how to implement doubly linked lists using only one pointer?

[tonywinslow1986]

I'm reading MIT's book "Introducti on to Algorithms".
The following is one of the excercises from it:
<
10.2-8
Explain how to implement doubly linked lists using only one pointer
value np[x] per item instead of the usual two (next and prev). Assume
that all pointer values can be interpreted as k-bit integers, and
define np[x] to be np[x] = next[x] XOR prev[x], the k-bit
"exclusive-or" of next[x] and prev[x]. (The value NIL is represented by

0.) Be sure to describe what information is needed to access the head
of the list. Show how to implement the SEARCH, INSERT, and DELETE
operations on such a list. Also show how to reverse such a list in O(1)

time.
/>

Could anybody tell me how to solve it? Thank you!!!

[Joe Seigh]

to************* @hotmail.com wrote:

I'm reading MIT's book "Introducti on to Algorithms".
The following is one of the excercises from it:
<
10.2-8
Explain how to implement doubly linked lists using only one pointer
value np[x] per item instead of the usual two (next and prev). Assume
that all pointer values can be interpreted as k-bit integers, and
define np[x] to be np[x] = next[x] XOR prev[x], the k-bit
"exclusive-or" of next[x] and prev[x]. (The value NIL is represented by

0.) Be sure to describe what information is needed to access the head
of the list. Show how to implement the SEARCH, INSERT, and DELETE
operations on such a list. Also show how to reverse such a list in O(1)

time.
/>

Could anybody tell me how to solve it? Thank you!!!

The book all but told you how to solve it.
If anything XOR'd with itself is zero, solve
the equation

np[x] = next[x] XOR prev[x]

for next[x].

To reverse the list, how would you revers a regular doubly
linked list with next and prev pointers?


--
Joe Seigh

When you get lemons, you make lemonade.
When you get hardware, you make software.

[Michal Nazarewicz]

"to************ *@hotmail.com" <to************ *@hotmail.comwr ites:

Explain how to implement doubly linked lists using only one pointer
value np[x] per item instead of the usual two (next and prev). Assume
that all pointer values can be interpreted as k-bit integers, and
define np[x] to be np[x] = next[x] XOR prev[x], the k-bit
"exclusive-or" of next[x] and prev[x]. (The value NIL is represented by

0.) Be sure to describe what information is needed to access the head
of the list. Show how to implement the SEARCH, INSERT, and DELETE
operations on such a list. Also show how to reverse such a list in O(1)
time.

http://en.wikipedia.org/wiki/XOR_linked_list

--
Best regards, _ _
.o. | Liege of Serenly Enlightened Majesty of o' \,=./ `o
..o | Computer Science, Michal "mina86" Nazarewicz (o o)
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[tonywinslow1986]

The book all but told you how to solve it.
If anything XOR'd with itself is zero, solve
the equation

np[x] = next[x] XOR prev[x]

for next[x].

That's the point! It seems that I know nothing about bit computations,
so silly questions are raised here.

[tonywinslow1986]

If anything XOR'd with itself is zero, solve

the equation

np[x] = next[x] XOR prev[x]

for next[x].

how to solve that equation? i'm confused!

[Jim Langston]

<to************ *@hotmail.comwr ote in message
news:11******** **************@ 73g2000cwn.goog legroups.com...

>If anything XOR'd with itself is zero, solve
the equation

np[x] = next[x] XOR prev[x]

for next[x].

how to solve that equation? i'm confused!

Play with bit math.

int X = 1;
int Y = 3;

now output X and Y, X or Y, X xor Y, etc... Find out why they come out the
way they do.

[tonywinslow1986]

Play with bit math.

>
int X = 1;
int Y = 3;

now output X and Y, X or Y, X xor Y, etc... Find out why they come out the
way they do.

i've known that:
if A ^ B = C, then B = A ^ C, A = B^C.
but anyway how to solve that equation with only np[x] given?

[Simon G Best]

Hello!

The original question is fun!

to************* @hotmail.com wrote:

>
i've known that:
if A ^ B = C, then B = A ^ C, A = B^C.

What's X ^ X? What's (X ^ X) ^ Y?

but anyway how to solve that equation with only np[x] given?

You won't have "only np[x] given".

The answer, as they say, is in the (original) question. It's all a
matter of knowing where to begin. You /will/ know where to begin. Just
use your head. Start at the beginning. After all, nothing will come
from nothing, as they say.

:-)

Simon

--
What happens if I mention Leader Kibo in my .signature?

[Risto Lankinen]

<to************ *@hotmail.comwr ote in message
news:11******** **************@ 48g2000cwx.goog legroups.com...

>
i've known that:
if A ^ B = C, then B = A ^ C, A = B^C.
but anyway how to solve that equation with only np[x] given?

I think this is an easy way to communicate the concept:

Instead of XOR, use addition and subtraction (they are
essentially equivalent when it comes to implementing the
trick of this method, but much easier to comprehend).

First, every node has a field to hold the distance from its
previous-node to its next-node. Note that this is exactly
equal to the sum of the distances from its previous-node
to itself, and from itself to its next-node.

When you iterate, you need pointers to two successive
nodes. Call them A and B. It is easy to compute the
distance between the nodes (B-A), and from there it is
easy to compute the address of A's previous-node [A's
sum-of-distances is reduced by (B-A) and the remainder
applied as an offset to A's address] or of B's next-node
[left as an exercise].

- Risto -