diff --git a/Lab/Lab#6/lab6.sql b/Lab/Lab#6/lab6.sql
index bd53776..e69de29 100644
--- a/Lab/Lab#6/lab6.sql
+++ b/Lab/Lab#6/lab6.sql
@@ -1,148 +0,0 @@
-WITH RECURSIVE tmp AS (
-  SELECT id, first_name, last_name, manager_id, 1 AS level
-  FROM bd6_employees
-  WHERE manager_id = 1 OR id = 1
-  UNION ALL
-  SELECT e.id, e.first_name, e.last_name, e.manager_id, t.level + 1
-  FROM bd6_employees e
-  JOIN tmp t ON e.manager_id = t.id
-)
-SELECT *
-FROM tmp;
-
-CREATE OR REPLACE PROCEDURE department_names() AS $$ 
-DECLARE d_attrs RECORD; 
-BEGIN
-
-FOR d_attrs IN 
-WITH RECURSIVE tmp AS ( 
-	SELECT 
-	 	id, 
-	 	first_name, 
-	 	last_name, manager_id, 
-	 	1 AS level 
-	 FROM bd6_employees 
-	 WHERE manager_id = 1 or id = 1 
-	UNION ALL 
-	SELECT 
-		e.id, 
-		e.first_name, 
-		e.last_name, 
-		e.manager_id, 
-		t.level + 1 
-	FROM bd6_employees e 
-	JOIN tmp t ON e.manager_id = t.id )
-	SELECT * FROM tmp LOOP RAISE INFO ' % % ',
-
-d_attrs.first_name, d_attrs.last_name; 
-END LOOP; 
-END $$ LANGUAGE plpgsql; 
-CALL department_names();
-
-DECLARE
-    prev_employee NUMBER := 0;
-    mod_salary NUMBER;
-BEGIN
-    FOR tmp IN (
-        SELECT first_name, last_name, salary,
-               ROW_NUMBER() OVER (ORDER BY salary) AS salary_rank
-        FROM employees
-    )
-    LOOP
-        IF emp.salary_rank = 1 THEN
-            modified_salary := FLOOR(emp.salary / 100) * 100; -- округляем до сотен в меньшую сторону
-        ELSE
-            modified_salary := FLOOR((emp.salary + previous_remainder) / 100) * 100;
-            previous_remainder := (emp.salary + previous_remainder) - modified_salary;
-        END IF;
-
-        DBMS_OUTPUT.PUT_LINE('Фамилия: ' || emp.last_name || ', Имя: ' || emp.first_name || ', Модифицированная заработная плата: ' || modified_salary);
-    END LOOP;
-END;
-
-
-CREATE OR REPLACE PROCEDURE f() AS $$ 
-DECLARE d_attrs RECORD; 
-BEGIN
-
-FOR d_attrs IN 
-    WITH RECURSIVE tmp AS ( 
-        SELECT i AS f1, i2 AS f2, i3 AS f3,i4 AS f4,i5 AS f5 
-        FROM generate_series(1,200) i)
-
-SELECT * FROM tmp LOOP RAISE INFO ' % % % % %',
-
-d_attrs.f1, d_attrs.f2, d_attrs.f3, d_attrs.f4, d_attrs.f5; END LOOP; 
-END $$ LANGUAGE plpgsql; 
-CALL f();
-
-WITH RECURSIVE tmp AS (
-  SELECT id, first_name, last_name, manager_id, 1 AS level
-  FROM bd6_employees
-  WHERE manager_id = 1 OR id = 1
-  UNION ALL
-  SELECT e.id, e.first_name, e.last_name, e.manager_id, t.level + 1
-  FROM bd6_employees e
-  JOIN tmp t ON e.manager_id = t.id
-)
-SELECT *
-FROM tmp;
-
-CREATE OR REPLACE PROCEDURE department_names() AS $$ 
-DECLARE d_attrs RECORD; 
-BEGIN
-
-FOR d_attrs IN 
-WITH RECURSIVE tmp AS ( 
-	SELECT 
-	 	id, 
-	 	first_name, 
-	 	last_name, manager_id, 
-	 	1 AS level 
-	 FROM bd6_employees 
-	 WHERE manager_id = 1 or id = 1 
-	UNION ALL 
-	SELECT 
-		e.id, 
-		e.first_name, 
-		e.last_name, 
-		e.manager_id, 
-		t.level + 1 
-	FROM bd6_employees e 
-	JOIN tmp t ON e.manager_id = t.id )
-	SELECT * FROM tmp LOOP RAISE INFO ' % % ',
-
-d_attrs.first_name, d_attrs.last_name; 
-END LOOP; 
-END $$ LANGUAGE plpgsql; 
-CALL department_names();
-
-CREATE OR REPLACE PROCEDURE modified_salary() AS $$ 
-DECLARE
-    prev_employee numeric := 0;
-    mod_salary numeric := (
-	SELECT FLOOR(min(salary_in_euro) / 100) * 100
