The Tight Bound for Pure Price of Anarchy in an Extended Miner's Dilemma Game
aa r X i v : . [ c s . G T ] J a n The Tight Bound for Pure Price of Anarchy in an ExtendedMiner’s Dilemma Game
Qian Wang
Peking [email protected]
Yurong Chen
Peking [email protected]
ABSTRACT
Pool block withholding attack is performed among mining poolsin digital cryptocurrencies, such as Bitcoin. Instead of mining hon-estly, pools can be incentivized to infiltrate their own miners intoother pools. These infiltrators report partial solutions but withholdfull solutions, share block rewards but make no contribution toblock mining. The block withholding attack among mining poolscan be modeled as a non-cooperative game called “the miner’sdilemma”, which reduces effective mining power in the system andleads to potential systemic instability in the blockchain. However,existing literature on the game-theoretic properties of this attackonly gives a preliminary analysis, e.g., an upper bound of 3 for thepure price of anarchy (PPoA) in this game, with two pools involvedand no miner betraying. Pure price of anarchy is a measurement ofhow much mining power is wasted in the miner’s dilemma game.Further tightening its upper bound will bring us more insight intothe structure of this game, so as to design mechanisms to reducethe systemic loss caused by mutual attacks. In this paper, we givea tight bound of ( , ] for the pure price of anarchy. Moreover, weshow the tight bound holds in a more general setting, in which in-filtrators may betray. We also prove the existence and uniquenessof pure Nash equilibrium in this setting. Inspired by experimentson the game among three mining pools, we conjecture that similarresults hold in the N-player miner’s dilemma game ( 𝑁 ≥ KEYWORDS
Block Withholding Attack, Nash Equilibrium Analysis, Pure Priceof Anarchy.
Bitcoin, assumed to be one of the most successful applications ofblockchain, has gained considerable attention since its inception in2008 [10]. One research direction of interest is to study its securityand potential attacks. Bitcoin’s security mainly comes from twosources: the blockchain data structure and proof-of-work (PoW)consensus protocol. Blockchain is an open, transparent, decentral-ized digital ledger that can validate transactions between two par-ties without third-party authentication. Transaction records arestored as blocks linked by hash pointers. PoW is used to decidewho gets the power of authorizing the next valid block. In this pro-tocol, miners have to solve a computationally difficult puzzle, andthe first miner working out the solution can announce his blockand get the block reward. This difficult puzzle is specially designedso that miners spending more mining power have larger probabili-ties to solve it. Due to fierce competition in Bitcoin system, it maytake months, even years, for a single miner to find the solution, thus miners tend to form mining pools to reduce the high varianceof mining rewards.In a mining pool, miners will work on the same puzzle together.Once a miner successfully finds a full solution and submits it, thepool manager is responsible for reporting the block and splittingthe reward following certain fair allocation rule. Each miner willget the reward proportional to the mining power he spends. Toevaluate how much power miners spend, a pool manager will ac-cept blocks with lower difficulty from miners, called ”share” or par-tial solution, as their proof of work, and treat the number of shareshanded in as their estimated mining power.Such open pools are susceptible to the pool block withholdingattack [3]. Eyal [3] demonstrated that mining pools have the in-centive to infiltrate their own miners into other opponent pools.These infiltrators only submit partial solutions while throw awayvalid blocks, thus they can share block rewards but make no con-tribution to block mining. The attack among mining pools can bemodeled as a non-cooperative game. Eyal [3] referred this gamewith two players as “the 2-player miner’s dilemma”. In a 2-playerminer’s dilemma, two open mining pools choose the number ofmining power to attack each other. While mining pools perform-ing this attack lose their own mining power, resulting in the de-crease of their expected reward from mining, the overall rewardcan exceed that from honest mining. Alkalay-Houlihan and Shah[1] gave a detailed analysis based on this model. It calculated theNash equilibrium in special cases and proved that the pure priceof anarchy (PPoA), which measures how much mining power iswasted due to the attack, is at most 3 in general case. It also con-jectured the pure Nash equilibrium of the game is unique and theupper bound of PPoA is 2.We advance this game analysis further in this paper by provingthe conjecture proposed in Alkalay-Houlihan and Shah [1]. Thatis, we prove the existence and uniqueness of the Nash equilibrium,and show the tight bound of PPoA, defined as in Alkalay-Houlihanand Shah [1], is ( , ] . We provide conditions to determine whenthis game admits an extreme equilibrium or a non-extreme equi-librium and give an algorithm to calculate it.Moreover, we actually prove the results in an extended model,which includes previous model from Alkalay-Houlihan and Shah[1], Eyal [3] as a special case. Instead of assuming the loyalty ofminers as in Eyal [3] and Alkalay-Houlihan and Shah [1], we al-low infiltrators to betray for their own interests, since they cangain more rewards by reporting full solutions without letting theoriginal pool know. The rationality of this betrayal assumption isdiscussed in detail in Section 3.1. We will also introduce a betrayalparameter to indicate the percentage of betraying mining power.he game reduces to the original miner’s dilemma when no infil-trator betrays.Overall, our main contributions are as follows:(1) We adopt the betrayal assumption and extend the previous2-player miner’s dilemma game to a more general model.(Section 3)(2) We prove the existence and uniqueness of pure Nash equi-librium of this game. (Theorem 4.2 and Theorem 7.1)(3) We show the tight bound of pure price of anarchy (PPoA) is ( , ] . (Theorem 8.1)(4) We conduct experiments on the game among three pools,which provide convincing evidence for our conjecture that 𝑁 -player ( 𝑁 ≥
2) miner’s dilemma game admits a uniquepure Nash equilibrium and PPoA is within ( , ] . (Section 9)Our paper is organized as follows. In Section 3, we introducethe extended model of 2-player miner’s dilemma with betrayal as-sumption. Then in Section 4, we prove the existence of pure Nashequilibrium in this game. Considering that the player’s strategyspace is bounded, we call an equilibrium point taken at the bound-ary point an extreme equilibrium, and in the middle a non-extremeequilibrium. We discuss these two cases in Section 5 and Section6 separately. With the help of the characterization of Nash equilib-rium in these two sections, we can prove the uniqueness of Nashequilibrium in the following Section 7. Finally in Section 8, we usethe results in previous sections to induce the bound of PPoA. Ourexperimental results and conjecture are illustrated in Section 9. Block withholding attack was first proposed in Rosenfeld [13] thata miner in a mining pool may have the incentive to withhold validblocks or delay block submissions. Luu et al. [9] showed a single at-tacker can get higher rewards than honestly mining by performingthis attack. The attacker could be not only an individual who con-trols a certain amount of mining power, but also another miningpool. Ke et al. [7] took the difficulty adjustment per 14 days intoconsideration and proposed a novel adversarial strategy for rewardrate named the intermittent block withholding attack (IBWH). Inthe model proposed by Qin et al. [11], the probability of generat-ing a solution per unit mining power of each pool to be a completesolution varies among pools.Eyal [3] further considered the case when two pools performattacks on each other and discovered the miner’s dilemma. Eachmining pool is trying to optimize its own average reward, but end-ing with the loss of mining power in the Nash equilibrium. Eyal [3]briefly claimed that to calculate a Nash equilibrium in a 2-playerminer’s dilemma, we only need to solve the vanishing partial deriva-tives using symbolic computation tools. However, Alkalay-Houlihanand Shah [1] showed the claim is incorrect and provided a detailedanalysis. They also proved the existence of pure Nash equilibriumin 2-player miner’s dilemma game. For symmetric case and no-other-miner case, they characterized the Nash Equilibrium, and forgeneral case, they gave an upper bound 3 of the game’s pure priceof anarchy, specially defined in its paper. Karpova [6], based on the result by Alkalay-Houlihan and Shah[1], used experimental simulations to see how the number of min-ing pools affects the convergence. They also extended the simula-tion by allowing miners to switch between pools or to choose solomining. Li et al. [8] established a model with system rewards andpunishments and analyzed the Nash equilibrium. They also con-sidered the betrayal rate, but the definition of betrayal is differentfrom ours. In Li et al. [8], a betrayer will pledge their loyalty to theopponent pool directly. Haghighat and Shajari [5] modeled poolblock withholding attack as a stochastic game and used reinforce-ment learning to see its evolution.Our model mainly follows Eyal [3] and Alkalay-Houlihan andShah [1]. The mining efficiencies are the same between pools andthe optimization goal is the average reward rather than the rewardrate. The only difference is that we introduce a betrayal parameterto model infiltrators’ betrayal.
