[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 476
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For those who like to solve problems:
A traffic cop who collects speeding tickets gains 11kg a year..,
and a traffic cop who collects fines for turning in a wrong place - only 6.5 kg.
1. Calculate the total annual weight of the traffic policemen in a 15-member squad,
if 7 of them are charged with speeding tickets
and 8 for making a U-turn in a wrong place.
Draw the weight gain curve as a graph. )))
2. How long will it take for traffic policemen 1 and 2 to starve to death if motorists stop breaking the rules?
Осталось доказать, что расстановка символов в закольцованной ленте 00111 - единственная. Ну например, ни при каких сдвигах и ни при каких поворотах нам не встречается последовательность - 01011
There are only three possible combinations of a looped tape: 1) 00111, 2) 01011 and 3) 11010. The third and second are mirrored, so they can be combined into one by formulating the rule: In a true looped tape, two zeros must stand in adjacent positions. The other three are occupied by three subordinate units.
Suppose that in a looped tape it is acceptable to have a single zero between the pair 11 and 1. For example, it is the combination 01011.
It is clear that to build a correct matrix, the initial top line should be shifted sequentially, position by position, cyclically. It's not hard to get to that point. If there is no such positional cyclical shift, we will get unordered (read: uncontrollable) chaos. Let's build exactly the same matrix with a shift that we get from line 01011. If it leads us to a contradiction in the problem condition, then our rule "In a true looped ribbon, two zeros must stand on adjacent positions. The other three are occupied by three subordinate ones" will be the only correct one. Let us construct a matrix
0 1 0 1 1
1 0 1 0 1
1 1 0 1 0
0 1 1 0 1
1 0 1 1 0
The matrix does not contradict the condition of the problem. It means that we have another 100 combinations to build the Karno map and that our rule is not true. The total is 200 ways.
A fun problem about arranging the units in a matrix. Well, we have to start somewhere. Trying to match at least one such matrix leads to this result:
1 0 0 1 1
1 1 0 0 1
1 1 1 0 0
0 1 1 1 0
0 0 1 1 1
Comparison of the first upper horizontal row with the second leads us to the conclusion that the second row is nothing else but the first one shifted by one position to the right. The rightmost character (the last one in the row) goes out of the matrix and we just put it in the first position, in the vacated place of the first character. Comparing all subsequent lines with previous ones leads to the same conclusion: each subsequent line is the previous one shifted by one position to the right. It is the same for the columns, only shifted vertically. So every line is a looped ribbon and every column is a looped ribbon. It turns out that this is not just a matrix - it is a Karno map. So the problem is not how many ways you can build such a matrix, but how many ways you can build such Karno maps.
Frankly, it seems to me that the ribbon has a single sequence of symbols, namely 00111, where the first zero and the last one are two adjacent symbols of the looped ribbon. If this assumption is correct (about the uniqueness of the sequence), the number of combinations is not difficult to calculate.
It is clear that if the upper ribbon is shifted horizontally, then all other horizontal ribbons should be shifted in the same direction and by the same number of positions. So we have 5 vertical and 5 horizontal shifts of the whole map field. For every vertical shift, there are 5 horizontal ones. The total is 5*5. But we can rotate the box. Let's paint the top line blue. How many positions will the square have? Blue top, blue right, blue bottom, blue left. In total there are 4 positions. Therefore we have 5*5*4 = 100 ways to build the given Karno map.
It remains to prove that the arrangement of symbols in the looped tape 00111 is the only one. For example, at no shifts and no turns do we meet the sequence - 01011
You have got one of the variants of filling the matrix. Now you can swap any columns and the result will also meet the conditions of the problem. You can also swap any rows. So here we have:
<number of column permutations> * <number of row permutations
You have obtained one of the options for filling the matrix. Now you can swap any columns and the result will also satisfy the conditions of the problem. You can also swap any rows. Thus we have:
<number of column permutations> * <number of row permutations>
No - look more closely - I've couaal 4 more matrix square rotation positions. Total <number of column permutations> * <number of row permutations> * <number of matrix square rotations
Plus I have found the second possible arrangement of symbols in the looped tape. So, the total number of combinations = <Number of column permutations> * <number of row permutations> * <number of matrix square rotations> * <2> = 200
drknn:
It remains to prove that the arrangement of characters in the looped tape 00111 is the only one. For example, at no shifts and no turns do we encounter the sequence - 01011
You can't prove it. There are many more permutations. For example, permutation of arbitrary columns or rows of a "proper" matrix creates a proper matrix.
An example off the top of my head:
zy: ))
PapaYozh is ahead of the curve.
You can't prove it. There are many more permutations. For example, rearranging arbitrary columns or rows of a "proper" matrix creates a proper matrix.
An example off the top of my head:
zy: ))
PapaYozh is ahead of the curve.
You have refuted the thesis "You can't prove it" with your own example. Look at your matrix - loop it horizontally - you will always have 111 and 00 in a row. It is the same if you loop it vertically. This leaves you with the only option to build a ribbon - to set a zero between 11 and 1