Avoidance Coupling Ohad N. Feldheim Institute of Mathematics and - PowerPoint PPT Presentation

Simple Examples - I - Tori On a 2 n cycle graph - maximal SAC is a least of size n . Lower bound: move all agents in 1 the same direction in each round. Upper bound: neighbours must keep moving in the same direction. This is a minimum-entropy 0 2 coupling. The same principle works for Z d / n Z d , Ohad N. Feldheim Avoidance Coupling

Simple Examples - II - loop triangle On a K ∗ 3 - maximal SAC is of size 2. Ohad N. Feldheim Avoidance Coupling

Simple Examples - II - loop triangle On a K ∗ 3 - maximal SAC is of size 2. Strategy: a 1 makes a random walk. 0 1 If a 1 stays - a 0 moves, otherwise - a 0 follows its only viable option. 0 1 0 1 Ohad N. Feldheim Avoidance Coupling

Simple Examples - II - loop triangle On a K ∗ 3 - maximal SAC is of size 2. Strategy: a 1 makes a random walk. 0 1 If a 1 stays - a 0 moves, otherwise - a 0 follows its only viable option. 0 1 0 1 This walk is: minimum-entropy coupling, invariant to time reversal. Ohad N. Feldheim Avoidance Coupling

Results for K n , K ∗ n Theorem (Angel, Holroyd, Martin, Wilson & Winkler) Let n = 2 d +1 . There exists a Markovian, minimum-entropy SAC of 2 d agents on K ∗ n , K ∗ n +1 and K n +1 . Ohad N. Feldheim Avoidance Coupling

Results for K n , K ∗ n Theorem (Angel, Holroyd, Martin, Wilson & Winkler) Let n = 2 d +1 . There exists a Markovian, minimum-entropy SAC of 2 d agents on K ∗ n , K ∗ n +1 and K n +1 . AC ( G ) := maximum SAC on G . Theorem AC ( K ∗ n ) is monotone in n . (AHMWW) AC ( K n ) is monotone in n . (F) Ohad N. Feldheim Avoidance Coupling

Results for K n , K ∗ n Theorem (Angel, Holroyd, Martin, Wilson & Winkler) Let n = 2 d +1 . There exists a Markovian, minimum-entropy SAC of 2 d agents on K ∗ n , K ∗ n +1 and K n +1 . AC ( G ) := maximum SAC on G . Theorem AC ( K ∗ n ) is monotone in n . (AHMWW) AC ( K n ) is monotone in n . (F) Corollary There exists a SAC of ⌈ n / 4 ⌉ agents on both K ∗ n and K n . Ohad N. Feldheim Avoidance Coupling

Results for K n , K ∗ n Theorem (Angel, Holroyd, Martin, Wilson & Winkler) Let n = 2 d +1 . There exists a Markovian, minimum-entropy SAC of 2 d agents on K ∗ n , K ∗ n +1 and K n +1 . AC ( G ) := maximum SAC on G . Theorem [Bernoulli SAC] AC ( K ∗ n ) is monotone in n . (AHMWW) [POSAC] AC ( K n ) is monotone in n . (F) Corollary There exists a SAC of ⌈ n / 4 ⌉ agents on both K ∗ n and K n . Ohad N. Feldheim Avoidance Coupling

Results for K n , K ∗ n Theorem (Angel, Holroyd, Martin, Wilson & Winkler) Let n = 2 d +1 . There exists a Markovian, minimum-entropy SAC of 2 d agents on K ∗ n , K ∗ n +1 and K n +1 . AC ( G ) := maximum SAC on G . Theorem [Bernoulli SAC] AC ( K ∗ n ) is monotone in n . (AHMWW) [POSAC] AC ( K n ) is monotone in n . (F) Corollary There exists a SAC of ⌈ n / 4 ⌉ agents on both K ∗ n and K n . These couplings are hidden Markovian. Ohad N. Feldheim Avoidance Coupling

Constructing an Avoidance Coupling on K d 2 +1 1 0 3 2

2 d agents SAC on K 2 d +1 +1 Write n = 2 d , V = { 0 , . . . , 2 n } , Ohad N. Feldheim Avoidance Coupling

2 d agents SAC on K 2 d +1 +1 Write n = 2 d , V = { 0 , . . . , 2 n } , assume WLOG a n ( t − 1) = 0. Ohad N. Feldheim Avoidance Coupling

2 d agents SAC on K 2 d +1 +1 Write n = 2 d , V = { 0 , . . . , 2 n } , assume WLOG a n ( t − 1) = 0. Let m < n and write m := � m i 2 i . We now define a m ( t ) | a n ( t − 1). Ohad N. Feldheim Avoidance Coupling

