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| -rw-r--r-- | challenge-257/pokgopun/python/ch-1.py | 60 | ||||
| -rw-r--r-- | challenge-257/pokgopun/python/ch-2.py | 173 |
2 files changed, 233 insertions, 0 deletions
diff --git a/challenge-257/pokgopun/python/ch-1.py b/challenge-257/pokgopun/python/ch-1.py new file mode 100644 index 0000000000..895ac3857b --- /dev/null +++ b/challenge-257/pokgopun/python/ch-1.py @@ -0,0 +1,60 @@ +### https://theweeklychallenge.org/blog/perl-weekly-challenge-257/ +""" + +Task 1: Smaller than Current + +Submitted by: [44]Mohammad Sajid Anwar + __________________________________________________________________ + + You are given a array of integers, @ints. + + Write a script to find out how many integers are smaller than current + i.e. foreach ints[i], count ints[j] < ints[i] where i != j. + +Example 1 + +Input: @ints = (5, 2, 1, 6) +Output: (2, 1, 0, 3) + +For $ints[0] = 5, there are two integers (2,1) smaller than 5. +For $ints[1] = 2, there is one integer (1) smaller than 2. +For $ints[2] = 1, there is none integer smaller than 1. +For $ints[3] = 6, there are three integers (5,2,1) smaller than 6. + +Example 2 + +Input: @ints = (1, 2, 0, 3) +Output: (1, 2, 0, 3) + +Example 3 + +Input: @ints = (0, 1) +Output: (0, 1) + +Example 4 + +Input: @ints = (9, 4, 9, 2) +Output: (2, 1, 2, 0) + +Task 2: Reduced Row Echelon +""" +### solution by pokgopun@gmail.com + +def stc(tup: tuple): + return tuple( + sum(x < y for x in tup) for y in tup + ) + +import unittest + +class TestStc(unittest.TestCase): + def test(self): + for inpt,otpt in { + (5, 2, 1, 6): (2, 1, 0, 3), + (1, 2, 0, 3): (1, 2, 0, 3), + (0, 1): (0, 1), + (9, 4, 9, 2): (2, 1, 2, 0), + }.items(): + self.assertEqual(stc(inpt),otpt) + +unittest.main() diff --git a/challenge-257/pokgopun/python/ch-2.py b/challenge-257/pokgopun/python/ch-2.py new file mode 100644 index 0000000000..31f932500e --- /dev/null +++ b/challenge-257/pokgopun/python/ch-2.py @@ -0,0 +1,173 @@ +### https://theweeklychallenge.org/blog/perl-weekly-challenge-257/ +""" + +Task 2: Reduced Row Echelon + +Submitted by: [45]Ali Moradi + __________________________________________________________________ + + Given a matrix M, check whether the matrix is in reduced row echelon + form. + + A matrix must have the following properties to be in reduced row + echelon form: +1. If a row does not consist entirely of zeros, then the first + nonzero number in the row is a 1. We call this the leading 1. +2. If there are any rows that consist entirely of zeros, then + they are grouped together at the bottom of the matrix. +3. In any two successive rows that do not consist entirely of zeros, + the leading 1 in the lower row occurs farther to the right than + the leading 1 in the higher row. +4. Each column that contains a leading 1 has zeros everywhere else + in that column. + + For example: +[ + [1,0,0,1], + [0,1,0,2], + [0,0,1,3] +] + + The above matrix is in reduced row echelon form since the first nonzero + number in each row is a 1, leading 1s in each successive row are + farther to the right, and above and below each leading 1 there are only + zeros. + + For more information check out this wikipedia [46]article. + +Example 1 + + Input: $M = [ + [1, 1, 0], + [0, 1, 0], + [0, 0, 0] + ] + Output: 0 + +Example 2 + + Input: $M = [ + [0, 1,-2, 0, 1], + [0, 0, 0, 1, 3], + [0, 0, 0, 0, 0], + [0, 0, 0, 0, 0] + ] + Output: 1 + +Example 3 + + Input: $M = [ + [1, 0, 0, 4], + [0, 1, 0, 7], + [0, 0, 1,-1] + ] + Output: 1 + +Example 4 + + Input: $M = [ + [0, 1,-2, 0, 1], + [0, 0, 0, 0, 0], + [0, 0, 0, 1, 3], + [0, 0, 0, 0, 0] + ] + Output: 0 + +Example 5 + + Input: $M = [ + [0, 1, 0], + [1, 0, 0], + [0, 0, 0] + ] + Output: 0 + +Example 6 + + Input: $M = [ + [4, 0, 0, 0], + [0, 1, 0, 7], + [0, 0, 1,-1] + ] + Output: 0 + __________________________________________________________________ + + Last date to submit the solution 23:59 (UK Time) Sunday 25th February + 2024. + __________________________________________________________________ + +SO WHAT DO YOU THINK ? +""" +### solution by pokgopun@gmail.com + +def areAllZeroes(tup: tuple): ### check if all members are zeros + return sum( x!=0 for x in tup )==0 + +def genNZ(tot: tuple): ### remove bottom all-zeroes rows, return others rows in reverse + c = 0 + l = len(tot) + for i in range(l): + if areAllZeroes(tot[l-i-1]): + c += 1 + if c > i: + continue + yield tot[l-i-1] + +def lopos(tup: tuple): ### position of leading 1, return -1 if there is no leadnig 1 + for i in range(len(tup)): + if tup[i]==1: + return i + return -1 + +def isRRE(tot: tuple): ### check if matrix is Reduced Row Echelon + l = len(tot[0]) ### to store previous leading 1 position, as we will do it in backward, initial value will be the row length which is always greater than any position + tot = tuple(genNZ(tot)) ### matrix with bottom all-zeroes rows removed, but the rows will be reversed + rc = len(tot) ### row count, this will be used by list comprehension that check if same columns as the leading 1 are all zeros or not + for t in tot: + lp = lopos(t) ### find out the poistion of leading 1 + if lp < 0 or lp >= l or sum(tot[r][lp]==0 for r in range(rc))!=rc-1: ### False if leading 1 cannot be found or its position is not less than the previous one or other values in its postions are not all zeros + return False + l = lp + return True + +import unittest + +class TestIsRRE(unittest.TestCase): + def test(self): + for inpt,otpt in { + ( + (1, 1, 0), + (0, 1, 0), + (0, 0, 0) + ): 0, + ( + (0, 1,-2, 0, 1), + (0, 0, 0, 1, 3), + (0, 0, 0, 0, 0), + (0, 0, 0, 0, 0) + ): 1, + ( + (1, 0, 0, 4), + (0, 1, 0, 7), + (0, 0, 1,-1) + ): 1, + ( + (0, 1,-2, 0, 1), + (0, 0, 0, 0, 0), + (0, 0, 0, 1, 3), + (0, 0, 0, 0, 0) + ): 0, + ( + (0, 1, 0), + (1, 0, 0), + (0, 0, 0) + ): 0, + ( + (4, 0, 0, 0), + (0, 1, 0, 7), + (0, 0, 1,-1) + ): 0, + }.items(): + self.assertEqual(isRRE(inpt),otpt) + +unittest.main() |