回答編集履歴
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修正と検証コード追記
test
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@@ -1,7 +1,181 @@
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- 1～K*(K-1) (= 1～12)の数値から相異なるK(=4)個を選ぶ。
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- 上記のK個の相異なる2つの差がすべて =
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- 上記のK個の相異なる2つの差がすべて( =4!=12種類)異なる。
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ものを選べばいけると思います。
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a...dの制約条件がうまく決めきれませんでしたが
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以下pythonコードで全探索と上記の手法とで結果が一致したので正しいと思います。
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K=2で2個, K=3で12個, K=4で96個, K=5で240個
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```Python
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import itertools
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K = 4 # 正方行列 K * K
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NUM_CNT = K*(K-1) # 対角線以外を埋める数値の個数
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NUMS = [i+1 for i in range(NUM_CNT)] # a...nの走査範囲値 1...NUM_CNT (= K*(K-1) )
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#NUMS = [i for i in range(NUM_CNT+1)]
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# 循環数を返す -1 = NUM_CNT, -2 = NUM_CNT-1, ...
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def circleNum( n):
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if n < 0:
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n += NUM_CNT+1
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return n
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# 数値リストから行列を作成
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def getMat( nums):
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return [[circleNum( nums[i]-nums[j]) for j in range(K)] for i in range(K)]
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# 重複判定 : 行列から
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def isMatDup( mat):
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dup = set()
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for i,j in itertools.product([k for k in range(K)], repeat=2): # 直積
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if i != j:
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if mat[i][j] in dup:
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return True
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dup.add(mat[i][j])
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return False
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# 重複判定 : 数値リストから
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def isNumDup( nums):
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dup = set()
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for n in itertools.permutations(nums, 2): # 順列
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diff = circleNum(n[0] - n[1])
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if diff in dup:
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return True
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dup.add(diff)
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return False
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# 計算 : 全探索＋行列から判定
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def calc1( path):
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print(path)
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with open(path, 'w') as f:
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dup = set() # 解（行列）の重複チェック用
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lopCnt,dupCnt,okCnt = 0,0,0
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for nums in itertools.product(NUMS, repeat=K): # 直積
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mat = getMat( nums)
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if not isMatDup(mat):
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key = str(mat)
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if key not in dup:
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f.write( 'nums[%s] mat[%s]\n'%(nums,mat))
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okCnt += 1
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dup.add(key)
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else:
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dupCnt += 1
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lopCnt += 1
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f.write( 'lop[%d] dup[%d] ok[%d]\n'%(lopCnt,dupCnt,okCnt))
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# 計算 : 順列＋数値リストから判定
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def calc2( path):
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print(path)
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with open(path, 'w') as f:
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dup = set() # 解（行列）の重複チェック用
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lopCnt,dupCnt,okCnt = 0,0,0
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for nums in itertools.permutations(NUMS, K): # 順列
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if not isNumDup(nums):
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mat = getMat(nums)
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key = str(mat)
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if key not in dup:
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f.write( 'nums[%s] mat[%s]\n'%(nums,mat))
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okCnt += 1
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dup.add(key)
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else:
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dupCnt += 1
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lopCnt += 1
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f.write( 'lop[%d] dup[%d] ok[%d]\n'%(lopCnt,dupCnt,okCnt))
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if __name__ == '__main__':
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calc1('calc1.txt')
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calc2('calc2.txt')
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```
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