Solution needed..urgent!!

[Ankur Arora]

Hi all,

The following post is probably not appropriate for this group but I
know there are brilliant minds active at this place (trust me, this
puzzle will involve some time and lots of grey matter), so would
appreciate if you guys can jump in with any solutions/suggestions.
Thanks!

You are given a deck containing n cards. While holding the deck:

1. Take the top card off the deck and set it on the table
2. Take the next card off the top and put it on the bottom of the deck
in your hand.
3. Continue steps 1 and 2 until all cards are on the table. This is a
round.
4. Pick up the deck from the table and repeat steps 1-3 until the deck
is in the original order.

Write a program to determine how many rounds it will take to put a
deck back into the original order. This will involve creating a data
structure to represent the
order of the cards. This program should be written in C or C++. Do
not use STL. It should
take a number of cards in the deck as a command line argument and
write the result to stdout.

 

[Juha Nieminen]

Do not use STL.

Any rational reason for this?

 

[Richard Herring]

Any rational reason for this?

It motivates homework question 2:

"Why didn't your first design work?
(a) dangling pointers;
(b) uninitialised pointers;
(b) double deletion;
(c) incorrect container insertion/deletion algorithm;
(d) all of the above"

:-/

 

[red floyd]

Hi all,
["do my homework" redacted]

You can find a good solution here:

http://www.parashift.com/c++-faq-lite/how-to-post.html#faq-5.2

 

[osmium]

The following post is probably not appropriate for this group but I
know there are brilliant minds active at this place (trust me, this
puzzle will involve some time and lots of grey matter), so would
appreciate if you guys can jump in with any solutions/suggestions.
Thanks!

You are given a deck containing n cards. While holding the deck:

1. Take the top card off the deck and set it on the table
2. Take the next card off the top and put it on the bottom of the deck
in your hand.
3. Continue steps 1 and 2 until all cards are on the table. This is a
round.
4. Pick up the deck from the table and repeat steps 1-3 until the deck
is in the original order.

Write a program to determine how many rounds it will take to put a
deck back into the original order. This will involve creating a data
structure to represent the
order of the cards. This program should be written in C or C++. Do
not use STL. It should
take a number of cards in the deck as a command line argument and
write the result to stdout.

The question is phrased in a way to start you on a wild goose chase.
Identify that bit and go on from there. I think that is the only part that
requires gray (or grey) matter.

 

[acehreli]

This program should be written in C or C++.

How about that? Any rationale for that?

Ali

 

[Ankur Arora]

It motivates homework question 2:

"Why didn't your first design work?
(a) dangling pointers;
(b) uninitialised pointers;
(b) double deletion;
(c) incorrect container insertion/deletion algorithm;
(d) all of the above"

:-/

Wow!!sorry about that, I can see its very offending.'will make it a
point to follow the guidelines.Thanks.

Here's what I have come up with so far (very basic, algorithmic
steps):-
- use of pointers to manipulate the deck since we will be doing a lot
of shuffling. A link list would be a good choice with each node
representing a card and the link list chain representing a particular
ordering.
- At least two loops, outer representing the condition till the deck
is in the original order and the inner representing a complete round.
- The inner loop will have its auxiliary list i.e. a particular
ordering after the completion of a complete round.
- At the end of each inner loop iteration we will check the auxiliary
list with the original list to see whether its the same as the
original list (this would be done in the conditional statement of the
outer loop).

Need to find out the implementation details for the inner loop (guess
that would easy) and a way to check whether the auxiliary list is same
as the original list at the end of each inner loop iteration.

Any other design/ideas ?

 

[Kai-Uwe Bux]

Ankur Arora wrote:

[snip]

Here's what I have come up with so far (very basic, algorithmic
steps):-
- use of pointers to manipulate the deck since we will be doing a lot
of shuffling. A link list would be a good choice with each node
representing a card and the link list chain representing a particular
ordering.
- At least two loops, outer representing the condition till the deck
is in the original order and the inner representing a complete round.
- The inner loop will have its auxiliary list i.e. a particular
ordering after the completion of a complete round.
- At the end of each inner loop iteration we will check the auxiliary
list with the original list to see whether its the same as the
original list (this would be done in the conditional statement of the
outer loop).

