Division of Labor in Ant Colonies

Transcription

1 Division of Labor in Ant Colonies Alex Cornejo, Anna Dornhaus, Nancy Lynch and Radhika Nagpal October 13, /19

2 What is division of labor? Division of labor is the process by which individual ants in a colony decide which task to perform to ensure the survival of the colony. 2/19

3 What is division of labor? Division of labor is the process by which individual ants in a colony decide which task to perform to ensure the survival of the colony. 2/19

4 What is division of labor? Division of labor is the process by which individual ants in a colony decide which task to perform to ensure the survival of the colony. 2/19

5 What is division of labor? Division of labor is the process by which individual ants in a colony decide which task to perform to ensure the survival of the colony. 2/19

6 What is division of labor? (cont.) Much of the existing work focuses on exploring the correlation between temporal and morphological variations and worker behavior. 3/19

7 What is division of labor? (cont.) Much of the existing work focuses on exploring the correlation between temporal and morphological variations and worker behavior. 3/19

8 What is division of labor? (cont.) Much of the existing work focuses on exploring the correlation between temporal and morphological variations and worker behavior. 3/19

9 Contributions A very general mathematical formulation for the phenomenon of division of labor in ant colonies. 4/19

10 Contributions A very general mathematical formulation for the phenomenon of division of labor in ant colonies. A distributed algorithm that quickly reaches a near optimal task allocation and imposing only minimal assumptions on the capabilities of individual ants. 4/19

11 A model for division of labor 5/19

12 A model for division of labor A set A of ants. 5/19

13 A model for division of labor A set A of ants. A set T of tasks. 5/19

14 A model for division of labor A set A of ants. A set T of tasks. For task τ T, ant a A and time t R 0 : 5/19

15 A model for division of labor A set A of ants. A set T of tasks. For task τ T, ant a A and time t R 0 : d(τ, t) is the total energy demand for task τ at time t. 5/19

16 A model for division of labor A set A of ants. A set T of tasks. For task τ T, ant a A and time t R 0 : d(τ, t) is the total energy demand for task τ at time t. e(τ, a, t) is the energy that can be supplied by ant a if engaged at task τ at time t. 5/19

17 Task Assignment 6/19

18 Task Assignment A task assignment is a function y : A R 0 T { }. 6/19

19 Task Assignment A task assignment is a function y : A R 0 T { }. Given y we define Y (τ, t) as the set of ants assigned to task τ at time t, and I(t) as the set of ants assigned to no task at time t. Y (τ, t) = {a A : y(a, t) = τ} I(t) = {a A : y(a, t) = } 6/19

20 Task Assignment A task assignment is a function y : A R 0 T { }. Given y we define Y (τ, t) as the set of ants assigned to task τ at time t, and I(t) as the set of ants assigned to no task at time t. Y (τ, t) = {a A : y(a, t) = τ} I(t) = {a A : y(a, t) = } Observe that by definition A = Y (τ, t) I(t). 6/19

21 What is a good task allocation? 7/19

22 What is a good task allocation? w(τ, t) = a Y (τ,t) e(τ, a, t) is the energy supplied to task τ at time t. 7/19

23 What is a good task allocation? w(τ, t) = a Y (τ,t) e(τ, a, t) is the energy supplied to task τ at time t. q(τ, t) = d(τ, t) w(τ, t), if negative represents a surplus of energy at task τ, if positive represents a deficit of energy at task τ, and if zero then the task τ is in equilibrium. 7/19

24 What is a good task allocation? w(τ, t) = a Y (τ,t) e(τ, a, t) is the energy supplied to task τ at time t. q(τ, t) = d(τ, t) w(τ, t), if negative represents a surplus of energy at task τ, if positive represents a deficit of energy at task τ, and if zero then the task τ is in equilibrium. An optimal task assignment is one that minimizes τ T q(τ, t)2 7/19

