### [2015-03-06] Challenge #204 [Hard] Addition chains

The bonus problem is actually very easy if you take in one optimization: the minimum amount of doubling values. For example, the largest number you can arrive at in 19 steps is 219 = 524288. Notice that it is not much larger than 123456. This means that, most of the time, the next number has to be double the previous number. In fact, you can calculate the minimum number of times this has to be the case as log2(target).

My solution is in C. All challenge inputs, including bonus problem, finish in less than 1 second.

``````#include <stdlib.h>
#include <stdio.h>
#include <stdbool.h>
#include <string.h>

int getMinDouble(target)
{
int retVal = 0;
while((target /= 2) > 1) retVal++;
return retVal;
}

bool getNext(int* chain, int nextIndex, int maxSize, int target, int reqDouble)
{
int addsLeft = maxSize - nextIndex;
int lastVal = chain[nextIndex-1];

return lastVal == target;
return false;
if( lastVal >= target )
return false;

// Try double first
chain[nextIndex]=lastVal*2;
if( getNext(chain, nextIndex+1, maxSize, target, reqDouble-1) )
return true;

// If double failed but haven't achieved minimum number of doubles,
// no need to try anything else
{
int i;
for(i = 0; i < nextIndex; i++)
{
int j;
for( j = i; j < nextIndex; j++)
{
int newVal = chain[i]+chain[j];
if( newVal <= lastVal || (i == j && i == nextIndex-1))
continue;
chain[nextIndex]=newVal;
if( getNext(chain, nextIndex+1, maxSize, target, reqDouble) )
return true;
}
}
chain[nextIndex]=0;
}
return false;
}

int main(int argc, char** argv)
{
if( argc != 3 )
{
printf("USAGE: %s sizeOfChain target\n",argv[0]);
return 1;
}

int size = atoi(argv[1]);
int target = atoi(argv[2]);
if( size <= 0 || target <= 0 )
{
printf("addition size and target must be positive integers\n");
return 1;
}
size++;

int* chain = malloc(size*sizeof(int));
memset(chain, 0, size*sizeof(int));

int minDouble = getMinDouble(target)-1;

chain[0]=1;
chain[1]=2;
if( getNext(chain, 2, size, target, minDouble) )
{
int i;
for(i = 0; i < size; i++ )
{
printf("%d\n", chain[i]);
}
}
else
{