The duplicity of hugs | THE GAME OF SCIENCE | EUROtoday
The drawback of nicely -avenue households posed final week admits a number of options, because the assertion is interpreted. Of course, in each households there are women and men, because the context suggests, and that when two males hug them it’s a single hug (it appears probably the most cheap, as a result of we are saying that Fulano and Mengano gave themselves a hug, not two). However, for the kiss that x The DA A y be the identical as y The DA A xit must be a kiss on the mouth, which isn’t often the case within the interfamily salutations, and fewer if two kisses are given per individual. So, if we think about that in every Besucón match there are 4 kisses and we name A to the variety of males Hernández, B to the variety of ladies Hernández, E to the variety of males Fernández and C To the variety of ladies Fernández, we could have:
A X E = 24
A x C + E x B + B x C = 132/4 = 33
The “Fermian” strategy of the issue would encompass beginning the believable decompositions of 24 in two components: 2 x 12, 3 x 8 and 4 x 6, which supplies rise to what number of options? But, I insist, different interpretations are potential (see feedback final week).
There is an easier model of this traditional by which the hugs are 21, which, on the one hand, is dominated out the chance that every hug is price two, and, on the opposite, there’s solely a possible decomposition in two components: 21 = 3 x 7.
As for the issue of formal greetings, it is extremely easy if we understand that in every handshake – even being one, such because the hug – two particular person handbook actions converge (which to simplify we’ll name “swipes”), that’s, 66 x 2 = 132. If there are n People on the assembly and each shakes all of the others, there can be N (N-1) “swipes”, in order that n (n-1) = 132. And it’s not crucial to resolve the second diploma equation to see that n = 12.
And with regard to the quantity of people that all through their lives have narrowed an odd variety of palms, Bretos Bursó presents an ingenious resolution very within the CVA line (outdated -fashioned outdated):
“Suppose that in each handshake someone separates the two sticks from a cross that takes from the sack of the crosses and gives one to each person part of the squeeze. All the crosses and all the sticks are the same. As soon as a person brings together two sticks it forms a cross and the strip to the well of the crosses. At any time, each person is without sticks (if he has given a pair number of squeezes) But until that moment a couple of sticks of the sack has come out, another even number has fallen to the well, and the difference, which is even, is the number of people who have a stick. ”
The rectangular board
They had been pending, for a few weeks, some points associated to the issue of the horse’s journey on rectangular boards (approached in his day by Euler himself), particularly within the 3×4. Here is the detailed Save Fuster evaluation:
“When drawing the graph associated with the problem of the horse’s journey for the 3×4 board, two cycles are obtained (1-7-9-2-8-10-1 and 3-5-11-4-6-12-3) with a couple of additional edges that connect them: 2-11 and 3-10.
If you start from the upper left corner (box 1) and the arrival box is not restricted, we will have two possibilities:
1-7-9-2-8-10-3-5-11-4-6-12
1-7-9-2-8-10-3-12-6-4-11-5
If we chose as the start box the first or last of the second row, we will have various possibilities. For example, if we choose box 5, the complete routes will be:
5-11-6-12-3-10-8-2-9-7-1 (coincides with 1-7-9-2-8-10-3-12-6-4-11-5 but traveled in reverse)
5-11-4-6-12-3-10-1-7-9-2-8
5-3-12-6-4-11-2-9-7-1-10-8 (symmetric to the previous one)
5-3-12-6-4-11-2-8-10-1-7-9 (inverted and symmetric of 1-7-9-2-8-10-3-12-6-4-11-5)
If we choose as a start box that is not a corner or any end of the second row, we will not be able to take a complete tour.
Therefore, if I’m not mistaken, there will be only 3 different solutions if we discount symmetries, turns or tours in reverse. ”

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