You are not logged in.
Pages: 1
This puzzle is from the book "Boolean Algebra and its Applications" by J. Eldon Whitesitt, but there is no solution given, so I'm interested in what anyone proposes. I tried 3 major AI platforms (Google Gemini, claude.ai and ChatGPT) and only ChatGPT came up with the answer which matched mine.
Here is the problem:
Simplify to a single rule the following set of rules regarding dress to be worn to a certain party for married couples:
1. If a man wears either a tie or a coat, then his wife must wear neither high heels nor a hat.
2. If a woman wears slacks or a hat, then her husband must wear either a tie or a hat, but not both.
3. No man and wife may both wear hats unless either the wife wears high heels and slacks or the man wears a tie and does not wear a coat.
4. For every couple, either the man or his wife must wear a hat and either the man wears a coat and the wife wears slacks.
5. If a man wears a hat or his wife wears slacks, then the man must wear a tie and his wife must not wear high heels.
I think this is an example of the coupon collector's problem (see https://en.wikipedia.org/wiki/Coupon_collector%27s_problem).
The formula for the expectation (average number of draws until you have the whole set) is n x (1/1 + 1/2 + 1/3 + ... + 1/n), and in this case n = 6 x 40 = 240 "coupons". Therefore at the rate of 1 draw/second it will take 240 x (1/1 + 1/2 + 1/3 + ... + 1/240) = 1455 seconds, or 24 mins 15 seconds.
Hi Bob,
Thanks for the reply. I hadn't thought of doing a Venn diagram, but that's probably the easiest and most intuitive solution.
For the record, here is my algebraic solution.
1) A = BC
2) A'C = {5,6}
3) (A + B)C' = {7}
Taking the complement of both sides of (1) gives A' = (BC)' = B'+C' (by de Morgans' rule)
Substituting this result into (2) gives (B'+C')C = B'C = {5,6}.
Substituting (1) into (3) gives (BC + B)C' = {7}, but BC + B = B, therefore BC' = {7}.
Now, B = BC + BC'
So B = A + {7} = {1,2,3,4} + {7} = {1,2,3,4,7}
and C = BC + B'C
So C = A + {5,6} = {1,2,3,4} + {5,6} = {1,2,3,4,5,6}
Given that A = {1,2,3,4}, calculate the elements of the sets B andC, given the following facts:
1) A = BC
2) A'C = {5,6}
3) (A + B)C' = {7}
I know I could have used LaTeX for the set notation but I think the boolean algebra notation is easier to read.
So juxtaposition is set intersection, "+" is set union, and a primed symbol (') is set complement.
I think I've solved it but the solution seems a bit long-winded, so I would like to see other solutions which may be simpler.
Hi phrontister,
It took me a little while to figure out how to use goal seek as I'd never used it before, but it worked and gave the correct results. This is cool, and it's nice to have another problem solving tool in my armoury.
Thanks again.
Hi phrontister,
I'm confused. dizzy Did you deliberately omit the brackets?
Not deliberately, but I guess I should have included them, although arguably "subtract M - A" isn't ambiguous.
I used the same method: see post #4's "Note: Their age difference at each stage is a constant, which in the first stage is established as being 2c (ie, 3c - c)."
Yes, I see now. Looking for the invariant is often a good strategy.
Thanks for the goal seek screenshot. Haven't tried it yet with Libreoffice Calc, but I will.
Hi phrontister,
Thanks for the links - very interesting. I solved the problem using the algebraic method as described in the Edna's age problem.
m = Mary's present age
a = Ann's present age
x = some years ago
y = some other years ago
z = some years hence
1. m + a = 44 Mary and Ann's combined age is 44 and
2. m = 2(a - x) Mary is twice as old as Ann was (x years ago)
3. m - x = (a + z)/2 when Mary was half as old as Ann will be (z years hence)
4. a + z = 3(m - y) when Ann is three times as old as Mary was (y years ago)
5. m - y = 3(a - y) when Mary was three times as old as Ann was. (y years ago)
Maple solution:
solve({m+a=44,m=2*(a-x),m-x=(a+z)/2,a+z=3*(m-y),m-y=3*(a-y)},[m,a,x,y,z]);
[[m = 55/2, a = 33/2, x = 11/4, y = 11, z = 33]]
Although it's not really much harder than solving the simple age problems you see in elementary algebra (at least, in terms of setting up the equations), what confused me was demarcating the different times correctly.
I don't have Excel, but I believe there is a similar solver in Libreoffice Calc, which I do have. I'll try to replicate the method you posted above using it.
Hi phrontister,
Thanks for your input. I should have perhaps mentioned that the text I got this from gives the answer, which is 27.5, but not a detailed solution. I posted it on another site and some bright spark came up with this:
M+A=44.
k years ago, Mary's age was M−k and Ann's age was A−k
You know that M−k=3(A−k)
So, now you have 2 linear equations, in the 3 unknowns, M,A,k
When Ann is 3(M−k), and Mary is (1/2) that age, Mary is (3/2)(M−k)
At that point, Ann was (3/2)(M−k)−(M−A).
Mary is now twice that age.
Therefore,
M=3(M−k)−2(M−A)⟹2A=3k.
The other two equations are
M+A=44
M−k=3(A−k)⟹[M−(2/3)A]=3A−2A⟹
M=(5/3)A.
From which, M = 27.5 and A = 16.5.
Like you, he started from right to left, but I must admit I don't quite understand this line :
At that point, Ann was (3/2)(M−k)−(M−A).
Not quite sure why you subtract M - A?
The combined ages of Mary and Ann are 44 years and Mary is twice as old as Ann was when Mary was half as old as Ann will be when Ann is three times as old as Mary was when Mary was three times as old as Ann was.
How old is Mary?
Translating into equations seems to be the way to go with this but the difficulty is where to start after the first one, which is clearly : m + a = 44.
Any suggestions?
Pages: 1