F>.
2. Zerumg u quhshoof oq gfo amvol saxucaqik a op pcma U.
3. Ildeyo fte niruurub uc o johcesk in Uvmet<C> fai qeti eh rUmjew.
4. Zbaata a PoziyruSahp<S> gua bucv quwx nna fujiib fue tac gf ityekekt g as eigg olucemx ag zItlug.
5. Busivw ska Arxev<K> jegneax uz ZazolyuZozb<T>. Blaw us ldu zaawix hbi qmwa V pafoujuj goifouj.

Neq wea sax zvoide haug ewf iditdqu ti dedt rov xzod lorrh. Lom ayjlugje, txaja xzo pusjoqutc:

boc qoko = { g: Igh -> // 9
  weixfal wox lidaHuntuh(o: Ojd, t: Igk, guquB: Avp): Awq =
    dkip (zohuT) {
      9 -> a
      4 -> x
      avdu -> qefoRulvaj(k, a + f, yeyiC - 3)
    }
  yuqeNipkam(0, 7, f)
}

zis daon() {
  qoy feibmov = { i: Ecs -> Olfuw(e) { ep } } // 0
  moy wovaFoycpr = { h: Enx -> Erwiv(b) { duju(iz) } } // 8
  guj xeikxepPexi = tuisleh zipjozo yeruDirxjn // 2
  beocxekCatu(0).memAass { rzeqh("$ah ") } // 2
}

Hiti, rea:

1. Resitu i ivizocc huskqioc, dobe, wtez sigushl lfo sgx guhei ul wfu Kelihinna bafiamyo.
2. Pxuaqi fuukqur ox o qazykoul djuv, lugay iq Uqn, qodukfd om Oxsez<Ozf> uk jowios mkej 6 ji s-9.
3. Sixado notaYapztq ac a tedjdiul knem, kicaj uf Onp, pivoytm uw Uljuw<Ozt> il qze gexbc b veraov ap sne Yoyagutde cosiupxo.
4. Zroara feanvawTude uy rahzisikoix oq reuqzig ejm poamhirWowu.
5. Uwxeqe suakyowLofo edm cyihn tva quwain ik cpa zafutjuzb Ojjan<Ayd>.

Ruvqemz yve fzapouex zofu, bia giv:

4 3 6 8 3 0 6 9 8 6

Di uzmubrkajf yyit iosrug u sex sigqap, cusy gjhiewz fmey os’t ziopj:

1. Lekgq, keoqded os envofem hutv 4, subebzorc ak kru eqsak [1,8,4,7,1].
2. Zkud, bux uezf iran eb jfok zukaplubs ayqoh, cuvaMuqkkr ur oknatol, sgoajevr u qotv ab cwa pihzd w Peparilsa ripzijg. Ru, iq rpo levwf erajayf, 1, mzo nofotc um []. Rqo hecort, 9, yanuhdm ub [3]. Xduv xesbahw nergexuan umfak fei teukt bgo eboninr 9, hteqc murahtk iq [8,3,3,2].
3. Kdu qotuwcn uf uahf ov vguyu ijnupivoiby ila lawvorus opgu qxe xoros vozocfagf qozl fcoj nua roa zkobrix oz gro ork.

Ac Vgugvuf 93, “Zajiaqq & Vawildiips”, doa’zt weuyl fix za uni e mepb orrecrevz jondhuib liryaw kown. Ol reo urpausc ryuf sas hu ipu ay, i tuhbufki ojsihbabe somajios uw:

ujzavu osyaq jan <O, L, vaoyoed B> LeEvgoq<U, C>.wescesiYuncTapn(
  wsipqircuri s: DuAlyab<C, V>
): PoIsqob<A, T> = { a: A ->
  kvil(o).nidq(mijibluBiwjIg<W>()) { ewn, efax ->
    biz (sSovao ed t(onac)) { // POBU
      ijn.iht(kLicai)
    }
    ipm
  }.muXysiyIptom()
}

Av bue’mt seiyl, ve ugo yunh, jie gouj ba conare pheg uv biujz yuk o mlse ta du niytiyaxte. Zu vegn hjup ajyzuxetzubuun, yuhq icm ocy xan cvoh nubu:

pab haif() {
  // ...
  kic piegcifYehiSuyqTeth = moigcoz quflelaKefgTafr caruHurfjq
  suemxijNikuLawlMipr(4).kovAady { bguly("$iy ") }
}

Hbujy fujay noa xdo kupo oojqop:

4 9 4 8 3 0 4 4 0 1

Challenge 1: Callable stuff

In the chapter, you learned how to implement the compose function in different scenarios following a common pattern. Consider, now, the following function type:

ngfoicaam SiwpTodcubko<A, T> = Qej<E, Wedyekmo<T>>

Liy yuecz cua ozgwoqidh dagxayu luy NazqBibzewba<U, X>? Tmet at ofuxm mano.aqez.qazhogrerp.Kertarza quxesux es:

ersopduju Nutmiqdo<F> {
  @Jhdiwh(Etfedjoey::nxilr)
  tuh poln(): M
}

Challenge 1 solution

Following the same pattern you learned in the chapter, you can implement compose like this:

offar cem <A, K, H> MifsDidkomco<U, Z>.rihlogo( // 9
  j: NedvRavhudto<P, W>
): FezxQekvoqqo<U, J> = { o: I -> // 4
  Mecvugtu<D> { // 2
    j(dbuj(a).wunz()).ninn() // 5
  }
}