2 d agents SAC on K 2 d +1 +1 Write n = 2 d , V = { 0 , . . . , 2 n } , assume WLOG a n ( t − 1) = 0. Let m < n and write m := � m i 2 i . We now define a m ( t ) | a n ( t − 1). t . . . ε d − 1 Let ε 0 be i.i.d. uniform { -1,1 } variables, and let δ t be an t independent uniform { 0 , 1 } variable. Ohad N. Feldheim Avoidance Coupling

2 d agents SAC on K 2 d +1 +1 Write n = 2 d , V = { 0 , . . . , 2 n } , assume WLOG a n ( t − 1) = 0. Let m < n and write m := � m i 2 i . We now define a m ( t ) | a n ( t − 1). t . . . ε d − 1 Let ε 0 be i.i.d. uniform { -1,1 } variables, and let δ t be an t independent uniform { 0 , 1 } variable. we set d − 1 a m ( t ) = 2 d + δ t + � m i ε i t 2 i , i =0 1 0 2 3 Ohad N. Feldheim Avoidance Coupling

2 d agents SAC on K 2 d +1 +1 Write n = 2 d , V = { 0 , . . . , 2 n } , assume WLOG a n ( t − 1) = 0. Let m < n and write m := � m i 2 i . We now define a m ( t ) | a n ( t − 1). t . . . ε d − 1 Let ε 0 be i.i.d. uniform { -1,1 } variables, and let δ t be an t independent uniform { 0 , 1 } variable. we set d − 1 a m ( t ) = 2 d + δ t + � m i ε i t 2 i , i =0 δ t determines a 0 ( t ) = a 00 ( t ). 1 0 2 3 Ohad N. Feldheim Avoidance Coupling

2 d agents SAC on K 2 d +1 +1 Write n = 2 d , V = { 0 , . . . , 2 n } , assume WLOG a n ( t − 1) = 0. Let m < n and write m := � m i 2 i . We now define a m ( t ) | a n ( t − 1). t . . . ε d − 1 Let ε 0 be i.i.d. uniform { -1,1 } variables, and let δ t be an t independent uniform { 0 , 1 } variable. we set d − 1 a m ( t ) = 2 d + δ t + � m i ε i t 2 i , i =0 0 δ t determines a 0 ( t ) = a 00 ( t ). Then ε 0 t determines a 1 ( t ) = a 01 ( t ), 1 2 3 Ohad N. Feldheim Avoidance Coupling

2 d agents SAC on K 2 d +1 +1 Write n = 2 d , V = { 0 , . . . , 2 n } , assume WLOG a n ( t − 1) = 0. Let m < n and write m := � m i 2 i . We now define a m ( t ) | a n ( t − 1). t . . . ε d − 1 Let ε 0 be i.i.d. uniform { -1,1 } variables, and let δ t be an t independent uniform { 0 , 1 } variable. we set d − 1 a m ( t ) = 2 d + δ t + � m i ε i t 2 i , i =0 0 1 δ t determines a 0 ( t ) = a 00 ( t ). Then ε 0 t determines a 1 ( t ) = a 01 ( t ), and ε 1 t determines a 2 ( t ) = a 10 ( t ). 2 3 Ohad N. Feldheim Avoidance Coupling

2 d agents SAC on K 2 d +1 +1 Write n = 2 d , V = { 0 , . . . , 2 n } , assume WLOG a n ( t − 1) = 0. Let m < n and write m := � m i 2 i . We now define a m ( t ) | a n ( t − 1). t . . . ε d − 1 Let ε 0 be i.i.d. uniform { -1,1 } variables, and let δ t be an t independent uniform { 0 , 1 } variable. we set d − 1 a m ( t ) = 2 d + δ t + � m i ε i t 2 i , i =0 0 1 δ t determines a 0 ( t ) = a 00 ( t ). Then ε 0 t determines a 1 ( t ) = a 01 ( t ), and ε 1 t determines a 2 ( t ) = a 10 ( t ). 2 a 3 ( t ) = a 11 ( t ) is fixed by ε 0 t , ε 1 t . 3 Ohad N. Feldheim Avoidance Coupling

2 d agents SAC on K 2 d +1 +1 Write n = 2 d , V = { 0 , . . . , 2 n } , assume WLOG a n ( t − 1) = 0. Let m < n and write m := � m i 2 i . We now define a m ( t ) | a n ( t − 1). t . . . ε d − 1 Let ε 0 be i.i.d. uniform { -1,1 } variables, and let δ t be an t independent uniform { 0 , 1 } variable. we set d − 1 a m ( t ) = 2 d + δ t + � m i ε i t 2 i , i =0 0 1 3 δ t determines a 0 ( t ) = a 00 ( t ). Then ε 0 t determines a 1 ( t ) = a 01 ( t ), and ε 1 t determines a 2 ( t ) = a 10 ( t ). 2 a 3 ( t ) = a 11 ( t ) is fixed by ε 0 t , ε 1 t . Ohad N. Feldheim Avoidance Coupling

