Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 16
You are missing trading opportunities:
- Free trading apps
- Over 8,000 signals for copying
- Economic news for exploring financial markets
Registration
Log in
You agree to website policy and terms of use
If you do not have an account, please register
OK, here's a problem that I've put into permanent subconscious thinking mode (weight - 3, not solved):
A megabrain in a very long text needs to replace all the letters "A" with "B" and all the letters "B" with "A". The text editor allows one arbitrary set of characters to be replaced by another in the whole text. For example, replacing "AA"->"BSAA" will cause the string "AAAAAL" to become "BSAABSAAL". How does he complete the task?
Explanation: The alphabet of the text is unknown, so no other characters besides A, B are known to us and cannot be used for substitution in the left part. Furthermore, no other characters in the right part of the replacement should be there either. This is some kind of bugger, but that's exactly the problem condition.
I already wrote several variants of "solution", but all of them turned out to be wrong.
As an option (probably silly, but no energy to check now):
1) "A" -> "AB".
2) "B" -> "BA".
3) "AB" -> "BA"
4) "AB" -> "A"
5) "BA" -> "B".
I'll have to prove it, I don't take your word for it.
P.S. Another one (weight 4, scored):
A brick falls on a perfectly elastic ball from a height of 1 metre and bounces back almost 1 metre. To what height will the ball bounce?
Explanation: the answer has to be numerical. That is the beauty of the problem.
I'll have to prove it, I don't take your word for it.
P.S. Another one (weight 4, scored):
A brick falls on a perfectly elastic ball from a height of 1 metre and bounces back almost 1 metre. To what height will the ball bounce?
Explanation: The answer has to be numerical. That is the beauty of the problem.
0 ?
Nah, bigger.
Have you ever tried to lift a tennis ball off the ground by hitting it with a racket?
As an option (probably silly, but no energy to check now):
1) "A" -> "AB"
2) "B" -> "BA"
3) "AB" -> "BA"
4) "AB" -> "A"
5) "BA" -> "B".
Yes, that's right silly, I forgot that already in the second step "BA" will replace all the "B", which is also the result of replacing "A" to "AB" from the first step.
The whole point of my option is just to exclude that sort of thing.
Another option:
1) "A" -> "AAA"
2) "B" -> "BBB".
3) "AAA" -> "BBB".
4) "BBB" -> "ABA"
5) "BAB" -> "B"
6) "ABA" -> "A"
0 ?
I'll have to prove it, I don't take your word for it.
P.S. Another one (weight 4, scored):
A brick falls on a perfectly elastic ball from a height of 1 metre and bounces back almost 1 metre. To what height will the ball bounce?
Explanation: The answer has to be numerical. That's the beauty of the problem.
The brick doesn't bounce... You can get it if the ball is placed on some kind of elastic/spring base. The ball will bounce off the brick together.
The answer is "almost 1 metre" minus the size of the brick ?
Yes, that's right silly, I forgot that already in the second step "BA" will replace all "B", which is also in the results of replacing "A" with "AB" from the first step.
The whole point of my option is just to exclude that sort of thing.
Another option:
1) "A" -> "AAA"
2) "B" -> "BBB".
3) "AAA" -> "BBB".
4) "BBB" -> "ABA"
5) "BAB" -> "B"
6) "ABA" -> "A".
Step 1 and 2 only postpones the mixing of letters in step 4 (AABBBB->BABBBB->BAABB->BAABB...
About letter substitution, here's what comes to mind:
1) A -> BAA
2) B -> ABB
3) ABBAA -> B
4) ABB -> A
It seemed to me that each A should be replaced by two (or three, ten...) AA, while shielding (marking) B. Same for B, only vice versa. Then the substitutions won't get lost.
What do you think?almost
Explanation: The answer has to be numerical. That's the beauty of the problem.
No. The brick bounces back at, say, 99.5cm. That's "almost".
The size of the brick does not play a role here. The important thing is to understand the physical processes that occur after the brick hits the ball, and then turn them into concrete numbers - say, centimetres.
Nothing complicated there really.
I.e. the answer is 0.