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 Mr. Wolfingham's Fiancée (Posted on 2023-03-01)
J. Weldon Wolfingham, Jr. will marry one of these women: Adrianna, Bianca, and Claire.

Some clues are furnished hereunder as follows:

1. Of Adrianna and Bianca:
• (a) Either they are both have blue eyes, or neither has blue eyes.
• (b) One has red hair and the other does not.

2. Of Adrianna and Claire:
• (a) Either they both have red hair, or neither has red hair.
• (b) One is 5'11" and the other is not.

3. Of Bianca and Claire:
• (a) One has blue eyes and the other does not.
• (b) One is 5'11" and the other is not.

4. Of the three characteristics - blue eyes, red hair and 5'11":
• (a) If any of the three women has exactly two of the three characteristics, Mr. Wolfingham will marry the one with the least number of characteristics.
• (b) If any of the three women has exactly one of the three characteristics, Mr. Wolfingham will marry the one with the greatest number of characteristics.

Who will Mr. Wolfingham marry?

 See The Solution Submitted by K Sengupta Rating: 5.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 computer solution | Comment 1 of 5
traitList= logical.empty(0,3);
for blue=[true false]
for red=[true false]
for five11=[true false]
traits=[blue red five11];
traitList(end+1,:)=traits;
end
end
end

for a=1:8
traitSet=traitList(a,:);
for b=1:8
traitSet(2,:)=traitList(b,:);
for c=1:8
traitSet(3,:)=traitList(c,:);

good=true;
if traitSet(1,1)~=traitSet(2,1) good=false; continue
end
if xor(traitSet(1,2),traitSet(2,2))==0 good=false; continue
end

if traitSet(1,2)~=traitSet(3,2) good=false; continue
end
if xor(traitSet(1,3),traitSet(3,3))==0 good=false; continue
end

if xor(traitSet(2,1),traitSet(3,1))==0 good=false; continue
end
if xor(traitSet(2,3),traitSet(3,3))==0 good=false; continue
end

disp(traitSet)
disp(' ')
end
end
end

finds  (annotated output -- 1 represents having the feature)
`       blue red 5'11"       eyes hair       A    1   1   1    B    1   0   1       Makes 4 inconsistent    C    0   1   0     A    1   1   0    B    1   0   0       Makes 4 inconsistent    C    0   1   1     A    1   0   1    B    1   1   1       Claire is the fiancee    C    0   0   0     A    1   0   0    B    1   1   0       Makes 4 inconsistent    C    0   0   1     A    0   1   1    B    0   0   1       Makes 4 inconsistent    C    1   1   0     A    0   1   0    B    0   0   0       Claire is the fiancee    C    1   1   1     A    0   0   1    B    0   1   1       Makes 4 inconsistent    C    1   0   0     A    0   0   0    B    0   1   0       Makes 4 inconsistent    C    1   0   1`

So Claire is the fiancee.

Edited on March 1, 2023, 9:30 am
 Posted by Charlie on 2023-03-01 09:29:43

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