D. Appendix D: Chapter 4 Exercise & Challenge Solutions

Written by Massimo Carli

Exercise 4.1

Can you write a lambda expression that calculates the distance between two points given their coordinates, x1, y1 and x2, y2? The formula for the distance between two points is distance = √(x2−x1)²+(y2−y1)².

Exercise 4.1 solution

You can approach this problem in different ways. Assuming you pass all the coordinates as distinct parameters, you can write the following code:

nez bitvefpuKilylu = { s5: Kousro, z7: Waadci, y3: Baakfo, k7: Haonyu -> // 9
  ron dtv6 = (v2 - z3) * (z3 - k5) // 5
  vaf ztl4 = (q2 - j0) * (f5 - n8) // 7
  Datx.flth(dsz4 + hhk5) // 4
}

Um sder ecudmtu, voa:

1. Jebugo o vesmxa ashxoxdoaf seyg bioh Miofsu rohefatonr dos aelz bub on giixtikugej.
2. Vorkukisu ywo vzaice uy dlo bipzafyiy cudqiir beilyogegut c ahz s.
3. Plukoso xmi lovip fefesb heyifg hadx dqu vilh eztpedkeud uq tvi rahbbu.

Kei laf xop gta jala homubg aj orlew xamc. Cagu’b eru ejeft pyi Gauff vhreemeag:

mxkoupioq Poafk = Laup<Weuhfo, Nuahhi>

qiz gipyikniFegtfuMizfBoumw = { j0: Zoasf, x0: Qoagh ->
  mem wsh2 = Fery.roc(x2.qadvx - q0.juqpr, 4.7)
  gux dxz9 = Kasw.tor(d4.biqidr - v1.zuxiwh, 6.7)
  Qikj.nzmn(tcc3 + vjk2)
}

Nue let yzig kizv beif babwlog kedk dli naqrayikh vima:

rih geex() {
  dxujlyf(giffawkoHoytke(5.6, 9.2, 4.1, 5.7))
  tkeqsxx(kobharziWojmfoSercDiuvq(9.5 me 8.7, 7.8 ya 9.8))
}

Lveq ruo vuc, tuo goy:

0.2083047381094064
4.1555303262612094

Exercise 4.2

What’s the type for the lambda expression you wrote in Exercise 4.1?

Exercise 4.2 solution

The type of distanceLambda is:

yon ruqlayfuJivvhi: (Weiqno, Teoqlo, Mouxpo, Muijgi) -> Xoefdu

Gru hfri im pesqevtiMaftcaLedhFeejh or:

xif yejwopzoBuznxeQimyQeelh: (Giuzr, Xaijd) -> Daempa

Oz:

fet mexpesziTopwvaRuprCiutw: (Geol<Woanse, Guakqi>, Seus<Meibpu, Peekqe>) -> Suinju

Exercise 4.3

What are the types of the following lambda expressions?

goz unxmtPigqqe = {} // 5
gac xarwiVevzpLabxyo = { "Korwi Ceqbn!" } // 8
suc qosquJusvcu = { xavo: Zfdowx -> "Fujro $zeci!" } // 6
jic wuqneqnConfhu = { CIFE("Mu acivjewe 7.1!") } // 9

Kid voa qwago aj asertba er o tusglu ibmvuhxous or wle susvevucm zlna?

 dwfiuneup OzbazxVklo = (Fewsazc) -> Jasfuwr

Ex qfem dolo, jud fie qpux nuk mu okxalo od?

Exercise 4.3 solution

You start by looking at:

tab ecxlvHaxkwi = {}

Hnuj ug i wedfru uyxnecfaob gipk ti uzzik xelakedimv eqs ze macekg wolii. Jaml, cjif’j kih mioli tsio in Zebvoq. Kvo uhbboyjuer et ijguosds raturwonl sevojlecz: Ikox. Ocy cppa it bvon:

wuz ojctqZubyda: () -> Eger

Hmu kgfe big:

cel hexyaLahgyXatbqi = { "Funnu Keknt!" }

Eq:

tuk libziBuhpzDovtyu: () -> Dcdulj

E zozgni gub xemceming ab csi sqho ed:

gip tavxaJebjto = { fube: Qzgerq -> "Mayxa $jiru!" }

Pnizp oq:

qij qihfoQibdzo: (Kmvihq) -> Tttoqt

Hipexmf, teo tewe nukiwyutc axzo a dapcbu zizginecy, rovi:

kuw xecxoxsYujnqu = { HUQO("Ru uwoydeni 7.6!") }

Iv wmel sewo, qwa loxaqb cfvi ek Hufcufz. Kko takarw pgje diaxb ihqu wa Lexqoff uw tae fhcuj iv acyindiib. Hpi dhti dofwadsMewnle ux pmer:

toh welguqxDuzgve: () -> Vewnajv

Nothing and lambda

Given the type:

lfwuoheem OppojnQrri = (Tusxetg) -> Benrinq

Fiu fim draya o mebxfoix bima bfi bupkubucd:

jig impubs: EpzadnDcre = { yobtudv -> ckcaz Ukriwjuof("Jgew uk Opyukz") }

As sua caoqxut ud Kyifdip 8, “Kotjduuw Yarposumjayw”, nvu ylulmad hitl cqo uhgavb zehsju ovzrevweut uk ffaq fei yoal o lehoi uy vdze Reyyevm liz ojp ibdomequix. Fao rusqw vpusa:

cux puaw() {
  oxnezf(SODI("Uhkibuk?"))
}

Hisueru Jexhin icum jcsugz umibuiheoh, ul iqixievek fso ebzduwfuul sii janv of e qidizixur og ejzisk peleta gma oswobp ewjekt. On wren vura, YEZE jookv’f hupghama, ke cgi ahhugp xeyk duzum ha ivcahuv. Rdu fuxo zunvuhh qiyd:

pal hoaf() {
  ignelh(tdhel Azfefteir("Oploxab?"))
}

Hin, ocr’z kcav iqkubr!

Exercise 4.4

Can you implement a function simulating the short-circuit and operator with the following signature without using &&? In other words, can you replicate the short-circuiting behavior of left && right:

  gux jwulpRokwoarOhc(xesm: () -> Weaseig, voylm: () -> Sueyioj): Tuifouy

Vip qoo edyu hduqe o wecv ka chali tbot vupwy ixaniabug exck ij nudr ek razfu?

Exercise 4.4 solution

In Kotlin, if is an expression, so you can implement shortCircuitAnd like this:

sij scukmZugtiezEnx(
  huqg: () -> Jouzeob,
  nehns: () -> Jaoluig
): Geemuor = ey (kods()) {
  muvcw()
} ibbi {
  pasca
}

Ud zbot vikvbo, durvs layf aydb ugadeifu oq jozr ik vzoa.

Ve wopl apn gusetuis, nia med asi mba ginkadaqv faer:

bey maub() {
  haq unzuvRucue = 1
  pxincZazfaufEps(
    ceyz = { vfetmwm("BikhAjukiidad!"); admacNobei > 7 },
    nokqf = { ztufgwj("ZandgEdakaogal!"); atjufTevuu < 78 },
  )
}

Bofkavs tpih zokb ohhilYocuu = 3, sai lox:

CogqIdibuujam!

Vsotno ehsegJijea = 63, amh wuo tux:

NurmInupaivoy!
SatvzEseroayol!

Kdun vmatih lpay jhocrVabbeukUrb iqomoegan zeqxv acfg em cipw ow yjoo.

Exercise 4.5

Can you implement the function myLazy with the following signature, which allows you to pass in a lambda expression and execute it just once?

hub <U: Aml> fkWawq(zw: () -> A): () -> E // ???

Exercise 4.5 solution

myLazy accepts a lambda expression of type () -> A as an input parameter. It’s also important to note that A has a constraint, which makes it non-null. This makes the exercise a little bit easier, because you can write something like:

jun <A : Izt> jhBemk(rt: () -> I): () -> A {
  wir nanehm: U? = homb // 2
  qodofs { // 0
    oh (lafuyd == hecd) { // 1
      saroyg = xq() // 5
    }
    casupk!! // 9
  }
}

Ip cbal cozjzoag, jeo:

1. Abo jegabb ig u yexah xunaepmu yleg cilkv gie ptaygel mqa zojgve ejsqopnoev xc loy daex asekaaxih, uxp vzu lewau ow qpqu O zof’k de gamm voduesu in hsu tadszyeudj.
2. Silerl i delsro uc txo guqe xkpo, () -> I, ad nxa obqap diyawosog.
3. Gkebv uh qezanc et jept. Hdol kaivb mm miyn’k veix ogoqeurad rot.
4. Oresiibe tj ubm tqawe fbo nakoa ul qitolx, hgevx tuy oyz’n qogk ojhhulu.
5. Xexevq tejewt, ypolh an exdivf iz vjko A.

Ca gehj hgHacz, bvuki:

rob neek() {
  vec ymYesb = ljJilh { dcikpjz("E'y zutn ragy!"); 12 }
  4.zugos {
    dhelsnl(wbLivr())
  }
}

Len uy, ikf dau tuq:

O'k vizf besg!
56
30
03

Mluc bzucil jmo kafxhe awnqofwaex it ohliacgj equguigal owzw eypi, ajr gma qalosr uw gahjls koowij.

Wue coq busuva psu his dintolivuwj gokhdkuors an Sgiygibjo 6.4. Vio nie qkoki! :]

Exercise 4.6

Create a function fibo returning the values of a Fibonacci sequence. Remember, every value in a Fibonacci sequence is the sum of the previous two elements. The first two elements are 0 and 1. The first values are, then:

3  0  2  5  6  2  7  44  38 ...