2 d agents SAC on K 2 d +1 +1 - cont. n = 2 d , V = { 0 , . . . , 2 d +1 } , a n ( t − 1) = 0. m i := i -th binary digit of m . t . . . ε d − 1 ε 0 uniform { -1,1 } , δ t uniform { 0 , 1 } . t d − 1 a m ( t ) = 2 d + δ t + � m i ε i t 2 i , i =0 We need to show: No collision in the same round Each agent performs simple random walk No collisions between rounds Ohad N. Feldheim Avoidance Coupling

2 d agents SAC on K 2 d +1 +1 - cont. n = 2 d , V = { 0 , . . . , 2 d +1 } , a n ( t − 1) = 0. m i := i -th binary digit of m . t . . . ε d − 1 ε 0 uniform { -1,1 } , δ t uniform { 0 , 1 } . t d − 1 a m ( t ) = 2 d + δ t + � m i ε i t 2 i , i =0 We need to show: No collision in the same round - straightforward. Each agent performs simple random walk No collisions between rounds Ohad N. Feldheim Avoidance Coupling

2 d agents SAC on K 2 d +1 +1 - cont. n = 2 d , V = { 0 , . . . , 2 d +1 } , a n ( t − 1) = 0. m i := i -th binary digit of m . t . . . ε d − 1 ε 0 uniform { -1,1 } , δ t uniform { 0 , 1 } . t d − 1 a m ( t ) = 2 d + δ t + � m i ε i t 2 i , i =0 We need to show: No collision in the same round - straightforward. Each agent performs simple random walk - we show this first. No collisions between rounds Ohad N. Feldheim Avoidance Coupling

2 d agents SAC on K 2 d +1 +1 - cont. n = 2 d , V = { 0 , . . . , 2 d +1 } , a n ( t − 1) = 0. m i := i -th binary digit of m . t . . . ε d − 1 ε 0 uniform { -1,1 } , δ t uniform { 0 , 1 } . t d − 1 a m ( t ) = 2 d + δ t + � m i ε i t 2 i , i =0 a m ( t − 1) ≡ a n ( t − 2) + 2 d + δ t − 1 + � d − 1 i =0 m i ε i t − 1 2 i , Ohad N. Feldheim Avoidance Coupling

2 d agents SAC on K 2 d +1 +1 - cont. n = 2 d , V = { 0 , . . . , 2 d +1 } , a n ( t − 1) = 0. m i := i -th binary digit of m . t . . . ε d − 1 ε 0 uniform { -1,1 } , δ t uniform { 0 , 1 } . t d − 1 a m ( t ) = 2 d + δ t + � m i ε i t 2 i , i =0 a m ( t − 1) ≡ a n ( t − 2) + 2 d + δ t − 1 + � d − 1 i =0 m i ε i t − 1 2 i , a n ( t − 1) ≡ a n ( t − 2) + 2 d + δ t − 1 + � d − 1 i =0 1 · ε i t − 1 2 i Ohad N. Feldheim Avoidance Coupling

2 d agents SAC on K 2 d +1 +1 - cont. n = 2 d , V = { 0 , . . . , 2 d +1 } , a n ( t − 1) = 0. m i := i -th binary digit of m . t . . . ε d − 1 ε 0 uniform { -1,1 } , δ t uniform { 0 , 1 } . t d − 1 a m ( t ) = 2 d + δ t + � m i ε i t 2 i , i =0 a m ( t − 1) ≡ a n ( t − 2) + 2 d + δ t − 1 + � d − 1 i =0 m i ε i t − 1 2 i , a n ( t − 1) ≡ a n ( t − 2) + 2 d + δ t − 1 + � d − 1 t − 1 2 i = 0 , i =0 1 · ε i Ohad N. Feldheim Avoidance Coupling

2 d agents SAC on K 2 d +1 +1 - cont. n = 2 d , V = { 0 , . . . , 2 d +1 } , a n ( t − 1) = 0. m i := i -th binary digit of m . t . . . ε d − 1 ε 0 uniform { -1,1 } , δ t uniform { 0 , 1 } . t d − 1 a m ( t ) = 2 d + δ t + � m i ε i t 2 i , i =0 a m ( t − 1) ≡ a n ( t − 2) + 2 d + δ t − 1 + � d − 1 i =0 m i ε i t − 1 2 i , a n ( t − 1) ≡ a n ( t − 2) + 2 d + δ t − 1 + � d − 1 t − 1 2 i = 0 , i =0 1 · ε i i =0 ( m i − 1) ε i a m ( t − 1) ≡ � d − 1 t − 1 2 i . Ohad N. Feldheim Avoidance Coupling

