## Forming matrix in a simple way

``````N1(n_) := Sqrt((NN - n)*(n + 1));

N2(n_) := Sqrt(n*(NN - n + 1));

N3(n_) := ((NN - n)*(NN - n - 1)) + (n*(n - 1));

FT(n_) := J*N1(n)

ST(n_) := J*N2(n)

TT(n_) := S*N3(n)

DCnt(n_) := (-I*FT(n)*C(n + 1)) + (-I*ST(n)*C(n - 1)) + (-I*TT(n)*C(n))
``````

Write now to form a matrix using the coefficients of above relations I am using like below:

Is there a simpler code for below?

``````MatA = {{Coefficient(DCnt(0), C(0)), Coefficient(DCnt(0), C(1)),
Coefficient(DCnt(0), C(2)), Coefficient(DCnt(0), C(3)),
Coefficient(DCnt(0), C(4)), Coefficient(DCnt(0), C(5)),
Coefficient(DCnt(0), C(6)), Coefficient(DCnt(0), C(7)),
Coefficient(DCnt(0), C(8))}, {Coefficient(DCnt(1), C(0)),
Coefficient(DCnt(1), C(1)), Coefficient(DCnt(1), C(2)),
Coefficient(DCnt(1), C(3)), Coefficient(DCnt(1), C(4)),
Coefficient(DCnt(1), C(5)), Coefficient(DCnt(1), C(6)),
Coefficient(DCnt(1), C(7)),
Coefficient(DCnt(1), C(8))}, {Coefficient(DCnt(2), C(0)),
Coefficient(DCnt(2), C(1)), Coefficient(DCnt(2), C(2)),
Coefficient(DCnt(2), C(3)), Coefficient(DCnt(2), C(4)),
Coefficient(DCnt(2), C(5)), Coefficient(DCnt(2), C(6)),
Coefficient(DCnt(2), C(7)),
Coefficient(DCnt(2), C(8))}, {Coefficient(DCnt(3), C(0)),
Coefficient(DCnt(3), C(1)), Coefficient(DCnt(3), C(2)),
Coefficient(DCnt(3), C(3)), Coefficient(DCnt(3), C(4)),
Coefficient(DCnt(3), C(5)), Coefficient(DCnt(3), C(6)),
Coefficient(DCnt(3), C(7)),
Coefficient(DCnt(3), C(8))}, {Coefficient(DCnt(4), C(0)),
Coefficient(DCnt(4), C(1)), Coefficient(DCnt(4), C(2)),
Coefficient(DCnt(4), C(3)), Coefficient(DCnt(4), C(4)),
Coefficient(DCnt(4), C(5)), Coefficient(DCnt(4), C(6)),
Coefficient(DCnt(4), C(7)),
Coefficient(DCnt(4), C(8))}, {Coefficient(DCnt(5), C(0)),
Coefficient(DCnt(5), C(1)), Coefficient(DCnt(5), C(2)),
Coefficient(DCnt(5), C(3)), Coefficient(DCnt(5), C(4)),
Coefficient(DCnt(5), C(5)), Coefficient(DCnt(5), C(6)),
Coefficient(DCnt(5), C(7)),
Coefficient(DCnt(5), C(8))}, {Coefficient(DCnt(6), C(0)),
Coefficient(DCnt(6), C(1)), Coefficient(DCnt(6), C(2)),
Coefficient(DCnt(6), C(3)), Coefficient(DCnt(6), C(4)),
Coefficient(DCnt(6), C(5)), Coefficient(DCnt(6), C(6)),
Coefficient(DCnt(6), C(7)),
Coefficient(DCnt(6), C(8))}, {Coefficient(DCnt(7), C(0)),
Coefficient(DCnt(7), C(1)), Coefficient(DCnt(7), C(2)),
Coefficient(DCnt(7), C(3)), Coefficient(DCnt(7), C(4)),
Coefficient(DCnt(7), C(5)), Coefficient(DCnt(7), C(6)),
Coefficient(DCnt(7), C(7)),
Coefficient(DCnt(7), C(8))}, {Coefficient(DCnt(8), C(0)),
Coefficient(DCnt(8), C(1)), Coefficient(DCnt(8), C(2)),
Coefficient(DCnt(8), C(3)), Coefficient(DCnt(8), C(4)),
Coefficient(DCnt(8), C(5)), Coefficient(DCnt(8), C(6)),
Coefficient(DCnt(8), C(7)), Coefficient(DCnt(8), C(8))}};
``````

## hash – What exactly happens to leaves when forming nodes?

Concatenation means the joining of the second sequence at the end of the first sequence. For example, with strings, the concatenation of `"hello, "` and `"world"` produces the string `"hello, world"`.

