How to determine the age of children? - MQL4 and MetaTrader 4

[Deleted] 2010.09.16 17:27 #3971

Hmmm... no, I won't be using today.

Alexey Subbotin 2010.09.16 18:08 #3972

dadi:
Hmmm... no, I won't be using today.

Get off the branch, don't spoil the fun!

Sceptic Philozoff 2010.09.16 19:54 #3973

Vita, will you write me the answer in a private message? Let's adjust the problem condition to the known answer...

P.S. I have answers to both interpretations; personally I like the strict one better ("all ages of children are different"). The reason is the singularity of the solution at a reasonable age of a friend.

P.S. It would be better if they went to a brothel - that way the natural age limit would be more obvious.

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Vitali 2010.09.17 08:42 #3974

Mathemat:

Vita, will you write me the answer in a private message? Let's adjust the problem condition to the known answer...

P.S. I have answers to both interpretations; personally I like the strict one better ("all ages of children are different"). The reason is the singularity of the solution at a reasonable age of a friend.

P.S. It would be better if they went to a brothel - that way the natural age restriction would be more obvious.

I was matching the answer to the non-strict interpretation.

My gut feeling is that if you write "the youngest is a redhead", does such wording rule out that there is no younger non-redhead, but the same age as the redhead? For example twins, a girl and a boy? They're both juniors. Neither of them is "the youngest". The younger one's a redhead and the younger one's a brunette, for example.

Generally, I assume that "youngest" is a characteristic of the redhead in relation to the remaining children, not a characteristic of the remaining two children or the entire row.

The characteristic of the redhead is that her age is not only extreme, but exceptional. I have found that the word 'most' does just that job.

Perhaps if you reduce the degree of sexism in the task and write "the youngest child is a redhead", then, that would be correct, but then the colouring disappears and the awl from the bag (the youngest is only one) sticks out more clearly.

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Ihor 2010.09.17 10:27 #3975

Vita:

I was picking up the answer under a loose interpretation.

My gut feeling is that if you write "the youngest is a redhead", does such wording rule out that there is no younger non-redhead, but the same age as the redhead? For example twins, a girl and a boy? They're both juniors. Neither of them is "the youngest". The younger one's a redhead and the younger one's a brunette, for example.

Generally, I assume that "the youngest" is a characteristic of the redhead in relation to the remaining children, not a characteristic of the remaining two children or the whole row.

The characteristic of the redhead is that her age is not only extreme, but exceptional. I have found that the word 'most' does just that job.

Perhaps if you reduce the degree of sexism in the task and write "the youngest child is a redhead", then, that would be correct, but then the colouring disappears and the awl out of the bag (the youngest is only one) sticks out more clearly.

If "the youngest is only one" and the oldest can be of the same age, then the problem has several solutions.

Sceptic Philozoff 2010.09.17 23:30 #3976

For those who want to check, there is a script:

//+------------------------------------------------------------------+
//|                                                      taskmq4.mq4 |
//|                                                   Mthmt (c) 2010 |
//+------------------------------------------------------------------+
#property copyright "Mthmt (c) 2010"
#define _MAX 30

int start( )
{
   for( int i = 1; i < _MAX; i ++ )   
      for ( int j = i; j < _MAX; j ++ )
         for ( int k = j; k < _MAX; k ++ )
            for( int i1 = 1; i1 < _MAX; i1 ++ )   
               for ( int j1 = i1; j1 < _MAX; j1 ++ )
                  for ( int k1 = j1; k1 < _MAX; k1 ++ )

                  {
                     if( 
                         ( i + j + k == i1 + j1 + k1 ) &&
                         ( i * j * k == i1 * j1 * k1 ) &&
                         ( ( i - i1 ) * ( i - i1 ) + ( j - j1 ) * ( j - j1 ) + ( k - k1 ) * ( k - k1 ) != 0 ) &&
                         ( i * j * k < 100 ) &&
                         ( 
// строгое условие
//                          ( ( i - j ) * ( i - k ) * ( j - k )       != 0 ) ||
//                          ( ( i1 - j1 ) * ( i1 - k1 ) * ( j1 - k1 ) != 0 ) 
                  
// нестрогое условие                            
                            ( ( i - j ) * ( i - k )     != 0 ) ||
                            ( ( i1 - j1 ) * ( i1 - k1 ) != 0 ) 
                         ) 
                       )
                        Print( "Age = " + i*j*k + ":  " + i + ", " + j + ", " + k + "; " + i1 + ", " + j1 + ", " + k1 );
                  }
   return( 0 );
}

The interpretation of task conditions (strict/unstrict) can be switched by commenting out/uncommenting the relevant lines. The structure of the output lines are: "Age of friend, first choice of ages, second choice". Reasonable solutions are highlighted in blue (there are always two identical solutions in this algorithm). I didn't make the algorithm perfect and flawless: it was enough for me to get solutions.

1. Strict:

03:23:44 taskmq4 USDCHF,Daily: Age = 40: 1 , 5, 8; 2 , 2, 10
03:23:44 taskmq4 USDCHF,Daily: Age = 96: 1, 8, 12; 2, 3, 16
03:23:44 taskmq4 USDCHF,Daily: Age = 90: 1, 9, 10; 2, 3, 15
03:23:51 taskmq4 USDCHF,Daily: Age = 40: 2, 2, 10; 1, 5, 8
03:23:51 taskmq4 USDCHF,Daily: Age = 90: 2, 3, 15; 1, 9, 10
03:23:51 taskmq4 USDCHF,Daily: Age = 96: 2, 3, 16; 1, 8, 12
03:23:51 taskmq4 USDCHF,Daily: Age = 90: 2, 5, 9; 3, 3, 10
03:23:51 taskmq4 USDCHF,Daily: Age = 90: 3, 3, 10; 2, 5, 9