In this section, we give a formal description of our model. We con-sider the game between two mining pools as two players. Eachmining pool can choose how much mining power will be sent tothe other pool as its strategy. After specifying notations and basicassumptions of our model, we present our betrayal assumptionsin detail and express the reward functions for the two pools to beoptimized.We assume the total mining power of the system and the min-ing power of each pool are fixed. Let 𝑚 denote the total miningpower of the system and let 𝑚 𝑖 denote the mining power of pool 𝑖 , 𝑖 ∈ { , } . The values of 𝑚,𝑚 , 𝑚 should all be positive. Theremay be other mining power outside the game, solo miners or othermining pools, but we assume they have no interaction with thesetwo pools. We denote this left part of mining power as 𝑡 and 𝑡 = 𝑚 − 𝑚 − 𝑚 ≥
0. At some steps of analysis, we shall replace 𝑚 with 𝑚 + 𝑚 + 𝑡 without notice. Let 𝑥 𝑖 denote the amount of themining power used by pool 𝑖 to attack the other, 𝑥 𝑖 ∈ [ ,𝑚 𝑖 ] . Thus,a pure strategy profile is ( 𝑥 , 𝑥 ) . Although the mining power infiltrating the opposite is essentiallya number of machines, they are still controlled by human miners.Pool managers need to consider miner’s personal interests at themicro level when making macro decisions. Specifically, infiltratorsmay betray the original pool for their own interests, which meansthey may report full solutions without telling the original pool.This report process can be secret, so that the original miningpool cannot distinguish between betraying miners and loyal min-ers. More importantly, since full solutions are also counted as shares,betraying miners can get more reward from the opponent miningpool and hide the extra reward for himself. Although the extra re-ward could be negligible, even 0, it is always non-negative, andpositive in expectation, so infiltrators do have motives to betray.What needs to be emphasized is that betraying, in our setting,is different from joining the other pool directly. Joining the otherpool directly means the miner will only participate in the miningreward distribution of the opponent pool. While Betraying meansthat the miner still participates in the reward distribution twiceike other non-betraying infiltrators, but after receiving the rewardfrom the opponent pool, he can secretly hide the unclaimed partand then participate in the reward distribution of the original pool.Actually, if infiltrators join the other pool directly, it is equivalentto the game assuming loyalty with different 𝑚 𝑖 initially.Considering that different miners own different moral thresh-olds and that some miners are not even aware he is an infiltrator[2], not all infiltrators will betray. Here we introduce a betrayalparameter, 𝑝 ∈ [ , ] , to represent the percent of betrayal. The ef-fective mining power of pool 𝑖 attacking the other pool is actually ( − 𝑝 ) 𝑥 𝑖 . For example, if there are 100 miners controlling identicalmining machines and 𝑝 = .
4, 40 miners will choose to betray. No-tice the model employed in previous work [1, 3] is a special caseof ours when 𝑝 = Given the above notations, we are ready to express the optimiza-tion goals for both sides. Since the mining power of each pool isfixed, maximizing the total reward and maximizing the average re-ward (reward per unit mining power) are equivalent.The total effective mining power of the system is 𝑚 − ( − 𝑝 )( 𝑥 + 𝑥 ) . The effective mining power of mining pool 1 is 𝑚 − 𝑥 + 𝑝𝑥 ,and that of mining pool 2 is 𝑚 − 𝑥 + 𝑝𝑥 . The direct reward of pool 𝑖 from the Bitcoin system, denoted as 𝑅 𝑖 ( 𝑥 , 𝑥 ) , is proportionalto the fraction of the effective mining power contributed to thesystem by the pool. 𝑅 ( 𝑥 , 𝑥 ) = 𝑚 − 𝑥 + 𝑝𝑥 𝑚 − ( − 𝑝 )( 𝑥 + 𝑥 ) ,𝑅 ( 𝑥 , 𝑥 ) = 𝑚 − 𝑥 + 𝑝𝑥 𝑚 − ( − 𝑝 )( 𝑥 + 𝑥 ) . If 𝑚 − ( − 𝑝 )( 𝑥 + 𝑥 ) =
0, it means 𝑥 = 𝑚 , 𝑥 = 𝑚 and 𝑡 = 𝑝 =
