A problem posed by Tony Corso (an APL user).
m comb n computes all the size m combinations
of i.n:
comb=: 4 : 0
c=. 1 {.~ - d=. 1+y-x
z=. i.1 0
for_j. (d-1+y)+/&i.d do. z=. (c#j) ,. z{~;(-c){.&.><i.{.c=. +/\.c end.
)
3 comb 5
0 1 2
0 1 3
0 1 4
0 2 3
0 2 4
0 3 4
1 2 3
1 2 4
1 3 4
2 3 4
Each row of the result specifies a boolean
vector of length n with m 1s. What is the
number of distinct Hamming distances
between the rows of m comb n? (The Hamming
distance between two boolean vectors is +/x~:y.)
n can be up to 200.
bcomb=: i.@] e."1 comb
hd =: ~. @ , @ (+/@:~:"1/~) @ bcomb
hdc =: i.@>: #@hd&> ]
hdc1 =: >:@i.@>: <. ] - <:@i.@>:
If m hdc n is the Hamming distance count
for size m combinations of iota n, it looks
like (m hdc n) = (m+1)>.n-(m+1) , but I
have not yet proven it.
hdc 10
1 2 3 4 5 6 5 4 3 2 1
hdc1 10
1 2 3 4 5 6 5 4 3 2 1
(hdc -: hdc1)"0 i.10
1 1 1 1 1 1 1 1 1 1
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