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hands_test.py
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hands_test.py
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import numpy as np
def hands_test(cnt):
childl = np.zeros(9, dtype='int')
childr = np.zeros(9, dtype='int')
l3 = np.nan
r3 = np.nan
# disconnect
for i in range(1, 8):
if cnt[i-1] and (not cnt[i]) and np.sum(cnt[i+1:]):
childl[:i] += cnt[:i]
childr[i:] += cnt[i:]
break
if any(childl):
l3, l2, lm, li = hands_test(childl)
r3, r2, rm, ri = hands_test(childr)
lrm = []
for lmi in lm:
for rmi in rm:
lrm.append(lmi + rmi)
return l3 + r3, l2 + r2, lrm, li * ri
# leaf
cntsum = np.sum(cnt)
cntmax = np.max(cnt)
cntpos = np.argmax(cnt)
if cntsum == 0:
return 0, 0, [[]], 1
if cntsum == 2 and cntmax == 2:
return 0, 1, [[str(cntpos+1) * 2]], 1
if cntsum == 3:
if cntmax == 3:
return 1, 0, [[str(cntpos+1)*3]], 1
if cntmax == 1:
return 1, 0, [[str(cntpos+1)+str(cntpos+2)+str(cntpos+3)]], 1
# decomposition
i = list(cnt > 0).index(True)
if i > 6: # only 8 or 9, can not form meld
melds = []
cntcnt = np.zeros(5)
for tile, num in enumerate(cnt):
cntcnt[num] += 1
if num:
melds.append(str(tile+1) * num)
if not (cntcnt[1] or cntcnt[4] or cntcnt[2] > 1):
return cntcnt[3], cntcnt[2], [melds], 1
elif cnt[i] == 1 or cnt[i] == 4: # 一定有至少一个顺子
if cnt[i+1] and cnt[i+2]:
# we can use childl here because if any(childl) has returned
childl += cnt
childl[i:i+3] -= 1
l3, l2, lm, li = hands_test(childl)
if not np.isnan(l3):
lm = [lmi + [str(i+1)+str(i+2)+str(i+3)] for lmi in lm]
return l3 + 1, l2, lm, li
elif cnt[i] == 2:
# 做将或有一杯口形状
childl += cnt
childl[i] -= 2
l3, l2, lm, li = hands_test(childl)
if cnt[i+1] > 1 and cnt[i+2] > 1:
childr += cnt
childr[i:i+3] -= 2
r3, r2, rm, ri = hands_test(childr)
if not np.isnan(l3):
lm = [lmi + [str(i+1)*2] for lmi in lm]
l2 += 1
if not np.isnan(r3):
if l3 == r3 + 2 and l2 == r2:
rm = [rmi + [str(i+1)+str(i+2)+str(i+3)]*2 for rmi in rm]
lm += rm
li += ri
return l3, l2, lm, li
elif not np.isnan(r3):
rm = [rmi + [str(i+1)+str(i+2)+str(i+3)]*2 for rmi in rm]
r3 += 2
return r3, r2, rm, ri
else: # 做刻或做将或有三杯口形状
# 虽然三杯口是三暗刻，但是在最本格的规则里需要区分，因为三暗刻未必是三杯口
childl += cnt
childl[i] -= 3
l3, l2, lm, li = hands_test(childl)
if cnt[i+1] and cnt[i+2]: # 做将或三杯口形状一定有至少一个顺子
childr += cnt
childr[i:i+3] -= 1
r3, r2, rm, ri = hands_test(childr)
if not np.isnan(l3):
lm = [lmi + [str(i+1)*3] for lmi in lm]
l3 += 1
if not np.isnan(r3):
rm = [rmi + [str(i+1)+str(i+2)+str(i+3)] for rmi in rm]
lm += rm
li += ri
return l3, l2, lm, li
elif not np.isnan(r3):
rm = [rmi + [str(i+1)+str(i+2)+str(i+3)] for rmi in rm]
r3 += 1
return r3, r2, rm, ri