2 d agents SAC on K 2 d +1 +1 - cont. n = 2 d , V = { 0 , . . . , 2 d +1 } , a n ( t − 1) = 0. m i := i -th binary digit of m . t . . . ε d − 1 ε 0 uniform { -1,1 } , δ t uniform { 0 , 1 } . t d − 1 a m ( t ) = 2 d + δ t + � m i ε i t 2 i , i =0 a m ( t − 1) ≡ a n ( t − 2) + 2 d + δ t − 1 + � d − 1 i =0 m i ε i t − 1 2 i , a n ( t − 1) ≡ a n ( t − 2) + 2 d + δ t − 1 + � d − 1 t − 1 2 i = 0 , i =0 1 · ε i i =0 ( m i − 1) ε i a m ( t − 1) ≡ � d − 1 t − 1 2 i . Thus a m ( t ) − a m ( t − 1) ≡ 2 d + δ t + � d − 1 t 2 i + � d − 1 i =0 m i ε i i =1 (1 − m i t − 1 ) ε i t − 1 2 i Ohad N. Feldheim Avoidance Coupling

2 d agents SAC on K 2 d +1 +1 - cont. n = 2 d , V = { 0 , . . . , 2 d +1 } , a n ( t − 1) = 0. m i := i -th binary digit of m . t . . . ε d − 1 ε 0 uniform { -1,1 } , δ t uniform { 0 , 1 } . t d − 1 a m ( t ) = 2 d + δ t + � m i ε i t 2 i , i =0 a m ( t − 1) ≡ a n ( t − 2) + 2 d + δ t − 1 + � d − 1 i =0 m i ε i t − 1 2 i , a n ( t − 1) ≡ a n ( t − 2) + 2 d + δ t − 1 + � d − 1 t − 1 2 i = 0 , i =0 1 · ε i i =0 ( m i − 1) ε i a m ( t − 1) ≡ � d − 1 t − 1 2 i . Thus a m ( t ) − a m ( t − 1) ≡ 2 d + δ t + � d − 1 t 2 i + � d − 1 i =0 m i ε i i =1 (1 − m i t − 1 ) ε i t − 1 2 i ≡ 2 d + δ t + � d − 1 i =0 b i ( t )2 i where b i are i.i.d. Bernoulli {− 1 , 1 } , Ohad N. Feldheim Avoidance Coupling

2 d agents SAC on K 2 d +1 +1 - cont. n = 2 d , V = { 0 , . . . , 2 d +1 } , a n ( t − 1) = 0. m i := i -th binary digit of m . t . . . ε d − 1 ε 0 uniform { -1,1 } , δ t uniform { 0 , 1 } . t d − 1 a m ( t ) = 2 d + δ t + � m i ε i t 2 i , i =0 a m ( t − 1) ≡ a n ( t − 2) + 2 d + δ t − 1 + � d − 1 i =0 m i ε i t − 1 2 i , a n ( t − 1) ≡ a n ( t − 2) + 2 d + δ t − 1 + � d − 1 t − 1 2 i = 0 , i =0 1 · ε i i =0 ( m i − 1) ε i a m ( t − 1) ≡ � d − 1 t − 1 2 i . Thus a m ( t ) − a m ( t − 1) ≡ 2 d + δ t + � d − 1 t 2 i + � d − 1 i =0 m i ε i i =1 (1 − m i t − 1 ) ε i t − 1 2 i ≡ 2 d + δ t + � d − 1 i =0 b i ( t )2 i where b i are i.i.d. Bernoulli {− 1 , 1 } , ≡ Unif { 1 . . . 2 d +1 } . Ohad N. Feldheim Avoidance Coupling

2 d agents SAC on K 2 d +1 +1 - cont. n = 2 d , V = { 0 , . . . , 2 d +1 } , a n ( t − 1) = 0. m i := i -th binary digit of m . t . . . ε d − 1 ε 0 uniform { -1,1 } , δ t uniform { 0 , 1 } . t d − 1 d − 1 a m ( t ) = 2 d + δ t + � � m i ε i t 2 i , ( m i − 1) ε i t − 1 2 i . a m ( t − 1) ≡ i =1 i =1 We need to show: No collision in the same round - Done. Each agent performs simple random walk - Done. No collisions between rounds Ohad N. Feldheim Avoidance Coupling

