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World consists of objects/components and their
interactions with each other |
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Goal is to be able to have proven, tested
software components that are reusable, wont have to write all code from
scratch speed software development |
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Came out of Simulation Simula first OO
language |
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Encapsulation, Inheritance, Polymorphism |
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Why save time, dont reinvent the wheel! |
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Three generic entities: |
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Container a data structure that organizes
objects a specific way, e.g. stack, queue, vector, list |
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Iterators object used to reference an element
in |
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a container, like a pointer, just for some
containers like list |
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Algorithms frequently used functions on
containers e.g. make-heap, find_first, sort ( Appendix B Drozdek) |
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Used to look at time and space complexity now
just time |
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As the size of the data set grows, how does the
time the algorithm takes to run grow? |
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In a formula, take fastest growing term, ignore
constant multipliers |
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Usually use n as the variable to denote size
of the data set |
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If have expression for run-time complexity |
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F(n) = 7 e**n + 100 n **2 + 250 n + 1,000,000 |
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Dominant term above, as n grows large is? |
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STL vectors |
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Lists |
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Stacks |
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Queues |
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#include <list> |
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list<int>
iList ; iList.push_front(5); iList.pop_back(); |
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list<char> cList;
cList.clear(); if(cList.empty())
. |
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list<anyClass> acList; int
size=acList.size(); |
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#include <stack> |
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stack<int> iStack iStack.push(5); |
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stack<char> cStack; cStack.pop(); |
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stack<anyClass> acStack; |
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A problem solving tool as well as a feature
incorporated in programming languages |
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Recursive problems have two features |
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Solutions are easy to give for special states
(stopping states) |
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For a given state (not a stopping state) there
are clear rules for how to proceed to a new state. This new state is either |
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a) a stopping state b) leads to a
stopping state |
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What goes on in memory with a recursive call? |
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With each recursive call a local environment (or
stack frame) is stacked which includes enough storage for: |
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1) LOCAL VARIABLES |
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2) VALUE PARAMETERS |
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3) REFERENCES (POINTERS) BACK TO
CALL-BY- REFERENCE
PARAMETERS |
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4) CELL FOR PASSING BACK THE VALUE OF
THE FUNCTION (FOR
NON-VOID FUNCTIONS) |
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5) THE ADDRESS TO RETURN TO AFTER
COMPLETING THE EXECUTION
OF THE CURRENT RECURSIVE CALL |
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Recursion exemplifies DIVIDE AND CONQUER problem-solving strategies --- |
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Example: TOWERS OF HANOI - 19th
century parlor game with 64 disks |
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Object: Move all disks from peg1 to peg3 using
peg2 for intermediate storage |
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Rules: One disk moved at a time, and a disk can
never rest on a smaller disk |
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Notes