feat: add reduction

main.typ

@@ -12,6 +12,7 @@
 #let TM = $T M$
 #let rej = $q_"rej"$
 #let acc = $q_"acc"$
+#let halt = $H A L T I N G$
 
 
 = Tjūringa Mašīnas
@@ -202,7 +203,7 @@ $ F(1)=0, F(2)=-1, F(3)=1, F(4)=-2, ... $
 Viens no veidiem, kā izveidot bijekciju pāra kopai, ir izmantot metodi, ko sauc
 par Kantoro pārošanas funkciju.
 Kantora pārošanas funkcija ir definēta sekojoši:
-$f(k_1, k_2) := 1/2(k_1 + k_2)(k_1 + k_2 + 1) + k_2$, kur $k_1,k_2 in NN$
+$f(k_1, k_2) := 1/2(k_1 + k_2)(k_1 + k_2 + 1) + k_2$, kur $k_1,k_2 in NN = {0, 1, 2, ...}$
 
 #figure(
   image("assets/img/cantors-pairing-function.png", width: 50%),
@@ -231,3 +232,52 @@ attēlots uz $n$, izmantojot Kantora pārošanas funkcijas apgriezto funkciju.
 Pielietojot apgriezto funkciju uz $n$, varam atgūt sākotnējo pāri $(k_1, k_2)$.
 Tādējādi funkcija $f$ ir surjektīva.
 
+= Redukcijas
+Given a problem $halt 2(M, x, y) = 1$ where turing machine $M$ halts on at least
+one of the inputs $x$ or $y$, prove and show that it can or can't be reduced to
+$halt(halt <= halt 2)$.
+
+To prove that the problem $halt 2(M, x, y)$ can be reduced to $halt$, we need to
+show that we can construct a Turing machine that solves $halt 2$ using a
+subroutine for solving $halt$.
+
+Let's assume we have a Turing machine $H$ that solves the $halt$ problem.
+We will construct a new Turing machine $H 2$ that solves the $halt 2$ problem
+using H as a subroutine.
+
+The Turing machine $H 2$ works as follows:
++ Given inputs $M$, $x$, and $y$.
++ Run $H$ on the input $(M, x)$.
++ If $H$ accepts $(M, x)$, halt and accept.
++ If $H$ rejects $(M, x)$, run $H$ on the input $(M, y)$.
++ If $H$ accepts $(M, y)$, halt and accept.
++ If $H$ rejects $(M, y)$, halt and reject.
+
+By constructing $H 2$ in this way, we are simulating the behavior of $H$ on
+both inputs $x$ and $y$.
+If $H$ accepts either $(M, x)$ or $(M, y)$, $H 2$ will also accept and halt.
+If $H$ rejects both $(M, x)$ and $(M, y)$, $H 2$ will also reject and halt.
+
+Now, let's analyze the reduction:
+
+- If $halt 2(M, x, y) = 1$, it means that Turing machine $M$ halts on at least
+  one of the inputs $x$ or $y$.
+  In this case, $H 2$ will also halt and accept, because it successfully
+  simulates $H$ on both inputs and accepts if $H$ accepts either of them.
+  Thus, $halt 2$ is reduced to $halt$.
+- If $halt 2(M, x, y) = 0$, it means that Turing machine $M$ does not halt on
+  both inputs $x$ and $y$.
+  In this case, $H 2$ will also not halt and will reject, because it simulates
+  $H$ on both inputs and rejects if $H$ rejects both of them.
+  Thus, $halt 2$ is reduced to $halt$.
+
+Therefore, we have shown that the problem $halt 2$ can be reduced to $halt$ by
+constructing a Turing machine $H 2$ that uses $H$ as a subroutine.
+This reduction demonstrates that $halt 2$ is computationally no harder than
+$halt$, implying that $halt 2$ is at least as undecidable as $halt$.
+
+= Daļēja atrisināmība
+= Algoritmiskā sanumurējamība
+= TM darbības laiks
+= NP (neatrisināmas problēmas)
+= Sarežģītības klases