return np.nan, np.nan, [[]], 0
# npless version
NOTAMELD = -100
argmax = np.argmax
def hands_test_(cnt):
cntsum = sum(cnt)
if not (cntsum % 3 == 1):
childl = [0] * 9
childr = [0] * 9
l3 = NOTAMELD
r3 = NOTAMELD
# disconnect
for i in range(1, 8):
if cnt[i-1] and sum(cnt[i+1:]) and not cnt[i]:
childl = cnt[:i] + [0] * (9 - i)
childr = [0] * i + cnt[i:]
break
if any(childl):
l3, l2, lm, li = hands_test_(childl)
r3, r2, rm, ri = hands_test_(childr)
lrm = []
for lmi in lm:
for rmi in rm:
lrm.append(lmi + rmi)
return l3 + r3, l2 + r2, lrm, li * ri
# leaf
cntmax = max(cnt)
cntpos = argmax(cnt)
if cntsum == 0:
return 0, 0, [[]], 1
if cntsum == 2 and cntmax == 2:
return 0, 1, [[str(cntpos+1) * 2]], 1
if cntsum == 3:
if cntmax == 3:
return 1, 0, [[str(cntpos+1)*3]], 1
if cntmax == 1:
return 1, 0, [[str(cntpos+1)+str(cntpos+2)+str(cntpos+3)]], 1
# decomposition
i = [j > 0 for j in cnt].index(True)
if i > 6: # only 8 or 9, can not form meld
melds = []
cntcnt = [0] * 5
for tile, num in enumerate(cnt):
cntcnt[num] += 1
if num:
melds.append(str(tile+1) * num)
if not (cntcnt[1] or cntcnt[4] or cntcnt[2] > 1):
return cntcnt[3], cntcnt[2], [melds], 1
elif cnt[i] == 1 or cnt[i] == 4: # 一定有至少一个顺子
if cnt[i+1] and cnt[i+2]:
childl = [k-(i-1<j and j<i+3) for j, k in enumerate(cnt)]
l3, l2, lm, li = hands_test_(childl)
if l3 >= 0:
lm = [lmi + [str(i+1)+str(i+2)+str(i+3)] for lmi in lm]
return l3 + 1, l2, lm, li
elif cnt[i] == 2:
# 做将或有一杯口形状
childl = [k-(i==j)*2 for j, k in enumerate(cnt)]
l3, l2, lm, li = hands_test_(childl)
if cnt[i+1] > 1 and cnt[i+2] > 1:
childr = [k-(i-1<j and j<i+3)*2 for j, k in enumerate(cnt)]
r3, r2, rm, ri = hands_test_(childr)
if l3 >= 0:
lm = [lmi + [str(i+1)*2] for lmi in lm]
l2 += 1
if r3 >= 0:
if l3 == r3 + 2 and l2 == r2:
rm = [rmi + [str(i+1)+str(i+2)+str(i+3)]*2 for rmi in rm]
lm += rm
li += ri
return l3, l2, lm, li
elif r3 >= 0:
rm = [rmi + [str(i+1)+str(i+2)+str(i+3)]*2 for rmi in rm]
r3 += 2
return r3, r2, rm, ri
else: # 做刻或做将或有三杯口形状
# 虽然三杯口是三暗刻，但是在最本格的规则里需要区分，因为三暗刻未必是三杯口
childl = [k-(i==j)*3 for j, k in enumerate(cnt)]
l3, l2, lm, li = hands_test_(childl)
if cnt[i+1] and cnt[i+2]: # 做将或三杯口形状一定有至少一个顺子
childr = [k-(i-1<j and j<i+3) for j, k in enumerate(cnt)]
r3, r2, rm, ri = hands_test_(childr)
if l3 >= 0:
lm = [lmi + [str(i+1)*3] for lmi in lm]
l3 += 1
if r3 >= 0:
rm = [rmi + [str(i+1)+str(i+2)+str(i+3)] for rmi in rm]
lm += rm
li += ri
return l3, l2, lm, li
elif r3 >= 0:
rm = [rmi + [str(i+1)+str(i+2)+str(i+3)] for rmi in rm]
r3 += 1
return r3, r2, rm, ri
return NOTAMELD, NOTAMELD, [[]], 0
print(hands_test_(np.array([0, 3, 1, 2, 1, 1, 0, 0, 0], dtype='int')))