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balatro-gba-chinese-jocker-…/source/hand_analysis.c
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MeirGavish 34889988a9 Added documentation to get_played/hand_distribution (#343)
* Added documentation to get_played/hand_distribution

* Documented get_played_distribution() behavior
2025-12-28 11:20:30 -08:00

457 lines
13 KiB
C

#include "hand_analysis.h"
#include "card.h"
#include "game.h"
void get_hand_distribution(u8 ranks_out[NUM_RANKS], u8 suits_out[NUM_SUITS])
{
for (int i = 0; i < NUM_RANKS; i++)
ranks_out[i] = 0;
for (int i = 0; i < NUM_SUITS; i++)
suits_out[i] = 0;
CardObject** cards = get_hand_array();
int top = get_hand_top();
for (int i = 0; i <= top; i++)
{
if (cards[i] && card_object_is_selected(cards[i]))
{
ranks_out[cards[i]->card->rank]++;
suits_out[cards[i]->card->suit]++;
}
}
}
void get_played_distribution(u8 ranks_out[NUM_RANKS], u8 suits_out[NUM_SUITS])
{
for (int i = 0; i < NUM_RANKS; i++)
ranks_out[i] = 0;
for (int i = 0; i < NUM_SUITS; i++)
suits_out[i] = 0;
CardObject** played = get_played_array();
int top = get_played_top();
for (int i = 0; i <= top; i++)
{
/* The difference from get_hand_distribution() (not checking if card is selected)
* is in line Balatro behavior,
* see https://github.com/GBALATRO/balatro-gba/issues/341#issuecomment-3691363488
*/
if (!played[i])
continue;
ranks_out[played[i]->card->rank]++;
suits_out[played[i]->card->suit]++;
}
}
// Returns the highest N of a kind. So a full-house would return 3.
u8 hand_contains_n_of_a_kind(u8* ranks)
{
u8 highest_n = 0;
for (int i = 0; i < NUM_RANKS; i++)
{
if (ranks[i] > highest_n)
highest_n = ranks[i];
}
return highest_n;
}
bool hand_contains_two_pair(u8* ranks)
{
bool contains_other_pair = false;
for (int i = 0; i < NUM_RANKS; i++)
{
if (ranks[i] >= 2)
{
if (contains_other_pair)
return true;
contains_other_pair = true;
}
}
return false;
}
bool hand_contains_full_house(u8* ranks)
{
int count_three = 0;
int count_pair = 0;
for (int i = 0; i < NUM_RANKS; i++)
{
if (ranks[i] >= 3)
{
count_three++;
}
else if (ranks[i] >= 2)
{
count_pair++;
}
}
// Full house if there is:
// - at least one three-of-a-kind and at least one other pair,
// - OR at least two three-of-a-kinds (second "three" acts as pair).
// This accounts for hands with 6 or more cards even though
// they are currently not possible and probably never will be.
return (count_three >= 2 || (count_three && count_pair));
}
// This is mostly from Google Gemini
bool hand_contains_straight(u8* ranks)
{
if (!is_shortcut_joker_active())
{
int straight_size = get_straight_and_flush_size();
// This is the regular case of detecting straights
int run = 0;
for (int i = 0; i < NUM_RANKS; ++i)
{
if (ranks[i])
{
if (++run >= straight_size)
return true;
}
else
{
run = 0;
}
}
// Check for ace low straight
if (straight_size >= 2 && ranks[ACE])
{
// With A as low, the highest rank you can use is FIVE.
