Problem 970

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Let f(m,n) represent the minimum number of steps to jump from the origin to (m,n).
Obviously,
f(±m,±n)=f(±n,±m).
Therefore, we only need to consider the case where m≥n≥0, as shown in Figure 2.
Note that for the knight's move, there is a special path
to reach each grid point on a certain line, this line divides the region
into two parts, A and B:
Region B: 2n>m>n>0.
For convenience, let T(1,2) represent a single step from any integer point
(x,y) to (x+1,y+2), and similarly defined,
Thus, f(m,n)=n+2k+1.
(iii) If r=2, then
(m,n)=(2n+4k+2,n).
From [m+1/2]=n+2k+1, we know this number has a different parity from m.
Therefore, f(m,n)≥n+2k+2.
On the other hand, take n+k steps T(2,1), k steps T(2,-1),
1 step T(1,2), 1 step T(1,-2). Then
(n+k)(2,1)+k(2,-1)+(1,2)+(1,-2)
= (2n+4k+2,n)=(m,n).
Thus, f(m,n)≤(n+k)+k+1+1=n+2k+2.
Therefore, f(m,n)=n+2k+2.
(iv) If r=3, then
the parity is different, so,
f(m,n)≥n+2k+3.
(n+k+2)(2,1)+k(2,-1)+(-1,-2)
= (2n+4k+3,n)=(m,n).
Thus, f(m,n)≤(n+k+2)+k+1
=n+2k+3.
Therefore, f(m,n)=n+2k+3.
Combining the above formulas, when (m,n)∈A, i.e., m≥2n
≥0, we have
f(1,0)=3;
and when (m,n)≠(1,0),
f(m,n)=m-n-2[(m-2n)/4].
f(m,n)=n+2k+r
=n+2k+(m-2n-4k)
(iii) If m+n=3k+2, since (m+n)/3>k,
Therefore, jumping to (m,n) requires at least
But since k+1 has a different parity from m+n (note
(m+n)-(k+1)=2k+1 is odd), then
f(m,n)≥k+2.
On the other hand, take m-1-k steps T(2,1), n-1-k
(2), 1 T T(2,-1), 1 step T(-1,2). Then
(m-1-k)(2,1)+(n-1-k)(1,2)+
=(2m+n-3k-2,2n+m-3k-2)
=(m,n),
n-1-k=(2n-m-1)/3≥0.
Thus, f(m,n)
≤(m-1-k)+(n-1-k+1+1)
=m+n-2k=k+2.
Combining the formulas, when (m,n)∈B, i.e., 2n>m
n>0, f(2,2)=4.
And when (m,n)≠(2,2),
f(m,n)=m+n-2[(m+n)/3].
Combining all the above cases, when m+n=3k+r,
f(m,n)=k+r (r=0,1,2).
And k=[(m+n)/3], r=m+n-3k, then
f(m,n)=k+(m+n-3k)=m+n-2k
=m+n-2[(m+n)/3].
Define T(2,1), T(2,-1), T(-1,2), etc.
For any point (m,n), if m+n is even, then the
point (m,n) is called an even point; if m+n is odd, then the point
(m,n) is called an odd point. Since the knight changes the sum of its
coordinates by an odd number (1 or 3) with each move,
the parity of the landing point changes with each move.
Therefore, to jump from the origin to (m,n) along any path, if
the point is odd, then an odd number of steps is required.
Thus, f(m,n) has the same parity as m+n.
(1) If (m,n) is in region A, i.e., m≥2n≥0,
Let m-2n=4k+r (k>0, 0≤r≤3). Since the knight needs at least [m+1/2] steps, where [x] represents the greatest integer not exceeding the real number x.
(i) If r=0, then
(m,n)=(2n+4k,n),
f(m,n)≥[m+1/2]=n+2k.
On the other hand, take n steps T(2,1), k steps T(2,-1).
Since (n+k)(2,1)+k(2,-1)
i.e., it only takes n+2k steps to jump to (m,n), so,
f(m,n)=n+2k
(ii) If r=1, then
f(m,n)≥[m+1/2]=n+2k+1.
Also, it is easy to know f(1,0)=3.
Thus, n+k-1≥0.
Take n+k-1 steps T(2,1), k+1 steps T(2,-1)
1 step T(1,2).
By (n+k-1)(2,1)+(k+1)(2,-1)+(1,2)
=(2n+4k+1,n)=(m,n),
Thus, f(m,n)≤(n+k-1)+(k+1)+1
=n+2k+1.
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