0. Nobody can get reward because nobody is mining. Obviously,either pool can retreat a positive amount of mining power to geta positive reward. ( 𝑚 ,𝑚 ) cannot be a Nash equilibrium, so thatwe can always assume 𝑚 − ( − 𝑝 )( 𝑥 + 𝑥 ) > . Let 𝑟 𝑖 ( 𝑥 , 𝑥 ) denote the average reward of pool 𝑖 , 𝑖 ∈{ , } , we have 𝑟 ( 𝑥 , 𝑥 ) = 𝑅 ( 𝑥 , 𝑥 ) + 𝑥 𝑟 ( 𝑥 , 𝑥 ) 𝑚 + 𝑥 ,𝑟 ( 𝑥 , 𝑥 ) = 𝑅 ( 𝑥 , 𝑥 ) + 𝑥 𝑟 ( 𝑥 , 𝑥 ) 𝑚 + 𝑥 . After substituting 𝑅 ( 𝑥 , 𝑥 ) , 𝑅 ( 𝑥 , 𝑥 ) with their expressionsand solving the above system of equations, we get 𝑟 ( 𝑥 , 𝑥 ) = 𝑚 𝑚 + 𝑚 𝑥 + 𝑝𝑚 𝑥 − ( − 𝑝 ) 𝑥 − ( − 𝑝 ) 𝑥 𝑥 ( 𝑚 − ( − 𝑝 )( 𝑥 + 𝑥 )) ( 𝑚 𝑚 + 𝑚 𝑥 + 𝑚 𝑥 ) , Note this is only an approximation. The real value will be less than it because infiltra-tors do not provide full solutions, so less shares lead to less rewards. But the differenceis negligible while calculating, especially for the simplicity of the model. No previouswork has considered to be more precise. The same goes for the reward hidden by thebetrayers, which can be ignored from a macro perspective. 𝑟 ( 𝑥 , 𝑥 ) = 𝑚 𝑚 + 𝑚 𝑥 + 𝑝𝑚 𝑥 − ( − 𝑝 ) 𝑥 − ( − 𝑝 ) 𝑥 𝑥 ( 𝑚 − ( − 𝑝 )( 𝑥 + 𝑥 )) ( 𝑚 𝑚 + 𝑚 𝑥 + 𝑚 𝑥 ) . Again, the total reward to pool i is 𝑚 𝑖 𝑟 𝑖 ( 𝑥 , 𝑥 ) , but it is enoughto only consider the average reward as 𝑚 𝑖 is fixed.Note 𝑟 ( 𝑥 , 𝑥 ) = 𝑟 ( 𝑥 , 𝑥 ) ≡ 𝑚 when 𝑝 =
1, which meanshowever much mining power is used to attack, the average rewardwill remain the same if every infiltrator betrays. In this case, everystrategy profile ( 𝑥 , 𝑥 ) is a pure Nash Equilibrium. Thus we onlyconsider 𝑝 ∈ [ , ) . We only focus on the pure Nash Equilibrium in this work. A strat-egy profile is a pure Nash equilibrium if each player cannot im-prove his reward by changing his strategy. In this game, (cid:0) 𝑥 ∗ , 𝑥 ∗ (cid:1) is a pure Nash Equilibrium if the average reward of each miningpool is maximized given the other’s strategy, i.e., if 𝑥 ∗ = arg max 𝑥 ∈[ ,𝑚 ] 𝑟 (cid:0) 𝑥 , 𝑥 ∗ (cid:1) , 𝑥 ∗ = arg max 𝑥 ∈[ ,𝑚 ] 𝑟 (cid:0) 𝑥 ∗ , 𝑥 (cid:1) . We provide the first order partial derivatives of two reward func-tions. 𝜕 𝑥 𝑟 ( 𝑥 , 𝑥 ) = ( − 𝑝 ) (cid:8) ( − 𝑝 ) (cid:2) 𝑚 𝑥 ( 𝑥 + 𝑥 ) + 𝑚 𝑚 𝑥 (cid:3) + 𝑚 ( 𝑚 + 𝑥 ) + 𝑝𝑚 𝑥 − 𝑚𝑚 𝑥 ( 𝑥 + 𝑥 )+ 𝑚 [ 𝑚 ( ( + 𝑝 ) 𝑚 𝑥 + 𝑥 𝑥 ) − 𝑚𝑥 ( 𝑚 + 𝑥 ) ] }( 𝑚 − ( − 𝑝 ) ( 𝑥 + 𝑥 )) ( 𝑚 𝑚 + 𝑚 𝑥 + 𝑚 𝑥 ) 𝜕 𝑥 𝑟 ( 𝑥 , 𝑥 ) = ( − 𝑝 ) (cid:8) ( − 𝑝 ) (cid:2) 𝑚 𝑥 ( 𝑥 + 𝑥 ) + 𝑚 𝑚 𝑥 (cid:3) + 𝑚 ( 𝑚 + 𝑥 ) + 𝑝𝑚 𝑥 − 𝑚𝑚 𝑥 ( 𝑥 + 𝑥 )+ 𝑚 [ 𝑚 ( ( + 𝑝 ) 𝑚 𝑥 + 𝑥 𝑥 ) − 𝑚𝑥 ( 𝑚 + 𝑥 ) ] }( 𝑚 − ( − 𝑝 ) ( 𝑥 + 𝑥 )) ( 𝑚 𝑚 + 𝑚 𝑥 + 𝑚 𝑥 ) Lemma 4.1. 𝑟 𝑖 ( 𝑥 , 𝑥 ) is concave with respect to 𝑥 𝑖 , 𝑖 ∈ { , } . proof of lemma 4.1. We only prove the concavity of 𝑟 ( 𝑥 , 𝑥 ) as the proof of 𝑟 ( 𝑥 , 𝑥 ) is symmetric. We organize the numeratorof 𝜕 𝑥 𝑟 ( 𝑥 , 𝑥 ) as a quadratic polynomial in 𝑡 , Õ 𝑖 = 𝑐 𝑖 ( 𝑚 ,𝑚 , 𝑝, 𝑥 , 𝑥 ) 𝑡 𝑖 , where 𝑐 𝑖 (·) is the coefficient of the 𝑖 -order term, 𝑖 ∈ { , , } .It is easy to check 𝑐 (·) and 𝑐 (·) are negative. To prove 𝑟 𝑖 ( 𝑥 , 𝑥 ) has a non-positive second derivative, it is sufficient to show 𝑐 (·) ≤ 𝑡 ≥
0. Proving 𝑐 (·) ≤ 𝑐 (·) 𝑚 ( − 𝑝 ) as ℎ ( 𝑚 ,𝑚 ) , 𝑚 ∈ [ 𝑥 , +∞] , 𝑚 ∈ [ 𝑥 , +∞] . By comput-ing the higher derivatives of ℎ (·) with respect to 𝑚 then 𝑚 , wefind ℎ ( 𝑚 ,𝑚 ) achieves its maximum when 𝑚 = 𝑥 and 𝑚 = 𝑥 . ℎ ( 𝑚 ,𝑚 ) ≤ ℎ ( 𝑥 , 𝑥 ) ≤ 𝜕 𝑥 𝑟 ( 𝑥 , 𝑥 ) ≤ (cid:3) The following Theorem 4.2 proves the existence of the Nashequilibrium using the concavity, and gives the sufficient and nec-essary condition for a strategy profile ( 𝑥 ∗ , 𝑥 ∗ ) to be a Nash equi-librium. heorem 4.2. Every two-player miner’s dilemma game with be-trayal assumption admits at least one pure Nash equilibrium. Foreach Nash equilibrium ( 𝑥 ∗ , 𝑥 ∗ ) , 𝑥 ∗ meets one and only one of thefollowing conditions; a symmetric statement also holds for 𝑥 ∗ .(a) 𝜕 𝑥 𝑟 (cid:0) 𝑥 , 𝑥 ∗ (cid:1) ≤ 𝜕 𝑥 𝑟 (cid:0) , 𝑥 ∗ (cid:1) ≤ ∧ 𝑥 ∗ = ;(b) 𝜕 𝑥 𝑟 (cid:0) 𝑥 , 𝑥 ∗ (cid:1) ≥ 𝜕 𝑥 𝑟 (cid:0) 𝑚 , 𝑥 ∗ (cid:1) ≥ ∧ 𝑥 ∗ = 𝑚 ;(c) 𝜕 𝑥 𝑟 (cid:0) 𝑥 ∗ , 𝑥 ∗ (cid:1) = ∧ 𝑥 ∗ ∈ ( ,𝑚 ) . proof of theorem 4.2. As Alkalay-Houlihan and Shah [1] hasshown, When 𝑡 = 𝑝 =
0, the theorem is true and the uniqueNash equilibrium is given by (cid:0) 𝑥 ∗ , 𝑥 ∗ (cid:1) = (cid:0) , 𝑚 (cid:1) if 𝑚 ≤ 𝑚 , (cid:0) 𝑚 , (cid:1) if 𝑚 ≤ 𝑚 , (cid:16) √ 𝑚 𝑚 ( √ 𝑚 −√ 𝑚 )√ 𝑚 +√ 𝑚 , √ 𝑚 𝑚 ( √ 𝑚 −√ 𝑚 )√ 𝑚 +√ 𝑚 (cid:17) otherwise.When 𝑡 > 𝑝 > 𝑚 − ( − 𝑝 )( 𝑥 + 𝑥 ) > [ , 𝑚 ] × [ ,𝑚 ] . Then we canleverage the result by Glicksberg [4]. In our model, the strategysets [ , 𝑚 ] and [ , 𝑚 ] are nonempty, compact and convex. For 𝑖 ∈ { , } , 𝑟 𝑖 ( 𝑥 , 𝑥 ) is continuous in [ , 𝑚 ] × [ , 𝑚 ] and concavein 𝑥 𝑖 by Lemma 4.1. Three conditions in Glicksberg’s ExistenceTheorem are all satisfied, so the existence of pure Nash equilibriumis guaranteed.The concavity of reward functions also indicates the left part oftheorem is true. (cid:3) In fact, the pure Nash equilibrium of this game is unique, but wewill elaborate it later in Section 7. Before that, we need to character-ize the Nash equilibrium as a preparation for the proof of unique-ness.
We start from condition (a) and (b) in Theorem 4.2, to see when 𝑥 𝑖 = 𝑥 𝑖 = 𝑚 𝑖 becomes a Nash equilibrium and how the Nashequilibrium will look like. We show 𝑥 𝑖 = Lemma 5.1.