// -1 for inclusive integer distance and another -1 for the Ace e.g. need=5 -> need 2..5
int last_needed = TWO + (straight_size - 2);
if (last_needed <= FIVE)
{
bool ok = true;
for (int r = TWO; r <= last_needed; ++r)
{
if (!ranks[r])
{
ok = false;
break;
}
}
if (ok)
return true;
}
}
return false;
}
else
{
// Shortcut Joker is active, we have to detect straights where any card may "skip" 1 rank
// We do this with a dynamic programming algorithm that calculates
// the longest possible straight that can end on each rank
// and stopping when we find one that is {straight-size} cards long
u8 longest_short_cut_at[NUM_RANKS] = {0};
// A low ace can start a sequence. 'ace_low_len' is 1 if an ace is present,
// acting as a potential predecessor for TWO and THREE.
int ace_low_len = ranks[ACE] ? 1 : 0;
// Iterate through all ranks from TWO up to ACE.
for (int i = 0; i < NUM_RANKS; i++)
{
// No cards in this rank, no straight can end here, continue
if (ranks[i] == 0)
{
longest_short_cut_at[i] = 0;
continue;
}
int prev_len1 = 0;
int prev_len2 = 0;
// This logic handles the special connections for ace-low straights.
if (i == TWO)
{
// A TWO can be preceded by a low ACE (no skip).
prev_len1 = ace_low_len;
}
else if (i == THREE)
{
// A THREE can be preceded by a TWO (no skip) or a low ACE (skip).
prev_len1 = longest_short_cut_at[TWO];
prev_len2 = ace_low_len;
}
else if (i == ACE)
{
// An ACE (as the highest card) can be preceded by a KING or a QUEEN.
prev_len1 = longest_short_cut_at[KING];
prev_len2 = longest_short_cut_at[QUEEN];
}
else // For all other cards (FOUR through KING).
{
// A card can be preceded by the rank directly below or two ranks below.
prev_len1 = longest_short_cut_at[i - 1];
prev_len2 = longest_short_cut_at[i - 2];
}
// The length of the straight ending at rank 'i' is 1 (for the card itself)
// plus the length of the longest valid preceding straight.
longest_short_cut_at[i] = 1 + max(prev_len1, prev_len2);
// If we've formed a sequence of {straight-size} or more cards, we have a straight.
if (longest_short_cut_at[i] >= get_straight_and_flush_size())
{
return true;
}
}
}
return false;
}
bool hand_contains_flush(u8* suits)
{
for (int i = 0; i < NUM_SUITS; i++)
{
if (suits[i] >= get_straight_and_flush_size())
{
return true;
}
}
return false;
}
// Returns the number of cards in the best flush found
// or 0 if no flush of min_len is found, and marks them in out_selection.
/**
* Finds the largest flush (set of cards with the same suit) in the given array of played cards.
* Marks the cards belonging to the best flush in the out_selection array.
*
* @param played Array of pointers to CardObject representing played cards.
* @param top Index of the top of the played stack.
* @param min_len Minimum number of cards required for a flush.
* @param out_selection Output array of bools; set to true for cards in the best flush, false
* otherwise.
* @return The number of cards in the best flush found, or 0 if no flush meets min_len.
*/
int find_flush_in_played_cards(CardObject** played, int top, int min_len, bool* out_selection)
{
if (top < 0)
return 0;
for (int i = 0; i <= top; i++)
out_selection[i] = false;
int suit_counts[NUM_SUITS] = {0};
for (int i = 0; i <= top; i++)
{
if (played[i] && played[i]->card)
{
suit_counts[played[i]->card->suit]++;
}
}
int best_suit = -1;
int best_count = 0;
for (int i = 0; i < NUM_SUITS; i++)
{
if (suit_counts[i] > best_count)
{
best_count = suit_counts[i];
best_suit = i;
}
}
if (best_count >= min_len)
{
for (int i = 0; i <= top; i++)
{
if (played[i] && played[i]->card && played[i]->card->suit == best_suit)
{
out_selection[i] = true;
}
}
return best_count;
}
return 0;
}
// Returns the number of cards in the best straight or 0 if no straight of min_len is found, marks
// as true them in out_selection[]. This is mostly from Google Gemini
int find_straight_in_played_cards(
CardObject** played,
int top,
bool shortcut_active,
int min_len,
bool* out_selection
)
{
if (top < 0)
return 0;
for (int i = 0; i <= top; i++)
out_selection[i] = false;
// --- Setup for Backtracking DP ---
u8 longest_straight_at[NUM_RANKS] = {0};
int parent[NUM_RANKS];
for (int i = 0; i < NUM_RANKS; i++)
parent[i] = -1;
u8 ranks[NUM_RANKS] = {0};
for (int i = 0; i <= top; i++)
{
if (played[i] && played[i]->card)
{
ranks[played[i]->card->rank]++;
}
}
// --- Run DP to find longest straight ---
// This is nearly identical to hand_contains_straight() logic
// TODO: Consolidate functions to avoid code duplication?
