Re: [algogeeks] Maximum height/depth of tree
should be 2. -Cheers, Moheed "I am who I am, no matter where I am or who I am with." * * On Sun, Mar 11, 2012 at 11:04 AM, rahul sharma wrote: > http://www.geeksforgeeks.org/archives/646 > > Plz tell that the link i have given in this mail for tree height...isn't > the tree height is 2 in this???height is the longest path...n in path we > include the number of edges or nodes??acc. to me its number of edges in > longest path...plz tell..thnx in advance > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to algogeeks@googlegroups.com. > To unsubscribe from this group, send email to > algogeeks+unsubscr...@googlegroups.com. > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] A strange doubt with cpp class. Why compile error ?
Isn't Abc() a constructor which means to construct the object. No need for
new..
-Cheers
Moheed
"I am who I am, no matter where I am or who I am with."
*
*
On Fri, Mar 9, 2012 at 9:16 PM, atul anand wrote:
> may u r confused with java
>
> Abc a ; // this itself is creating an object of type ABC in C++ , no need
> to use new keyword to allocate m/m.
>
> where as in C++ new keyword allocate m/m and return an address ,
> so to make it work do.
> Abc *a=new ABC();
>
> On Fri, Mar 9, 2012 at 8:28 PM, rahul sharma wrote:
>
>> on left side of new there should be pointer that should collect the
>> address of the memory allocated..
>>
>> write.
>>
>> ABC *a=new ABC();
>>
>>
>> correct if wrng...
>>
>>
>> On Fri, Mar 9, 2012 at 8:19 PM, sanjiv yadav wrote:
>>
>>> write..
>>>
>>> Abc a=Abc()
>>>
>>> it will execute...
>>>
>>>
>>> On Fri, Mar 9, 2012 at 8:15 PM, ~*~VICKY~*~ wrote:
>>>
#include
using namespace std;
class Abc
{
public :
int i;
Abc(){ i = 0;}
};
int main()
{ Abc a = new Abc();
cout<>>> }
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Vicky
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>>>
>>>
>>>
>>> --
>>> Regards
>>>
>>> Sanjiv Yadav
>>>
>>> MobNo.- 8050142693
>>>
>>> Email Id- sanjiv2009...@gmail.com
>>>
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Re: [algogeeks] Doubt in removing loop from linked list
@Rahul: As Dev said: Now start the fast pointer at the head and take m single steps with both pointers. The fast pointer is at the beginning of the cycle, and the slow pointer has traversed the cycle (2*t - u) times and is back at the beginning of the cycle. k= (2*t-u)p -m When fast pointer is being started from the beginning, the slow pointer has not yet completed 2*t-u round, but m nodes less. That means if fast pointer and slow pointer both are moved m nodes further, they will meet at the beginning of the cycle. Following is just a reasoning why both pointer can't meet sooner than m step: The pointers couldn't be equal sooner since the fast pointer is not in the cycle after less than m steps and the slow pointer is in the cycle. " -Cheers Moheed "I am who I am, no matter where I am or who I am with." * * On Sat, Mar 10, 2012 at 4:02 PM, rahul sharma wrote: > 1-2-3-4-5-6-7-8-9-10-11 > fast and slow meet at 11 > > m=6; > k=4 > > > ...i cant get last two lineswhen k= sometimes around the circle - > m.. > > then after that taking fast at begining and slow within circle ..i cant > get this...@ dave plz explain with this example...will b of gr8 > help..thnx in advance.. > > > On Sat, Mar 10, 2012 at 3:02 PM, Kumar Vishal wrote: > >> Hi >> On Mar 9, 2012 3:48 PM, "rahul sharma" wrote: >> >>> i have 2 pointers fast and slow.now if tehy meet there is a loop... >>> >>> now keep one ptr at meeting point and take other one to the begining of >>> listmove both at speed of one..they will meet at start of loophow >>> this happens???why they meet at start..plz tell logic behind this???thnx in >>> advance >>> >>> -- >>> You received this message because you are subscribed to the Google >>> Groups "Algorithm Geeks" group. >>> To post to this group, send email to algogeeks@googlegroups.com. >>> To unsubscribe from this group, send email to >>> algogeeks+unsubscr...