[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 477
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Actually, for a 5^5 counter
No it isn't. A counter is a tuple. If the counter has only two disks with digits from 0 to 9, the total number of combinations is 10 to the power of two. 10 disc elements to the power of 2 - to the power of the number of discs.
But we have a different situation here - we can't swap two adjacent rows - we have to shift all five rows at once. Otherwise the matrix will contradict the condition. So we have two disks with 5 elements in each. Therefore the number of combinations = 5 to the power of two. Just think: we shift the horizontal line by one position only and go through all the combinations of vertical line shifts for that shift. This is equivalent to the counter having a new one in the high digit and going through all combinations of the digits of the disk that displays the low digit for it.
P.S.
Actually, the statement "5 to the power of 5" would be true if each disk of the counter contained 5 digits and there were 5 disks as well.
No it isn't. The counter is a tuplet. If the counter has only two disks with digits from 0 to 9, the total number of combinations is 10 to the power of two. 10 disc elements to the power of two is to the power of the number of discs.
But we have a different situation here - we can't swap two adjacent rows - we have to shift all five rows at once. Otherwise the matrix will contradict the condition. So we have two discs with 5 elements in each. Therefore the number of combinations = 5 to the power of two. Just think: we shift the horizontal line by one position only and go through all the combinations of vertical line shifts for that shift. This is equivalent to the counter having a new one in the high digit and going through all combinations of the digits of the disk that displays the low digit for it.
P.S.
Actually, the statement "5 to the power of 5" would be true if each counter disk contained 5 digits and there were 5 disks as well.
Take a close look at the bottom 2 lines:
So?
And what?
Where is the looped "11100" in them?
Perhaps this should explain why 5 to the power of 5 won't work.
Imagine that the vertical columns of the matrix are the vertical-standing discs of the meter. Let's set the counter to the zero position, where the top row shows the slot where we see the meter reading. Our matrix will thus take the form:
00000
00000
11111
11111
11111
Thus in the bottom three horizontals we observe the contradiction of the problem's condition: there are 5 units in the rows instead of 3.
It means that we cannot go through the vertical discs the way the electricity meter does. We have to move the whole matrix at once, but only in one plane at a time. Thus we have 2 planes of 5 elements each. Hence the total number of combinations is 5 to the power of 2.
What's "and"?
Where's the looped "11100" in them?
Take a strip of paper. Divide it into 5 cells. Write the combination 00111 in them. Ring the strip so that the first zero and the last one are side by side. Now do the same with the second stripe. Now place one strip above the other so that 00 of the top strip is above 01 of the bottom strip.
This is the principle by which the edges of the Carnot card are glued together. You probably have never dealt with them - that's why you have not been able to understand me half-heartedly.
P.S.
Concerning the combination 10110 I have already proved, that setting zero between 1 and 11 is also a variant of the solution. Well, I just explained that it will work too. And I showed that we have only 2 ways of making up a strip - when standing together 111 and 00 and the second way - when standing between 11 and 1 is zero.
Take a strip of paper. Divide it into five cells. Write the combination 00111 into them. Ring the strip so that the first zero and the last one stand side by side. Now do the same with the second stripe. Now place one strip above the other so that the top 10 is above the bottom 01.
This is the principle by which the edges of the Carnot card are glued together. You have not dealt with them, so you have not been able to understand me.
You talk about Thomas and you talk about Yeroma.
There are the conditions of the problem. Your solution is a special case.
Take a strip of paper. Divide it into 5 cells. Write the combination 00111 in them. Ring the strip so that the first zero and the last one stand side by side. Now do the same with the second stripe. Now place one strip above the other so that the top 10 is above the bottom 01.
This is the principle by which the edges of the Carnot card are glued together. You've probably never dealt with them - that's why you can't understand me half-heartedly.
You've got it right on the nose. There's no way you can do it. :) I'll try again.
Very carefully analyze this matrix for (1) consistency with your theory and (2) with the conditions of the problem.
Then think about it further.
MetaDriver has already proved it to you.
Well, his comment makes a difference - I admit it. Well, you had to start somewhere. An error is a result. So the circle of search is widening, that's all.
Well, his comment makes a difference - I admit it. Well, you had to start somewhere. A mistake is a result. So the search is widening, that's all.
So the problem is now formulated differently. There are only 2 possible sequences of characters in the strip: 1) when 111 and 00 are side by side and 2) when there is zero between 1 and 11.
MetaDriver has already shown us the combination where three lines consist of characters from the first sequence and 2 from the other one. What remains to be determined is whether the 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?