local

Author: mostafa-saad

18-Programming-4kids/05_homework1.cpp

@@ -1,3 +1,5 @@
+// Homework 1: Guess the output
+
 #include<iostream>
 using namespace std;
 

18-Programming-4kids/05_homework2_answer.cpp

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+/*
+ Homework 2: Create logic!
+
+ Write a program that reads 3 integers about the class room
+ Number of boys (nb), number of girls (ng), number of teachers (nt)
+
+ Prepare and print a boolean variable for these cases:
+	 nb greater than 25
+	 ng less than or equal to 30
+	 nb > 20 and nt > 2 or ng > 30 and nt > 4
+	 Either nb < 60 or ng < 70
+	 Neither nb >= 60 nor ng >= 70
+	 nb is 10 more students than ng
+	 Difference between nb and ng is more than 10 or nt > 5
+	 Either nb is 10 more students than ng or ng is 15 more students than nb
+ */
+
+#include<iostream>
+using namespace std;
+
+int main() {
+	int nb, ng, nt;
+	cin >> nb >> ng >> nt;
+
+	// nb greater than 25
+	cout << (nb > 25) << "\n";
+
+	// ng less than or equal to 30
+	cout << (ng <= 30) << "\n";
+
+	// nb > 20 and nt > 2 or ng > 30 and nt > 4
+	cout << (nb > 20 && nt > 2 || ng > 30 && nt > 4) << "\n";
+
+	// Either nb < 60 or ng < 70
+	cout << (nb < 60 || ng < 70) << "\n";
+
+	// Neither nb >= 60 nor ng >= 70
+	cout << (  !(nb >= 60) && !(ng >= 70) ) << "\n";
+
+	// nb is 10 more students than ng
+	cout << (nb > ng + 10) << "\n";
+
+	// Difference between nb and ng is more than 10 or nt > 5
+	cout << (nb - ng > 10 || nt > 5) << "\n";
+
+	// Either nb is 10 more students than ng or ng is 15 more students than nb
+	cout << (nb > ng + 10 || ng > nb + 15) << "\n";
+
+	return 0;
+}
+

18-Programming-4kids/05_homework3_answer.txt

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+Homework 3: Simplify expressions
+
+
+
+T && T && F && T
+- F        [as all of them are && and one of them is F]
+========================================================================
+
+T && T && F && T || T && T
+- T && T && F && T ==> Reduce to F
+- F || T ==> T
+========================================================================
+
+T && T && T && T || T && (T || F)
+- (T || F) ==> Reduce to T
+- T && T && T && T || T && T
+- T && T && T && T ==> reduce to T
+- T || T && T
+- T && T  ==> Reduce to T
+- T || T ==> T
+========================================================================
+
+T && T && T || T && (F || (T && (T && T)))
+- (T && T) ==> reduce to T
+- T && T && T || T && (F || (T && T))
+- (T && T) ==> reduce to T
+- T && T && T || T && (F || T)
+- (F || T) ==> reduce to T
+- T && T && T || T && T
+- T && T && T  ==> reduce to T
+- T || T && T
+- T && T ==> reduce to T
+- T || T ==> T
+========================================================================
+
+T && T || T && F && T || T && T && F || (T && (T || F))
+- (T || F) ==> reduce to T
+- T && T || T && F && T || T && T && F || (T && T)
+- (T && T) ==> reduce to T
+- T && T || T && F && T || T && T && F || T
+- T && T ==> reduce to T
+- T || T && F && T || T && T && F || T
+- T && F && T ==> Reduce to F
+- T || F || T && T && F || T
+- T && F && T ==> Reduce to F
+- T || F || T && T && F || T
+- T && T && F ==> Reduce to F
+- T || F || F || T ==> T
+========================================================================
+
+T && T || T && F && T || T && T && F || (T && (T || F))
+- (T || F) ==> reduce to T
+- T && T || T && F && T || T && T && F || (T && T)
+- (T && T) ==> reduce to T
+- T && T || T && F && T || T && T && F || T
+- (T && T) ==> reduce to T
+- T || whatever = T     [smart reduction]
+========================================================================
+
+T && T || T && F && T || (T && T && F || (T && (T || F)))
+- Notice T && T || something = T, so we know the answer T
+========================================================================
+
+(T && T || T && F && T || T) && T && F || (T && (T || F))
+- (T && (T || F)) ==> reduced in 2 steps to: T
+- (T && T || T && F && T || T) ==> (T || F || T) ==> T
+- So now is:
+- T && T && F || T ==> F || T ==> T
+========================================================================
+
+
+T && T || T && (F && T || T && T) && F || (T && (T || F))
+- (T && (T || F)) ==> reduced in 2 steps to: T
+- (F && T || T && T) ==> (F || T) ==> T
+- T && T || T && T && F || T ==> T || F || T = T
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