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SuicideFuel Math thread problem (official)

Ahnfeltia

Ahnfeltia

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Is this computational complexity?
While computational complexity is more broad than Turing machines, the second question is indeed concerned with the computational complexity of a certain Turing machine.
 
trying to ascend

trying to ascend

Oldcel KHHV
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F(x,x) = x.
F(x,y) = F(y,x).
(x + y)f(x, y) = yf(x, x + y).

Find F(14, 52)
 
Ahnfeltia

Ahnfeltia

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F(x,x) = x.
F(x,y) = F(y,x).
(x + y)f(x, y) = yf(x, x + y).

Find F(14, 52)
F(14,52) = F(14,38)*52/38 = F(14,24)*38/24*52/38 = F(14,10)*24/10*52/24 = F(10,14)*52/10 = F(10,4)*14/4*26/5 = F(4,10)*7*13/5 = F(4,6)*10/6*7*13/5 = F(4,2)*6/2*10/6*7*13/5 = F(2,4)*10/2*7*13/5 = F(2,2)*4/2*7*13 = 2*2*7*13 = 364
 
trying to ascend

trying to ascend

Oldcel KHHV
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F(14,52) = F(14,38)*52/38 = F(14,24)*38/24*52/38 = F(14,10)*24/10*52/24 = F(10,14)*52/10 = F(10,4)*14/4*26/5 = F(4,10)*7*13/5 = F(4,6)*10/6*7*13/5 = F(4,2)*6/2*10/6*7*13/5 = F(2,4)*10/2*7*13/5 = F(2,2)*4/2*7*13 = 2*2*7*13 = 364
Correct :feelsokman:
 
trying to ascend

trying to ascend

Oldcel KHHV
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Find S(n), given that S(1) = 0. S(2) = 1.2, S(3) = 1.2 + 1.3 + 2.3. S(4) = 1.2 + 1.3 + 1.4 + 2.3 + 2.4 + 3.4
 
Grim_Reaper

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Find S(n), given that S(1) = 0. S(2) = 1.2, S(3) = 1.2 + 1.3 + 2.3. S(4) = 1.2 + 1.3 + 1.4 + 2.3 + 2.4 + 3.4
S(n) = S(n-1) + Σ (i = 1 to n-1) i+(n/10) for n3
 
Grim_Reaper

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What is that summation adding up?

Also, it has to be as a function of n, not S(n).
Like for S(4) for example, the summation will be 1 + (4/10) + 2 + (4/10) + 3 + (4/10) = 1.4+2.4+3.4
 
trying to ascend

trying to ascend

Oldcel KHHV
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Like for S(4) for example, the summation will be 1 + (4/10) + 2 + (4/10) + 3 + (4/10) = 1.4+2.4+3.4
. as in a multiplication sign, it's 1 times 2, 1 times 3 and so on
 
trying to ascend

trying to ascend

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@trying to ascend ?
Is it a doule summation? Assuming it is, then it's not correct, provided that S(3) = 11 and yours gives 14.

We will have sigma (upper equals 3 and lower 2) from (i)(i + 1) + Sigma (upper 3 and lower 3).

1.2 + 2.3 + 2.3.

You won't have pair of terms whose difference is higher than 1 in that summation.

Hint: The problem has a relation with arithmetic progressions
 
Grim_Reaper

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Is it a doule summation? Assuming it is, then it's not correct, provided that S(3) = 11 and yours gives 14.

We will have sigma (upper equals 3 and lower 2) from (i)(i + 1) + Sigma (upper 3 and lower 3).

1.2 + 2.3 + 2.3.

You won't have pair of terms whose difference is higher than 1 in that summation.

Hint: The problem has a relation with arithmetic progressions
Wdym my summation does give me 11.
 
Ahnfeltia

Ahnfeltia

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Find S(n), given that S(1) = 0. S(2) = 1.2, S(3) = 1.2 + 1.3 + 2.3. S(4) = 1.2 + 1.3 + 1.4 + 2.3 + 2.4 + 3.4
(1 + 2 + ... + n)^2 = 1^2 + 2^2 + ... + n^2 + 2S(n). Since 1 + 2 + ... + n = n(n+1)/2 and 1^2 + 2^2 + ... + n^2 = n(n+1)(2n+1)/6, one can easily solve for S(n) and find that S(n) = (n-1)n(n+1)(3n+2)/24.
S(n) = Σ (i = 1 to n-1) ( Σ (j= i+1 to n) i*j)
You have merely rewritten his function in terms of summation notation, which hardly counts as solving the problem.
 
trying to ascend

trying to ascend

Oldcel KHHV
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(1 + 2 + ... + n)^2 = 1^2 + 2^2 + ... + n^2 + 2S(n). Since 1 + 2 + ... + n = n(n+1)/2 and 1^2 + 2^2 + ... + n^2 = n(n+1)(2n+1)/6, one can easily solve for S(n) and find that S(n) = (n-1)n(n+1)(3n+2)/24.

