Dynamic programming algorithm - c++

Question

Hi I have a question on an existing algo problem.
Existing problem description: Generate 10-digit number using a phone keypad
1 2 3
4 5 6
7 8 9
0

Answer 1

Though this question has a tag of C++, consider this pseudo-code to express the algorithm (which conveniently happens to be written in ruby.)
# Where the knight can jump to
$m = {
0 => [4,6], 1 => [6,8], 2 => [7,9], 3 => [4,8], 4 => [0,3,9],
5 => [], 6 => [0,1,7], 7 => [2,6], 8 => [1,3], 9 => [2,4]
}
$cache = Hash.new
# return count
def nseq( k, n, e=0 )
e += 1 if k.even?
return 0 if 3 < e
return 1 if n == 1
key = "#{k}:#{n}:#{e}" # for the memoization
return $cache[key] if $cache.has_key? key
# Sum nseq(j,n-1,e) for j in $m[k]
return $cache[key] = $m[k].inject(0) { |sum,j| sum + nseq( j, n-1, e ) }
end
0.upto(9) do |k|
2.upto(8) do |n|
count = nseq(k,n)
puts "k=#{k},n=#{n}: #{count}"
break if count.zero?
end
end
This outputs
k=0,n=2: 2
k=0,n=3: 6
k=0,n=4: 8
k=0,n=5: 16
k=0,n=6: 0
k=1,n=2: 2
k=1,n=3: 5
k=1,n=4: 10
k=1,n=5: 24
k=1,n=6: 32
k=1,n=7: 64
k=1,n=8: 0
k=2,n=2: 2
k=2,n=3: 4
k=2,n=4: 10
k=2,n=5: 16
k=2,n=6: 32
k=2,n=7: 0
k=3,n=2: 2
k=3,n=3: 5
k=3,n=4: 10
k=3,n=5: 24
k=3,n=6: 32
k=3,n=7: 64
k=3,n=8: 0
k=4,n=2: 3
k=4,n=3: 6
k=4,n=4: 14
k=4,n=5: 16
k=4,n=6: 32
k=4,n=7: 0
k=5,n=2: 0
k=6,n=2: 3
k=6,n=3: 6
k=6,n=4: 14
k=6,n=5: 16
k=6,n=6: 32
k=6,n=7: 0
k=7,n=2: 2
k=7,n=3: 5
k=7,n=4: 10
k=7,n=5: 24
k=7,n=6: 32
k=7,n=7: 64
k=7,n=8: 0
k=8,n=2: 2
k=8,n=3: 4
k=8,n=4: 10
k=8,n=5: 16
k=8,n=6: 32
k=8,n=7: 0
k=9,n=2: 2
k=9,n=3: 5
k=9,n=4: 10
k=9,n=5: 24
k=9,n=6: 32
k=9,n=7: 64
k=9,n=8: 0
The result is the number of all n-length sequences starting on key k, which have no more than 3 even digits in them. For example, the last entry is k=9,n=8: 0. This means that all sequences of length 8 starting on key 9 include more than 3 even digits.
EDIT: Here it is translated into C++. It produces identical output as above.
#include<iostream>
#include<map>
using namespace std;
const int MAX_EVENS = 3; // Assume < 8
// Where the knight can jump to
const int jumpto[][3] = { {4,6}, // 0
{6,8}, {7,9}, {4,8}, // 1 2 3
{0,3,9}, {}, {0,1,7}, // 4 5 6
{2,6}, {1,3}, {2,4} }; // 7 8 9
const int jumpto_size[] = { 2, // 0
2, 2, 2, // 1 2 3
3, 0, 3, // 4 5 6
2, 2, 2 }; // 7 8 9
typedef map<unsigned,int> cachetype;
cachetype cache;
int nseq( int k, int n, int e=0 )
{
e += k&1^1; // increment e if k is even.
if( MAX_EVENS < e ) return 0;
if( n <= 1 ) return 1;
unsigned key = (n << 4 | k) << 3 | e; // n is left with 32-7=25 bits
cachetype::const_iterator it = cache.find(key);
if( it != cache.end() ) return it->second;
int sum = 0;
for( int i=0 ; i<jumpto_size[k] ; ++i ) sum += nseq( jumpto[k][i], n-1, e );
return cache[key] = sum;
}
int main()
{
for( int k=0 ; k<=9 ; ++k )
for( int n=2 ; n<=8 ; ++n )
{
int count = nseq(k,n);
cout << "k="<<k<<",n="<<n<<": "<<count<<endl;
if( count == 0 ) break;
}
return 0;
}