4-rings or 4-squares puzzle: Difference between revisions
add resursion solution
(clean up) |
(add resursion solution) |
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Line 7,211:
4 unique solutions in 3 to 9
[[7, 8, 3, 4, 5, 6, 9], [8, 7, 3, 5, 4, 6, 9], [9, 6, 4, 5, 3, 7, 8], [9, 6, 5, 4, 3, 8, 7]]
2860 non-unique solutions in 0 to 9
</pre>
Recursion solution
<syntaxhighlight lang="v">import math.bits
fn main ( ) {
mut n := 0
mut t := []u8{len:7}
n = solve ( 0, mut t, 1, 7, true )
println ( "$n unique solutions in 1 to 7" )
n = solve ( 0, mut t, 3, 9, true )
println("$n unique solutions in 3 to 9")
n = solve ( 0, mut t, 0, 9, false )
println("$n non-unique solutions in 0 to 9")
}
fn solve ( lvl u8, mut t []u8, low u8, high u8,unique bool ) int {
if lvl==7 {
return int(suma(t) && (!unique || isunique( t )))
}
mut cnt:=0
for v := low; v <= high; v++ {
t[lvl] = v
cnt += solve ( lvl + 1, mut t, low, high, unique )
}
return cnt
}
@[inline]
fn isunique ( t []u8 ) bool {
return bits.ones_count_32(u32(1)<<t[0] | 1<<t[1] | 1<<t[2] | 1<<t[3] | 1<<t[4] | 1<<t[5] | 1<<t[6]) == 7
}
fn suma ( t []u8 ) bool {
s1 := t[0] + t[1]
s2 := t[1] + t[2] + t[3]
s3 := t[3] + t[4] + t[5]
s4 := t[5] + t[6]
return s1 == s2 && s2 == s3 && s3 == s4
}
</syntaxhighlight>
{{out}}
<pre>
8 unique solutions in 1 to 7
4 unique solutions in 3 to 9
2860 non-unique solutions in 0 to 9
</pre>
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