coded_binary_logic

Example Problem

Example: Three people have a 3-bit code where 1=T, 0=L. A says: 'My bit is 1.' B says: 'A's bit is 0.' C says: 'The code has exactly two 1's.' Find the code. Solution: Step 1: Let code = (a,b,c) where a,b,c ∈ {0,1}. Step 2: A's statement: a=1. Truth of A's statement = (a==1). A's type = a, so a = (a==1). This is true for a=1 (1=1), false for a=0 (0=0? 0=(0==1)=0 works too? Let's solve: a = (a==1). If a=1, RHS=1, works. If a=0, RHS=0, works. So A's statement gives no constraint on a alone? Actually, it's an identity. So A can be 0 or 1. Step 3: B's statement: 'A's bit is 0' means a=0. Truth of B's statement = (a==0). B's type = b, so b = (a==0). Step 4: C's statement: 'The code has exactly two 1's' means a+b+c = 2. Truth of C's statement = (a+b+c==2). C's type = c, so c = (a+b+c==2). Step 5: Test possibilities. If a=1, then b = (1==0)=0. Then a+b+c=1+0+c=1+c. For c=(1+c==2), if c=1, RHS=(2==2)=1, so c=1 works. Then code (1,0,1) sum=2. Works. If a=0, then b=(0==0)=1. Then a+b+c=0+1+c=1+c. For c=(1+c==2), if c=1, RHS=(2==2)=1, so c=1 works. Code (0,1,1) sum=2. Works. Two solutions: (1,0,1) and (0,1,1). Answer: Two possible codes.