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Never explain
Radix/bucket sort..
won't that help?
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Fri, May 27, 2011 at 7:15 PM, adityasir...@gmail.com wrote:
how about this case:
9, 100 - 9100
100 9
9100
2, 3, 9, 78 --
78 9 3 2
9 78 3 2
I guess
Can this work...
Lets say I have following numbers
8 9 7 4 2 121 23 21 24 27 35 79 2334 6785
Now repeat the last number to make all the number of equal length...
1211 2333 2111 2444 2777 3555 7999 2334 6785
Sort the following numbers in descending order.
7999
check this sol frends
#includeiostream
using namespace std;
class A
{
private:
int a;
public:
A(int m)
{
a=m;
}
};
int main()
{
coutEnter no of objects in arrayendl;
int p,n;
cinn;
coutEnter value of a for first object of arrayendl;
cinp;
Hi Saurabh,
Can you try it for 10? Could not really understand, what are you gonna
communicate?
10 = 2*5 (2^2 + 1^2 )*(1*2 + 1^2)... If with this logic you are saying 10 is
prime then all numbers divisible by 5 should be prime.
Could you elaborate your answer more?
Thanks Regards
Vishal Jain
here in this part
if bitmap[bi] matrix[i][j]) 0x1) == 0x0)
(((bitmap[bj] matrix[i][j]) 0x1) == 0x0)
(((bitmap[bk] matrix[i][j])
0x1) == 0x0)) {
bitmap[bi] |= 1 matrix[i][j];
yes, you are right. bitmap will be filled in the process of solving
the grid. in verify routine, if the expression evaluates to false, it
mean an element is encountered which is already present in row, col
and 3x3 cube. this way you an tell that the solution is wrong.
hope that helps.
On Sun,
no... what i mean to say is you have filled the bitmap in the process
of solving the grid. The way you are filling the bitmap is - you are
taking values from the matrix and placing them in the bitmaps using bi
bj bk.
now in the verification you are checking the same matrix value against
the same
I think ur verification function is correct but it works only if user
is entering the values one by one. as soon as he enters a duplicate
value it will show a error. It will work for the case of ur solver as
ur code itself is generating the values.
But here in this verification problem u r given a
sorry, i made a mistake in explaining. when you are solving the gird,
you will have the matrix[][] array partially filled. the bitmap used
for solving the grid is different than the one being used for
verifying. in case of verifying, you will have a completely blank
bitmap, with not bits set. you
what i understood from your problem is, when the configuration is
given by user, it can be directly stored into matrix[][] array. that
will solve your problem in o(n^2), where n=9.
On Sun, May 29, 2011 at 5:50 PM, Dumanshu duman...@gmail.com wrote:
I think ur verification function is correct but
for eg 10, 9
most signifacnt digit of 10 is 1 and 9 is 9
so after sorting based on most significant digit is
10,9
output 910
2nd ex 2,3,5,78
most significant digit is 2,3,5,7
so out put is 78532
On May 29, 12:59 am, Vishal Jain jainv...@gmail.com wrote:
Can this work...
Lets say I have
oh yeah... now it works. thanx a lot!
On May 29, 5:20 pm, Vishal Thanki vishaltha...@gmail.com wrote:
sorry, i made a mistake in explaining. when you are solving the gird,
you will have the matrix[][] array partially filled. the bitmap used
for solving the grid is different than the one being
use long long int instead of int for X and all except t.and type cast the
sqrt returned value to integer..like int(sqrt(X))
On Sun, May 29, 2011 at 1:55 PM, Vishal Jain jainv...@gmail.com wrote:
Hi Saurabh,
Can you try it for 10? Could not really understand, what are you gonna
communicate?
There are n persons.
You are provided with a list of ppl which each person does not like.
Determine the minm no. of houses required such that, in no house
2 people should dislike each other.
Is there a polynomial time solution exist for this? Or is this not
solvable at all?
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#includestdio.h
int main(void)
{
float a=0.08;
if(a0.08)
printf(Hello\n);
else
printf(Hii\n);
return 0;
}
The o/p is: *Hello * why
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To
a 0.08f will make it compare to float. by default 0.08 is
considered as double.
On Sun, May 29, 2011 at 9:51 PM, Ankit Agarwal ankitgeniu...@gmail.com wrote:
#includestdio.h
int main(void)
{
float a=0.08;
if(a0.08)
printf(Hello\n);
else
printf(Hii\n);
@vishal,Sanjeev..
for the inputs 18,187.. apply ur method..
18 -- 188
187-- 187
18187 - ur method
18718 - actual
@Sunny...
i agree that your algorithm takes the *O(N logN)* time.. but again..
the problem is it* doesn't get* the exact solution.
Do we really have a polynomial solution for this
and, I read it long time back that.. the value of 0.8 alone will be stored
as 0.7995 (not sure on the number of 9's but.. the last digit in the
precision will be a 5)
that could be a reason.
may be what vishal said is correct.
