The Duke of Densmore was killed by an explosion that damaged his castle. His testament, evidentally destroyed, was said to have displeased one of his 7 ex-wives. However, each of these women had come in the castle by invitation shortly before the crime. Each of the 7 ex-wifes swears that the invitation was the only time she went there. All of them could be guilty, but because of the careful preparation of the bomb, which was hidden and adjusted to fit in a knight's armor in the Duke's bedroom, the murderess must have come in the castle more than one time. So the guilty woman lied: she came in the castle several times. The women do not remember exactly the date when they went there, but they remember who they met:

Ann met Betty, Charlotte, Felicia, Georgia.
Betty met Ann, Charlotte, Edith, Felicia, Helen.
Charlotte met Ann, Betty, Edith.
Edith met Betty, Charlotte, Felicia.
Felicia met Ann, Betty, Edith, Helen.
Georgia met Ann, Helen.
Helen met Betty, Felicia, Georgia.

Which one was lying? Who is therefore the murderess? ________________________________________________

The original problem was posed in 1980 by the French mathematician Claude Berge (1926 - 2002) to demonstrate a solution by graph theory.

The puzzle is more a mathematical problem than a logical one. We can also see that every statement of an ex-wife confirms the other ones. (I put the problem in the category of 'logic' only because of its flavor.)

CLAUDE BERGE (1980):
"L'énigme policière", Regards sur la théorie des graphes,
Actes du Colloque de Cérisy, Presses polytechniques romandes, Lausanne.

CLAUDE BERGE (1994): "Qui a tué le duc de Densmore?",
Bibliothèque Oulipienne no. 67.

CLAUDE BERGE (2000): "Qui a tué le duc de Densmore?", Une nouvelle policière où le meurtrier est confondu grâce à l'utilisation d'un théorème de combinatoire.
(= Réédition du 1994 par Ed. Castor Astral)

The version of the problem given above is based on the following source:

http://mathafou.free.fr/pbm_en/pb205.html

I would like to thank Brian Smith for his inspiring solution in which he unmasked the guilty wife and even reconstructed a possible chronological form, showing that Ann stayed three times for sure (instead of only one time, as she said).

Of the four cycles AGHB, AGHF, ACEF, ABFH, the last one is not entirely chordless: ABFH has the chord BH. Whether we take it as ABFH or as AFHB or as ABHF, there is always at least one chord (BH or FB or BF). So the first three cycles (all chordless) that Brian Smith found are sufficient for the solution.

Posted by ollie on 2017-05-01 16:55:32