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Generation of constructible numbers in Mathematica
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README.md
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ConstructibleNumbers.nb
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KonstruktibilnaStevila.nb
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Raw
predavanje.md

Koliko je konstruktibilnih točk četrtega reda?

Povzetek: Konstruktibilne točke tvorimo tako, da začnemo z dvema točkama na ravnini. V naslednjem koraku z ravnilom narišemo vse premice, ki potekajo skozi vsaj dve od predhodno konstruiranih točk, s šestilom pa vse krožnice, ki imajo središče v eni od točk in vsebujejo še eno od točk. V drugem koraku tako dobimo 6 točk (prvotni dve in še štiri presečišča med dvema krožnicama in premico). V tretjem koraku dobimo 203 točke. Katero število dobimo v čertem koraku? Hitro vidimo, da je število zelo veliko in da ga ne moremo kar tako stresti iz rokava. Pomagamo si lahko z računalniki, a moramo pred tem spoznati še konstruktibilna števila in postopke za računanje z njimi.

Konstruktibilne točke

V klasični geometriji uporabljamo (neoznačeno) ravinolo in šestilo. Z ravnilom lahko povlečemo premico skozi dve točki, s šestilom pa narišemo krožnico z danim središčem in polmerom. Mi se bomo omejili le na risanje krožnic z danim središčem in dano točko na krožnici, tako bodo razmisleki nekoliko bolj preprosti, a še vedno zanimmivi. (Se pravi, da ne moremo odmeriti polmera na daljici in ga prenesti v neko tretjo točko.)

Točko lahko konstruiramo kot sečišče dveh premic, krožnica, ali krožnice in premice.

Definicija: Točka je konstruktibilna, če jo lahko načrtamo z ravnilom in šestilom kot presečišče premic in krožnic, začenši z dvema danima točkama, ki določata enotsko daljico.

Posvetili se bomo naslednjemu vprašanju:

Koliko točk lahko skonstruiramo na ta način v nekaj korakih?

Prvi korak: začnimo z dvema točkama.

Drugi korak: narišemo eno premico in dve krožnici, skupaj dobimo 6 točk.

Tretji korak? Kaj dobimo v tretjem koraku? Na roko bomo težko vse narisal. Sicer imamo simetrijo, to bo malce pomagalo, a je še vedno preveč. Pomagali si bomo z računalnikom.

Konstruktibilna števila

Pri delu z računalnikom bomo delali analitično, se pravi, da uvedemo koordinatni sistem in računamo koordinate točk.

Najprej premislimo, kako zapleteni bodo izračuni.

Premica je podana z enačbo oblike a x + b y = c

Krožnica s središčem (x₁,y₁) in polmerom r je podana z enačbo oblike (x - x₁)² + (y - y₁)² = r².

Če želimo izračunati koordinate presečišč, moramo rešiti sistem enačb:

  • premica-premica: dve linearni enačbi z dvema neznankama
  • premica-krožnica: ena linearna in ena kvadratna enačba z dvema neznankama
  • krožnica-krožnica: dve kvadratni enačbi z dvema neznankama

Paziti je treba tudi na število rešitev (od 0 do 4). Poleg tega bomo dobivali rešitve v komplesnih številih, ki jih zavržemo. Seveda takih enačb ne bomo reševali na roko, pomagamo si z Mathematico.

S kakšnimi števili imamo opravka pri reševanju enačb? Geometrijsko gledano, gre za števila, ki jih lahko "konstruiramo" z ravnilom in šestilom:

Definicija: Število r je konstruktibilno, če lahko začenši z daljico dolžine 1 z ravnilom in šestilom načrtamo daljico dolžine r.

Iz stališča algebre gre za števila, ki jih lahko tvorimo iz 0 in 1 z operacijami +, ×, -, / in √. (Kvadratni koreni izhajajo iz reševanja kvadratnih enačb.) Če bi imeli le prve štiri operacije, bi dobili natanko vsa racionalna števila. Kvadratni koren pa nabor števil razširi, na primer: √2, √(√2), 2 + √(3 + √5) itd. Konstruktibilna števila tvorijo obseg K, ki je hkrati zaprt še za kvadratne korene.