2 d agents SAC on K 2 d +1 +1 - cont. n = 2 d , V = { 0 , . . . , 2 d +1 } , a n ( t − 1) = 0. m i := i -th binary digit of m . t . . . ε d − 1 ε 0 uniform { -1,1 } , δ t uniform { 0 , 1 } . t d − 1 d − 1 a m ( t ) = 2 d + δ t + � � m i ε i t 2 i , ( m i − 1) ε i t − 1 2 i . a m ( t − 1) ≡ i =1 i =1 Ohad N. Feldheim Avoidance Coupling

2 d agents SAC on K 2 d +1 +1 - cont. n = 2 d , V = { 0 , . . . , 2 d +1 } , a n ( t − 1) = 0. m i := i -th binary digit of m . t . . . ε d − 1 ε 0 uniform { -1,1 } , δ t uniform { 0 , 1 } . t d − 1 d − 1 a m ( t ) = 2 d + δ t + � � m i ε i t 2 i , ( m i − 1) ε i t − 1 2 i . a m ( t − 1) ≡ i =1 i =1 Let m < q and recall that: a q ( t − 1) ≡ � d − 1 i =1 ( q i − 1) ε i t − 1 2 i , Ohad N. Feldheim Avoidance Coupling

2 d agents SAC on K 2 d +1 +1 - cont. n = 2 d , V = { 0 , . . . , 2 d +1 } , a n ( t − 1) = 0. m i := i -th binary digit of m . t . . . ε d − 1 ε 0 uniform { -1,1 } , δ t uniform { 0 , 1 } . t d − 1 d − 1 a m ( t ) = 2 d + δ t + � � m i ε i t 2 i , ( m i − 1) ε i t − 1 2 i . a m ( t − 1) ≡ i =1 i =1 Let m < q and recall that: a q ( t − 1) ≡ � d − 1 i =1 ( q i − 1) ε i t − 1 2 i , and thus: a m ( t ) − a q ( t − 1) ≡ 2 d + δ t + � d − 1 � m i ε i t + (1 − q i ) ε i � 2 i t − 1 i =1 Ohad N. Feldheim Avoidance Coupling

2 d agents SAC on K 2 d +1 +1 - cont. n = 2 d , V = { 0 , . . . , 2 d +1 } , a n ( t − 1) = 0. m i := i -th binary digit of m . t . . . ε d − 1 ε 0 uniform { -1,1 } , δ t uniform { 0 , 1 } . t d − 1 d − 1 a m ( t ) = 2 d + δ t + � � m i ε i t 2 i , ( m i − 1) ε i t − 1 2 i . a m ( t − 1) ≡ i =1 i =1 Let m < q and recall that: a q ( t − 1) ≡ � d − 1 i =1 ( q i − 1) ε i t − 1 2 i , and thus: a m ( t ) − a q ( t − 1) ≡ 2 d + δ t + � d − 1 � m i ε i t + (1 − q i ) ε i � 2 i t − 1 i =1 Write ∆ i := m i ε i t + (1 − q i ) ε i t − 1 . Ohad N. Feldheim Avoidance Coupling

2 d agents SAC on K 2 d +1 +1 - cont. n = 2 d , V = { 0 , . . . , 2 d +1 } , a n ( t − 1) = 0. m i := i -th binary digit of m . t . . . ε d − 1 ε 0 uniform { -1,1 } , δ t uniform { 0 , 1 } . t d − 1 d − 1 a m ( t ) = 2 d + δ t + � � m i ε i t 2 i , ( m i − 1) ε i t − 1 2 i . a m ( t − 1) ≡ i =1 i =1 Let m < q and recall that: a q ( t − 1) ≡ � d − 1 i =1 ( q i − 1) ε i t − 1 2 i , and thus: a m ( t ) − a q ( t − 1) ≡ 2 d + δ t + � d − 1 � m i ε i t + (1 − q i ) ε i � 2 i t − 1 i =1 Write ∆ i := m i ε i t − 1 . Taking k = max i ( m i � = q i ), we have t + (1 − q i ) ε i m k = 0, q k = 1, Ohad N. Feldheim Avoidance Coupling