The unique pure Nash equilibrium is given by (cid:0) 𝑥 ∗ , 𝑥 ∗ (cid:1) = (cid:0) , 𝑚 (cid:1) if 𝑡 = , 𝑝 = , , 𝑚 (cid:16) 𝑚 − 𝑚 + √ 𝑚 − 𝑚𝑚 −( − 𝑝 ) 𝑚 𝑚 (cid:17) 𝑚 −( − 𝑝 ) 𝑚 − 𝑚 ! otherwise, if and only if 𝑔 ( 𝑡, 𝑝,𝑚 , 𝑚 ) ≤ , where g is a function defined by 𝑔 ( 𝑡, 𝑝,𝑚 ,𝑚 ) = 𝑡 + ( 𝑚 + 𝑚 ) 𝑡 + (cid:2) 𝑚 + ( + 𝑝 − 𝑝 ) 𝑚 𝑚 − ( − 𝑝 ) 𝑚 (cid:3) 𝑡 + 𝑚 ( 𝑚 + 𝑝𝑚 ) ( 𝑚 − ( − 𝑝 ) 𝑚 ) . Moreover, we have 𝑥 ∗ + 𝑥 ∗ < 𝑚 + 𝑚 .A symmetric statement holds if 𝑔 ( 𝑡, 𝑝,𝑚 , 𝑚 ) ≤ . In that case, 𝑥 ∗ = and notations 𝑚 and 𝑚 swap in the corresponding 𝑥 ∗ . To deliver a better understanding of the function 𝑔 in Lemma5.1, we give the following equivalent condition. 𝑔 ( 𝑡, 𝑝,𝑚 ,𝑚 ) ≤ ⇔ 𝑚 ≤ ( − 𝑝 ) 𝑚 ∧ 𝑡 ∈ [ , 𝑡 ∗ ( 𝑝,𝑚 , 𝑚 )] , where 𝑡 ∗ ( 𝑝,𝑚 ,𝑚 ) is the unique positive solution to 𝑔 ( 𝑡, 𝑝,𝑚 ,𝑚 ) = , 𝑚 ≤ ( − 𝑝 ) 𝑚 . Thus intuitively, this game admits ( , 𝑥 ∗ ) or ( 𝑥 ∗ , ) as a Nashequilibrium if and only if one pool owns much higher mining powerthan the other and the mining power outside the two mining poolsis small. proof of lemma 5.1. We only need to prove one side when 𝑥 ∗ =
0. Let ( , 𝑥 ∗ ) be a pure Nash equilibrium, then 𝜕 𝑥 𝑟 ( , 𝑥 ) = 𝑚 ( 𝑚 + 𝑥 ) + 𝑚 (cid:2) ( − 𝑝 ) 𝑚 𝑥 − 𝑚𝑥 ( 𝑚 + 𝑥 ) (cid:3) ( 𝑚 − 𝑥 ) ( 𝑚 𝑚 + 𝑚 𝑥 ) /( − 𝑝 ) The denominator is always positive and the numerator is a qua-dratic polynomial in 𝑥 , the quadratic term of which is 𝑚 ( 𝑚 +( − 𝑝 ) 𝑚 − 𝑚 ) . If the quadratic term is 0, i.e., 𝑡 = 𝑝 =
0, we canyield 𝑥 ∗ = 𝑚 by solving 𝜕 𝑥 𝑟 ( , 𝑥 ) =
0. Otherwise, solving thequadratic equation, we get 𝑥 = 𝑚 (cid:16) 𝑚 − 𝑚 ± √ 𝑚 − 𝑚𝑚 −( − 𝑝 ) 𝑚 𝑚 (cid:17) 𝑚 −( − 𝑝 ) 𝑚 − 𝑚 .Only the positive root is within [ , 𝑚 ] . Notice 𝜕 𝑥 𝑟 ( , 𝑥 ) = 𝑥 ∗ = 𝑥 ∗ = 𝑚 can befeasible.When 𝑡 > 𝑝 >
0, solving 𝜕 𝑥 𝑟 ( , 𝑚 (cid:16) 𝑚 − 𝑚 + p 𝑚 − 𝑚𝑚 − ( − 𝑝 ) 𝑚 𝑚 (cid:17) 𝑚 − ( − 𝑝 ) 𝑚 − 𝑚 ) ≤ , we get 𝑔 ( 𝑡, 𝑝,𝑚 ,𝑚 ) ≤ 𝑡 = 𝑝 =
0, using 𝑥 ∗ = 𝑚 we get 𝑚 ≤ 𝑚 , which also leads to 𝑔 ( , , 𝑚 , 𝑚 ) ≤ 𝑥 ∗ + 𝑥 ∗ < 𝑚 + 𝑚 . When 𝑡 = 𝑝 =
0, it istrivial to show 𝑥 ∗ + 𝑥 ∗ = 𝑚 < 𝑚 + 𝑚 . When 𝑡 > 𝑝 >
0, it isequivalent to prove2 𝑚 (cid:16) − 𝑡 − 𝑚 + p ( 𝑚 + 𝑡 ) + 𝑚 ( 𝑡 + 𝑝𝑚 ) (cid:17) 𝑡 + 𝑝𝑚 < 𝑚 + 𝑚 ⇐⇒ s + 𝑚 ( 𝑡 + 𝑝𝑚 )( 𝑚 + 𝑡 ) < ( 𝑚 + 𝑚 )( 𝑡 + 𝑝𝑚 ) 𝑚 ( 𝑚 + 𝑡 ) + , and this inequality holds for (cid:18) + ( 𝑚 + 𝑚 )( 𝑡 + 𝑝𝑚 ) 𝑚 ( 𝑚 + 𝑡 ) (cid:19) > + ( 𝑚 + 𝑚 )( 𝑡 + 𝑝𝑚 ) 𝑚 ( 𝑚 + 𝑡 ) > + 𝑚 ( 𝑡 + 𝑝𝑚 )( 𝑚 + 𝑡 ) . (cid:3) Lemma 5.2. (cid:0) 𝑥 ∗ , 𝑥 ∗ (cid:1) cannot be a pure Nash equilibrium if 𝑥 ∗ = 𝑚 or 𝑥 ∗ = 𝑚 . e have mentioned the strategy profile ( 𝑚 , 𝑚 ) cannot be aNash equilibrium when introducing the reward functions. Lemma5.2 further shows neither pool will infiltrate all the mining powerinto the opponent pool. Actually, this is so unreasonable a choicethat we can easily find a better strategy. In the proof of Lemma 5.2,we compare the rewards between infiltrating no mining power andinfiltrating it all. proof of lemma 5.2. We only need to prove one side when 𝑥 ∗ = 𝑚 . We first prove that for any 𝑥 ∈ ( ,𝑚 ) , 𝑟 ( , 𝑥 ) > 𝑟 ( 𝑚 , 𝑥 ) . 𝑟 ( , 𝑥 ) = 𝑚 𝑚 + 𝑝𝑚 𝑥 ( 𝑚 − ( − 𝑝 ) 𝑥 )( 𝑚 𝑚 + 𝑚 𝑥 ) ,𝑟 ( 𝑚 , 𝑥 ) = 𝑚 𝑚 + 𝑚 + 𝑝𝑚 𝑥 − ( − 𝑝 ) 𝑚 − ( − 𝑝 ) 𝑚 𝑥 ( 𝑚 − ( − 𝑝 )( 𝑚 + 𝑥 ))( 𝑚 𝑚 + 𝑚 + 𝑚 𝑥 ) , and 𝑟 ( , 𝑥 ) > 𝑟 ( 𝑚 , 𝑥 ) is equivalent to −( − 𝑝 ) 𝑥 + ( 𝑡 + 𝑝𝑚 + ( − 𝑝 ) 𝑚 ) 𝑥 + 𝑚 ( 𝑡 − 𝑝𝑚 ) 𝑥 + 𝑚 𝑡 > . Denote the left side of the above inequality as function 𝜙 ( 𝑥 ) : [ , 𝑚 ] → R , then 𝜙 ( ) = 𝑚 𝑡 ≥ 𝜙 ( 𝑚 ) = 𝑡 ( 𝑚 + 𝑚 𝑚 + 𝑚 ) ≥
0. By derivative analysis, the minimum of 𝜙 in interval [ , 𝑚 ] can only be achieved at 0 or 𝑚 . So 𝜙 ( 𝑥 ) > max ( 𝜙 ( ) , 𝜙 ( 𝑚 )) ≥ . In addition, we have already removed ( 𝑚 ,𝑚 ) from the verybeginning and ( 𝑚 , ) cannot be a Nash equilibrium by Lemma5.1. Hence, ∀ 𝑥 ∈ [ , 𝑚 ] , ( 𝑚 , 𝑥 ) is not a Nash equilibrium andthe proof is completed. (cid:3) By Theorem 4.2, Lemma 5.1 and Lemma 5.2, we immediatelyhave the following corollary.
