  C RUBY-ON-RAILS MYSQL ASP.NET DEVELOPMENT RUBY .NET LINUX SQL-SERVER REGEX WINDOWS ALGORITHM ECLIPSE VISUAL-STUDIO STRING SVN PERFORMANCE APACHE-FLEX UNIT-TESTING SECURITY LINQ UNIX MATH EMAIL OOP LANGUAGE-AGNOSTIC VB6 MSBUILD # Algorithm for ‘spreading` sum in Hash

## Algorithm for ‘spreading` sum in Hash Tag : ruby , By : B3CFT Date : January 11 2021, 03:32 PM

I hope this helps you . This solution could be a starting point, I didn't test for avoiding infinite loop:
``````banknotes = { 5 => 6, 10 => 7, 20 => 8, 50 => 9 }
amount = 335

min = banknotes.keys.min
init_avg = amount / banknotes.keys.sum # calculate the initial average
result = banknotes.transform_values { init_avg } # initialize with average
delta = amount - result.sum { |k,v| k * v }

if delta > 0 # start distribution
loop do
banknotes.keys.reverse_each do |val|
if delta - val >= 0 and result[val] + 1 < banknotes[val]
delta -= val
result[val] += 1
end
end
break if delta < min
end
end

result
#=> {5=>3, 10=>4, 20=>4, 50=>4}
`````` Boards Message : You Must Login Or Sign Up to Add Your Comments .

## why should a good hash algorithm not allow attackers to find two messages producing the same hash?

Tag : security , By : warttack
Date : March 29 2020, 07:55 AM
may help you . Yes, it's inevitable that there will be collisions when mapping a long message onto a shorter hash, as the hash cannot contain all possible values of the message. For the same reason you cannot 'invert' the hash to uniquely produce either Beowulf or The Hobbit - but if you generated every possible text and filtered out the ones that had your particular hash value, you'd find both texts (amongst billions of others).
The article is saying that it should be hard for an attacker to find or construct a second message that has the same hash value as a first. Cryptographic hash functions are often used as proof that a message hasn't been tampered with - if even a single bit of data flips then the hash value should be completely different.

## Is it possible to use label spreading scikit algorithm on edgelist?

Tag : python , By : user171752
Date : March 29 2020, 07:55 AM
wish of those help To use Label Spreading you should follow these steps: 1. create a vector of labels (y), where all the unlabeled instances are set to -1. 2. fit the model using your feature data (X) and y. 3. create predict_entropies vector using stats.distributions.entropy(yourmodelname.label_distributions_.T) 4. create an uncertainty index by sorting the predict_entropies vector. 5. send the samples of lowest certainty for label query.

Tag : java , By : Tim
Date : March 29 2020, 07:55 AM
like below fixes the issue The yellow circle indicates the starting node and the red circle is indicating the goal node. I can't understand why my current node is spreading outwards instead of the image below where the node is just going straight into the goal. , I think you have a problem to get the lowestcost:
``````private Node getLowestFCost(List<Node> open) {
int lowestCost = 0;
int index = 0;
for (int i = 0; i < open.size(); i++) {
if (open.get(i).fCost <= lowestCost) {
lowestCost = open.get(i).fCost;
index = i;
}
}
return open.get(index);
}
``````

## Leveinshtein and hash - finding one hash algorithm that results in correlation (closer distance)

Tag : python , By : Zinovate
Date : March 29 2020, 07:55 AM 