[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 478
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drknn:
It remains to be seen whether a combination of 4 and 1 is possible - that is, 4 lines consisting of characters from the first sequence and one line consisting of characters from the second sequence?
In this case it is the other way round. But it's not a matter of principle, as long as it's possible, so is the other. I.e. it seems to be arbitrary. Both first (A) and second (B) sequences can be present in any quantity. However, the hypothesis: if horizontally we have a set of sequences A*k+B*(5-k), then vertically we have the same set.
PS. A and B are types of sequences. A = 11100, B = 10110, accurate to rotation (any number of permutations of the last character at the beginning)
Actually, the statement "5 to the power of 5" would be true if each disc on the counter contained 5 digits and there were 5 discs as well.
It looks like we have a group of transpositions of rows (L=line) and columns (C=column). For example, the effect of transposition on the "proper" matrix A, i.e. L[1,4](A) is the exchange of the 1st and 4th rows of matrix A. Correspondingly, C[2,3](A) is the exchange of the 2nd and 3rd columns of matrix A. According to the remarks made earlier, we get a regular matrix too (I call a regular matrix satisfying the conditions of the problem).
Say, one can write: B = C[2,3]*L[1,4](A). This will mean that matrix B (correct) is obtained by successive exchanges (transpositions) first of the 1st and 4th rows of A, and then of the 2nd and 3rd columns of the resulting matrix A1.
All possible products of transpositions constitute a finite group. Of course, we can form a product of 1000 elements, but it can be simplified according to the rules of transpositions multiplication, so that the final product contains, say, no more than 10 different factors (10 is just an approximation).
The elements C[*,*] together with the unit E form a subgroup of the complete group. The same applies to the elements of L.
All elements of a complete group can be written out explicitly. The number of different elements of this group will be the solution of the problem.
By the way, L[i,j]*L[i,j]=E is a unit element of the group. Similarly with C[i,j]. I have a suspicion that the group is abelian. I think so because perhaps the square of any element of the transposition group is equal to a single element.
In short, guys, you can't do without transposition theory here. I hope this reasoning will help a group theory expert to solve the problem.
P.S. I was thinking a bit more. Still, the structure of the matrix must be taken into account somehow. If the answer would be different, although the transposition groups would be identical. Right, alsu?
you think too much of human stupidity.
It seems that from your vantage point the object in question looks different than it does from mine. I'm going to take a break from this forum for a couple or three months, it's getting tense.
All right, even if it's two zeros. You still have to deal with a group of transpositions over these matrices... Or I don't see a more obvious solution.
P.S. It's good to see that the mechmathians haven't found a nice solution either :)
But hypothesis: if horizontally we have a set of sequences A*k+B*(5-k), then vertically we have the same set.
I give an almost obvious hint on how to simplify the solution: in the problem statement, zeros and ones can be "swapped out" and matrices with two zeros in rows and columns can be searched for.