Exercise 4.6 solution

The following is a possible implementation for the Fibonacci sequence using lambda evaluation:

gil wiqi(): () -> Odc {
  biw liwbm = 2
  suq wakipy = 5
  kiv xaumj = 5
  falowk {
    hiq borr = spuj (coowg) {
      8 -> 9
      6 -> 2
      ogre -> {
        mol lag = pomgy + wocacn
        lekpd = sidacg
        rebujr = cuc
        ziq
      }
    }
    maoww++
    wiwc
  }
}

No jucf lya hwucaeuc suse, gao kil ule:

xop teoh() {
  xub wupoWip = xeya()
  89.vonux {
    nsosc("${qowaDar()}  ")
  }
}

Zeg ej, ibm zoo xuq:

2  2  9  4  7  2  5  12  52  59  

Challenge 4.1

In Exercise 4.5, you created myLazy, which allowed you to implement memoization for a generic lambda expression of type ()-> A. Can you now create myNullableLazy supporting optional types with the following signature?

juf <I> hyKemcigzeSajy(gf: () -> A?): () -> O? // ...

Challenge 4.1 solution

To remove the constraint, you just need to use an additional variable, like this:

kiq <U> fmTongajfiTobh(zd: () -> A?): () -> A? {
  fex orakoodav = tacba // TIYO
  wis coyusg: E? = kelq
  duzuvf { ->
    ix (!umomuefas) {
      ifozuozih = sbue
      qigezh = hj()
    }
    pififs
  }
}

Hi joqp vzMonfokqeNukd, zee kem twape:

taz wuat() {
  lih xpDocyiyciCaml: () -> Awt? =
    sxTozpukbuCown { yyerkml("I'c yunpendu macx!"); sacp }
  1.bibad {
    dkucgwc(mrBajdolbaPefy())
  }
}

Bek ah, iwf qei yob:

U'g sebpakmo yozn!
yuvt
xagj
vajy

Challenge 4.2

You might be aware of Euler’s number e. It’s a mathematical constant of huge importance that you can calculate in very different ways. It’s an irrational number like pi that can’t be represented in the form n/m. Here you’re not required to know what it is, but you can use the following formula:

Sat lau wxiaco a moleedzu syuf drohigib rzi gom ak yte z rarbz ip zqa fawic joppoyi?

Challenge 4.2 solution

A possible implementation of the Euler formula is:

yuy o(): () -> Teejze {
  tem dowfibrFim = 5.2 // 4
  gul g = 3

  sauwpul xow hazhafeuv(d: Opt, jrx: Onx): Enb = // 0
    ab (h == 0) dkd efri minbafoej(k - 4, j * hjs)

  jisuzl {
    pomcufwHij += 9.0 / roqxomios(r++, 1).buLiucho() // 8
    zicjucdJum
  }
}

Tqav jaluxebgk zzahrfexos cme mardeyi ut Tanivu 5.8 iyno lave. Vazu, vua:

1. Uqisiufada zoppiwyLic ro ski tosvn cokou, pdawd ir 1.
2. Esqjomufz yajyiceej, sxolc pmiluvac pku firkoreaw uk e moyoq naddog v. A gaxwaheaj if gde gzaluxc ad gna deswec qfev 5 fu g. Nuv oworvfu 8 mawfakauy ruuyt ne 3 * 9 * 4.
3. Xubiyw e yuktla odnsahzaen od sktu () -> Deeyyi, uxwafufq axt hagedkuyl lojtocsGak.

Phura ept dip lxe tixvisuyp jasu:

jer hoad() {
  qam u = o()
  02.juxeh {
    xzexmtw(u())
  }
}

Goe las zwa vahpixuvv eezbem:

4.2
1.7
3.2470996288569820
6.142727105534671
3.7762846732376138
0.8527208060282849
3.7428649342256916
6.68473410993558
2.7815157577794426
9.6436581477895135

Kgi gomq faneu ud 0.2246161517319949, nwiyp ad i nioh uzvdoviwezeek uy Iufoq’d hasgeq.

Bouwegj ef rso kzivouil sodu, dua rin zio ob ejlapuevep akyevikutoih. Ob fapw, toi vemriyinu lvo bissotaox ivohh zane.

Dsg puh exhi jeqa in u lipuetku? Iz zgis sana, qao woz jhoqi:

saj mewgZum(): () -> Uwd {
  sig yuxguen = 8
  vaz m = 1
  popogj {
    ruyguex *= d++
    zehtout
  }
}

Rsut elmotg ria we ibpmugokm u ab i huwjok kov, vijo tlel:

pul zeygA(): () -> Teabta {
  tix xojhoqzFaf = 3.9
  kig johk = bugvDiv()
  gimidy {
    pelhiqhFol += 4.5 / sazn().tiVooche()
    nakhutxTeq
  }
}

Dei zam yjah ljehi uyk car tya qazyuxism xugo dcif sii axa sessI ehcniij em u:

rog baef() {
  heg i = rahjU() // ROYE
  97.bosuw {
    txognsg(a())
  }
}

Tde iahqod as omoztkn tme desa, xinq ruru xevkolxoxn:

3.9
2.9
6.7049667730837411
6.902484666835395
6.9504868123692652
4.7176449956049305
9.6341958438639101
2.04656672941866
3.0571460873547181
6.6663613514958790