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Is it possible to create a Fibonacci Heap that has exactly 5 nodes: one root node and 4 children of that root?. If yes please explain the sequence of operations to do so. If it is not possible then why?

I faced this question in an interview but couldn't answer it. Hopefully can learn it from here.

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    $\begingroup$ Welcome to Computer Science! The title you have chosen is not well suited to representing your question. Please take some time to improve it; we have collected some advice here. Thank you! $\endgroup$ – Raphael Mar 6 at 8:11
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    $\begingroup$ What have you tried and where did you get stuck? $\endgroup$ – Raphael Mar 6 at 8:11
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    $\begingroup$ Choose and cite a definition of Fibonacci Heap, and start arguing. (There are data structures where certain configurations can not be reached by additions to empty, only: you may need deletions, too.) $\endgroup$ – greybeard Mar 6 at 8:13
  • $\begingroup$ @greybeard Hi, for me the definition is Fibonacci Heap is a collection of trees with min-heap or max-heap property. In Fibonacci Heap, trees can can have any shape even all trees can be single nodes . $\endgroup$ – Anuj Mittal Mar 6 at 8:31
  • $\begingroup$ I think I see where your problem started. I am not happy with Fredman&Tarjan's We impose no explicit constraints on the number or structure of the trees, there are constraints [implied by] the way the trees are manipulated. $\endgroup$ – greybeard Mar 6 at 19:05
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Corollary 1 in Fredman & Tarjan's paper states:

A node of rank $k$ in an F-heap has at least $F_{k+2} \ge \phi^k$ descendants, including itself; where $F_k$ is the $k$th Fibonacci number ($F_0 = 0$, $F_1 = 1$, $F_k = F_{k-2} + F_{k-1}$ for $k \ge 2$), and $\phi = (1 + \sqrt{5})/2$ is the golden ratio.

Any node of rank 4 must have at least $F_{4+2} = 8$ descendants including itself, but the root node in the example has only 5. Therefore it could not have been created using Fibonacci heap operations.

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