Need to find out the implementation details for the inner loop (guess
that would easy) and a way to check whether the auxiliary list is same
as the original list at the end of each inner loop iteration.

Any other design/ideas ?

Yes: The shuffle rule describes a fixed permutation of [1...n]. All you need
to do is find the order of that permutation. That can be accomplished by
finding the cycle representation and computing the LCM of the lengths of
all cycles. This approach has the advantage of being much faster in those
cases where the counting would take forever (and there are such n).


Best

Kai-Uwe Bux

 

[Erik Wikström]

Wow!!sorry about that, I can see its very offending.'will make it a
point to follow the guidelines.Thanks.

Here's what I have come up with so far (very basic, algorithmic
steps):-
- use of pointers to manipulate the deck since we will be doing a lot
of shuffling. A link list would be a good choice with each node
representing a card and the link list chain representing a particular
ordering.
- At least two loops, outer representing the condition till the deck
is in the original order and the inner representing a complete round.
- The inner loop will have its auxiliary list i.e. a particular
ordering after the completion of a complete round.
- At the end of each inner loop iteration we will check the auxiliary
list with the original list to see whether its the same as the
original list (this would be done in the conditional statement of the
outer loop).

Need to find out the implementation details for the inner loop (guess
that would easy) and a way to check whether the auxiliary list is same
as the original list at the end of each inner loop iteration.

A tip: represent each card with a number and let the original deck be
sorted (so a deck of 4 cards would start as 1, 2, 3, 4). Then checking
if the deck is back to the original state is easy.

 

[James Kanze]

By not allowing you to use STL, they encourage you to write
your own classes to do the same thing - the process of
reinventing the wheel will convey to you (a) a fair amount
about the STL, (b) an understanding of why reinventing the
wheel is generally to be avoided and (c) a healthy
appreciation of the hard work put in by the people who have
worked on the STL over the years.

I'm not so sure about that. Writing a simple array class which
is usable in one particular context is actually very simple.
Designing the interface so that it can be generally usable in
the widest number of different contexts is another matter, not
to mention implementing it in a manner so that it will have
adequate performance for all possible uses.

It's worth implementing some simple containers and algorithms
yourself, to understand the issues better, but doing so will not
help you better understand the STL.

 

[gw7rib]

Here's what I have come up with so far (very basic, algorithmic
steps):-
- use of pointers to manipulate the deck since we will be doing a lot
of shuffling. A link list would be a good choice with each node
representing a card and the link list chain representing a particular
ordering.

This sounds good. I'd recommend the type of linked list where you keep
track of both the first and the last element in the list. This makes
adding to the end much quicker.

Hope that helps.
Paul.

 

[Ankur Arora]

When you're manipulating a deck of 52 cards, making append fast is not
important. An array is a much more natural data structure for this
purpose, and it's much easier to use.

--
  Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of "The
Standard C++ Library Extensions: a Tutorial and Reference
(www.petebecker.com/tr1book)

Guys..Thanks for the tips!
A circular link list could also be considered..i think.

 

[yog.mittal]

I will really appreciate if anybody could please post the solution ?
Thanks

 

[Kai-Uwe Bux]

to restore context:

You are given a deck containing n cards. While holding the deck:

1. Take the top card off the deck and set it on the table
2. Take the next card off the top and put it on the bottom of the deck
in your hand.
3. Continue steps 1 and 2 until all cards are on the table. This is a
round.
4. Pick up the deck from the table and repeat steps 1-3 until the deck
is in the original order.

Write a program to determine how many rounds it will take to put a
deck back into the original order.

I will really appreciate if anybody could please post the solution ?

I shall ignore the requirement that STL not be used. I hope that my solution
therefore will not qualify for the homework (also, some time has passed).
Also, I shall not use the algorithm outlined in the description of the
problem but use an entirely different idea that will give the same output.

