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Alex. Gerbessiotis  CS 610 FALL   2025
(c) All rights reserved.      SEPTEMBER    2025.
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PROBLEM 1:  FALSE, TRUE, FALSE,  FALSE,  FALSE, FALSE.
PROBLEM 2                                
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 1. O(n).     It is O(lg(2^n -1 +1)) = O(n).
 2. O(n^4).   It is O((n^2)^2) = O(n^4).
 3. (b).      (a) is $\Theta(nlgn)$ by Stirling's formula and thus 
    $nlgn = o(n^2)$ concluding (b) is asymptotically larger.
 4.  f(n) is asymptotically larger.
     2**(4**n)= 2**(2**n x 2**n)
     4**(2**n)= 2**(2 x 2**n)
     Thus taking limits f(n)/g(n) = 2**( 2**n (2**n -2)) -> oo

     However, I am going to accept as ok the misinterpretation
     (a) is ${16}^n$ and (b) is  ${16}^n$, 
     None; they are asymptotically equal.

PROBLEM 3
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 omega, Theta, o, omega.

PROBLEM 4 
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Merge(X1,X2,X3,X4,n,3n,n,n)
1. Y1 = X3 merge X4 ; // 2n keys in  <= 2n-1 comps.
2. Y2 = Y1 merge X1 ; // 3n keys in  <= 3n-1 comps.
3. Y3 = Y2 merge X2 ; // 6n keys in  <= 6n-1 comps
   return(Y3)         // 6n keys total; 11n-3 = 11n+o(n) comps total.

PROBLEM 5  
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<3 4 5 ... n-1   n   1   2>

The subsequence of length n-2  < 3 4 5 ... n-1 n> is sorted so
InsertionSort would take n-3 comparisons to sort it!
Then key 1 is to be inserted to the far left thus requiring 
n-2 comparisons for the n-2 keys of the subsequence.
Then we get n-1 keys of the input sorted < 1 3 4 5 ... n-1 n>
Moreover key 2 will need to be compared to all those keys 
and deposited to the right of 1 and left of 3. This
requires n-1 comparisons.
The comparison total is (n-3)+(n-2)+(n-1) = 3n - 6.

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Alex. Gerbessiotis  CS 610 FALL   2025
(c) All rights reserved.      SEPTEMBER    2025.
Time Created:09/25/2025@16:01 Last Update: 10/02/2025@10:48 
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