2. Unstressed:

03:15:20 taskmq4 USDCHF,Daily: Age = 40: 1 , 5, 8; 2, 2, 10
03:15:20 USDCHF,Daily: Age = 36: 1 , 6, 6; 2, 2, 9
03:15:20 USDCHF,Daily: Age = 96: 1, 8, 12; 2, 3, 16
03:15:20 taskmq4 USDCHF,Daily: Age = 90: 1, 9, 10; 2, 3, 15
03:15:20 taskmq4 USDCHF,Daily: Age = 36: 2, 2, 9; 1, 6, 6
03:15:20 taskmq4 USDCHF,Daily: Age = 40: 2, 2, 10; 1, 5, 8
03:15:20 USDCHF,Daily: Age = 90: 2, 3, 15; 1, 9, 10
03:15:20 USDCHF,Daily: Age = 96: 2, 3, 16; 1, 8, 12
03:15:20 USDCHF,Daily: Age = 90: 2, 5, 9; 3, 3, 10
03:15:20 USDCHF,Daily: Age = 72: 2, 6, 6; 3, 3, 8
03:15:20 USDCHF,Daily: Age = 72: 3, 3, 8; 2, 6, 6
03:15:20 USDCHF,Daily: Age = 90: 3, 3, 10; 2, 5, 9

As you can see, the only sensible option is to go to a brothel when interpreted strictly. It is 1, 5, 8.

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Vitali 2010.09.18 17:57 #3977

Mathemat:

For those who want to check it out, here's a script:

Interpretation of problem conditions (strict/unstrict) can be switched by commenting/uncommenting the appropriate lines. The structure of the output lines: "Friend's age, first variant of children's age, second variant. Reasonable solutions are highlighted in blue (there are always two identical solutions in this algorithm). I didn't make the algorithm perfect and flawless: it was enough for me to get solutions.

1. Strict:

03:23:44 taskmq4 USDCHF,Daily: Age = 40: 1 , 5, 8; 2 , 2, 10
03:23:44 taskmq4 USDCHF,Daily: Age = 96: 1, 8, 12; 2, 3, 16
03:23:44 taskmq4 USDCHF,Daily: Age = 90: 1, 9, 10; 2, 3, 15
03:23:51 taskmq4 USDCHF,Daily: Age = 40: 2, 2, 10; 1, 5, 8
03:23:51 taskmq4 USDCHF,Daily: Age = 90: 2, 3, 15; 1, 9, 10
03:23:51 taskmq4 USDCHF,Daily: Age = 96: 2, 3, 16; 1, 8, 12
03:23:51 taskmq4 USDCHF,Daily: Age = 90: 2, 5, 9; 3, 3, 10
03:23:51 taskmq4 USDCHF,Daily: Age = 90: 3, 3, 10; 2, 5, 9

2. Unstressed:

03:15:20 taskmq4 USDCHF,Daily: Age = 40: 1 , 5, 8; 2, 2, 10
03:15:20 USDCHF,Daily: Age = 36: 1 , 6, 6; 2, 2, 9
03:15:20 USDCHF,Daily: Age = 96: 1, 8, 12; 2, 3, 16
03:15:20 taskmq4 USDCHF,Daily: Age = 90: 1, 9, 10; 2, 3, 15
03:15:20 taskmq4 USDCHF,Daily: Age = 36: 2, 2, 9; 1, 6, 6
03:15:20 taskmq4 USDCHF,Daily: Age = 40: 2, 2, 10; 1, 5, 8
03:15:20 USDCHF,Daily: Age = 90: 2, 3, 15; 1, 9, 10
03:15:20 USDCHF,Daily: Age = 96: 2, 3, 16; 1, 8, 12
03:15:20 USDCHF,Daily: Age = 90: 2, 5, 9; 3, 3, 10
03:15:20 USDCHF,Daily: Age = 72: 2, 6, 6; 3, 3, 8
03:15:20 USDCHF,Daily: Age = 72: 3, 3, 8; 2, 6, 6
03:15:20 USDCHF,Daily: Age = 90: 3, 3, 10; 2, 5, 9

As you can see, the only sensible option is to go to a brothel when interpreted strictly. It is 1, 5, 8.

Thank you, Mathemat, I've forgotten something about the condition over the years. I need to refactor the problem.

Considering that at 40 it's safe to say "You know, honey, I don't feel like it anymore", it's better to leave it about beer rather than brothels :). In general, the script is not running on the right periods and currency. Every day (Daily), and even for Swiss francs (USDCHF) - definitely only beer. :)

Maybe in the problem was iteration: Two of my twins - little info, and only after that - the younger redhead. But I don't remember exactly.

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Сергей 2010.09.23 07:02 #3978

Now the granddaughter of a school employee called and asked for help with a physics problem. :)

There are 8 metal balls of the same size. One of them is hollow.

The question is how to find the hollow ball with the help of 2 weighings on a lever scale.

I had to think about it! The grandfather was happy when we solved it.

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Neutron 2010.09.23 07:32 #3979

First option: Place three balls each on the scales (first weighing); if the scales are balanced, the hollow ball is one of the remaining two, which will reveal the second weighing.

Second option: We put three balls on the scales each (first weighing); if the scales are out of balance, the hollow ball is one of the three. We take two of them at random and carry out the second weighing; if the scales are in equilibrium, the hollow ball is the one remaining alone. If the scales are not balanced, the lighter one is.

Both options are two-way and there are no more options.

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Neutron 2010.09.23 07:38 #3980

We all spit on the range.

It is known that if you spit from the window of a speeding train, the range of the "projectile" (without taking into account the drift along the carriage) will be much less than if you do the same operation with a stationary train.

Question: How is this possible?

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