@googlegroups.com. >>> For more options, visit this group at >>> http://groups.google.com/group/algogeeks?hl=en. >>> >> -- >> You received this message because you are subscribed to the Google Groups >> "Algorithm Geeks" group. >> To post to this group, send email to algogeeks@googlegroups.com. >> To unsubscribe from this group, send email to >> algogeeks+unsubscr...@googlegroups.com. >> For more options, visit this group at >> http://groups.google.com/group/algogeeks?hl=en. >> > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to algogeeks@googlegroups.com. > To unsubscribe from this group, send email to > algogeeks+unsubscr...@googlegroups.com. > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Practical way to check the primality in efficent time
There are deterministic polyomial-time algorithm. [AKS-primality] Miller-Robin primality test is a non-deterministic primality test with almost zero chance of error. [All above based on some modification/improvement on fermat's theorem] GNU Multi Precission library can be used to handle large integers/primes. -Moheed "I am who I am, no matter where I am or who I am with." * * On Sat, Feb 11, 2012 at 8:32 PM, shady wrote: > There are many, but the one i know how to code is fermat's primality test. > How to calculate all prime numbers between a given range efficiently i > read somewhere that we can do bit-masking to store whether a number is > prime or not, thus saving space ? > I generally use double prime sieve. > > On Sat, Feb 11, 2012 at 8:24 PM, rspr wrote: > >> what are the efficient ways to check that a given number is primer >> assuming the numbers can be large. >> >> -- >> You received this message because you are subscribed to the Google Groups >> "Algorithm Geeks" group. >> To post to this group, send email to algogeeks@googlegroups.com. >> To unsubscribe from this group, send email to >> algogeeks+unsubscr...@googlegroups.com. >> For more options, visit this group at >> http://groups.google.com/group/algogeeks?hl=en. >> >> > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to algogeeks@googlegroups.com. > To unsubscribe from this group, send email to > algogeeks+unsubscr...@googlegroups.com. > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] [Combinatorics] count possible number of binary search trees, given number of nodes
Yeah, its BST(given that each node's data is different from others.) -Moheed "I am who I am, no matter where I am or who I am with." * * On Fri, Feb 10, 2012 at 5:07 PM, Manni mbd wrote: > are you sure u want to ask BINARY SEARCH treees and not Binary trees.. > > On 1/29/12, Moheed Moheed Ahmad wrote: > > I know how to solve it programatically, can anybody pls help me to solve > it > > using combinatorics. > > -Moheed > > > > -- > > You received this message because you are subscribed to the Google Groups > > "Algorithm Geeks" group. > > To post to this group, send email to algogeeks@googlegroups.com. > > To unsubscribe from this group, send email to > > algogeeks+unsubscr...@googlegroups.com. > > For more options, visit this group at > > http://groups.google.com/group/algogeeks?hl=en. > > > > > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to algogeeks@googlegroups.com. > To unsubscribe from this group, send email to > algogeeks+unsubscr...@googlegroups.com. > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Quick Sort
A typical implementation of quick sort works on quick sorting subarray and comparison is done in a linear manner that is a[i] is compared with pivot and then a[i+1] and so on. Virtual memory systems employ some sort of caching. Caching works on the principle of spatial locality of reference. That said, once a comparison begins, after a miss in the cache, cache is fetched for the miss(a[i]) as well as for next few more memory cells (a[i+1], a[i+2], etc). This ensures that next 'few'(typically 3-5) iteration of the loop will be a cache hit, there by speeding up the algorithm. -Moheed "I am who I am, no matter where I am or who I am with." * * On Sun, Jan 29, 2012 at 12:38 AM, karthikeya s wrote: > How QuickSort is good in Virtual Memory Enviroment ? > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to algogeeks@googlegroups.com. > To unsubscribe from this group, send email to > algogeeks+unsubscr...@googlegroups.com. > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] [Combinatorics] count possible number of binary search trees, given number of nodes
