Question

Hello, i am not sure on how to work this out, can you explain and show how you get the answer? Cheers

The lecture notes show the top 5 levels of the call graph for a call coins_reqd(30, 11, 10, 25]) with a purely recursive implementation. Here is a similar call graph for a call to coins_reqd(11,[1, 5, 10]) but this time showing all levels except calls with negative values. How many times is the yellow node "5" called? 10

Answer 1

Detailed explanation of how to do such problems is as follows:

Trace This problem using a stack: call to coins_read(11, [1,5,10]) is made, Create a stack and put all neccesary information in it. Kall-read (11) 1 [1, 5, 10 take one path, lets say first choose 1, and go to node la me fal 7010) possible paths on top of the stack we have 10 now, again choose any one path first, go to that node, and remore that path from list of possible paths as follows: Lett say we again choosed feath I, which goes to node 9. | 10 T (10) 119) (X, 5, 10, 5, 10] [15 I don't include lo in this list because from node 9, there's no possible path 1o. (You can include it, but its of no use for our analysis).

TWITTER follow same procedure till we reach node 5. The time we reach node 5, our stack will look like as follows: (10) T110) 119 18 17 16 T5 | [1,5, 10] [x,5,10] [1,5 M [15] [15] [115] [5] we have reached node 5 once, Therefore till now node "5" is called once only. follow same procedure and move further down the tree, But we can observe that we cannot reach ñode 5 from any keres Neste of Node having Lesser number [44]." so we can assume we have seen all possible paths from Node '5' and we pop off the node "S" from stack. |(7) 162 Ton apo 46 (1,5, 10] 1,5, 10] [15] [85] [85/5 y Now we have node 6 at top of the stack, havene the next possible input of node 6, which is beth "S" I w. 1(10) 119 18 17 16) Tu 100,5,103 101,5,6] [15] [195] 95] 51] [] we reached node one, follow the same procedure, untill, we have seen all possible paths for each node. stack updates for the complete procedure is as follows. 119) 1

au fob poop in this direction, 5,10] - Assume all falls. travessed as shown above, I now, just pop 15% off one more call for "5" here, till now = 2. Total calls sa Evento all paths paresed Tekia Výs] 0,1 Eva EL (5) same assume [15] Z one more call total calls till now=3.

stack is empty now, we have paversed our complete free, In whole baversal, we get Node is 3 times. Hence, It is called 3 times.

Conclusion: Node "5" (yellow node) is called 3 times in this procedure.

Given only tree this was the most visualized way to solve this problems, it we would have given the recursion, then there are other ways to solve such kind of problems.

But this is the best way if you want to visualize things.

I have drawn stack horizontally, just to save space, don't confuse it with the queue. It's a stack.

Pardon my handwriting, i would have typed this answer if possible, but i needed to draw these stack instances.

If you still got any doubt in the explanation, lemme know in the comment section.