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CSE 2050 Exam 2 - Recursion, Sorting, and Hashing Questions with 100% Actual correct answers | verified | latest update | Graded A+ | Already Passed | Complete Solution £6.37   Add to cart
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CSE 2050 Exam 2 - Recursion, Sorting, and Hashing Questions with 100% Actual correct answers | verified | latest update | Graded A+ | Already Passed | Complete Solution

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CSE 2050 Exam 2 - Recursion, Sorting, and Hashing Questions with 100% Actual correct answers | verified | latest update | Graded A+ | Already Passed | Complete Solution

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  • June 24, 2024
  • 9
  • 2023/2024
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  • cse 2050 exam 2 recursion sorting and hashing

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CSE 2050 Exam 2 - Recursion, Sorting,
and Hashing
Recursion - ANS-Problem-solving strategy that involves breaking a problem into smaller
instances of the same problem (also called subproblems) until we get a small enough
subproblem with a trivial solution.

First X Numbers - ANS-def f(x):
if x > 0:
return f(x-1) + x
return 0

Proofs by Induction - ANS-Base Step: prove the predicate is true when n = 1.
Hypothesis Step: assume it is true for n = K.
Inductive Step: show it is true for n = K + 1.

Rules of Recursion - ANS-Base Case: there must be at least one base case at which,
when such condition is met, the function stops calling itself.
Recursive Step: must move towards the base case.

The Function Call Stack - ANS-Calling a function pushes onto the stack.
Returning pops off the stack.

Factorial - ANS-def fact(x):
if x == 1:
return 1
else:
return x * fact(x-1)

Euclid's Algorithm - ANS-def GCD(a, b):
if a > b:
a, b = b, a
if a == 0:
return b
return GCD(a, b % a)

Fibonacci Numbers - ANS-def fib(n):
if n in [0, 1]:

, return 1
return fib(n-1) + fib(n-2)

Golden Ratio - ANS-Ratio of adjacent Fibonacci numbers.
Often called the most beautiful number in the universe.
Applications from geometry to the proportions of the human body itself.

Problems - ANS-Some recursive algorithms are inefficient because they repeatedly call
to a function with the same parameters, wasting time.

Greedy Algorithms - ANS-Strategy that makes the most optimal choice at each small
stage with the goal of eventually leading to a globally optimal solution.
Picks the best solution at the moment without regard for consequences (hence, it is
greedy).
In many problems, a greedy strategy does not produce an optimal solution.

Dynamic Programming - ANS-Stores previously calculated values to prevent repetitive
calculations and allow for quicker retrievals as well as using smaller values to solve for
even larger ones.
Created subproblems are overlapping, that is, we see the same subproblems repeated.
Two Approaches: Memoization and Tabulation.

Memoization - ANS-Write the recursive function top-down.
Alter the function to check if we've already calculated the value.
If so, use the pre-calculated value,
If not, do the recursive call.

Fib - Memo - ANS-def fib(n, fib_cache):
if n in fib_cache:
return fib_cache[n]
fib_cache[n] = fib(n-1, fib_cache) + fib(n-2, fib_cache)
return fib_cache[n]

Memoization Pros - ANS-Can be more intuitive to code / understand.
Can be memory saving if you don't need answers to all subproblems.

Memoization Cons - ANS-Depending on implementation, larger overhead due to
recursion.

Tabulation - ANS-Bottom-Up.

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