7 people are arguing what the current day of the week might be. Each states what he believes to know:
All of them, except one, is wrong. What day is it it?
Rewording their statements:
We know that exactly one of these is right. It can't be Wednesday (since then 2 and 6 would both be right), nor can it be Thursday, Friday, or Saturday (since then 4 and 7 would both be right), nor can it be Monday or Tuesday (since then 7 would be right and so would 1 or 3). So today is
Sunday
and the
4th
speaker is the only correct one.
7 says that it's not Sunday, which agrees with 1,2,3,5,6. therefor proof not only that all but 4 is wrong, but also that since the 7th statement is wrong, it means that today IS Sunday. All can be proven with just that 1 statement.
The answer is
Sunday
The best way to visualize it is by creating a table with values:
$\begin{array}{c|c|c|c|c|c|c|c}\underset{(Statement~\#)}{\text{Speaker}}&\text{Mon}&\text{Tue}&\text{Wed}&\text{Thu}&\text{ Fri }&\,\text{Sat}\,&\text{Sun}\\\hline1&\text{X}\\\hline2&&&\text{X}\\\hline3&&\text{X}\\\hline4&&&&\text{X}&\text{X}&\text{X}&\color{red}{\text{X}}\\\hline5&&&&&\text{X}\\\hline6&&&\text{X}\\\hline7&\text{X}&\text{X}&\text{X}&\text{X}&\text{X}&\text{X}\end{array}$
Filling in the rows of the table:
Statement 1 is true only if today is Monday.
Statement 2 is true only if today is Wednesday.
Statement 3 is true only if today is Tuesday.
Statement 4 is true only if today is in the range from Thursday to Sunday.
Statement 5 is true only if today is Friday.
Statement 6 is true only if today is Wednesday.
Statement 7 says that yesterday was not Saturday. Then yesterday could be Monday, Tuesday, Wednesday, Thursday, Friday or Sunday. So today is Tuesday, Wednesday, Thursday, Friday, Saturday or Monday – any day except Sunday.
Finally, reading down the columns of the table:
On Monday, statements 1 and 7 are true.
On Tuesday, statements 3 and 7 are true.
On Wednesday, statements 2, 6, and 7 are true.
On Thursday, statements 4 and 7 are true.
On Friday, statements 4, 5, and 7 are true.
On Saturday, statements 4 and 7 are true.
On Sunday, only statement 4 is true.
The only day when only one statement is true is the correct day. That is Sunday.
A computer program can be used to solve it (following is in Racket language):
; SUN M T W TH F SAT
; 0 1 2 3 4 5 6
(define (f)
; assume today is x;
(for ((x 7)) ; check x for 0 to 6
(printf "x=~a; count=~a ~n"
x
(count
(lambda(x) x)
(list (= 3 (+ x 2)) ; statements are listed here
(= x 3)
(= x 2)
(and (not(= x 1)) (not(= x 2)) (not(= x 3)))
(= x 5)
(= x 3)
(not (= 0 x))
)))))
(f)
It takes values of 0 to 6 for Sun to Sat and checks how many statements are correct for each of them. The output is:
x=0; count=1
x=1; count=2
x=2; count=2
x=3; count=3
x=4; count=2
x=5; count=3
x=6; count=2
Hence, only 1 statement is correct only for Sunday (x = 0), hence that is the answer.
Using SymPy:
>>> from sympy import *
>>> sunday, monday, tuesday, wednesday, thursday, friday, saturday = symbols('sunday monday tuesday wednesday thursday friday saturday')
Since only one of the $7$ Boolean variables can be true:
>>> Sun = sunday & Not(monday) & Not(tuesday) & Not(wednesday) & Not(thursday) & Not(friday) & Not(saturday)
>>> Mon = Not(sunday) & monday & Not(tuesday) & Not(wednesday) & Not(thursday) & Not(friday) & Not(saturday)
>>> Tue = Not(sunday) & Not(monday) & tuesday & Not(wednesday) & Not(thursday) & Not(friday) & Not(saturday)
>>> Wed = Not(sunday) & Not(monday) & Not(tuesday) & wednesday & Not(thursday) & Not(friday) & Not(saturday)
>>> Thu = Not(sunday) & Not(monday) & Not(tuesday) & Not(wednesday) & thursday & Not(friday) & Not(saturday)
>>> Fri = Not(sunday) & Not(monday) & Not(tuesday) & Not(wednesday) & Not(thursday) & friday & Not(saturday)
>>> Sat = Not(sunday) & Not(monday) & Not(tuesday) & Not(wednesday) & Not(thursday) & Not(friday) & saturday
>>> Today = Sun | Mon | Tue | Wed | Thu | Fri | Sat
Translating the $7$ statements:
>>> Phi1 = monday
>>> Phi2 = wednesday
>>> Phi3 = tuesday
>>> Phi4 = Not(monday) & Not(tuesday) & Not(wednesday)
>>> Phi5 = friday
>>> Phi6 = wednesday
>>> Phi7 = Not(sunday)
Since $6$ out of $7$ are false:
>>> Psi1 = (Phi1 & Not(Phi2) & Not(Phi3) & Not(Phi4) & Not(Phi5) & Not(Phi6) & Not(Phi7))
>>> Psi2 = (Not(Phi1) & Phi2 & Not(Phi3) & Not(Phi4) & Not(Phi5) & Not(Phi6) & Not(Phi7))
>>> Psi3 = (Not(Phi1) & Not(Phi2) & Phi3 & Not(Phi4) & Not(Phi5) & Not(Phi6) & Not(Phi7))
>>> Psi4 = (Not(Phi1) & Not(Phi2) & Not(Phi3) & Phi4 & Not(Phi5) & Not(Phi6) & Not(Phi7))
>>> Psi5 = (Not(Phi1) & Not(Phi2) & Not(Phi3) & Not(Phi4) & Phi5 & Not(Phi6) & Not(Phi7))
>>> Psi6 = (Not(Phi1) & Not(Phi2) & Not(Phi3) & Not(Phi4) & Not(Phi5) & Phi6 & Not(Phi7))
>>> Psi7 = (Not(Phi1) & Not(Phi2) & Not(Phi3) & Not(Phi4) & Not(Phi5) & Not(Phi6) & Phi7)
>>> Psi = Psi1 | Psi2 | Psi3 | Psi4 | Psi5 | Psi6 | Psi7
Simplifying:
>>> simplify(Today & Psi)
And(Not(friday), Not(monday), Not(saturday), Not(thursday), Not(tuesday), Not(wednesday), sunday)
Hence, today is Sunday.
