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concept check please

Posted on 2004-09-26
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Last Modified: 2011-09-20
1. A stack can be used to permute numbers. For example, assume you have '1
2 3' as an input stream (reading from the right) into your stack. Assume
you can push or pop in any order at any time (with the restriction that
you cannot pop from an empty stack). Thus, if you push all three and
pop_top all three, your output is '1 2 3'. What are all the possible
permutations for output you can achieve for this input stream?

ANSWER: 5

321   312   231   213   123

The choices in this case are limited by the fact that you cannot execute
a pop on an empty stack and you cannot push from an empty input stream.
Thus allowable choices can be shown by using a binary decision tree.

<EXAMPLE>

                        PUSH
                       /      \
                    POP       PUSH
                   /          /   \
                PUSH        POP    PUSH
               /  \        /   \      \
           POP    PUSH   POP  PUSH    POP
           /        \     /      \       \
       PUSH        POP  PUSH     POP     POP
        |           |    |        |       |
       POP         POP   POP     POP     POP
      321          312   231     213     123

2. Prove by induction that for any n >= 1, 1^3 + 2^3 + 3^3 + ... + n^3 =
(1 + 2 + 3 + ... + n)^2.

Call this summation SUM3.

*1. Establish a base case:     SUM3(1) = 1^3 = 1.

*2. Now assume SUM3 true for some k.  Recall that we proved by induction
    (in class) that 1 + 2 + 3 + ... + k =  2 | k(k+1) (meaning 2 divides
    k(k+1) )   thus:

    SUM3(k) =  ( 2 | k(k+1) ) ^2

*3. Show that if true for k, true for k+1.

    SUM3(k + 1) =  ( 2 | k(k+1) ) ^2  + (k+1)^3
                       =  ( 2 | k^2 + k ) ^2 + (k+1)^3
                       =  ( 2 | k+1(k+2) ) ^2 =  ( 2 | k^2 + 3k + 2) ^2


    SUM3(k+1) = SUM3(k) + (k+1)^3

    ( 2 | k(k+1) ) ^2


3. Prove by induction that the sum 1 + 3 + 5 + 7 + ... + 2n-1 is n^2; e.g. // not sure on this one...please advise.
for n=3, 1 + 3 + 5 = 3^2

        1 + 3 + 5 + ... + (2n - 1) = n^2
        Assume true for sample n;

        1 + 3 + 5 + ... + (2n - 1) - 2(n - 1) = (n - 1)^2

        n^2 - 2(n - 1) = (n - 1)^2
        n^2 - 2n + 2 - 1 = n^2 - 2n +1
        n^2 - 2n + 1 = n^2 - 2n + 1

        Since true for n + 1,..true for any n.
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Question by:edelossantos
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by:edelossantos
ID: 12157932
Please advise on #3.  Del
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jhshukla earned 500 total points
ID: 12158005
>> ANSWER: 5
>> 321   312   231   213   123

NO! Answer: 6
321   312   231   213   123   132
this may b because i did not understand how you permute using stack.

and this is not a math room. anyways since most comp sci people like math this is a good place to seek answers.
1 + 3 + 5 + 7 + ... + 2n-1 is n^2
next number in the series is 2(n+1)-1; substituting (1 + 3 + 5 + 7 + ... + 2n-1) = n^2 the series becomes
n^2 + 2(n+1)-1
= n^2 + 2n + 1
= (n+1)^2
THE END
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Author Comment

by:edelossantos
ID: 12162712
sorry... and thanks again for your help.  Truly appreciate it.  Regards.  Del
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Expert Comment

by:jhshukla
ID: 12166068
sorry? for what? you haven't hurt anyone or done anything harmful to this world. plus I love induction. so i thank you for asking.
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Expert Comment

by:jhshukla
ID: 12166536
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