2 d agents SAC on K 2 d +1 +1 - cont. n = 2 d , V = { 0 , . . . , 2 d +1 } , a n ( t − 1) = 0. m i := i -th binary digit of m . t . . . ε d − 1 ε 0 uniform { -1,1 } , δ t uniform { 0 , 1 } . t d − 1 d − 1 a m ( t ) = 2 d + δ t + � � m i ε i t 2 i , ( m i − 1) ε i t − 1 2 i . a m ( t − 1) ≡ i =1 i =1 Let m < q and recall that: a q ( t − 1) ≡ � d − 1 i =1 ( q i − 1) ε i t − 1 2 i , and thus: a m ( t ) − a q ( t − 1) ≡ 2 d + δ t + � d − 1 � m i ε i t + (1 − q i ) ε i � 2 i t − 1 i =1 Write ∆ i := m i ε i t − 1 . Taking k = max i ( m i � = q i ), we have t + (1 − q i ) ε i m k = 0, q k = 1, and so,  1 i > k    | ∆ i | ≤ 0 i = k   2 i < k  Ohad N. Feldheim Avoidance Coupling

2 d agents SAC on K 2 d +1 +1 - cont. n = 2 d , V = { 0 , . . . , 2 d +1 } , a n ( t − 1) = 0. m i := i -th binary digit of m . t . . . ε d − 1 ε 0 uniform { -1,1 } , δ t uniform { 0 , 1 } . t d − 1 d − 1 a m ( t ) = 2 d + δ t + � � m i ε i t 2 i , ( m i − 1) ε i t − 1 2 i . a m ( t − 1) ≡ i =1 i =1 Let m < q and recall that: a q ( t − 1) ≡ � d − 1 i =1 ( q i − 1) ε i t − 1 2 i , and thus: a m ( t ) − a q ( t − 1) ≡ 2 d + δ t + � d − 1 � m i ε i t + (1 − q i ) ε i � 2 i t − 1 i =1 Write ∆ i := m i ε i t − 1 . Taking k = max i ( m i � = q i ), we have t + (1 − q i ) ε i m k = 0, q k = 1, and so,  1 i > k    i =2 2 i < 2 d ⇒ | δ t + � d − 1 i =1 ∆ i 2 i | < 1 + � d − 1 | ∆ i | ≤ 0 i = k   2 i < k  Ohad N. Feldheim Avoidance Coupling

2 d agents SAC on K 2 d +1 +1 - cont. n = 2 d , V = { 0 , . . . , 2 d +1 } , a n ( t − 1) = 0. m i := i -th binary digit of m . t . . . ε d − 1 ε 0 uniform { -1,1 } , δ t uniform { 0 , 1 } . t d − 1 d − 1 a m ( t ) = 2 d + δ t + � � m i ε i t 2 i , ( m i − 1) ε i t − 1 2 i . a m ( t − 1) ≡ i =1 i =1 Let m < q and recall that: a q ( t − 1) ≡ � d − 1 i =1 ( q i − 1) ε i t − 1 2 i , and thus: a m ( t ) − a q ( t − 1) ≡ 2 d + δ t + � d − 1 � m i ε i t + (1 − q i ) ε i � 2 i t − 1 i =1 Write ∆ i := m i ε i t − 1 . Taking k = max i ( m i � = q i ), we have t + (1 − q i ) ε i m k = 0, q k = 1, and so,  1 i > k    i =2 2 i < 2 d ⇒ | δ t + � d − 1 i =1 ∆ i 2 i | < 1 + � d − 1 | ∆ i | ≤ 0 i = k   2 i < k  Ohad N. Feldheim Avoidance Coupling

2 d agents SAC on K 2 d +1 +1 - cont. n = 2 d , V = { 0 , . . . , 2 d +1 } , a n ( t − 1) = 0. m i := i -th binary digit of m . t . . . ε d − 1 ε 0 uniform { -1,1 } , δ t uniform { 0 , 1 } . t d − 1 d − 1 a m ( t ) = 2 d + δ t + � � m i ε i t 2 i , ( m i − 1) ε i t − 1 2 i . a m ( t − 1) ≡ i =1 i =1 Let m < q and recall that: a q ( t − 1) ≡ � d − 1 i =1 ( q i − 1) ε i t − 1 2 i , and thus: a m ( t ) − a q ( t − 1) ≡ 2 d + δ t + � d − 1 � m i ε i t + (1 − q i ) ε i � 2 i t − 1 i =1 Write ∆ i := m i ε i t − 1 . Taking k = max i ( m i � = q i ), we have t + (1 − q i ) ε i m k = 0, q k = 1, and so,  1 i > k    i =2 2 i < 2 d ⇒ | δ t + � d − 1 i =1 ∆ i 2 i | < 1 + � d − 1 | ∆ i | ≤ 0 i = k   2 i < k  ⇒ a m ( t ) − a q ( t − 1) � = 0 . Ohad N. Feldheim Avoidance Coupling