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@sravanreddy001
i don't find any cases for which my algo fails and its O(nlgn)
i may be missing something
can you tell any case where it fails
On Sun, May 29, 2011 at 10:15 PM, sravanreddy001
sravanreddy...@gmail.comwrote:
@vishal,Sanjeev..
for the inputs 18,187.. apply ur method..
18 --
it is exactly a graph coloring problem. so it has no polynomial order
solution.
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you may want to check how the floats and doubles are stored into
memory using ieee notation.
i tried to print 0.08 and 0.08f in hex format and got the following result.
vishal@ubuntu:~/progs/c\ 10:03:56 AM $ cat fl.c
#include stdio.h
int main()
{
float f=0.08;
if (f 0.08f)
@anshu mishra:
Yeah. Thanks! :)
On May 30, 8:26 am, anshu mishra anshumishra6...@gmail.com wrote:
it is exactly a graph coloring problem. so it has no polynomial order
solution.
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what about using a hash function?
On Mon, May 30, 2011 at 10:18 AM, ross jagadish1...@gmail.com wrote:
Given a matrix, you need to find the number of blocks in it.
A block has the same numbers.
EG:
1 1 3
1 2 3
2 2 4
has 4 blocks namely,
1 1
1
2
2 2
3
3
4
1 2 3
4 5 6
7 8 9
write these three lines in ur function, it will bind that particular thread
to (id)th processor wher
void func(int id)
{
unsigned long mask;
mask = 1id;
pthread_setaffinity_np(pthread_self(), sizeof(mask), mask);
}
main()
{
pthread_t *thread;
thread =
PS dont forget to bind ith thread with ith processor
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@vishal
Hi,
I do not get you.
Can you please elaborate a little more how you ll use hash?
On May 30, 8:50 am, Vishal Thanki vishaltha...@gmail.com wrote:
what about using a hash function?
On Mon, May 30, 2011 at 10:18 AM, ross jagadish1...@gmail.com wrote:
Given a matrix, you need to
Okay, i thought it this way: iterate through the whole matrix, and
take the value as a key to a hash table. and the value
corresponding to the key would be the count. increment the count
everytime you encounter the same key. it is very easy to implement
this in python (using dictionary) but i am
when u store 0.08 on the float variable a then it does not store exactly
0.08 on 'a'
actually it store 0.7999
that's why after comparition it make less quantity with 0.08
i.e 'if ' condition is true and print it HELLO
On Sun, May 29, 2011 at 9:21 AM, Ankit Agarwal
@ross...
I have adoubt regarding this question...
what should be output if the matrix is like this..
1 1 2
1 3 1
2 3 4
U reply I will tell you then what doubt I have.
On Mon, May 30, 2011 at 10:36 AM, Vishal Thanki vishaltha...@gmail.comwrote:
Okay, i thought it this way: iterate through
it is a simple graph problem. travese the whole matrix using BFS. it will be
O(n^2).
for (i = 0; i n; i++)
{
for (j = 0; j n; j++)
{
if (flag[i][[j])
{
bfs(mat, flag, i, j);
count++;
}
}
}
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@vishal ur sol wil give wrong answer for this
1 1 2
1 3 1
2 3 4
answer should be 6 but ur sol wil give 4.
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@anshucan u plz elaborate ur solution for a particular case??
On Mon, May 30, 2011 at 10:49 AM, anshu mishra anshumishra6...@gmail.comwrote:
@vishal ur sol wil give wrong answer for this
1 1 2
1 3 1
2 3 4
answer should be 6 but ur sol wil give 4.
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@anshu, i got you. my bad!!
On Mon, May 30, 2011 at 10:49 AM, anshu mishra
anshumishra6...@gmail.com wrote:
@vishal ur sol wil give wrong answer for this
1 1 2
1 3 1
2 3 4
answer should be 6 but ur sol wil give 4.
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Made snake and ladder in c using graphics some time ago. Maybe it could help
http://algoritmus.in/blog/2011/03/snakes-and-ladder-in-cgraphics/
On Sat, May 28, 2011 at 2:58 PM, utkarsh srivastav usdilogi...@gmail.comwrote:
sorry snake and ladder
On Sat, May 28, 2011 at 2:27 AM, Dilogical King
@piyush
void bfs(int mat[][n], bool flag[][n], int i, int j)
{
queue.push(mat[i][j]);
while (!q.empty())
{
x = q.top();
q.pop();
add top bottom, left right element in qeuue if their flag is true and their
value is equal to x and mark their flag false;
}
}
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Thank u guys :) :)
On Mon, May 30, 2011 at 10:40 AM, arjoo kumar 2009ar...@gmail.com wrote:
when u store 0.08 on the float variable a then it does not store exactly
0.08 on 'a'
actually it store 0.7999
that's why after comparition it make less quantity with 0.08
i.e 'if ' condition is
At the each level, traversed by BFS, you will have to check whether the
vertex in this level has the element same as it found in the previous level.
If it is different, then count it.
On Sun, May 29, 2011 at 10:43 PM, anshu mishra anshumishra6...@gmail.comwrote:
@piyush
void bfs(int mat[][n],
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