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Page 320 Lei et al. Intell Robot 2022;2(4):313­32 I http://dx.doi.org/10.20517/ir.2022.18

( ) 2
2L L − 2 L
L = D + D = (4)
2L
where is the total length of spiral motion planning. However, the start and end points of the robot coverage
path are different based on the two traditional methods, which cannot meet Requirement 5. The total path
length should include the returning path towards the start point, which could be suboptimal under the two
planning methods.
Lemma 1. The length of returning path under zigzag planning in the workspace (L ) is constrained as

√
( ) 2 ( ) 2
L − L ≤ L ≤ L − L + L − L (5)
√
( ) 2 ( ) 2
Proof. Since L ≥ L > L , we have L − L < L − L + L − L .

The equality holds on the left of Equation (5) when L = 2× ×L for = 1, 2, 3, .... Then, the L = L −L .

The equality is satisfied on the right of Equation (5) when L = (2 × − 1) × L for = 2, 3, 4, .... Then, the
L is
√
2
L = (L − L ) + (L − L ) 2 (6)
This completes the proof of Lemma 1.

Lemma 2. The length of returning path under the spiral planning in the workspace (L ) is given
√ √
L L s − 2L L − L 2L − L + L
2
2
2
( ) + ( ) ≤ L ≤ ( ) + ( ) 2 (7)
2 2 2 2
√
( ) 2 ( ) 2 ( ) 2 ( ) 2 ( ) 2 ( ) 2
Proof. Since L ≥ L > L , −2 < L −L and L < 2L l −L +L , we have L + L −2L
2 2 2 2 2 2
√
( ) 2 ( ) 2
< L − + 2L −L +L .
2 2
The equality holds on the left of Equation (7) when L = 2 × × L for = 1, 2, 3, ... Then, the L is
√
( ) 2 ( ) 2
L L s − 2L
L = + (8)
2 2
The equality holds on the right of Equation (7) when L = (2 × − 1) × L for = 2, 3, 4, ...
Then, the L is
√
( ) 2 ( ) 2
L − L 2L − L + L
L = + (9)
2 2
This completes the proof of Lemma 2.

The robot follows the first two steps of spiral motion planning, then repeats the zigzag motion planning, and
finally returns to the start point. The robot is assumed to run along the feeding and drinking lines without
moving across them. The DCPP finally provides the robot with overall trajectories from the start to end points.

The total path length of DCPP (L ) is calculated through Equation (10):

( ) ( ) L − 2L
L = 2 × L + L − 2L + L − 2L × (10)
L