}
+ /* If card_one < card_two, then return value will be negative
+ if they are equal, 0 will be returned,
+ if card_one > card_two, then return value will be positive */
static int card_compare(card_t *card_one, card_t *card_two)
{
return card_one->number - card_two->number;
}
- static int card_group_is_run(card_group_t *card_group)
+ static int card_group_is_run_one(card_group_t *card_group)
+ {
+ int i;
+ qsort (&card_group->cards[0], card_group->num_cards,
+ sizeof (card_t), card_compare);
+
+ if (card_group->num_cards > 13 || card_group->num_cards < 3)
+ {
+ return 0;
+ }
+ for (i = 0; i < card_group->num_cards - 1; ++i)
+ if(card_group->cards[i].color != card_group->cards[i + 1].color)
+ {
+ return 0;
+ }
+ if(card_group->cards[i].number != card_group->cards[i + 1].number -1)
+ {
+ return 0;
+ }
+ return 1;
+ }
+
+
+ static int card_group_is_run_two(card_group_t *card_group)
{
int i;
int lowest = 14, highest = 0;
color_t run_color;
- if (card_group->num_cards > 13 || card_group->num_cards < 3)
+ /* By definition, a run must have at least 3 cards. Also, it's
+ * impossible for any group of cards with more than 13 cards to be
+ * a run, (there are only 13 unique numbers so a group with more
+ * than 13 cards must have some duplicates).
+ */
+ if (card_group->num_cards < 3 || card_group->num_cards > 13)
{
return 0;
}
- run_color = card_group->cards[i].color;
+ /* Loop through all cards in the group, ensuring that they are all
+ * the same color and finding the highest and lowest number in the
+ * group. */
+ run_color = card_group->cards[0].color;
+
for (i = 0; i < card_group->num_cards; i++)
{
if (card_group->cards[i].color != run_color)
return 0;
-
if (card_group->cards[i].number > highest)
{
highest = card_group->cards[i].number;
lowest = card_group->cards[i].number;
}
}
+
+ /* For a run, the difference between the highest and lowest cards
+ * will always be one less than the number of cards in the
+ * group. If not then we know it's not a run.
+ */
if (highest - lowest != card_group->num_cards - 1)
{
return 0;
}
+
+ /* XXX: There's a bug here. We're guessing that at this point
+ * anything we're looking at must be a run. This would be correct
+ * if there were no duplicate cards, but since there are
+ * duplicates this us quite broken. For example consider two
+ * sequences of entirely red cards:
+ *
+ * This is a run: 1, 2, 3, 4
+ * But this is not: 1, 3, 4, 4
+ *
+ * As currently written, this function will consider both of these
+ * groups to be a run. One possible fix is to throw away the
+ * highest - lowest heuristic and instead simply sort the cards up
+ * front and ensure the difference between each adjacent pair is
+ * exactly 1.
+ */
return 1;
}