2 d agents SAC on K 2 d +1 +1 - cont. n = 2 d , V = { 0 , . . . , 2 d +1 } , a n ( t − 1) = 0. m i := i -th binary digit of m . t . . . ε d − 1 ε 0 uniform { -1,1 } , δ t uniform { 0 , 1 } . t d − 1 d − 1 a m ( t ) = 2 d + δ t + � � m i ε i t 2 i , ( m i − 1) ε i t − 1 2 i . a m ( t − 1) ≡ i =1 i =1 Let m < q and recall that: a q ( t − 1) ≡ � d − 1 i =1 ( q i − 1) ε i t − 1 2 i , and thus: a m ( t ) − a q ( t − 1) ≡ 2 d + δ t + � d − 1 � m i ε i t + (1 − q i ) ε i � 2 i t − 1 i =1 Write ∆ i := m i ε i t − 1 . Taking k = max i ( m i � = q i ), we have t + (1 − q i ) ε i m k = 0, q k = 1, and so,  1 i > k    i =2 2 i < 2 d ⇒ | δ t + � d − 1 i =1 ∆ i 2 i | < 1 + � d − 1 | ∆ i | ≤ 0 i = k   2 i < k  ⇒ a m ( t ) − a q ( t − 1) � = 0 . Ohad N. Feldheim Avoidance Coupling

2 d agents SAC on K 2 d +1 - cont. And there is also an applet! (by David Wilson) http://dbwilson.com/avoidance.svg Ohad N. Feldheim Avoidance Coupling

Monotonicity of Avoidance coupling on K n 2 3 0 1

Partly Ordered Simple Avoidance Coupling (POSAC) Let G = ( V , E ) be a finite graph (loops and multi-edges are OK). m agents, a 0 , . . . , a m − 1 moving on V , are said to form a k - POSAC if: 2 1 0 3 Ohad N. Feldheim Avoidance Coupling

Partly Ordered Simple Avoidance Coupling (POSAC) Let G = ( V , E ) be a finite graph (loops and multi-edges are OK). m agents, a 0 , . . . , a m − 1 moving on V , are said to form a k - POSAC if: The agents move one at a time 2 but in changing order, 1 0 3 Ohad N. Feldheim Avoidance Coupling

Partly Ordered Simple Avoidance Coupling (POSAC) Let G = ( V , E ) be a finite graph (loops and multi-edges are OK). m agents, a 0 , . . . , a m − 1 moving on V , are said to form a k - POSAC if: The agents move one at a time 2 but in changing order, 1 Agents a 1 , . . . , a k are always moving in order , The agents never collide 0 3 Ohad N. Feldheim Avoidance Coupling

Partly Ordered Simple Avoidance Coupling (POSAC) Let G = ( V , E ) be a finite graph (loops and multi-edges are OK). m agents, a 0 , . . . , a m − 1 moving on V , are said to form a k - POSAC if: The agents move one at a time 2 but in changing order, 1 Agents a 1 , . . . , a k are always moving in order , The agents never collide The path of each agent is a 0 3 simple random walk. Ohad N. Feldheim Avoidance Coupling

Partly Ordered Simple Avoidance Coupling (POSAC) Let G = ( V , E ) be a finite graph (loops and multi-edges are OK). m agents, a 0 , . . . , a m − 1 moving on V , are said to form a k - POSAC if: The agents move one at a time 2 but in changing order, 1 0 Agents a 1 , . . . , a k are always moving in order , The agents never collide The path of each agent is a 3 simple random walk. Ohad N. Feldheim Avoidance Coupling

Partly Ordered Simple Avoidance Coupling (POSAC) Let G = ( V , E ) be a finite graph (loops and multi-edges are OK). m agents, a 0 , . . . , a m − 1 moving on V , are said to form a k - POSAC if: The agents move one at a time 2 3 but in changing order, 1 0 Agents a 1 , . . . , a k are always moving in order , The agents never collide The path of each agent is a simple random walk. Ohad N. Feldheim Avoidance Coupling

Partly Ordered Simple Avoidance Coupling (POSAC) Let G = ( V , E ) be a finite graph (loops and multi-edges are OK). m agents, a 0 , . . . , a m − 1 moving on V , are said to form a k - POSAC if: The agents move one at a time 2 3 but in changing order, 0 Agents a 1 , . . . , a k are always moving in order , 1 The agents never collide The path of each agent is a simple random walk. Ohad N. Feldheim Avoidance Coupling

Partly Ordered Simple Avoidance Coupling (POSAC) Let G = ( V , E ) be a finite graph (loops and multi-edges are OK). m agents, a 0 , . . . , a m − 1 moving on V , are said to form a k - POSAC if: The agents move one at a time 3 but in changing order, 0 2 Agents a 1 , . . . , a k are always moving in order , 1 The agents never collide The path of each agent is a simple random walk. Ohad N. Feldheim Avoidance Coupling