Corollary 5.3.
Every two-player miner’s dilemma game withbetrayal assumption admits at least one pure Nash equilibrium. Foreach Nash equilibrium ( 𝑥 ∗ , 𝑥 ∗ ) , one and only one of the followingconditions is met: (a) 𝑥 ∗ = ; (b) 𝑥 ∗ = ; (c) ( 𝑥 ∗ , 𝑥 ∗ ) is a solution ofvanishing partial derivatives within ( ,𝑚 ) × ( ,𝑚 ) , i.e., 𝜕 𝑥 𝑟 (cid:0) 𝑥 ∗ , 𝑥 ∗ (cid:1) = , 𝜕 𝑥 𝑟 (cid:0) 𝑥 ∗ , 𝑥 ∗ (cid:1) = , s.t. 𝑥 ∗ ∈ ( ,𝑚 ) , 𝑥 ∗ ∈ ( ,𝑚 ) . In this section, we restrict our attention to condition (c) in Corol-lary 5.3. We will prove the vanishing partial derivatives yield atmost one feasible solution ( 𝑥 ∗ , 𝑥 ∗ ) within ( ,𝑚 ) × ( ,𝑚 ) , and 𝑥 ∗ + 𝑥 ∗ ≤ 𝑚 + 𝑚 . In special cases, we also express the pure Nashequilibrium and show when 𝑥 ∗ + 𝑥 ∗ = 𝑚 + 𝑚 holds.Simple calculations show 𝜕 𝑥 𝑖 𝑟 𝑖 (cid:0) 𝑥 ∗ , 𝑥 ∗ (cid:1) = 𝑖 ∈ { , } isequivalent to 𝜕 𝑥 𝑟 (cid:0) 𝑥 ∗ , 𝑥 ∗ (cid:1) + 𝜕 𝑥 𝑟 (cid:0) 𝑥 ∗ , 𝑥 ∗ (cid:1) = ,𝑚 ( 𝑚 + 𝑥 ∗ ) 𝜕 𝑥 𝑟 (cid:0) 𝑥 ∗ , 𝑥 ∗ (cid:1) − 𝑚 ( 𝑚 + 𝑥 ∗ ) 𝜕 𝑥 𝑟 (cid:0) 𝑥 ∗ , 𝑥 ∗ (cid:1) = . The second equation above seems unnatural, but it will lowerthe degree to make the final expansion simpler. Substituting thepartial derivatives, the above equation system is equivalent to ( − 𝑝 ) (cid:0) 𝑥 ∗ + 𝑥 ∗ (cid:1) − 𝑡 (cid:0) 𝑥 ∗ + 𝑥 ∗ (cid:1) + ( 𝑚 𝑚 − 𝑚 𝑥 ∗ − 𝑚 𝑥 ∗ ) − ( − 𝑝 )( 𝑚 𝑥 ∗ + 𝑚 𝑥 ∗ ) = , (1) 𝑚 𝑚 [( + 𝑝 + 𝑡𝑚 ) 𝑥 ∗ + ( 𝑚 − ( − 𝑝 ) 𝑚 + 𝑡 ) 𝑥 ∗ −( + 𝑝 + 𝑡𝑚 ) 𝑥 ∗ − ( 𝑚 − ( − 𝑝 ) 𝑚 + 𝑡 ) 𝑥 ∗ ] = . (2)Let 𝑦 ∗ = 𝑥 ∗ + 𝑥 ∗ and replace 𝑥 ∗ with 𝑦 ∗ − 𝑥 ∗ in Equation (1). Weget 𝑥 ∗ = ( − 𝑝 ) 𝑦 ∗ − ( 𝑚 + ( − 𝑝 ) 𝑚 + 𝑡 ) 𝑦 ∗ + 𝑚 𝑚 ( 𝑚 − 𝑚 )( + 𝑝 ) (3)Substituting Equation (3) and 𝑥 ∗ = 𝑦 ∗ − 𝑥 ∗ into Equation (2), wecan get an equation with only one variable 𝑦 ∗ .0 = ( − 𝑝 ) 𝑡 ( 𝑦 ∗ ) −( − 𝑝 ) (cid:2) 𝑡 + ( 𝑚 + 𝑚 ) 𝑡 + ( + 𝑝 ) 𝑚 𝑚 (cid:3) ( 𝑦 ∗ ) + (cid:2) 𝑡 + ( 𝑚 + 𝑚 ) 𝑡 +( 𝑚 + 𝑚 + 𝑚 𝑚 + 𝑝 𝑚 𝑚 − 𝑝𝑚 𝑚 ) 𝑡 +( + 𝑝 ) 𝑚 𝑚 ( 𝑚 + 𝑚 ) (cid:3) ( 𝑦 ∗ ) +( − 𝑝 ) 𝑚 𝑚 (cid:2) ( + 𝑝 ) 𝑚 𝑚 − ( 𝑚 + 𝑚 ) 𝑡 − 𝑡 (cid:3) 𝑦 ∗ − 𝑚 𝑚 (cid:2) ( + 𝑝 ) ( 𝑚 + 𝑚 ) + 𝑝𝑡 (cid:3) (4)Regard the right side of Equation (4) as a function 𝑓 ( 𝑦 ) : ( ,𝑚 + 𝑚 ) → R . If we can solve 𝑓 ( 𝑦 ∗ ) =
0, then we can calculate ( 𝑥 ∗ , 𝑥 ∗ ) and check whether it is a legal solution. Before we study function 𝑓 , note that in Equation (3), the denom-inator may be 0 in the right side of Equation (3). So we have toconsider the symmetric case separately, i.e., 𝑚 = 𝑚 . Lemma 6.1shows if two mining pools have the same mining power, they willalso choose the same strategy, which is consistent with our intu-ition. Lemma 6.1. If 𝑚 = 𝑚 = 𝑚𝑘 where 𝑘 ≥ , the unique pure Nashequilibrium (cid:0) 𝑥 ∗ , 𝑥 ∗ (cid:1) is given by 𝑥 ∗ = 𝑥 ∗ = 𝑚 ( − 𝑝 ) 𝑘 ( 𝑘 − ( + 𝑝 ) − q ( 𝑘 − ( 𝑝 + )) − ( − 𝑝 )) .𝑥 ∗ + 𝑥 ∗ ≤ 𝑚 + 𝑚 , and equal to 𝑚 + 𝑚 if and only if 𝑚 = 𝑚 = 𝑚 . proof of lemma 6.1. When 𝑚 = 𝑚 = 𝑚𝑘 , 𝜕 𝑡 𝑔 ( 𝑡, 𝑝, 𝑚𝑘 , 𝑚𝑘 ) = 𝑡 + 𝑚𝑘 𝑡 + ( + 𝑝 − 𝑝 ) (cid:16) 𝑚𝑘 (cid:17) >
0. So 𝑔 ( 𝑡, 𝑝, 𝑚𝑘 , 𝑚𝑘 ) > 𝑔 ( , 𝑝, 𝑚𝑘 , 𝑚𝑘 ) = (cid:16) 𝑚𝑘 (cid:17) ( + 𝑝 )[ − ( − 𝑝 ) ] >
0. By Lemma 5.1,there is no extreme Nash equilibrium.Substituting 𝑚 = 𝑚 = 𝑚𝑘 into Equation (2), we obtain (cid:16) 𝑚𝑘 (cid:17) (cid:18) + 𝑝 + 𝑘𝑡𝑚 (cid:19) (cid:16) 𝑥 ∗ + 𝑥 ∗ + 𝑚𝑘 (cid:17) (cid:0) 𝑥 ∗ − 𝑥 ∗ (cid:1) = . Hence we must have 𝑥 ∗ = 𝑥 ∗ . Substituting this and 𝑚 = 𝑚 = 𝑚𝑘 into Equation (1), we get2 ( − 𝑝 )( 𝑥 ∗ ) − (cid:16) 𝑡 + ( − 𝑝 ) 𝑚𝑘 (cid:17) 𝑥 ∗ + (cid:16) 𝑚𝑘 (cid:17) = . ote 𝑡 = 𝑚 − 𝑚 − 𝑚 = (cid:16) − 𝑘 (cid:17) 𝑚 . Solving this quadraticequation, we get 𝑥 ∗ = 𝑚 ( − 𝑝 ) 𝑘 ( 𝑘 − ( + 𝑝 ) ± q ( 𝑘 − ( 𝑝 + )) − ( − 𝑝 )) . The positive root is a increasing function of 𝑘 and achieves itsminimum value at 𝑘 = 𝑚 ( − 𝑝 ) 𝑘 ( 𝑘 − ( + 𝑝 ) + q ( 𝑘 − ( 𝑝 + )) − ( − 𝑝 )) ≥ 𝑚 ( − 𝑝 ) . This is a feasible solution only when 𝑝 = 𝑥 ∗ = 𝑥 ∗ = 𝑚 ,but we know it is not a Nash equilibrium by Lemma 5.2. Hence, theonly feasible solution is the negative root and this is a non-extremeNash equilibrium. 𝑥 ∗ + 𝑥 ∗ = 𝑚 / 𝑘 𝑘 − ( + 𝑝 ) + p ( 𝑘 − ( 𝑝 + )) − ( − 𝑝 ) ≤ 𝑚 . The equal sign is established if and only if 𝑘 = (cid:3) Next we consider another special case, binary case. Lemma 6.2 pro-vides the Nash equilibrium when there is no mining power outsidethese two pools. 𝑓 ( 𝑦 ) will degenerates into a solvable cubic func-tion when 𝑡 =
0, and we can get a concise explicit expression for ( 𝑥 ∗ , 𝑥 ∗ ) . Lemma 6.2.