You have merely rewritten his function in terms of summation notation, which hardly counts as solving the problem.
Correct :feelsokman:
 
cripplecel

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f(x) = 2x^2 what is f’(x)
 
Ahnfeltia

Ahnfeltia

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Bruh I thought it was a decimal sign. You should've used the * sign for multplication.
@trying to ascend I concur. While using a "." for multiplication is fine when it comes to variables, using it this way between numbers is very confusing because virtually everyone will interpret it as a radix point. Moreover, I believe using a "." for multiplication is Indian notation, but Wikipedia suggests India uses the "." as the decimal separator as well, so I can't imagine using "." to indicate multiplication between numbers is common practice.
 
fukurou

fukurou

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if the amount of sentences is x, and each param list is marked as a1...an with y elements per list.
how many senstence options are there?
Java:
    public PerChanceTest() {
        super();
        sentences.add("here is a  salad vegi1 #, vegi2 # and herb #.");
        sentences.add("how about this salad vegi1 #, vegi2 # and herb #. it goes well with tuna fish");
        UniqueItemSizeLimitedPriorityQueue temp = new UniqueItemSizeLimitedPriorityQueue();
        temp.setLimit(3);
        wordToList.put("vegi1", temp);
        temp = new UniqueItemSizeLimitedPriorityQueue();
        temp.setLimit(3);
        wordToList.put("vegi2", temp);
        temp = new UniqueItemSizeLimitedPriorityQueue();
        temp.setLimit(3);
        wordToList.put("herb", temp);
    }

output example:
output:
here is a salad cucmber, purple onion and cpriander.
here is a salad tomato, purple onion and irit.
here is a salad lettuce, pickle and cpriander.
how about this salad tomato, purple onion and cpriander. it goes well with tuna fish
how about this salad lettuce, pickle and irit. it goes well with tuna fish
how about this salad lettuce, pickle and parcely. it goes well with tuna fish
how about this salad tomato, onion and cpriander. it goes well with tuna fish
how about this salad cucmber, pickle and parcely. it goes well with tuna fish
here is a salad cucmber, purple onion and parcely.
here is a salad cucmber, purple onion and irit.

Process finished with exit code 0
 
Ahnfeltia

Ahnfeltia

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if the amount of sentences is x, and each param list is marked as a1...an with y elements per list.
how many senstence options are there?
If I understand you correctly, you have x different sentences, each with n variable slots and each slot permitting y options, correct? In that case, the total number of possible sentences is x*(y^n).
 
trying to ascend

trying to ascend

Oldcel KHHV
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f(x) = 2x^2 what is f’(x)
4x

@trying to ascend I concur. While using a "." for multiplication is fine when it comes to variables, using it this way between numbers is very confusing because virtually everyone will interpret it as a radix point. Moreover, I believe using a "." for multiplication is Indian notation, but Wikipedia suggests India uses the "." as the decimal separator as well, so I can't imagine using "." to indicate multiplication between numbers is common practice.
Yes, it indeed looked confusing, though I was just copying the original problem
 
Grim_Reaper

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(1 + 2 + ... + n)^2 = 1^2 + 2^2 + ... + n^2 + 2S(n). Since 1 + 2 + ... + n = n(n+1)/2 and 1^2 + 2^2 + ... + n^2 = n(n+1)(2n+1)/6, one can easily solve for S(n) and find that S(n) = (n-1)n(n+1)(3n+2)/24.

You have merely rewritten his function in terms of summation notation, which hardly counts as solving the problem.
I thought that was okay:feelsjuice:.
 
Ahnfeltia

Ahnfeltia

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I thought that was okay:feelsjuice:.
Generally not (although there technically are exceptions). Here's a rather extreme example: would you say that sum from j=2 to 4 of j is a valid solution to "find the value of 2 + 3 + 4"?
 
Grim_Reaper

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Generally not (although there technically are exceptions). Here's a rather extreme example: would you say that sum from j=2 to 4 of j is a valid solution to "find the value of 2 + 3 + 4"?
No but the upper limit for your example is bounded.
 
Ahnfeltia

Ahnfeltia

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No but the upper limit for your example is bounded.
Does any of the limits being a variable change anything significant? In that case you could argue that summation notation is more formal as opposed to the ellipses, but that's about it, no? At the end of the day you have merely recast the given expression in a more formal form without really doing any kind of simplification.
 
fukurou

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If I understand you correctly, you have x different sentences, each with n variable slots and each slot permitting y options, correct? In that case, the total number of possible sentences is x*(y^n).
that looks like a lot when it's put into a formula, maybe it could be used to write visual novel type of games, scripts
 
Ahnfeltia

Ahnfeltia

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that looks like a lot when it's put into a formula, maybe it could be used to write visual novel type of games, scripts
it is a lot, esp. for large n
 
uranium235

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Consider 9 dots arranged as 3x3 what is the highest amount of closed figures that can be drawn by connecting the dots one by one in a single trial you can imagine this as drawing pattern lock in your phone i have achieved 10 closed figures so far
 
Ahnfeltia

Ahnfeltia

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Consider 9 dots arranged as 3x3 what is the highest amount of closed figures that can be drawn by connecting the dots one by one in a single trial you can imagine this as drawing pattern lock in your phone i have achieved 10 closed figures so far
I take it you want the number of essentially distinct ones? Mirror images counting as the same figure etc.
 
uranium235

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There are 10 closed figures in this image
 

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uranium235

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Let P(x) be a polynomial with positive real coefficients. Prove that
Sqrt(P(a)P(b) )≥ Sqrt(P( ab)),for all positive real numbers a and b.
 
Ahnfeltia

Ahnfeltia

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Let P(x) be a polynomial with positive real coefficients. Prove that
Sqrt(P(a)P(b) )≥ Sqrt(P( ab)),for all positive real numbers a and b.
I'm not convinced this is true. Take P(x) = ¼ + x. In that case your desideratum amounts to ¼ + ab ≤ (¼ + a)(¼ + b) for positive a & b, but this is evidently false for a = b = ¼.
 

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