[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 150
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How about starting with an octagon?
If we still assume that each edge is oriented, then at a glance the neighbouring nodes (vertices) should be marked with two triples with signs. Say,(-4, +2, +3) and (+1, +4, -5)). The numbers associated with the same edge are marked in bold. They naturally have different signs.
Anyway, we obtain a homogeneous system of 8 equations with 12 unknowns. In this system, each weight of an edge occurs in only two different equations, but with different signs.
But you still look for the problem, Alexander.
2 TheXpert: How about a tetrahedron?
Если все же считать, что каждое ребро ориентированное, то навскидку соседние узлы (вершины) должны быть размечены двумя тройками со знаками. Скажем, (-4, +2, +3) и (+1, +4, -5)). Жирным выделены числа, связанные с одним и тем же ребром. Они, естественно, имеют разные знаки.
Все равно выходит однородная система из 8 уравнений с 12 неизвестными. В этой системе каждый вес ребра встречается только в двух разных уравнениях, но с разными знаками.
Но ты все равно ищи задачку, Александр.
2 TheXpert: а, может, с тетраэдра?
there was also a drawing like this, I'm kind of sketching it out here
There are more parents than children.
Every boy has a sister.
There are more boys than girls.
There are no childless families in the house.
Every boy and girl has a father and a mother in the family.
This report was rejected. Why was it rejected?
Solution for the murahedron (top view).
sanyooooook, take your time, your cubic war ants have already diverted half the resources of the forum.
Here's an idea how to simplify the problem.
If you take two adjacent vertices together with their adjacent edges, you get a squaring of the central edge O and four edges A, B, C, D, departing in pairs from each vertex of the edge. Now I am going to "shorten" this misfit. After reduction only original A, B, C, D will remain with their weights and the same directions (the sum is also zero, if with signs), and the central O will disappear. The centre of this construction is the vertex O.
The main question is: can we unambiguously reconstruct the weight of O, knowing only the weights of edges A, B, C, D - of course, under the conditions of this problem?
MetaDriver, have you already coped with suspended ceilings?
1 Родителей больше, чем детей.
2 У каждого мальчика есть сестра.
3 Мальчиков больше, чем девочек.
4 Бездетных семей в доме нет.
5 У каждого мальчика и у каждой девочки есть в семье и папа и мама.
Since (1), (4) and (5), some (not all) families have 2 children and others have 1. There cannot be 2 boys in a two-child family, because each of them should have a sister (condition 2), hence in such a family there is at most one boy and at least one girl. Since condition (3) exists, to compensate for the shortage of boys some boys should be concentrated in at least a larger part of one-child families, but this is impossible due to (2).
Поскольку (1),(4) и (5), значит в некоторых (не всех) семьях детей 2, а в остальных 1. 2 мальчика в двухдетной семье быть не может, т.к. у каждого должна быть сестра (условие 2), следовательно в такой семье мальчик максимум один, а девочка соответственно минимум одна. Поскольку имеется условие (3), то чтобы скомпенсировать недостаток мальчиков необходимо, чтобы часть мальчиков была сконцентрирована по крайней мере в большей части однодетных семей, однако это невозможно в силу (2).
Look for the answer in Science and Life 1998 No.5
sanyooooook,
Есть идея, как задачку упростить.
I am a retard of course, but can you clarify the problem for me
Is it possible to say - 12 ants guard the cube, show the route ( the number will mean how many ants are on this face) or what ?
ответ ищите в журнале наука и жизнь за 1998 №5
will it be different from mine???