Partly Ordered Simple Avoidance Coupling (POSAC) Let G = ( V , E ) be a finite graph (loops and multi-edges are OK). m agents, a 0 , . . . , a m − 1 moving on V , are said to form a k - POSAC if: The agents move one at a time 2 3 but in changing order, 0 Agents a 1 , . . . , a k are always moving in order , 1 The agents never collide The path of each agent is a simple random walk. Ohad N. Feldheim Avoidance Coupling

* E F 3 E F 1 1 A D 2 A D 2 0 0 C B C B Theorem If there is a k -POSAC of m agents on K n , then there also is a k -POSAC of m + 1 agents on K n +1 . Ohad N. Feldheim Avoidance Coupling

* E F 3 E F 1 1 A D 2 A D 2 0 0 C B C B Add a special vertex ∗ with a special disordered agent. Ohad N. Feldheim Avoidance Coupling

* E F 3 E F 1 1 A D 2 A D 2 0 0 C B C B Add a special vertex with a special disordered agent. At start of a round flip the special vertex with another vertex. Ohad N. Feldheim Avoidance Coupling

B E F 3 E F 1 1 A D 2 A D 2 0 0 C B * C Add a special vertex with a special disordered agent. At start of a round flip the special vertex with another vertex. Continue the process respecting the new labels. Ohad N. Feldheim Avoidance Coupling

B E F 3 E F 1 1 A D 2 A D 2 0 0 C B * C Add a special vertex with a special disordered agent. At start of a round flip the special vertex with another vertex. Continue the process respecting the new labels. Ohad N. Feldheim Avoidance Coupling

B E F 3 E F 1 0 1 0 A D 2 A D 2 C B * C Add a special vertex with a special disordered agent. At start of a round flip the special vertex with another vertex. Continue the process respecting the new labels. As soon as ∗ clears - shift there the new agent. Ohad N. Feldheim Avoidance Coupling

B E F E F 1 0 1 0 A D 2 A D 2 3 C B * C Add a special vertex with a special disordered agent. At start of a round flip the special vertex with another vertex. Continue the process respecting the new labels. As soon as ∗ clears - shift there the new agent. Ohad N. Feldheim Avoidance Coupling

B 2 E F E F 1 0 1 0 A D A D 2 3 C B * C Add a special vertex with a special disordered agent. At start of a round flip the special vertex with another vertex. Continue the process respecting the new labels. As soon as ∗ clears - shift there the new agent. Ohad N. Feldheim Avoidance Coupling

B 2 E F E F 0 0 A D A D 1 2 1 3 C B * C Add a special vertex with a special disordered agent. At start of a round flip the special vertex with another vertex. Continue the process respecting the new labels. As soon as ∗ clears - shift there the new agent. Ohad N. Feldheim Avoidance Coupling

B 2 E F * F 0 0 A D A D 1 2 1 3 C B E C Add a special vertex with a special disordered agent. At start of a round flip the special vertex with another vertex. Continue the process respecting the new labels. As soon as ∗ clears - shift there the new agent. Ohad N. Feldheim Avoidance Coupling

B 2 E F * F 0 0 3 A D A D 1 2 1 C B E C Add a special vertex with a special disordered agent. At start of a round flip the special vertex with another vertex. Continue the process respecting the new labels. As soon as ∗ clears - shift there the new agent. Ohad N. Feldheim Avoidance Coupling

B E F * F 0 2 0 3 A D A D 1 1 2 C B E C Add a special vertex with a special disordered agent. At start of a round flip the special vertex with another vertex. Continue the process respecting the new labels. As soon as ∗ clears - shift there the new agent. Ohad N. Feldheim Avoidance Coupling

B E F * F 2 3 A D 0 A D 0 1 1 2 C B E C Add a special vertex with a special disordered agent. At start of a round flip the special vertex with another vertex. Continue the process respecting the new labels. As soon as ∗ clears - shift there the new agent. Ohad N. Feldheim Avoidance Coupling

B 1 E F * F 2 3 A D 0 A D 0 1 2 C B E C Add a special vertex with a special disordered agent. At start of a round flip the special vertex with another vertex. Continue the process respecting the new labels. As soon as ∗ clears - shift there the new agent. Ohad N. Feldheim Avoidance Coupling

What is there to show? No collisions occur. Each walker makes a simple random walk. * 3 E F E F 1 1 A D 2 A D 2 0 0 B C C B Ohad N. Feldheim Avoidance Coupling