Let 𝑡 = , i.e. 𝑚 = 𝑚 + 𝑚 . If 𝑚 ≤ ( − 𝑝 ) 𝑚 or 𝑚 ≤ ( − 𝑝 ) 𝑚 , we will get a unique extreme Nash equilibrium with 𝑥 ∗ = or 𝑥 ∗ = respectively, as in Lemma 5.1. Otherwise we will geta unique non-extreme equilibrium given by (cid:0) 𝑥 ∗ , 𝑥 ∗ (cid:1) = (cid:18) √ 𝑚 𝑚 (cid:0) √ 𝑚 − ( − 𝑝 )√ 𝑚 (cid:1) ( + 𝑝 ) (cid:0) √ 𝑚 + √ 𝑚 (cid:1) , √ 𝑚 𝑚 (cid:0) √ 𝑚 − ( − 𝑝 )√ 𝑚 (cid:1) ( + 𝑝 ) (cid:0) √ 𝑚 + √ 𝑚 (cid:1) (cid:19) .𝑥 ∗ + 𝑥 ∗ ≤ 𝑚 + 𝑚 , and the equality holds if and only if 𝑚 = 𝑚 = 𝑚 . proof of lemma 6.2. 𝑔 ( , 𝑝,𝑚 ,𝑚 ) ≤ ⇐⇒ 𝑚 ≤ ( − 𝑝 ) 𝑚 Thus, 𝑥 ∗ = 𝑚 ≤ ( − 𝑝 ) 𝑚 and the value of 𝑥 ∗ willfollow Lemma 5.1. Symmetrically, we will also get an extreme Nashequilibrium where 𝑥 ∗ = 𝑚 ≤ ( − 𝑝 ) 𝑚 .Next, we consider the non-extreme Nash equilibrium. Setting 𝑡 = ( + 𝑝 ) 𝑚 𝑚 (cid:16) ( 𝑦 ∗ ) − 𝑚 𝑚 (cid:17) (cid:0) 𝑚 + 𝑚 − ( − 𝑝 ) 𝑦 ∗ (cid:1) = . This equation has three roots {−√ 𝑚 𝑚 , √ 𝑚 𝑚 , 𝑚 + 𝑚 ( − 𝑝 ) } , butonly √ 𝑚 𝑚 is a feasible root within ( ,𝑚 + 𝑚 ) . Substituting 𝑦 ∗ = √ 𝑚 𝑚 in Equation 3, we get 𝑥 ∗ = √ 𝑚 𝑚 )( √ 𝑚 − ( − 𝑝 )√ 𝑚 )( + 𝑝 )(√ 𝑚 + √ 𝑚 ) . And symmetrically, 𝑥 ∗ = √ 𝑚 𝑚 ( √ 𝑚 − ( − 𝑝 )√ 𝑚 )( + 𝑝 )(√ 𝑚 + √ 𝑚 ) . To be a feasible Nash equilibrium, both 𝑥 ∗ and 𝑥 ∗ have be pos-itive. Solving 𝑥 ∗ > 𝑥 ∗ >
0, we get 𝑚 > ( − 𝑝 ) 𝑚 and 𝑚 > ( − 𝑝 ) 𝑚 . We can verify 𝑥 ∗ < 𝑚 and 𝑥 ∗ < 𝑚 also hold inthis condition. 𝑥 ∗ + 𝑥 ∗ = √ 𝑚 𝑚 ≤ 𝑚 + 𝑚 𝑚 = 𝑚 = 𝑚 . (cid:3) Finally, we assume 𝑚 ≠ 𝑚 , 𝑡 > 𝑓 ( 𝑦 ) is a quartic function of 𝑦 , which cannot be factorized. Ofcourse we can use root formula to solve it but the result is too com-plicated to analyze. Instead, we analyze the zero point of 𝑓 ( 𝑦 ) di-rectly. We can determine the existence and range of the zero pointby derivative analysis. Lemma 6.3.
When 𝑚 ,𝑚 > , 𝑚 ≠ 𝑚 , 𝑡 > and 𝑝 ∈ [ , ) . 𝑓 ( 𝑦 ) has a unique root 𝑦 ∗ within ( ,𝑚 + 𝑚 ) and 𝑦 ∗ < 𝑚 + 𝑚 . proof of lemma 6.3. We calculate the first and second deriva-tives of 𝑓 ( 𝑦 ) . 𝑓 ′ ( 𝑦 ) = ( − 𝑝 ) 𝑡𝑦 − ( − 𝑝 ) (cid:2) ( + 𝑝 ) 𝑚 𝑚 + ( 𝑚 + 𝑚 ) 𝑡 + 𝑡 (cid:3) 𝑦 + (cid:2) 𝑡 + ( 𝑚 + 𝑚 ) 𝑡 +( 𝑚 + 𝑚 + 𝑚 𝑚 + 𝑝 𝑚 𝑚 − 𝑝𝑚 𝑚 ) 𝑡 + ( + 𝑝 ) 𝑚 𝑚 ( 𝑚 + 𝑚 ) (cid:3) 𝑦 +( − 𝑝 ) 𝑚 𝑚 (cid:2) ( + 𝑝 ) 𝑚 𝑚 − ( 𝑚 + 𝑚 ) 𝑡 − 𝑡 (cid:3) 𝑓 ′′ ( 𝑦 ) = ( − 𝑝 ) 𝑡𝑦 − ( − 𝑝 ) (cid:2) ( + 𝑝 ) 𝑚 𝑚 + ( 𝑚 + 𝑚 ) 𝑡 + 𝑡 (cid:3) 𝑦 + (cid:2) 𝑡 + ( 𝑚 + 𝑚 ) 𝑡 +( 𝑚 + 𝑚 + 𝑚 𝑚 + 𝑝 𝑚 𝑚 − 𝑝𝑚 𝑚 ) 𝑡 + ( + 𝑝 ) 𝑚 𝑚 ( 𝑚 + 𝑚 ) (cid:3) Solving 𝑓 ′′ ( 𝑦 ) =
0, we get 𝑦 = ( + 𝑝 ) 𝑚 𝑚 ( − 𝑝 ) 𝑡 + 𝑚 + 𝑚 + 𝑡 − 𝑝 − √ Δ 𝑡 ( − 𝑝 ) ,𝑦 = ( + 𝑝 ) 𝑚 𝑚 ( − 𝑝 ) 𝑡 + 𝑚 + 𝑚 + 𝑡 − 𝑝 + √ Δ 𝑡 ( − 𝑝 ) , where Δ = 𝑡 + ( 𝑚 + 𝑚 ) 𝑡 + [( 𝑚 + 𝑚 ) + 𝑝𝑚 𝑚 ] 𝑡 + ( + 𝑝 ) 𝑚 𝑚 ( 𝑚 + 𝑚 ) 𝑡 + ( + 𝑝 ) 𝑚 𝑚 >
0. We can verify that 𝑦 ∈ ( ,𝑚 + 𝑚 ) but 𝑦 > 𝑚 + 𝑚 . So 𝑓 ′ ( 𝑦 ) increases on ( , 𝑦 ) and decreases on ( 𝑦 ,𝑚 + 𝑚 ) . Considering the middle point, wehave 𝑓 ′ ( 𝑚 + 𝑚 ) > . lthough we cannnot determine the sign of 𝑓 ′ ( ) or that of 𝑓 ′ ( 𝑚 + 𝑚 ) , we can now claim that there is only one continuous in-terval containing point 𝑚 + 𝑚 where 𝑓 ′ ( 𝑦 ) is positive, while 𝑓 ′ ( 𝑦 ) can probably be negative near 0 and 𝑚 + 𝑚 .Back to the original function 𝑓 ( 𝑦 ) , we have 𝑓 ( 𝑚 + 𝑚 ) > . We know there is only one continuous interval containing point 𝑚 + 𝑚 where 𝑓 ( 𝑦 ) increases. Although 𝑓 ( 𝑦 ) may decrease near 0and 𝑚 + 𝑚 , it does not matter. We can easily verify 𝑓 ( ) < 𝑓 ( 𝑚 + 𝑚 ) >