Also: I will assume that the compiler supports long long (you really want
some arbitrary length integral type). The program will silently wrap around
and not report the overflow. I attach some values that I found, which
illustrate that point. At n=280, 2^32 will not suffice and at n=1004, the
result is beyond 2^64. (These numbers also illustrate why I will not
determine the order of the permutation by counting: it would take forever.)


#include <cassert>
#include <algorithm>
#include <vector>
#include <set>
#include <iterator>
#include <iostream>

// #include <kubux/integer>

// typedef kubux::integer arithmetic_type;
typedef unsigned long long arithmetic_type;

typedef std::vector< unsigned long > permutation;

permutation id ( unsigned long n ) {
permutation result;
result.reserve( n );
for ( unsigned long i = 0; i < n; ++i ) {
result.push_back( i );
}
return ( result );
}

void shuffle ( permutation & perm ) {
permutation result;
result.reserve( perm.size() );
while ( ! perm.empty() ) {
result.push_back( perm.front() );
perm.erase( perm.begin() );
std::rotate( perm.begin(), perm.begin() + 1, perm.end() );
}
std::swap( result, perm );
}

arithmetic_type gcd ( arithmetic_type a, arithmetic_type b ) {
if ( b < a ) {
std::swap( a, b );
}
while ( a != 0 ) {
arithmetic_type dummy( b % a );
b = a;
a = dummy;
}
return( b );
}

arithmetic_type lcm ( arithmetic_type a, arithmetic_type b ) {
arithmetic_type quot = b / gcd(a,b);
return ( a * ( b / gcd(a,b) ) );
}

arithmetic_type order ( permutation const & p ) {
arithmetic_type l ( 1 );
std::set< unsigned long > still_free;
std::copy( p.begin(), p.end(),
std::inserter( still_free, still_free.end() ) );
while ( ! still_free.empty() ) {
unsigned long index = *(still_free.begin());
unsigned long where = index;
arithmetic_type length = arithmetic_type();
do {
++ length;
where = p[where];
still_free.erase( where );
} while ( where != index );
l = lcm( l, length );
}
return ( l );
}

arithmetic_type number ( unsigned long n ) {
permutation p ( id( n ) );
shuffle( p );
return ( order( p ) );
}

int main ( void ) {
for ( unsigned long n = 1; ; ++n ) {
std::cout << n << " --> " << number(n) << '\n';
}
}


/*
Some values:
1 --> 1
3 --> 2
5 --> 3
6 --> 5
7 --> 6
10 --> 9
12 --> 28
18 --> 70
23 --> 210
35 --> 308
38 --> 990
44 --> 2618
58 --> 13300
67 --> 27720
68 --> 203490
83 --> 360360
111 --> 1021020
112 --> 1511640
113 --> 6230070
162 --> 59254020
185 --> 1376845470
230 --> 2153331180
280 --> 56270739378
315 --> 591118103520
338 --> 2328245689770
441 --> 7600186994400
554 --> 247651817843430
689 --> 3993173946822060
839 --> 16415300091350160
907 --> 24858416865744000
944 --> 27303154083485280
947 --> 150103870204688400
949 --> 4073014663549605312
1004 --> 889351355342914944480
1307 --> 1423178772983737169400
1649 --> 709044136174739580415200
1705 --> 1384460057373814306578000
1799 --> 17843737077119301787398960
1934 --> 26473022063487459901268520
1948 --> 412108799554731513698624160
2591 --> 10132942664275877164628800800
2606 --> 492252055940052814523176659600
3011 --> 4061133619046911989389682594720
3368 --> 28987127583211271154658957499520
3472 --> 28786869259134681128048175521961000
4037 --> 13968231274398330387225456621929156160
4451 --> 44613968405036781772417636768076148600
4743 --> 47127723619287268204539720673201387200
4924 --> 1015809212163473864903934664023523839570
6387 --> 91321559704961673892472654357633041307427900
7772 --> 9027755125844404863192942627813586968870041001600
8921 --> 46137862364188321895572339367552812592575522848000
9749 --> 1514900913329751772087279522915115470030101944229120
*/


Best

Kai-Uwe Bux