I know how to solve it programatically, can anybody pls help me to solve it using combinatorics. -Moheed -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] LINKED LIST
http://cslibrary.stanford.edu/105/LinkedListProblems.pdf
-Moheed
On Sun, Dec 25, 2011 at 11:39 PM, atul anand wrote:
> @ashish : please provide link for that page
>
> On Sun, Dec 25, 2011 at 10:36 PM, Ashish Goel wrote:
>
>> refer stanford page
>> 1,2,3,4,5,6
>>
>> will become
>>
>> 5,3,1
>> 6,4,2
>>
>> void MoveNode(struct node** destRef, struct node** sourceRef) {
>> struct node* newNode = *sourceRef; // the front source node
>> assert(newNode != NULL);
>> *sourceRef = newNode->next; // Advance the source pointer
>> newNode->next = *destRef; // Link the old dest off the new node
>> *destRef = newNode; // Move dest to point to the new node
>> }
>>
>> void AlternatingSplit(struct node* source,
>> struct node** aRef, struct node** bRef) {
>> struct node* a = NULL; // Split the nodes to these 'a' and 'b' lists
>> struct node* b = NULL;
>> struct node* current = source;
>> while (current != NULL) {
>> MoveNode(&a, ¤t); // Move a node to 'a'
>> if (current != NULL) {
>> MoveNode(&b, ¤t); // Move a node to 'b'
>> }
>> }
>> *aRef = a;
>> *bRef = b;
>> }
>>
>>
>> Best Regards
>> Ashish Goel
>> "Think positive and find fuel in failure"
>> +919985813081
>> +919966006652
>>
>>
>>
>> On Sat, Dec 24, 2011 at 11:37 PM, atul anand wrote:
>>
>>> because you are doing odd=odd->next; and even=even->next;
>>> you will lose head pointers for the two linked list formed once you come
>>> out of the loop.
>>>
>>>
>>>
>>> void segregate(node* head)
>>> {
>>> node *temp,*odd,*even;
>>> toggle=1;
>>> if(head==NULL)
>>> {
>>> return;
>>> }
>>> odd=head;
>>>
>>> if(head->next==NULL)
>>> {
>>>
>>> return;
>>> }
>>> even=head->next;
>>>
>>> if(even->next==NULL)
>>> {
>>> return;
>>> }
>>> else
>>> {
>>> temp=even->next;
>>> }
>>> node *otemp,*etemp;
>>> otemp=odd;
>>> etemp=even;
>>>
>>> while(temp!=NULL)
>>> {
>>>
>>> if( toggle == 1)
>>> {
>>>
>>> otemp->next=temp;
>>> otemp=otemp->next;
>>>
>>> temp=temp->next;
>>> otemp->next=NULL;
>>>
>>> toggle=0;
>>> }
>>> else
>>> {
>>> etemp->next=temp;
>>> etemp=etemp->next;
>>>
>>> temp=temp->next;
>>> etemp->next=NULL;
>>>
>>> toggle=1;
>>>
>>> }
>>>
>>>
>>>
>>> }
>>>
>>>
>>> }
>>>
>>> On Sat, Dec 24, 2011 at 10:56 PM, Karthikeyan V.B
>>> wrote:
>>>
Segregate even and odd psoitioned nodes in a linked list
Eg:
2->3->1->9->7->5
The output should be two separate lists
2->1->7
3->9->5
*My code is:*
void segregate(node* head)
{
int i=1;
node* sec=head->next;
node *odd=head,*even=head->next;
while(even)
{
if(i%2)
{
odd->next=even->next;
even->next=NULL;
odd=odd->next;
if(!odd->next) break;
}
else
{
even->next=odd->next;
odd->next=NULL;
even=even->next;
if(!even->next) break;
}
i++;
}
}
Pls correct me if i'm wrong or suggest me a better approach
Regards,
KARTHIKEYAN.V.B
PSGTECH
CBE
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"Algorithm
Re: [algogeeks] Re: Nice question
Thanks Don for pointing it out. It will not work [the second and
consecutive abs-diffs are not mutually exclusive].
However, here are the 10 possible numbers that will work for n=3 and
absdiff={0,0}:
0 0 0
1 1 1
2 2 2
--
8 8 8
9 9 9
-Moheed
'If a man neglects education, he walks lame to the end of his life.'
On Wed, Dec 14, 2011 at 12:17 AM, Don wrote:
> Moheed,
> If n=3 and absdiff = {0,0}, your program says that there are 100
> possible numbers. Can you show me at least 10 of them?