For TXIDs, they are taken as a 32-byte array. The concatenation of the two TXIDs produces an array of 64 bytes, which then becomes the input for another round of double SHA256. The resulting hash becomes the parent hash of the two TXIDs in the Merkle tree.

Each pair of transactions is hashed in such way, and if the number of transactions is odd, the last TXID in the sequence is duplicated (concatenated with itself). The parent hashes are then recursively treated in the same way, concatenated pairwise, until a single hash remains, which is the Merkle root.

## combinatorics – Probability of forming an odd integer from a 5-permutation of integers in [0,9]

My intuition is telling me that the probability of forming an odd integer from a 5-element partial permutation of integers in (0,9) is one-half, because it should be the same probability as the integer being even, which makes those probabilities one-half.

However, I’m having trouble proving it.

I split the problem into five disjoint cases: the number consists of 1, 2, 3, 4, and 5 odd integers.

P(the number is odd and it uses only 1 odd integer) = $$frac{5*4*3*2*binom{5}{1}}{10*9*8*7*6} = frac{600}{30240}$$, because the last digit must be an odd integer, and there are five candidates, therefore 5 choose 1, and the rest are even integers where the first even integer has 5 candidates, the second, 4 candidates and so on.

P(the number is odd and it uses only 2 odd integer) = $$frac{5*4*3*binom{4}{1}binom{5}{1}}{10*9*8*7*6} = frac{1200}{30240}$$, because the last digit must be an odd integer, and there are five candidates, therefore 5 choose 1, and there is another odd integer, which has 4 possible candidates; the rest are even integers where the first even integer has 5 candidates, the second, 4 candidates and so on.

I see a problem with this approach is that I can continue to the case of 3, 4, 5, but the numerator would not add up to half of 30240. In fact, the numerator only adds up to 3720.

I feel it might be the placement of the odd integers other than the last, so I tried:

P(the number is odd and it uses only 2 odd integer) = $$frac{5*4*3*4*binom{4}{1}binom{5}{1}}{10*9*8*7*6} = frac{4800}{30240}$$, because the last digit must be an odd integer, and there are five candidates, therefore 5 choose 1, and there is another odd integer, which has 4 possible candidates but it can be in any of the 4 places, therefore multiplied by 4; the rest are even integers where the first even integer has 5 candidates, the second, 4 candidates and so on.

But this time the numbers are too large, and it’s way over half of 30240.

What am I doing wrong here?

## algorithm – Assessing BigO/small o for forming a dynamic dictionary based on searched key and values pair

I am trying to create a dictionary with a file containing text based on a matched pattern. Lines containing `key_str` should become keys and subsequent lines not matching `key_str` should become values and get associated with keys in the dictionary. so i have below code working: But I need help and getting the Big O analysis. How could i say my logic is worth case, Best Case or good case? Also, which case is small o and which is big O

File: `file2dict.result-soa1`

``````ml1
/var
/home

cpuml2
/var
/home

``````

Output

``````my_dict: {ml1: ('/var','/home'), cpuml2: ('/var','/home')}
``````

Code:

``````import os
homedir = os.environ.get('HOME')

key_str = "ml"
my_dict = {}
val_list = ()
key = ''
with open(homedir + '/backup/file2dict.result-soa1') as file2dict:
for line in file2dict:
words = line.split()
for aWord in words:
if key_str in aWord:
if key:
my_dict(key) = val_list
val_list = ()
key = aWord
else:
key = aWord
else:
val_list.append(aWord)
my_dict(key) = val_list
print(my_dict)
``````

## Inequality forming [closed]

a polo league renting a banquet hall for an end – of – season celebration. it costs 800 dollars to rent the hall and 40 dollars per person for prizes and food . the costs will be split equally among the attending players . how many players must attend the banquet for the cost per player to be at most 72 dollars?