25 Restricted System Model 8/19

26 Restricted System Model We assume e(τ, a, t) = c for all τ T, every ant a A, and every time t R 0. 8/19

27 Restricted System Model We assume e(τ, a, t) = c for all τ T, every ant a A, and every time t R 0. We consider synchronous model where time proceeds in lock-step rounds 1, 2,..., and for the duration of each round i N each ant a A works at task y(a, i). 8/19

28 Capabilities of Individuals 9/19

29 Capabilities of Individuals Recall that q(τ, i) = d(τ, i) w(τ, i) is the difference between the energy demand for task τ at round i and the energy supplied for task τ at round i. 9/19

30 Capabilities of Individuals Recall that q(τ, i) = d(τ, i) w(τ, i) is the difference between the energy demand for task τ at round i and the energy supplied for task τ at round i. For each task τ T ants can sense only a binary feedback function f(τ, i) where: f(τ, i) = { +1 q(τ, i) 0, 1 q(τ, i) < 0. 9/19

31 Algorithm Idea 10/19

32 Algorithm Idea Intuition: Given sufficient memory, ants could use a randomized binary-search-like strategy, guided by the function f, to reach the optimal division of labor. 10/19

33 Algorithm Idea Intuition: Given sufficient memory, ants could use a randomized binary-search-like strategy, guided by the function f, to reach the optimal division of labor. Idea 1: Offload the burden of memory from the individuals to the colony. 10/19

34 Algorithm Idea Intuition: Given sufficient memory, ants could use a randomized binary-search-like strategy, guided by the function f, to reach the optimal division of labor. Idea 1: Offload the burden of memory from the individuals to the colony. Idea 2: Let ants use a response-threshold strategy to pick tasks, and allow them to specialize. 10/19

35 Algorithm Details Each ant stores a task currentt ask T { } and a potential table ϱ[τ]. 11/19

36 Algorithm Details Each ant stores a task currentt ask T { } and a potential table ϱ[τ]. Initially ants start in the Resting state with currentt ask = and a potential of zero for every task τ T, ϱ[τ] = 0. 11/19

37 Algorithm Details Each ant stores a task currentt ask T { } and a potential table ϱ[τ]. Initially ants start in the Resting state with currentt ask = and a potential of zero for every task τ T, ϱ[τ] = 0. Ants can be in one of the five states Resting, FirstReserve, SecondReserve, TempWorker and CoreWorker. 11/19

38 Simplified State Machine +1 prob prob Resting FirstReserve SecondReserve TempWorker CoreWorker threshold 12/19

39 Resting State τ T, ϱ[τ] { 0 if f(τ, i) < 0 min(ϱ[τ] + 1, 3) if f(τ, i) > 0 13/19

40 Resting State { 0 if f(τ, i) < 0 τ T, ϱ[τ] min(ϱ[τ] + 1, 3) if f(τ, i) > 0 candidate {τ T ϱ[τ] = 3} 13/19

41 Resting State { 0 if f(τ, i) < 0 τ T, ϱ[τ] min(ϱ[τ] + 1, 3) if f(τ, i) > 0 candidate {τ T ϱ[τ] = 3} if candidate then 13/19

42 Resting State { 0 if f(τ, i) < 0 τ T, ϱ[τ] min(ϱ[τ] + 1, 3) if f(τ, i) > 0 candidate {τ T ϱ[τ] = 3} if candidate then with probability 1 do 2 τ T, ϱ[τ] 0 currentt ask random candidate task state TempWorker end with end if 13/19

43 Working States case state = TempWorker 14/19

44 Working States case state = TempWorker if f(currentt ask, i) < 0 then state FirstReserve 14/19

45 Working States case state = TempWorker if f(currentt ask, i) < 0 then state FirstReserve else state CoreWorker end if 14/19

46 Working States case state = TempWorker if f(currentt ask, i) < 0 then state FirstReserve else state CoreWorker end if case state = CoreWorker 14/19