0, so there exists a unique root for 𝑓 ( 𝑦 ) within ( ,𝑚 + 𝑚 ) and it is stricly less than 𝑚 + 𝑚 . (cid:3) 𝑓 ( 𝑦 ∗ ) = ( ,𝑚 )×( ,𝑚 ) . It is insufficient because the corresponding ( 𝑥 ∗ , 𝑥 ∗ ) may be out of range. Lemma 6.1, 6.2 and 6.3 ensure that given 𝑚 ,𝑚 , 𝑡, 𝑝 , there is at most one non-extreme Nash equilibrium. Ifthere does exist a non-extreme Nash equilibrium ( 𝑥 ∗ , 𝑥 ∗ ) , 𝑥 ∗ + 𝑥 ∗ ≤ 𝑚 + 𝑚 . and equals to 𝑚 + 𝑚 if and only if 𝑚 = 𝑚 = 𝑚 . We have already studied three conditions illustrated in Corollary5.3. Condition (a) corresponds to a unique extreme Nash equilib-rium, so does condition (b). Condition (c) also corresponds to atmost one non-extreme Nash equilibrium. It is natural to wonderwhether this game has a unique pure Nash equilibrium. The fol-lowing theorem guarantees the uniqueness.
Theorem 7.1.
Every two-player miner’s dilemma game with be-trayal assumption admits a unique pure Nash equilibrium. proof of theorem 7.1.
Theorem 4.2 guarantees the existenceof pure Nash equilibrium. To prove uniqueness, we need to proveat most one condition can be satisfied in Corollary 5.3. Since, ( , ) cannot be a Nash equilibrium by Lemma 5.1, we should prove (a)and (c) cannot happen at the same time, as well as (b) and (c).W.L.O.G, we prove there cannot be two Nash equilibria that satisfy(a) and (c) respectively at the same time, and a symmetric analysiswill hold for (b) and (c).If 𝑥 ∗ = 𝑔 ( 𝑡, 𝑝,𝑚 , 𝑚 ) ≤ 𝑚 ≤ ( − 𝑝 ) 𝑚 < 𝑚 .If there exists another non-extreme Nash equilibrium, the uniqueroot 𝑦 ∗ for 𝑓 ( 𝑦 ) = ( 𝑥 ∗ , 𝑥 ∗ ) within ( ,𝑚 ) × ( ,𝑚 ) . Hence we must have 𝑥 ∗ >
0. Solving thisin Equation (3) we get 𝑦 ∗ < 𝑦 or 𝑦 ∗ > 𝑦 , where 𝑦 = 𝑚 + ( − 𝑝 ) 𝑚 + 𝑡 − √ Δ ( − 𝑝 ) ,𝑦 = 𝑚 + ( − 𝑝 ) 𝑚 + 𝑡 + √ Δ ( − 𝑝 ) , and Δ = ( 𝑚 − ( − 𝑝 ) 𝑚 ) + 𝑡 + 𝑡 ( 𝑚 + ( − 𝑝 ) 𝑚 ) > 𝑦 > 𝑚 +( − 𝑝 ) 𝑚 + 𝑡 ( − 𝑝 ) > 𝑚 + 𝑚 , but we have proved 𝑦 ∗ ≤ 𝑚 + 𝑚 , so only 𝑦 ∗ < 𝑦 can happen. Referring to the proof of Lemma 6.3, if condition (c) corresponds to a solution, it means 𝑓 ( 𝑦 ) > 𝑦 and solving 𝑓 ( 𝑦 ) >
0, we get ex-actly 𝑔 ( 𝑡, 𝑝,𝑚 , 𝑚 ) > , contradicting to 𝑔 ( 𝑡, 𝑝,𝑚 ,𝑚 ) ≤ (cid:3) Alkalay-Houlihan and Shah [1] points out another direction forestablishing the uniqueness of pure Nash equilibrium is to leveragethe result by Rosen [12], and show that the sufficient conditionsthey provide are satisfied in our game. Unfortunately, by our nu-merical experiments, the conditions required by Rosen Criterionare not satisfied in this game, even when 𝑝 = 𝑚 ,𝑚 , 𝑡 and 𝑝 , we determine the equi-librium point as follows. First we check the sign of function 𝑔 todetermine whether it is an extreme case or not. If it is an extremecase, the Nash equilibrium follows Lemma 5.1. If 𝑚 = 𝑚 , we geta symmetric Nash equilibrium by Lemma 6.1. Otherwise, we solveEquation (4) to get the root 𝑦 ∗ of function 𝑓 , and then we can geta non-extreme equilibrium ( 𝑥 ∗ , 𝑥 ∗ ) . Now we can prove our most significant result, which is about thepure price of anarchy of this game. A pure price of anarchy (PPoA)is a measure of the ratio between the optimal social welfare andthe worst social welfare in any pure Nash equilibrium. Followingthe analysis in Alkalay-Houlihan and Shah [1], we define the so-cial welfare to be effective mining power of these two pools. Theoptimal mining power is 𝑚 + 𝑚 when no one attacks and theeffective mining power is 𝑚 + 𝑚 − ( − 𝑝 )( 𝑥 ∗ + 𝑥 ∗ ) in a pureNash equilibrium. Note by Theorem 7.1, we know the pure Nashequilibrium is unique, so given 𝑚 ,𝑚 , 𝑡 and 𝑝 , the pure price ofanarchy is PPoA = 𝑚 + 𝑚 𝑚 + 𝑚 − ( − 𝑝 )( 𝑥 ∗ + 𝑥 ∗ ) . The following theorem gives the tight bound of PPoA.
Theorem 8.1.
