CS 300 Assignment 2Due Friday Oct. 27 by midnightName: Email:Instructions: Fill in your name and email. Use webct to submit your worksheet. There are four questions, each worth 25 points. Assignments must be done individually; however, if you have questions or do not understand how to do something, you should ask or send your question to the class mailing list.Overview: The purpose of this assignment is to utilize Maple to further investigate permutations and to analyze the performance of two sorting algorithms. The exercises require empirical and theoretical analysis and will take some time to do well (do not wait until the last minute to do this assignment). All of the questions build on and are related to material presented in class. Successful completion of the assignment will help you become more comfortable using Maple and will help you develop skills related to algorithm analysis and combinatorics.
<Text-field style="Heading 1" layout="Heading 1">Question 1 (Quotient Distribution in the Euclidean Algorithm)</Text-field>In this question, you will investigate the distribution of quotients in the Euclidean algorithm. Recall that the Euclidean algorithm performs a sequence of integer divisions with remainder.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You will generate 1000 random inputs 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with 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for 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 and will look at the distribution of the values of the quotients LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYjLUklbXN1YkdGJDYlLUkjbWlHRiQ2JVEicUYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1GIzYjLUYvNiVRImlGJ0YyRjUvJS9zdWJzY3JpcHRzaGlmdEdRIjBGJw== in all of the NumSample calls to the Euclidean algorithm.First, modify the iterative form, gcdit, of the Euclidean algorithm below so that it returns a list of the quotients that occur in the call to the algorithm. Then perform the above experiment and show a histogram (use the Histogram command in the Statistics package) of quotients that occurred. How many (what percentage) of the quotients are equal to 1, 2, 3, 4, 5? What percentage are less than 10, what percentage are less than 100? Note that the range option to Histogram allows you to ignore values outside a given range. You will need to use this option if most of the quotients are in a given range while a few are a large distance from the range.restart;with(Statistics);gcdit := proc(a,b) local a1,a2,a3,q; a1 := a; a2 := b; while (a2 <> 0) do a3 := a1 mod a2; q := floor(a1/a2); a1 := a2; a2 := a3; od; return a1;end;
<Text-field style="Heading 1" layout="Heading 1">Question 2 (Population Count)</Text-field>In this question, you will investigate the cumulative population count of the 1 bits in the binary representation of positive integers.First use Maple's sum command to get a formula for the number of bits in the binary representation of the numbers less than LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEiTkYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JI21vR0YkNjBRIi5GJy9GM1Enbm9ybWFsRicvJSZmZW5jZUdRJmZhbHNlRicvJSpzZXBhcmF0b3JHRj0vJSlzdHJldGNoeUdGPS8lKnN5bW1ldHJpY0dGPS8lKGxhcmdlb3BHRj0vJS5tb3ZhYmxlbGltaXRzR0Y9LyUnYWNjZW50R0Y9LyUlZm9ybUdRJmluZml4RicvJSdsc3BhY2VHUSQwZW1GJy8lJ3JzcGFjZUdGTy8lKG1pbnNpemVHUSIxRicvJShtYXhzaXplR1EpaW5maW5pdHlGJw==The