Kaj še lahko povemo o obsegu K? Zagotovo vsebuje racionalna števila ℚ, je razširitev obsega racionalnih števil. Če bi na primer razširili ℚ samo z enim kvadratnim korenom √r, bi dobili razširitev ℚ(√r). Imamo dve množnosti:

  1. √r je racionalno število (če je r = a^2/b^2 za neki celi števili a in b), v tem primeru ℚ(√r) = ℚ.
  2. √r je iracionalno število, v tem primeru so elementi ℚ(√r) oblike p + q √r, kjer sta p, q ∈ ℚ.

Kako računamo v obsegu ℚ(√r)? Vprašajmo Mathematico!

V splošnem je obseg K bolj kompliciran. Vsako konstruktibilno število dobimo z nekim zaporedjem razširitev s kvadratnimi koreni:

ℚ(√r₁)(√r₂)⋯(√rᵢ)

To je r₁ ∈ ℚ, r₂ ∈ ℚ(√r₁), ... Pravila za računanje postanejo bolj zapletena in se vanje ne bomo spuščali.

Konstruktibilne točke v Mathematici

Oglejmo si, kako v Mathematici sestavimo program, ki za nas konstruira točke, jih prešteje in tudi nariše.

Dobimo odgovor:

Konstruktibilnih točk tretjea reda je 203.

Koliko je konstruktibilnih točk čertega reda?

Če poskusimo v Mathematici izračunati, koliko je konstruktibilnih točk četrtega, ne dobimo odgovora. Koliko jih sploh je? Ali lahko naredimo kakšno zgornjo oceno?

Naredimo zelo grobo oceno. Denimo, da imamo n točk. Iz njih lahko tvorimo:

  • največ n² premic, po eno za vsaki dve točki (pravzaprav n(n-1)/2, a poenostavimo)
  • največ n² krožnic, po dve za vsaki dve točki
  • največ n⁴ parov premic, ki dajo največ n⁴ presečišč
  • največ n⁴ parov premica-krožnica, ki dajo največ 2 n⁴ presečišč
  • največ n⁴ parov krožnica-krožnica, ki dajo največ 2 n⁴ presečišč

Skupaj torej ne več kot 5 n⁴ točk v naslednjem koraku. Dobimo torej naslednje ocene:

  • prvi red: največ 2 točki
  • drugi red: največ 5 · 2⁴ = 80 točk (v resnici 6)
  • tretji red: največ 5 · 80⁴ = 204.800.000 ≈ 200 · 10⁶ točk (v resnici 203)
  • četrti red: največ 5 · 204800000⁴ = 8796093022208000000000000000000000 ≈ 10³⁴ točk

A ta zgornja ocena je precej groba, čeprav nas opozarja, da bo število zares veliko.

Koliko je konstruktibilnih točk četrtega reda, je pvrašal Joel Hamkins na math.stackexchange.com.

Kar z nekaj truda je nanj odgovoril Teofil Camarasu. Poglejmo, kako je to naredil in kakšen je odgovor.

Napisal je program v Haskellu (Mathematica je prepočasna), uporabil je knjižnico za računanje s konstruktibilnimi števili. Ker je računanje s konstruktibilnimi števili precej počasnejše od vgrajenih števil s plavajočo vejico (ki jih ne smemo uporabiti!), je za pospešitev vsako konstruktibilno število izračunal še približno spodnjo iz zgornjo oceno, saj je lahko tako hitro ugotovil, ali sta števili različni. Za račuanje presečiš premic in krožnic je uporabil še nekatere algoritme iz računske geometrije. Uporabil je tudi simetrijo med kvadranti, da ni bilo treba računati vsega.

Šest vzporednih procesov je na treh računalnikih 6 dni računalo odgovor. Nikoli niso presegli uporabe več kot 5 GB pomnilnika. Odgovor je:

Konstruktibilnih točk četrtega reda je 1723816861.

Zaporedje je zavedeno v OEIS pod A333944.

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