47 Working States case state = TempWorker if f(currentt ask, i) < 0 then state FirstReserve else state CoreWorker end if case state = CoreWorker if f(currentt ask, i) < 0 then state TempWorker end if end case 14/19

48 Reserve States case state = FirstReserve 15/19

49 Reserve States case state = FirstReserve if f(currentt ask, i) < 0 then state Resting else 15/19

50 Reserve States case state = FirstReserve if f(currentt ask, i) < 0 then state Resting else with probability 1 2 do state TempWorker otherwise state SecondReserve end with end if 15/19

51 Reserve States case state = FirstReserve if f(currentt ask, i) < 0 then state Resting else with probability 1 2 do state TempWorker otherwise state SecondReserve end with end if case state = SecondReserve 15/19

52 Reserve States case state = FirstReserve if f(currentt ask, i) < 0 then state Resting else with probability 1 2 do state TempWorker otherwise state SecondReserve end with end if case state = SecondReserve if f(currentt ask, i) < 0 then state Resting 15/19

53 Reserve States case state = FirstReserve if f(currentt ask, i) < 0 then state Resting else with probability 1 2 do state TempWorker otherwise state SecondReserve end with end if case state = SecondReserve if f(currentt ask, i) < 0 then state Resting else state TempWorker end if 15/19

54 Analysis Outline 16/19

55 Analysis Outline Lemma 1: If during an interval tasks are not too oversatisfied, the probability there is a task with a deficit is exponentially small in the interval length. 16/19

56 Analysis Outline Lemma 1: If during an interval tasks are not too oversatisfied, the probability there is a task with a deficit is exponentially small in the interval length. ants 0 time 16/19

57 Analysis Outline Lemma 1: If during an interval tasks are not too oversatisfied, the probability there is a task with a deficit is exponentially small in the interval length. ants 0 time 16/19

58 Analysis Outline Lemma 1: If during an interval tasks are not too oversatisfied, the probability there is a task with a deficit is exponentially small in the interval length. ants 0 time 16/19

59 Analysis Outline Lemma 1: If during an interval tasks are not too oversatisfied, the probability there is a task with a deficit is exponentially small in the interval length. ants 0 time 16/19

60 Analysis Outline Lemma 1: If during an interval tasks are not too oversatisfied, the probability there is a task with a deficit is exponentially small in the interval length. ants 0 time 16/19

61 Analysis Outline Lemma 2: Once a task transitions from having a deficit to a surplus, it will keep oscillating every two or three rounds; moreover, with constant probability the number of ants involved in the oscillations is reduced by a constant fraction. 17/19

62 Analysis Outline Lemma 2: Once a task transitions from having a deficit to a surplus, it will keep oscillating every two or three rounds; moreover, with constant probability the number of ants involved in the oscillations is reduced by a constant fraction. ants 0 time 17/19

63 Analysis Outline Lemma 2: Once a task transitions from having a deficit to a surplus, it will keep oscillating every two or three rounds; moreover, with constant probability the number of ants involved in the oscillations is reduced by a constant fraction. ants 0 time 17/19

64 Analysis Outline Lemma 2: Once a task transitions from having a deficit to a surplus, it will keep oscillating every two or three rounds; moreover, with constant probability the number of ants involved in the oscillations is reduced by a constant fraction. ants 0 time 17/19

65 Analysis Outline Lemma 2: Once a task transitions from having a deficit to a surplus, it will keep oscillating every two or three rounds; moreover, with constant probability the number of ants involved in the oscillations is reduced by a constant fraction. ants 0 time 17/19

66 Analysis Outline Lemma 2: Once a task transitions from having a deficit to a surplus, it will keep oscillating every two or three rounds; moreover, with constant probability the number of ants involved in the oscillations is reduced by a constant fraction. ants 0 time 17/19

67 Analysis Outline Theorem: After O(log n) rounds, with high probability, ants reach an optimal task allocation. ants 0 time 18/19

68 19/19 Questions?