-	FROM bd6_employees
-		);
-	d_attrs RECORD;
-BEGIN
-    FOR d_attrs IN  WITH tmp AS (
-        SELECT first_name, last_name, salary_in_euro
-        FROM bd6_employees
-		ORDER BY salary_in_euro ASC
-    )
-	SELECT * FROM tmp
-	RAISE INFO ' % % % ',
-		d_attrs.last_name,
-		d_attrs.first_name,
-		mod_salary;
-    LOOP
-		mod_salary := FLOOR((d_attrs.salary_in_euro + prev_employee) / 100) * 100;
-        prev_employee := (d_attrs.salary_in_euro + prev_employee) - mod_salary;
-		RAISE INFO ' % % % ',
-		d_attrs.last_name,
-		d_attrs.first_name,
-		mod_salary;
-    END LOOP;
-END $$ LANGUAGE plpgsql;
-CALL modified_salary();
diff --git a/Lab/Lab#6/lab6_done.sql b/Lab/Lab#6/lab6_done.sql
new file mode 100644
index 0000000..bbce025
--- /dev/null
+++ b/Lab/Lab#6/lab6_done.sql
@@ -0,0 +1,132 @@
+/* Вариант 1 || Работа 6 || Антипенко, Дробышевский */
+
+
+-- Task 1 -- 
+/* a) Напишите запрос, используя конструкцию WITH, выбирающий
+рекурсивно сотрудника с идентификатором 1 и все его подчинённых,
+как прямых, так и подчинённых более низкого ранга*.
+b) Напишите программу на языке PL/SQL, печатающую на экран фамилию
+и имя сотрудника с идентификатором 1 и всех его подчинённых, как
+прямых, так и подчинённых более низкого ранга. */
+
+CREATE OR REPLACE PROCEDURE department_names() AS 
+DECLARE d_attrs RECORD; 
+BEGIN
+
+FOR d_attrs IN WITH RECURSIVE tmp AS ( 
+    SELECT id, first_name, last_name, manager_id, 1 AS level 
+    FROM bd6_employees 
+    WHERE manager_id = 1 or id = 1 
+    UNION ALL 
+    SELECT e.id, e.first_name, e.last_name, e.manager_id, t.level + 1 
+    FROM bd6_employees e JOIN tmp t ON e.manager_id = t.id ) 
+
+SELECT * FROM tmp 
+LOOP 
+    RAISE INFO ' % % ', d_attrs.first_name, d_attrs.last_name; 
+END LOOP; 
+END  LANGUAGE plpgsql; 
+CALL department_names();
+
+
+-- Task 2 --
+/* Напишите программу на языке PL/SQL, выбирающую строки из таблицы
+employees в порядке возрастания заработной платы (salary) и печатающие на
+экран следующие данные: фамилия, имя, модифицированная заработная
+плата. Модифицированная заработная плата получается следующим
+образом: у первого по порядку сотрудника она округляется до сотен в
+меньшую сторону, а у всех последующих сотрудников она сначала
+увеличивается на остаток от округления, полученный от предыдущего
+сотрудника, а затем округляется до сотен в меньшую сторону. */ 
+
+CREATE OR REPLACE PROCEDURE print_employees() RETURNS VOID AS $$
+DECLARE
+current_salary INTEGER;
+m_salary INTEGER;
+mod_salary INTEGER := 0;
+employee_rec RECORD;
+BEGIN
+FOR employee_rec IN (
+   SELECT last_name, first_name, salary_in_euro 
+   FROM bd6_employees 
+   ORDER BY salary_in_euro) 
+LOOP
+   current_salary := employee_rec.salary_in_euro;
+
+   IF mod_salary=0 THEN
+      m_salary := (current_salary / 100) * 100;
+   ELSE
+      m_salary := ((current_salary + mod_salary) / 100) * 100;
+END IF;
+
+RAISE NOTICE 'Employee: % %, Modified Salary: %', employee_rec.last_name, employee_rec.first_name, m_salary;
+mod_salary := current_salary - m_salary;
+END LOOP;
+END $$ LANGUAGE plpgsql;
+
+SELECT print_employees();
+
+-- Task 3 -- 
+/* Напишите программу на языке PL/SQL, удаляющую 10 сотрудников с
+самой маленькой заработной платой. При этом их заработная плата должна