// Might cost performance because this does a little more
int ace_low_len = ranks[ACE] ? 1 : 0;
for (int i = 0; i < NUM_RANKS; i++)
{
if (ranks[i] > 0)
{
int prev1 = 0, prev2 = 0;
int parent1 = -1, parent2 = -1;
if (shortcut_active)
{
if (i == TWO)
{
prev1 = ace_low_len;
parent1 = ACE;
}
else if (i == THREE)
{
prev1 = longest_straight_at[TWO];
parent1 = TWO;
prev2 = ace_low_len;
parent2 = ACE;
}
else if (i == ACE)
{
prev1 = longest_straight_at[KING];
parent1 = KING;
prev2 = longest_straight_at[QUEEN];
parent2 = QUEEN;
}
else
{
prev1 = longest_straight_at[i - 1];
parent1 = i - 1;
if (i > 1)
{
prev2 = longest_straight_at[i - 2];
parent2 = i - 2;
}
}
}
else
{
if (i == TWO)
{
prev1 = ace_low_len;
parent1 = ACE;
}
else if (i == ACE)
{
prev1 = longest_straight_at[KING];
parent1 = KING;
}
else
{
prev1 = longest_straight_at[i - 1];
parent1 = i - 1;
}
}
// Parallels longest_short_cut_at[i] = 1 + max(prev_len1, prev_len2);
// in hand_contains_straight()
if (prev1 >= prev2)
{
longest_straight_at[i] = 1 + prev1;
parent[i] = parent1;
}
else
{
longest_straight_at[i] = 1 + prev2;
parent[i] = parent2;
}
}
}
// --- Find best straight and backtrack ---
int best_len = 0;
int end_rank = -1;
for (int i = 0; i < NUM_RANKS; i++)
{
if (longest_straight_at[i] >= best_len)
{
best_len = longest_straight_at[i];
end_rank = i;
}
}
if (best_len >= min_len)
{
u8 needed_ranks[NUM_RANKS] = {0};
int current_rank = end_rank;
while (current_rank != -1 && best_len > 0)
{
needed_ranks[current_rank]++;
current_rank = parent[current_rank];
best_len--;
}
for (int i = 0; i <= top; i++)
{
if (played[i] && played[i]->card && needed_ranks[played[i]->card->rank] > 0)
{
out_selection[i] = true;
needed_ranks[played[i]->card->rank]--;
}
}
int final_card_count = 0;
for (int i = 0; i <= top; i++)
{
if (out_selection[i])
final_card_count++;
}
return final_card_count;
}
return 0;
}
// This is used for the special case in "Four Fingers" where you can add a pair into a straight
// (e.g. AA234 should score all 5 cards)
void select_paired_cards_in_hand(CardObject** played, int played_top, bool* selection)
{
// Build a set of ranks that are already selected
bool rank_selected[NUM_RANKS] = {0};
bool any_selected_rank = false;
for (int i = 0; i <= played_top; i++)
{
if (selection[i] && played[i] && played[i]->card)
{
rank_selected[played[i]->card->rank] = true;
any_selected_rank = true;
}
}
// If no ranks were selected initially, nothing to do
if (!any_selected_rank)
return;
// Add any unselected card to the selection if if shares a rank with the selected ranks
for (int i = 0; i <= played_top; i++)
{
if (played[i] && played[i]->card && !selection[i])
{
if (rank_selected[played[i]->card->rank])
{
selection[i] = true;
}
}
}
}