Extract
[...] We can then write the following heap sort function: def triParTas(L,LTriee): if return LTriee+L for i in range(1,len(L)): L = tamiser(L,i) LTriee = LTriee+[L[0]] return triParTas(L[1:],LTriee) The operation is as follows: initially, we provide an unsorted list and the empty list that will contain the sorted list. The termination condition is the case where the list contains only one element. Otherwise, we build a heap by successively considering the first two, then the first three (etc.) elements and using the previous 'sift' function. Once this is done, we can remove the first element and then rebuild a heap with the remaining terms. Question 27. [...]
[...] (Explanation: As the tree is not empty, it starts with the root » obligatorily. We are therefore interested in the 4ème element (so the index 3). If we have , this means that the left subtree is empty (or there is none) so we can return directly the else, otherwise, we count the parentheses to know when the tree is closed, that is, when we have crossed as many closing parentheses as opening ones). As it is incremented by 1 at the end of the loop, when we exit the loop we are on the comma so , or we don't want to return the last one (and in Python a range of the form a:b excludes the index b). [...]
[...] Binary Trees - Heap Sort Question 1 - First case : The numbers who are left without adoptive children are the numbers for which , soit . Thus, the numbers that end up without adopted children are the numbers between and . - Second case : The numbers are found without adopted children are the numbers for which , soit . Thus, the numbers that end up without adopted children are the numbers between and . - First case : The numbers who are left with a single adopted child are the numbers for which , soit . [...]
[...] So, according to the previous question, we have: and - Let us suppose . - Then, according to rule so then - From the previous inequality, we get, according to rule By summarizing the two previous points, we can write: et Suppose now that we have et - De , it is deduced, according to rule that: This ultimately gives: - De , it is deduced, according to rule that: This ultimately gives: In making the synthesis of the two previous points, we can write: et Finally, the two implications found are synthesized in: et Question 5 - Yes est pair, it is written with - Yes is odd is written with - The function here calculates depending on the following is - On a - First case: then so . [...]
[...] h It is a heap. Question 23. No not necessarily (as in the heap of the sentence or Question 24. We are building the following pile: Thus, as well as the step of removal:" We are building the following pile: Thus, as well as the step of removal: We are building the following pile: Thus, as well as the step of removal: Question 25. The sifting function is therefore as follows: def sifting(L,i): if L if L[iRac]>L[i]: L[iRac],L[i]=L[i],L[iRac] tamiser(L,iRac) else: return L Question 26. [...]