+добавиться 10 сотрудникам с самой большой заработной платой. Причём
+самая маленькая заработная плата должна добавиться к человеку с самой
+большой заработной платой. 2-я с конца заработная плата должна добавиться
+к человеку со второй по размеру заработной платой и т.д. */
+
+CREATE OR REPLACE PROCEDURE update_salary() AS $$
+DECLARE
+   emp_record bd6_employees%ROWTYPE;
+   counter integer := 0;
+   min_salary bd6_employees.salary_in_euro%TYPE;
+   max_salary bd6_employees.salary_in_euro%TYPE;
+BEGIN
+   SELECT MIN(salary_in_euro), MAX(salary_in_euro) INTO min_salary, max_salary FROM bd6_employees;
+   
+   FOR emp_record IN
+   SELECT * FROM (SELECT * FROM bd6_employees ORDER BY salary_in_euro LIMIT 3) AS min_salaries
+   LOOP
+
+      DELETE FROM bd6_employees WHERE id = emp_record.id;
+      counter := counter + 1;
+   END LOOP;
+   FOR emp_record IN
+   SELECT * FROM (SELECT * FROM bd6_employees ORDER BY salary_in_euro DESC LIMIT 3) AS max_salaries
+   LOOP
+      UPDATE bd6_employees SET salary_in_euro = salary_in_euro + min_salary WHERE id = emp_record.id;
+      counter := counter + 1;
+   END LOOP;
+   
+   RAISE NOTICE 'Successfully updated % bd6_employees.', counter;
+END $$ LANGUAGE plpgsql;
+CALL update_salary();
+
+-- Task 4 --
+/* Создайте таблицу spiral с 5 полями f1, f2, f3, f4, f5 – целые числа.
+Напишите программу на языке PL/SQL, заполняющую данную таблицу 1000
+строк по следующему принципу:
+1 2 3 4 5
+10 9 8 7 6
+11 12 13 14 15
+20 19 18 17 16
+21 22 23 24 25 */
+
+CREATE OR REPLACE PROCEDURE f() AS $$ 
+DECLARE d_attrs RECORD; 
+BEGIN
+
+FOR d_attrs IN WITH RECURSIVE tmp AS ( 
+    SELECT 1+i5 AS f1, 2+i5 AS f2, 3+i5 AS f3, 4+i5 AS f4,5+i*5 AS f5 
+    FROM generate_series(0,199) i)
+
+SELECT * FROM tmp 
+LOOP 
+case 
+when ( mod(d_attrs.f1,2)!=0) 
+then RAISE INFO ' % % % % %', d_attrs.f1, d_attrs.f2, d_attrs.f3, d_attrs.f4, d_attrs.f5; 
+when ( mod(d_attrs.f1,2)=0) 
+then RAISE INFO ' % % % % %', d_attrs.f5, d_attrs.f4, d_attrs.f3, d_attrs.f2, d_attrs.f1; 
+end case; 
+END LOOP; 
+END $$ LANGUAGE plpgsql; 
+call f();
\ No newline at end of file
diff --git a/Lab/Lab#6/test3.sql b/Lab/Lab#6/test3.sql
new file mode 100644
index 0000000..35ed016
--- /dev/null
+++ b/Lab/Lab#6/test3.sql
@@ -0,0 +1,26 @@
+CREATE OR REPLACE PROCEDURE update_salary() AS $$
+DECLARE
+   emp_record bd6_employees%ROWTYPE;
+   counter integer := 0;
+   min_salary bd6_employees.salary_in_euro%TYPE;
+   max_salary bd6_employees.salary_in_euro%TYPE;
+BEGIN
+   SELECT MIN(salary_in_euro), MAX(salary_in_euro) INTO min_salary, max_salary FROM bd6_employees;
+   
+   FOR emp_record IN
+   SELECT * FROM (SELECT * FROM bd6_employees ORDER BY salary_in_euro LIMIT 3) AS min_salaries
+   LOOP
+
+      DELETE FROM bd6_employees WHERE id = emp_record.id;
+      counter := counter + 1;
+   END LOOP;
+   FOR emp_record IN
+   SELECT * FROM (SELECT * FROM bd6_employees ORDER BY salary_in_euro DESC LIMIT 3) AS max_salaries
+   LOOP
+      UPDATE bd6_employees SET salary_in_euro = salary_in_euro + min_salary WHERE id = emp_record.id;
+      counter := counter + 1;
+   END LOOP;
+   
+   RAISE NOTICE 'Successfully updated % bd6_employees.', counter;
+END $$ LANGUAGE plpgsql;
+CALL update_salary();
\ No newline at end of file