following loop shows the binary represenation of all numbers less than 16. Observe that the total number of bits is equal to 1 + 2*2 + 3*4 + 4*8 = 49.restart;for i from 1 to 15 do convert(i,binary);end do;The function numbits, from class, counts the number of 1 bits in the binary representation of a given number. Use this function to write another function, CumulativeBits(N), which counts the number of one bits in all of the integers between 1 and LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEiTkYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JI21vR0YkNjBRIi5GJy9GM1Enbm9ybWFsRicvJSZmZW5jZUdRJmZhbHNlRicvJSpzZXBhcmF0b3JHRj0vJSlzdHJldGNoeUdGPS8lKnN5bW1ldHJpY0dGPS8lKGxhcmdlb3BHRj0vJS5tb3ZhYmxlbGltaXRzR0Y9LyUnYWNjZW50R0Y9LyUlZm9ybUdRJmluZml4RicvJSdsc3BhY2VHUSQwZW1GJy8lJ3JzcGFjZUdGTy8lKG1pbnNpemVHUSIxRicvJShtYXhzaXplR1EpaW5maW5pdHlGJw==What is the result of calling CumulativeBits when 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Plot the CumulativeBits(N) and then plot CumulativeBits(N)-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 in the ranges from 1 to 64 and 1 to 512.numbits := proc(n) local count, nb; count := 0; nb := n; while (nb <> 0) do if nb mod 2 = 1 then count := count + 1; fi; nb := floor(nb/2); od; return count;end;with(plots);
<Text-field style="Heading 1" layout="Heading 1">Question 3 (Generating Rightmost Binary Partition Trees)</Text-field>A binary partition tree is a binary tree such that the nodes are labeled by positive integers with the property that the sum of the values of the left and right nodes is equal to the value of the node. E.G. 5 / \ 2 3 / \ / \ 1 1 2 1A rightmost binary partition tree is a binary partition tree whose left nodes are leaf nodes. The previous example is not a rightmost tree. The following example is. 5 / \ 2 3 / \ 2 1You are to write a Maple procedure that generates a list of all rightmost binary trees with a specified root value, LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEibkYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JI21vR0YkNjBRIixGJy9GM1Enbm9ybWFsRicvJSZmZW5jZUdRJmZhbHNlRicvJSpzZXBhcmF0b3JHRjEvJSlzdHJldGNoeUdGPS8lKnN5bW1ldHJpY0dGPS8lKGxhcmdlb3BHRj0vJS5tb3ZhYmxlbGltaXRzR0Y9LyUnYWNjZW50R0Y9LyUlZm9ybUdRJmluZml4RicvJSdsc3BhY2VHUSQwZW1GJy8lJ3JzcGFjZUdRM3Zlcnl0aGlja21hdGhzcGFjZUYnLyUobWluc2l6ZUdRIjFGJy8lKG1heHNpemVHUSlpbmZpbml0eUYnand whose leaf nodes are less than or equal to a given constant LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEiY0YnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JI21vR0YkNjBRIixGJy9GM1Enbm9ybWFsRicvJSZmZW5jZUdRJmZhbHNlRicvJSpzZXBhcmF0b3JHRjEvJSlzdHJldGNoeUdGPS8lKnN5bW1ldHJpY0dGPS8lKGxhcmdlb3BHRj0vJS5tb3ZhYmxlbGltaXRzR0Y9LyUnYWNjZW50R0Y9LyUlZm9ybUdRJmluZml4RicvJSdsc3BhY2VHUSQwZW1GJy8lJ3JzcGFjZUdRM3Zlcnl0aGlja21hdGhzcGFjZUYnLyUobWluc2l6ZUdRIjFGJy8lKG1heHNpemVHUSlpbmZpbml0eUYnwhich is passed as a parameter. Note that nodes whose value is less than or equal to LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEiY0YnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JI21vR0YkNjBRIixGJy9GM1Enbm9ybWFsRicvJSZmZW5jZUdRJmZhbHNlRicvJSpzZXBhcmF0b3JHRjEvJSlzdHJldGNoeUdGPS8lKnN5bW1ldHJpY0dGPS8lKGxhcmdlb3BHRj0vJS5tb3ZhYmxlbGltaXRzR0Y9LyUnYWNjZW50R0Y9LyUlZm9ybUdRJmluZml4RicvJSdsc3BhY2VHUSQwZW1GJy8lJ3JzcGFjZUdRM3Zlcnl0aGlja21hdGhzcGFjZUYnLyUobWluc2l6ZUdRIjFGJy8lKG1heHNpemVHUSlpbmZpbml0eUYn are not further expanded. For example, when 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 and 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 possible trees are the following. 4 4 4 / \ / \ / \2 2 1 3 1 3 / \ / \ 2 1 1 2Hint. You can obtain this procedure by simple modifications of the routine, GenBinTrees, presented in class to generate all binary trees of a given size. For your convenience GenBinTrees and PrintBinTree, a routine for printing binary trees, are duplicated below.# PrintBinTree - print a binary tree. Use inorder traversal.# Input: T - a binary tree.# Side Effect: T is printed, in ASCII, rotated 90 degrees.PrintBinTree := proc(T) local PrintTreeIndent, PrintBinTreeAux; PrintBinTreeAux := proc(T,indent) local i, value, left, right; if T = [] then return; end if; value := T[1]; left := T[2]; right := T[3]; PrintBinTreeAux(right,indent || " "); printf("%s%d\n",indent,value); PrintBinTreeAux(left,indent || " "); end; PrintBinTreeAux(T,"");end;# GenBinTrees - Generate all binary trees with a given number of nodes.# Input: n - a positive integer.# Output: a list of all binary trees with n nodes.GenBinTrees := proc(n) option remember; local k,Tl,Tr,T,P,children,tree; if n = 0 then return [[]]; end if; if n = 1 then return [[1,[],[]]]; end if; T := []; for k from 0 to n-1 do Tl := GenBinTrees(k); Tr := GenBinTrees(n-k-1); P:=combinat[cartprod]([Tl,Tr]); while not P[finished] do children := P[nextvalue](); tree := [n,children[1],children[2]]; T := [tree,op(T)]; end do; end do; return T;end;S3 := GenBinTrees(3);for T in S3 do PrintBinTree(T); print();end do;
<Text-field style="Heading 1" layout="Heading 1">Question 4 (Counting Rightmost Binary Partition Trees)</Text-field>In this question, you will investigate the number of rightmost binary partition trees using different leaf nodes sizes. LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYjLUkjbW9HRiQ2MFEifkYnLyUsbWF0aHZhcmlhbnRHUSdub3JtYWxGJy8lJmZlbmNlR1EmZmFsc2VGJy8lKnNlcGFyYXRvckdGNC8lKXN0cmV0Y2h5R0Y0LyUqc3ltbWV0cmljR0Y0LyUobGFyZ2VvcEdGNC8lLm1vdmFibGVsaW1pdHNHRjQvJSdhY2NlbnRHRjQvJSVmb3JtR1EhRicvJSdsc3BhY2VHUSQwZW1GJy8lJ3JzcGFjZUdGRi8lKG1pbnNpemVHUSIxRicvJShtYXhzaXplR1EpaW5maW5pdHlGJw==First write a recursive Maple procedure to count the number of rightmost binary trees of a specified size LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEibkYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JI21vR0YkNjBRIn5GJy9GM1Enbm9ybWFsRicvJSZmZW5jZUdRJmZhbHNlRicvJSpzZXBhcmF0b3JHRj0vJSlzdHJldGNoeUdGPS8lKnN5bW1ldHJpY0dGPS8lKGxhcmdlb3BHRj0vJS5tb3ZhYmxlbGltaXRzR0Y9LyUnYWNjZW50R0Y9LyUlZm9ybUdRIUYnLyUnbHNwYWNlR1EkMGVtRicvJSdyc3BhY2VHRk8vJShtaW5zaXplR1EiMUYnLyUobWF4c2l6ZUdRKWluZmluaXR5Ric=and leaf size LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2JVEiY0YnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JI21vR0YkNjBRIi5GJy9GM1Enbm9ybWFsRicvJSZmZW5jZUdRJmZhbHNlRicvJSpzZXBhcmF0b3JHRj0vJSlzdHJldGNoeUdGPS8lKnN5bW1ldHJpY0dGPS8lKGxhcmdlb3BHRj0vJS5tb3ZhYmxlbGltaXRzR0Y9LyUnYWNjZW50R0Y9LyUlZm9ybUdRJmluZml4RicvJSdsc3BhY2VHUSQwZW1GJy8lJ3JzcGFjZUdGTy8lKG1pbnNpemVHUSIxRicvJShtYXhzaXplR1EpaW5maW5pdHlGJy1GNjYwUSJ+RidGOUY7Rj5GQEZCRkRGRkZIL0ZLUSFGJ0ZNRlBGUkZVRlg= The procedure is similar to the one from the lecture on binary trees. Confirm that the counts produced are the same as the number of trees generated in question 3. Show the sequences of counts for LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYuLUkjbWlHRiQ2JVEiY0YnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JI21vR0YkNjBRIj1GJy9GM1Enbm9ybWFsRicvJSZmZW5jZUdRJmZhbHNlRicvJSpzZXBhcmF0b3JHRj0vJSlzdHJldGNoeUdGPS8lKnN5bW1ldHJpY0dGPS8lKGxhcmdlb3BHRj0vJS5tb3ZhYmxlbGltaXRzR0Y9LyUnYWNjZW50R0Y9LyUlZm9ybUdRJmluZml4RicvJSdsc3BhY2VHUS90aGlja21hdGhzcGFjZUYnLyUncnNwYWNlR0ZPLyUobWluc2l6ZUdRIjFGJy8lKG1heHNpemVHUSlpbmZpbml0eUYnLUkjbW5HRiQ2JEZURjktRjY2MFEiLEYnRjlGOy9GP0YxRkBGQkZERkZGSEZKL0ZOUSQwZW1GJy9GUVEzdmVyeXRoaWNrbWF0aHNwYWNlRidGUkZVLUZZNiRRIjJGJ0Y5RmVuLUZZNiRRIjNGJ0Y5RmVuLUY2NjBRIn5GJ0Y5RjtGPkZARkJGREZGRkgvRktRIUYnRmluL0ZRRmpuRlJGVS1GWTYkUSM0LkYnRjlGY29GY28=What can you say about their growth rates? Based on your recursive procedure for counting rightmost binary partition trees, derive a recurrence relation for the number of such trees, and use Maple's rsolve command to find the solution to the recurrence for 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and LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYnLUkjbW5HRiQ2JFEjMy5GJy8lLG1hdGh2YXJpYW50R1Enbm9ybWFsRictSSNtb0dGJDYwUSJ+RidGLy8lJmZlbmNlR1EmZmFsc2VGJy8lKnNlcGFyYXRvckdGOC8lKXN0cmV0Y2h5R0Y4LyUqc3ltbWV0cmljR0Y4LyUobGFyZ2VvcEdGOC8lLm1vdmFibGVsaW1pdHNHRjgvJSdhY2NlbnRHRjgvJSVmb3JtR1EhRicvJSdsc3BhY2VHUSQwZW1GJy8lJ3JzcGFjZUdGSi8lKG1pbnNpemVHUSIxRicvJShtYXhzaXplR1EpaW5maW5pdHlGJ0YyRjJGMg==Try to use the solution from rsolve to generate the sequence of counts (see the worksheet from class on the Fibonacci numbers). Note that you may need to use evalf to generate numeric approximations as the solutions returned by rsolve may not be in a form that Maple can simplify.The recurrence relations for the number of rightmost binary partition trees are examples of linear recurrence with constant coefficients. In the worksheet on Fibonacci numbers it was shown how to solve these recurrences in terms of a linear combination of the roots of the characteristic polynomial of the recurrence relation. A linear recurrence with constant coefficients of order LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYjLUkjbWlHRiQ2JVEia0YnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJw== is of the form 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with base cases 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The characteristic polynomial for the recurrence is equal to LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzY0LUklbXN1cEdGJDYlLUkjbWlHRiQ2JVEieEYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1GIzYjLUYvNiVRImtGJ0YyRjUvJTFzdXBlcnNjcmlwdHNoaWZ0R1EiMEYnLUkjbW9HRiQ2MFEqJnVtaW51czA7RicvRjZRJ25vcm1hbEYnLyUmZmVuY2VHUSZmYWxzZUYnLyUqc2VwYXJhdG9yR0ZILyUpc3RyZXRjaHlHRkgvJSpzeW1tZXRyaWNHRkgvJShsYXJnZW9wR0ZILyUubW92YWJsZWxpbWl0c0dGSC8lJ2FjY2VudEdGSC8lJWZvcm1HUSZpbmZpeEYnLyUnbHNwYWNlR1EwbWVkaXVtbWF0aHNwYWNlRicvJSdyc3BhY2VHRlovJShtaW5zaXplR1EiMUYnLyUobWF4c2l6ZUdRKWluZmluaXR5RictSSVtc3ViR0YkNiUtRi82JVEiYUYnRjJGNS1GIzYlRjpGQC1JI21uR0YkNiRGaW5GRC8lL3N1YnNjcmlwdHNoaWZ0R0Y/LUZBNjBRJyZzZG90O0YnRkRGRkZJRktGTUZPRlFGU0ZVL0ZZUSQwZW1GJy9GZm5GXnBGZ25Gam4tRiw2JUYuRmNvRj1GQEZqb0Zqb0Zqb0ZALUZebzYlRmBvLUYjNiNGZW9GaG9Gam9GLkZALUZebzYlRmBvLUYjNiMtRmZvNiRRIjBGJ0ZERmhvLUZBNjBRIi5GJ0ZERkZGSUZLRk1GT0ZRRlNGVUZdcEZfcEZnbkZqbi1GQTYwUSJ+RidGREZGRklGS0ZNRk9GUUZTL0ZWUSFGJ0ZdcEZfcEZnbkZqbkZgcQ==If the characteristic polynomial has distinct roots, 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then the solution to the recurrence is of the form 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where the constants 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 can be determined from the base cases (see the worksheet on Fibonacci numbers). Solve the recurrence for the number of rightmost binary partition trees for LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYlLUkjbWlHRiQ2JVEiY0YnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JI21vR0YkNjBRIj1GJy9GM1Enbm9ybWFsRicvJSZmZW5jZUdRJmZhbHNlRicvJSpzZXBhcmF0b3JHRj0vJSlzdHJldGNoeUdGPS8lKnN5bW1ldHJpY0dGPS8lKGxhcmdlb3BHRj0vJS5tb3ZhYmxlbGltaXRzR0Y9LyUnYWNjZW50R0Y9LyUlZm9ybUdRJmluZml4RicvJSdsc3BhY2VHUS90aGlja21hdGhzcGFjZUYnLyUncnNwYWNlR0ZPLyUobWluc2l6ZUdRIjFGJy8lKG1heHNpemVHUSlpbmZpbml0eUYnLUkjbW5HRiQ2JFEiM0YnRjk= using this technique. Note that you should use fsolve to find the roots of the characteristic polynomial. You should solve the the constants in solution to the recurrence using Maple's solve command (note that since the roots are numeric approximations, you will get an approximate solution). Check to see that this is an approximate solution to the recurrence (how far off is your solution - for small inputs and large inputs). Note that you will need to use the option complex when using fsolve in order to get all of the roots. What happens if you only use the largest root in your approximate solution of the recurrence?For extra credit, experiment with the recurrence for the number of rightmost binary partition trees for larger values of LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2JVEiY0YnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JI21vR0YkNjBRIi5GJy9GM1Enbm9ybWFsRicvJSZmZW5jZUdRJmZhbHNlRicvJSpzZXBhcmF0b3JHRj0vJSlzdHJldGNoeUdGPS8lKnN5bW1ldHJpY0dGPS8lKGxhcmdlb3BHRj0vJS5tb3ZhYmxlbGltaXRzR0Y9LyUnYWNjZW50R0Y9LyUlZm9ybUdRJmluZml4RicvJSdsc3BhY2VHUSQwZW1GJy8lJ3JzcGFjZUdGTy8lKG1pbnNpemVHUSIxRicvJShtYXhzaXplR1EpaW5maW5pdHlGJy1GNjYwUSJ+RidGOUY7Rj5GQEZCRkRGRkZIL0ZLUSFGJ0ZNRlBGUkZVRlg=What do you observe? What patterns do you see for the characteristic polynomial and its roots? What can you say about the asymptotic growth of the number of trees as LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkjbWlHRiQ2JVEiY0YnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JI21vR0YkNjBRIn5GJy9GM1Enbm9ybWFsRicvJSZmZW5jZUdRJmZhbHNlRicvJSpzZXBhcmF0b3JHRj0vJSlzdHJldGNoeUdGPS8lKnN5bW1ldHJpY0dGPS8lKGxhcmdlb3BHRj0vJS5tb3ZhYmxlbGltaXRzR0Y9LyUnYWNjZW50R0Y9LyUlZm9ybUdRIUYnLyUnbHNwYWNlR1EkMGVtRicvJSdyc3BhY2VHRk8vJShtaW5zaXplR1EiMUYnLyUobWF4c2l6ZUdRKWluZmluaXR5Ric=increases?