Let x be the number of players.

I have tried making different equations to solve it but the answers are always imaginary so I posted it here for help.

## runtime analysis – Forming recurrence relations

I have this 2 examples in my textbook:

Example 1

``````public void f(int n) {
if (n = 1)
return 1;
else
return n * f (n-1);
}
``````

The textbook shows how the recurrence relation is being form from by the above code

``````T(0) = a           for some constant a
T(n) = T(n-1)+b    for some constant b and a recursive term
``````

Example 2

``````public int myFunction (int n) {
if (n == 1)
return 1;
else
return 2 * myFunction(n/2) + myFunction(n/2) + 1;
}
``````

The textbook shows how the recurrence relation is being form from by the above code

``````T(1) = c              for some constant c
T(n) = 2T(n/2) + b    for some constant b and a recursive term
``````

The problem

Despite reading the textbook multiple time (and failed attempt at searching online), I still do not understand how the +b come about for both examples. Is anyone able to enlighten on this?

Thanks.

## algorithms – Tight upper bound for forming an \$n\$ element Red-Black Tree from scratch

I learnt that in a order-statistic tree (augmented Red-Black Tree, in which each node $$x$$ contains an extra field denoting the number of nodes in the sub-tree rooted at $$x$$) finding the $$i$$ th order statistics can be done in $$O(lg(n))$$ time in the worst case. Now in case of an array representing the dynamic set of elements finding the $$i$$ th order statistic can be achieved in the $$O(n)$$ time in the worst case.( where $$n$$ is the number of elements).

Now I felt like finding a tight upper bound for forming an $$n$$ element Red-Black Tree so that I could comment about which alternative is better : “maintain the set elements in an array and perform query in $$O(n)$$ time” or “maintaining the elements in a Red-Black Tree (formation of which takes $$O(f(n))$$ time say) and then perform query in $$O(lg(n))$$ time”.

So a very rough analysis is as follows, inserting an element into an $$n$$ element Red-Black Tree takes $$O(lg(n))$$ time and there are $$n$$ elements to insert , so it takes $$O(nlg(n))$$ time. Now this analysis is quite loose as when there are only few elements in the Red-Black tree the height is quite less and so is the time to insert in the tree.

I tried to attempt a detailed analysis as follows (but failed however):

Let while trying to insert the $$j=i+1$$ th element the height of the tree is atmost $$2.lg(i+1)+1$$. For an appropriate $$c$$, the total running time,

$$T(n)leq sum_{j=1}^{n}c.(2.lg(i+1)+1)$$

$$=c.sum_{i=0}^{n-1}(2.lg(i+1)+1)$$

$$=c.left(sum_{i=0}^{n-1}2.lg(i+1)+sum_{i=0}^{n-1}1right)$$

$$=2csum_{i=0}^{n-1}lg(i+1)+cntag1$$

Now

$$sum_{i=0}^{n-1}lg(i+1)=lg(1)+lg(2)+lg(3)+…+lg(n)=lg(1.2.3….n)tag2$$

Now $$prod_{k=1}^{n}kleq n^n, text{which is a very loose upper bound}tag 3$$

Using $$(3)$$ in $$(2)$$ and substituting the result in $$(1)$$ we have $$T(n)=O(nlg(n))$$ which is the same as the rough analysis…

Can I do anything better than $$(3)$$?

All the nodes referred to are the internal nodes in the Red-Black Tree.

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## MSC Simufact Forming 16.0 (x64) – NulledTeam UnderGround

File size: 3.1 GB

Simufact Forming is a simulation tool for professional users who work with forming technology. Our goal is to tailor the software to the practical needs of the users. Our customers confirm that we have achieved a lot in this area. The use of simulation software often requires a lot of expertise to build and program complex models. This is not the case with Simufact Forming: the user can focus on the design-related details of the forming process. He does not have to deal with the physics of the forming process or simulation-specific details. Simufact Forming is application-oriented and easy and fast to master.
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Forming terminology
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All objects can be provided by databases
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