> Don
>
> On Dec 13, 12:24 pm, Moheed Moheed Ahmad wrote:
> > To get a abs difference of 0 there are 10 ways
> > similarly getting abs difference of 1 there are 9x2 ways(w1)
> > for 2 its 8x2 (say w(2)
> > for 3 its 7x2
> > .
> > for 9 its 1x2(w9)
> > let w(i) represents the number of ways to get abs diff of i.
> > So total numbers that are possible from the given abs diff i j k l m
> ...
> > (w(i) x w(j) x w(k) x w(l) x.)
> >
> > Now algo will be to scan the given abs diff and multiply the w(i) for
> each
> > absdiff .
> >
> > int calculate_possible_nums(int absdiff[], int len){
> > int ways[]={10, 18, 16, 14, 12, 10, 8, 6, 4, 2, 1};
> > int numways=1;
> > for ( i=0; i < len; i++){
> > numways = numways * ways[absdiff[i]];
> > }
> > return numways;}
> >
> > -Moheed
> > 'If a man neglects education, he walks lame to the end of his life.'
> >
> > On Tue, Dec 13, 2011 at 11:20 PM, Don wrote:
> > > There should be 39 combinations with that input. You are missing
> > > numbers which include the digit zero, such as 14610, 30278, and 52056.
> >
> > > Don
> >
> > > On Dec 13, 11:37 am, tech coder wrote:
> > > > I tried the problem and written the code for it . it is in java. it
> is
> > > > printing all the possible numbers
> > > > I am treating the differences ans an array of integers.
> >
> > > > here is the code
> >
> > > > public class Main {
> >
> > > > public static void main(String[] args)
> > > > {
> > > >int digit[]={3,2,5,1};// array of absolute differences
> >
> > > > int digit[]={3,2,5,1};
> > > >for(int num=1;num<=9;num++) // call with all possible
> initial
> > > > numbers
> > > >findNumber(digit,4,num,0,num);
> > > > }
> >
> > > > public static void findNumber(int digit[],int n,int num,int i,int
> > > > oldDigit)
> > > > {
> > > > if(i==n)
> > > > {
> > > > System.out.print(num+" ");
> > > > return;
> > > > }
> >
> > > > {
> > > > int o=digit[i]+oldDigit;
> > > > if(o<10)
> > > > findNumber(digit,n,10*num+o,i+1,o);
> > > > o=oldDigit-digit[i];
> > > > if(o>0)
> > > > findNumber(digit,n,10*num+o,i+1,o);
> >
> > > > }
> > > > }
> >
> > > > }
> >
> > > > and here is the output
> >
> > > > 14612 14278 14276 25723 25721 25389 25387 36834 36832 36498
> > > 47945
> > > > 47943 41389 41387 58612 52498 69723 69721 63167 63165 74612
> > > > 74278 74276 85723 85721 85389 85387 96834 96832 96498
> > > > BUILD SUCCESSFUL (total time: 0 seconds)
> >
> > > > On Tue, Dec 13, 2011 at 11:11 PM, Dave
> wrote:
> > > > > @Amir: Presumably, since these are digits in a number, they are
> > > > > bounded on the bottom by 0 and on the top by radix-1. So in
> decimal,
> > > > > if a digit is 7 and the absolute difference between it and the next
> > > > > digit is 3, there is only one possibility for the next digit, 7-3
> = 4,
> > > > > since 7+3 is too large. So only some subset of the 2^(n-1)
> > > > > combinations of addition and subtraction may be possible.
> >
> > > > > Dave
> >
> > > > > On Dec 13, 4:15 am, Amir hossein Shahriari
> > > > > wrote:
> > > > > > actually there are infinite number of sequences that match it
> > > > > > for example if the absolute differences are 3 2 5 1
> > > > > > one possible sequence is 6 3 5 0 1 one other is 7 4 6 1 2 or 8 5
>
Re: [algogeeks] Re: Nice question
To get a abs difference of 0 there are 10 ways
similarly getting abs difference of 1 there are 9x2 ways(w1)
for 2 its 8x2 (say w(2)
for 3 its 7x2
.
for 9 its 1x2(w9)
let w(i) represents the number of ways to get abs diff of i.
