Hi, Quintopia,
Maybe you misunderstood my algorithm above.
S_k is a set of ordered pair (i,j)
each time we pop the min(S_k) from S_k, we insert at most 2 element in
it
so its size is in O(k)
I think the following should be more clear:
heap: MIN_HEAP
visited[n][n] =
Hi, ljb,
Let me have a try to prove the correctness of it.
Let define a property A(S) :
S is a set of ordered pair (i,j), and every
pair (x,y) whose sum(x,y) is less than min{S} = sum(mi,mj) is already
output.
Initially, we have S_1 = {(1,1)}, A(S_1) holds.
Suppose that for n, A(S_n) holds,
Reading about hamming distance and clustering methods might help.
bullockbefriending bard wrote:
A single 6 Pick bet looks like this:
RACE1 RACE2RACE3RACE4 RACE5 RACE6
runner1 / runner 2 / runner 3 / runner4 / runner5 / runner6 - $amount
e.g. we might have:
5 / 3 / 11 / 7
I have the same question. It seems the S-G function doesn't work when
there is a draw... ?
Rajiv Mathews 写道：
Is there a way to leverage this solution to handle more than one pile
of matches?
Say there were k piles each containing n_1, n_2, ..., n_k matches,
with the same rules, that is
Hi
can any one help me to find this ebook free ( please help me ) thank you :
Horowitz, Sahni, Mehta: Fundamentals of Data Structures in C++
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Algorithm Geeks
wrb wrote:
in arrange 1..n there are n different numbers. how can you fill A[1..n]
without any one of them?
That occurred to me as well, but I assumed that it must be allowed that
for A[i] == A[j], i != j because otherwise it would be impossible for
there
to be any missing numbers.
you're right. I didn't understand it. your solution is correct. My
version would only work in cases in which we didn't need to print the
same minimum sum as many times as it appears. . .but it wouldn't even
be very good at this.
In other words, I solved a completely different problem, yes?
I think this one might have some optimal substructure, though it's not
exactly clear what that may be. I'll give it some thought.
On Dec 1, 6:26 am, smartdude [EMAIL PROTECTED] wrote:
Reading about hamming distance and clustering methods might help.
bullockbefriending bard wrote:
A single
Thanks. I'm going to try to see what happens if I throw my data at
py-cluster with some kind of nearness metric based initially upon
Hamming distance - perhaps later can include bet size in this and kill
two birds with one stone. Still going to be a lot of computational work
to then group
Any pointers much appreciated - I simply don't have the background or
experience to be able to visualise what this might be. I take it as a
given that the problem is NP-hard at the very least and that there is
no alternative but to go at it with some kind of greedy heuristic - and
any way of
I'm going to want to be able to recognise and merge the likes of:
5 + 7 / 3 / 11 / 7 + 14 / 1 / 9 - $50 ($200 total)
5 + 7 / 3 / 11 / 7 + 14 / 1 / 2 - $50 (ditto)
into
5 + 7 / 3 / 11 / 7 + 14 / 1 / 2 + 9 $50 ($400 total)
And so on, 'recursively' so that in a perfect world I'd end up with
each
Oops... I think I just did a silly thing and changed title of entire
thread due to unfamiliarity with Google Groups. Anyway, changing
subject back to '6 Pick Bet Grouping' just in case.
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