Lemma 1. For a simple polygon with vertices there exists a diagonal that separates or edges. 0 k Proof. Among the diagonals that separate 6 and more edges choose a diagonal that separates the smallest number of edges. Denote the number of separated edges to be . Set the labels to the vertices as on the picture: . Consider a triangle . s , otherwise a diagonal separates less than but equal or more than edges. The same argument for . Adding these two inequalities get . Lemma 2. Consider a simple polygon with or vertices. Assume that for every the number of periscopic guards needed to cover a -gon is (assumption 1). Claim that for every edge in polygon for there exist covering sets , and . For every edge in polygon for there exists a covering set or . Proof by induction: Base: . Any pentagon can be covered by one periscopic guard placed in any of vertices. For any edge of a hexagon a guard placed in one of two vertices can cover the hole polygon. Induction hypothesis: suppose that for any the lemma is proved. Induction step: a) ( -gon) Count the vertices from to starting from to . Consider a k-gon s m-gon triangle . i. If (or ) then a polygon ­gon has no more than vertices ( ­gon has no more than 0 5t - 1 A B vertices). Than can be covered from any of or vertices. The remaining -gon ( -gon) can be covered by ( ) guards (by assumption 1). Thus, the polygon can be covered by guards and one of these guards can be placed in any of or vertices. ii. If then and . Notice that . Consider the possible variants (the remaining variants are simmetrical to one of these): 1. , Thus -gon can be covered by guards and -gon can be covered by guards (by assumption 1). So can be covered by guards with the last one placed in any of A and B nodes. 2. , Thus -gon can be covered by guards and -gon can be covered by guards (by assumption 1). So can be covered by guards with the last one placed in any of A and B nodes. 3. , , Using induction hypothesis, -gon can be covered by guards with one guard in any of or vertices and -gon can be covered by guards with one guard placed in or . If -gon has a coverage set that contains then can be covered by guards with one guard places in any of and vertices. Otherwise -gon and -gon can be covered by guards, thus can be covered by guards with one guard places in any of and vertices. b) ( -gon) Count the vertices from to starting from to . Consider a k-gon s m-gon triangle . i. If (or ) then a polygon ­gon ( ­gon ) has vertices and can be covered by guards with one guard 0 5t A B placed in any of and nodes (any of B and nodes). Thus can be covered by guards with one guard placed in one of or nodes. ii. If (or ) then a polygon ­gon has no more than vertices ( ­gon has no more than vertices). Than can be covered from one of or vertices. The remaining -gon ( -gon) can be covered by ( ) guards (by assumption 1). Thus, the polygon can be covered by guards and one of these guards can be placed in any of or vertices. iii. If then and . Notice that . Consider the possible variants (the remaining variants are simmetrical to one of these): 1. , Thus -gon can be covered by guards (by assumption 1). And -gon can be covered by guards with one placed in . So can be covered by guards with one placed in one of A or B nodes. 2. , Thus -gon can be covered by guards and -gon can be covered by guards (by assumption 1). So can be covered by guards with the last one placed in or . 3. , , Using induction hypothesis, -gon can be covered by guards with one guard in or and -gon can be covered by guards with one guard placed in or . If -gon (or -gon) has a coverage set with vertex (vertex ) then can be covered by guards with one guard places in or vertices. If they both have coverage sets that contain then -gon and -gon can be covered by guards, thus can be covered by guards with one guard places in or . Theorem. Any simple polygon with vertices can be covered by periscopic guards. Proof by induction: e 0 k Base: . Any pentagon, hexagon, heptagon, octagon or nonagon can be covered by one periscopic guard. l-gon m-gon Induction hypothesis: suppose that for any the theorem is proved. s Induction step: Among the diagonals that separate 6 and more edges choose a diagonal that separates the smallest number of edges. Denote the number of edges . By lemma 1 . Denote a polygon built of vertices. has vertices. a) If , or then is a heptagon, octagon or nonagon. can be covered by guard. A polygon has vertices, thus can be covered by guards. So, can be covered by guards. b) If , consider a triangle . or , otherwise doesn’t separate the smallest number of edges. Without loss of generality . -gon has vertices, -gon has vertices. A polygon has vertices. Consider a polygon -gon , it has vertices. i. If then . So, can be covered by guards. -gon can be covered by guard, thus can be covered by guards. ii. If then a polygon has vertices. It can be covered by guards with one guard placed in vertex or (by lemma 2). 1. If it is placed in then -gon is covered by this guard. -gon can be covered by 1 guard. So, can be covered by guards. 2. If it is placed in then a. if it covers -gon, the rest -gon can be covered by 1 guard. b. if it doesn’t cover -gon, a guard placed in will cover -gon (pentagon) and -gon (by lemma 2). can be covered by guards. c) If , consider a triangle . , otherwise doesn’t separate the smallest number of edges. -gon and -gon have vertices. A polygon has vertices. Consider a polygon -gon , it has vertices. i. If then . So, can be covered by guards. -gon can be covered by guard, thus can be covered by guards. ii. If then a polygon has vertices. It can be covered by guards with one guard placed in any of vertices or (by lemma 2). -gon can be covered by guard placed in or . -gon can be covered by guard placed in or . 1. If -gon is covered by a guard placed in . -gon can be covered by guard, -gon can be covered by 1 guard. So, can be covered by guards. 2. If -gon is covered by a guard placed in . -gon can be covered by guard, -gon can be covered by 1 guard. So, can be covered by guards. 3. If -gon and -gon can be covered by a guard placed in then can be covered by guards. Thus, any -gon can be covered by periscopic guards. .