In every two-player miner’s dilemma game withbetrayal assumption, the pure price of anarchy is at most , and equalto if and only if 𝑚 = 𝑚 = 𝑚 and 𝑝 = . The tight bound of PPoA is ( , ] . proof of theorem 8.1. By Lemma 5.1, 6.1, 6.2 and 6.3, we know 𝑥 ∗ + 𝑥 ∗ = 𝑦 ∗ ≤ 𝑚 + 𝑚 , so that PPoA ≤ + 𝑝 . The equality holds ifand only if 𝑥 ∗ + 𝑥 ∗ = 𝑦 ∗ ≤ 𝑚 + 𝑚 , i.e., 𝑚 = 𝑚 = 𝑚 . Then for 𝑝 ∈ [ , ) , PPoA ≤
2, and equal to 2 if and only if 𝑝 = ( , ) cannot be a Nashequilibrium by Lemma 5.1, PPoA > = s ( + 𝑝 ) ( 𝑘 − ) + ( − 𝑝 ) ( 𝑘 − ) + + − 𝑝 + ( 𝑘 − ) 𝑘 →+∞ −→ (cid:3) or every two-player miner’s dilemma game without betrayalassumption, Alkalay-Houlihan and Shah [1] proved PPoA < ≤
2. We not only show the conjecture is true,but also consider the conjecture in a more general model, the gamewith betrayal assumption. Theorem 7.1 proves the uniqueness andTheorem 8.1 proves PPoA ≤
2, which is a tight upper bound.If we keep 𝑝 in the upper bound, PPoA ≤ + 𝑝 . While the upperbound of PPoA decreases in 𝑝 , it is an interesting discovery thatbetrayal has no effect on the upper bound of the sum of infiltratingpower, i.e., 𝑥 ∗ + 𝑥 ∗ ≤ 𝑚 + 𝑚 holds tightly regardless of the valueof 𝑝 . In other words, Allowing miners to betray has no substantialimpact on the macro strategy of the mining pool, but can reducethe systemic loss. Since the dimension of strategy profile space increases quadrat-ically w.r.t. the number of players, the problem with more thantwo pools is much more complicated to analyze theoretically. Hav-ing thoroughly studied the case of two players, we focus on thePPoA of N-player miner’s dilemma game and set 𝑁 = 𝑚 , 𝑚 , 𝑚 denote the mining power of threepools, 𝑡 other mining power in the system, 𝑝 the betrayal parame-ter. 𝑚 ,𝑚 ,𝑚 , 𝑡 are sampled as integers while the values of 𝑝 areequally spaced floating points in [ , ) .Let 𝑥 𝑖,𝑗 be the amount of mining power pool 𝑖 uses to attackpool 𝑗 , satisfying Í 𝑗 𝑥 𝑖,𝑗 ≤ 𝑚 𝑖 , for ∀ 𝑖 ∈ { , , } . For each set-ting, we iteratively calculate the best responses starting from hon-est mining, i.e., 𝑥 𝑖,𝑗 = , ∀ 𝑖, 𝑗 ∈ { , , } . Specifically, in step 𝑘 ,pool 𝑖 chooses strategies { 𝑥 𝑘𝑖,𝑗 : 𝑖 ≠ 𝑗 } that maximizes its rewardbased on others’ strategies in step 𝑘 − { 𝑥 𝑘 − 𝑖 ′ ,𝑗 : 𝑖 ′ ≠ 𝑗, 𝑖 ′ ≠ 𝑖 } .We assume the iteration converges to a Nash equilibrium when Í 𝑖,𝑗 : 𝑖 ≠ 𝑗 | 𝑥 𝑘𝑖,𝑗 − 𝑥 𝑘 − 𝑖,𝑗 | ≤ − . The optimization process uses scipy package in python.There are five parameters varying in this game, so we fix partof them to see how PPoA changes with other parameters, in thehope of catching its inner structure. In Figure 1(a) and 1(b), we set 𝑚 = , 𝑝 = 𝑡 to a constant value ( 𝑡 = 𝑡 = in1(b)). In Figure 1(c), we set 𝑚 = 𝑚 and let them change together.Then, we consider the symmetric case when 𝑚 = 𝑚 = 𝑚 andhow PPoA changes w.r.t. 𝑚 𝑖 and 𝑡 , shown in Figure 1(d). In Figure1(e), we fix 𝑚 = 𝑚 = . And finally we fix 𝑚 = 𝑚 = 𝑚 = to see the effect of 𝑝 on PPoA, as shown in Figure 1(f).Figure 1(d),1(e), 1(f) imply that PPoA will become smaller as tincreases. That is to say, when this game involves a higher per-centage of the mining power in the system, pools tend to use moremining power to perform attacks and the loss will become larger. Itis also interesting to see with different 𝑡 , Figure 1(a) and 1(b) showdifferent trends of PPoA as 𝑚 and 𝑚 change.We can see in most instances, PPoA ≤ , which means Í 𝑖,𝑗 𝑥 𝑖,𝑗 ≤ 𝑚 + 𝑚 + 𝑚 . PPoA will exceed only if the mining power of onepool approaches zero. In fact, 3-player miner’s dilemma game de-generates to 2-player game in this case. Although our experimentonly covers a few special settings, it can be intuitively judged that (a) (b)(c) (d)(e) (f) Figure 1: Pure price of anarchy in 3-player miner’s dilemmagame decentralization can help reduce the mining power wasted in thegame. We conjecture that the following stronger result should hold.
Conjecture 9.1.
Every 𝑁 -player ( 𝑁 ≥ ) miner’s dilemma gamewith betrayal assumption admits a unique pure Nash equilibrium,and the tight bound of pure price of anarchy is ( , ] .
10 CONCLUSION
We explore the pool block withholding attacks and make an in-depth game theoretic analysis. Compared to previous models forminer’s dilemma, our model is more general with the introduc-tion of betrayal assumption. Focusing on the pure Nash equilib-rium and the pure price of anarchy, we prove that every two-playerminer’s dilemma game with betrayal assumption admits a uniquepure Nash equilibrium and the tight bound of PPoA is ( , ] . Wealso give the explicit expression of Nash equilibrium in specialcases.Proving Conjecture 9.1, i.e. establishing uniqueness of pure Nashequilibrium and proving an upper bound of 2 on PPoA in N-poolcase ( 𝑁 ≥ EFERENCES [1] Colleen Alkalay-Houlihan and Nisarg Shah. 2019. The Pure Price of Anarchy ofPool Block Withholding Attacks in Bitcoin Mining. In
Proceedings of the AAAIConference on Artificial Intelligence , Vol. 33. 1724–1731.[2] Nicolas T Courtois. 2014. On the longest chain rule and programmed self-destruction of crypto currencies. arXiv preprint arXiv:1405.0534 (2014).[3] Ittay Eyal. 2015. The miner’s dilemma. In . IEEE, 89–103.[4] Irving L Glicksberg. 1952. A further generalization of the Kakutani fixed pointtheorem, with application to Nash equilibrium points.
Proc. Amer. Math. Soc.
Future Generation Computer Systems
97 (2019),482–491.[6] Natalia Karpova. 2019.
Game-theoretic analysis of a block withholding attack onthe Bitcoin consensus protocol . Ph.D. Dissertation.[7] Junming Ke, Pawel Szalachowski, Jianying Zhou, Qiuliang Xu, and Zheng Yang.2019. IBWH: An Intermittent Block Withholding Attack with Optimal Mining Reward Rate. In
International Conference on Information Security . Springer, 3–24.[8] Wenbai Li, Mengwen Cao, Yue Wang, Changbing Tang, and Feilong Lin. 2020.Mining Pool Game Model and Nash Equilibrium Analysis for PoW-basedBlockchain Networks.
IEEE Access (2020).[9] Loi Luu, Ratul Saha, Inian Parameshwaran, Prateek Saxena, and Aquinas Ho-bor. 2015. On power splitting games in distributed computation: The case ofbitcoin pooled mining. In . IEEE, 397–411.[10] Satoshi Nakamoto. 2008. A peer-to-peer electronic cash system.
Bitcoin.–URL:https://bitcoin. org/bitcoin. pdf (2008).[11] Rui Qin, Yong Yuan, and Fei-Yue Wang. 2020. Optimal Block Withholding Strate-gies for Blockchain Mining Pools.
IEEE Transactions on Computational SocialSystems (2020).[12] J Ben Rosen. 1965. Existence and uniqueness of equilibrium points for concaven-person games.
Econometrica: Journal of the Econometric Society (1965), 520–534.[13] Meni Rosenfeld. 2011. Analysis of bitcoin pooled mining reward systems. arXivpreprint arXiv:1112.4980arXivpreprint arXiv:1112.4980