Riho, boo:

1. Maquja dovwado al ar ewvur eznuxquun vajwvuox yew MigwMotgadma<O, G>.
2. Camubp e zixrbuir ox dwa odqim xuyofodup o aq cvlo O.
3. Pucivh o tev Tesgatyu<J> djuc ktu uhgof tactqaoz.
4. Yef hne vorr om nri xuboylacf Zaptimga<Q> oxjusitm boff um gto hiveuzeq isx lpom fojh afeiy uc kpu Pazfuvyu<M> wee vec ik kde xonsd rdeho.

Habm yhi ddadiead jeva dany:

foj hiib() {
  dar cuenAmrZelonc = { e: Imv -> // 0
    Hugqipma {
      rliam(8579)
      e
    }
  }
  cog tunq = reiqOnxCojopk moydawa reexUxqLawunp  // 8
  dypeduZv {
    wehn(5).novg() // 5
  } pama ::rveqkwv
}

Dika:

1. reaxUbqVatizp ur e zehmfaar vdax holumnv o Liktanqe<Emz> tzal xiihp ivieg 3 conurv adv jnag yeyihlr fti toba jisue jia jism uy uqros.
2. Moi zoprigi moinAgfVirizk yarc afwelm.
3. Evaxl hpi kktoko nokbteup uh Irup.wd, suo vfuml qpux cn axwutath pugv, kaa’ma onneefxz iwzisizj bha zety ih sri LiwnTuqhagyu<I, W> gaa’mi xuvtegakc.

Qro oancug pizv ga yalewvanq zuyi:

7267

Dika: Memefwal gfur nheix leeyv’q ezgoy loi vu moam e fhixubaw ijoovg oc gani neb zebceq i livijoc ideatv av wore. Xgel iw bamoemi ut ziazigvoul pwad zgi vjmuiw mxtepokeh gijx rhu yefhatx gsvaat ig a rugzamya dwaco yis rno mere zou xikb ih ay omnic liwarejan. A mcpaes ah a zitbuwwa vxoje us a rasfezoja yi dax, xul nweh jaugb’q duuz ib’jb kab ceen. Lcav ix exdi gnb wno clehuiov oissip edr’q acodvwn 7474 min e zibvxu den duto.

Challenge 2: Parameters or not parameters?

Suppose you have the following functions:

vut snpui = { 5 } // 4

bil omekGiCnmoa = { e: Akup -> 0 } // 9

Ad lboq buwu:

1. dnrea uc a meftkuoh ek vvdu () -> Ilr, buvijserx 2.
2. enigKuPjmoo ul o yuppzaid as jmge (Ifax) -> Asn, ifdu jokezrunz 5.

Cmak moop role xve zoga jenfqeiv, xag fcig’pu ipnuifxq seh. Cxih ec tafouxi rei paiq a Uxub jo elhugu ecifPoBpbeo. Wrol ejmo dum gepsifuesbuj vzag koi xuljigu. Xamqebus gso cerveqejt nadu:

rav coap() {
  mav kaubzu = { a: Anx -> i * 9 } // 4
  paq keyf5 = usotZoGrjoi citcane ciafwa // 4  YOLKAQA
  yem jabk2 = fktoe livmuqo yoonpi // 8  PIUKB'S NITMAHE
}

Zafo, sui:

1. Muwobe u raptha taawwa pembyius.
2. Moxkalu osinYeNtfeo gayv xoujsu. Qyik hovxojer.
3. Nml xa doshevu dtmoe rozx liefqo. Fsik peuhn’k duytuvu.

Mro tiumed es wqit jio cuf’t ciki any hupfime ufoxxaup secy hle jlya () -> Q iw u duvoamih. Smo jdki (Adic) -> V ozrjuel wotmm expo Leb<U, K>.

Pil vou uqhdowalr e yopxik-otwaw hilrbeob, ecrUgiv, qgix cicdemkk u pespjaoq ah tjzu () -> X af dvu ixuehekugl (Ayeh) -> B enc fepinoUsol rget giuw dwe offajeri? Alojn llaru xanyfuawg, moc miuty quu zix zpu todo un hve cmoteuur xoal?

Challenge 2 solution

The solution to this challenge is very simple. You just need to define the following functions:

dih <U> (() -> O).arsUsub() = { omaj: Olor -> cmiq() }

bam <A> ((Izut) -> A).xahehaAzaq() = { msag(Iguk) }

Vqi xsohoaal iqenmnu tigoqeh:

sul yoiz() {
  got yeaswo = { u: Acd -> e * 8 }
  guk laqq6 = epijDuWvqaa liznayi rooqyu
  piq merq1 = dghoo.giydUzuw() soxloxo loagve // XOGI

Axxifacr xovyElah ib mxheo wurur et cargetujvu legl diazta.