Discussion on Beautiful Quadruples Challenge

1 year ago+ 0 comments

I found a way mathmateically to calculate the number of ALL combinations where order does not matter. I meant to subtract the number of quadruples where each number XOR = 0 from that total, but it turns out it isn't easy to find that.

def sum_range(start,end):
    #integer skips
    start,end=sorted([start,end])
    return (end-start+1)*(start+end)//2

def findTotal(a,b,c,d):
    #because a,b,c,d is sorted, d is max
    base=d-c+1
    same_frequency_as_bass=d-(b-1)
    same_frequency_sum=sum_range(base,same_frequency_as_bass)*sum_range(b-a+1,b)
  
    staggered_sum=0
    freq=0
    for i in range(0,b-1): #remember that range is [).
        freq+=min([i+1,a])
        staggered_sum+=(d-i)*freq
  
    return same_frequency_sum+staggered_sum

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