So total numbers that are possible from the given abs diff i j k l m ...
(w(i) x w(j) x w(k) x w(l) x.)
Now algo will be to scan the given abs diff and multiply the w(i) for each
absdiff .
int calculate_possible_nums(int absdiff[], int len){
int ways[]={10, 18, 16, 14, 12, 10, 8, 6, 4, 2, 1};
int numways=1;
for ( i=0; i < len; i++){
numways = numways * ways[absdiff[i]];
}
return numways;
}
-Moheed
'If a man neglects education, he walks lame to the end of his life.'
On Tue, Dec 13, 2011 at 11:20 PM, Don wrote:
> There should be 39 combinations with that input. You are missing
> numbers which include the digit zero, such as 14610, 30278, and 52056.
>
> Don
>
> On Dec 13, 11:37 am, tech coder wrote:
> > I tried the problem and written the code for it . it is in java. it is
> > printing all the possible numbers
> > I am treating the differences ans an array of integers.
> >
> > here is the code
> >
> > public class Main {
> >
> > public static void main(String[] args)
> > {
> >int digit[]={3,2,5,1};// array of absolute differences
> >
> > int digit[]={3,2,5,1};
> >for(int num=1;num<=9;num++) // call with all possible initial
> > numbers
> >findNumber(digit,4,num,0,num);
> > }
> >
> > public static void findNumber(int digit[],int n,int num,int i,int
> > oldDigit)
> > {
> > if(i==n)
> > {
> > System.out.print(num+" ");
> > return;
> > }
> >
> > {
> > int o=digit[i]+oldDigit;
> > if(o<10)
> > findNumber(digit,n,10*num+o,i+1,o);
> > o=oldDigit-digit[i];
> > if(o>0)
> > findNumber(digit,n,10*num+o,i+1,o);
> >
> > }
> > }
> >
> > }
> >
> > and here is the output
> >
> > 14612 14278 14276 25723 25721 25389 25387 36834 36832 36498
> 47945
> > 47943 41389 41387 58612 52498 69723 69721 63167 63165 74612
> > 74278 74276 85723 85721 85389 85387 96834 96832 96498
> > BUILD SUCCESSFUL (total time: 0 seconds)
> >
> >
> >
> > On Tue, Dec 13, 2011 at 11:11 PM, Dave wrote:
> > > @Amir: Presumably, since these are digits in a number, they are
> > > bounded on the bottom by 0 and on the top by radix-1. So in decimal,
> > > if a digit is 7 and the absolute difference between it and the next
> > > digit is 3, there is only one possibility for the next digit, 7-3 = 4,
> > > since 7+3 is too large. So only some subset of the 2^(n-1)
> > > combinations of addition and subtraction may be possible.
> >
> > > Dave
> >
> > > On Dec 13, 4:15 am, Amir hossein Shahriari
> > > wrote:
> > > > actually there are infinite number of sequences that match it
> > > > for example if the absolute differences are 3 2 5 1
> > > > one possible sequence is 6 3 5 0 1 one other is 7 4 6 1 2 or 8 5 7 2
> 3
> > > > and you can add any integer value to all elements and the result will
> > > still
> > > > be valid
> > > > actually you can start with any number and and then the second number
> > > will
> > > > be equal to the first number that you chose plus/minus the first
> absolute
> > > > difference and so on
> >
> > > > so if we are given the first element of the sequence there are
> 2^(n-1)
> > > ways
> > > > to find a valid sequence because for each absolute difference we can
> > > either
> > > > add the absolute difference to the last sequence element or subtract
> the
> > > > absolute difference from it
> >
> > > > On Mon, Dec 12, 2011 at 9:01 PM, KAY
> > > wrote:
> > > > > If for a number n digits long, the absolute difference between
> > > > > adjacent digits is given, how to find out the number of different
> > > > > numbers with these absolute differences ?
> >
> > > > > for eg,
> > > > > if n=5
> > > > > and the absolute differences are
> > > > > 3 2 5 1
> > > > > then 1 possible number is
> > > > > 6 3 5 0 1(because |6-3|=3,|3-5|=2 and so on...)
> >
> > > > > How many such numbers will be there?
> >
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