The Byzantine Generals Problem

Transcription

1 The Byzantne Generals Problem A paper by: Lesle Lamport, Robert Shostak, and Marshall Pease. Summary by: Roman Kaplan. Every computer system must cope wth computer malfunctons, whereas a malfuncton does not necessarly mean that the component has stopped workng. It can also mean the component wll send a conflctng nformaton to dfferent parts of the system. The stuaton can be expressed abstractly n terms of a group of generals of the Byzantne army camped wth ther troops around an enemy cty. In order to accomplsh best achevements all generals must decde upon a common plan of acton. However there are trators among the generals that wsh to a falure of accomplshng an agreement. The communcaton s done usng messengers and t s relable therefore, a trator can confuse ts fellow generals by sendng msleadng nformaton that wll cause all loyal generals to decde upon dfferent plans of acton and consequently lead to a system falure. Our objectve s to fnd an algorthm to ensure that all loyal generals wll decde upon the same plan of acton, meanng they wll acheve an agreement. We wll show that usng oral messages, agreement can be acheved f and only f more than two-thrds of the generals are loyal. The proof wll be based on the fact that two loyal generals cannot reach an agreement n the presence of a sngle trator. If we wll add the assumpton that messages cannot be forged, the problem s solvable for any number of generals and possble trators. Later we wll dscuss the applcaton of the aforementoned results to relable computer systems. 1. Introducton Relable systems must know how to cope wth multple malfunctonng components. As prevously sad, these components may send conflctng nformaton to dfferent parts of the system. In order to ncrease reader's nterest n the subject, the problem s abstractly expressed as an army dealng wth treacherous generals rather than a CPU falng. We present the problem as the Byzantne army besegng a cty, whle each devson commanded by ts own general. Each of the generals can be a trator, whch has dfferent nterests from hs loyal fellows. Generals' communcaton s done only by a messenger and the loyal ones wll use t to reach fellows. Generals' communcaton s done only by a messenger and the loyal ones wll use t to reach unanmous plan of acton. Possble plans of acton are: "retreat", "attack", etc. Tratorous` generals man objectve s to prevent the loyal ones from reachng a consensus. The loyal generals are aware of the fact that there mght be trators among them so they must have an algorthm that wll ensure they wll reach a consensus. We wll now defne the wanted results from the loyal generals' algorthm. Frst wanted result s they all must reach an agreement regardng ther plan of acton. The agreement must be acheved regardless of trators' actons. Second wanted result s to ensure that all loyal generals wll adopt a "good" plan. Whle t s hard to defne what a good plan s, we wll try explan the meanng to remove some obscurty: If every one of the loyal generals` opnon s to retreat, after sharng ths nformaton wth every other general t`s not acceptable for them to reach the decson to attack, meanng the algorthm must use robust technque.

2 We can rephrase the condton: The decson that the loyal generals wll take must not be greatly nfluenced by the trators. To prevent a small number of trators to nterfere, the majorty method wll be used n the algorthm. In the smple form of ths method every loyal general wll share hs opnon usng messengers wth the rest of the generals and accept a decson regardng hs plan of acton based on the majorty vote among them. The smple form of the majorty method may work n some cases, but t s not fal proof and t s based on the assumpton that all loyal generals wll receve the same values, even from tratorous generals. It s a necessty snce a small number of trators can confuse loyal generals' majorty vote f they are almost equally dvded between a few dfferent opnons. Hence, we understand that a loyal general cannot use the values he receved drectly from the generals snce trators can send conflctng values to dfferent generals. Ths lmtaton must be handled wth care snce we may use a dfferent value from the one sent by some general whle calculatng the majorty functon, even though he s loyal. Therefore, we reach a concluson regardng every loyal general: If a specfc general s loyal then the value he sent must be used as hs value by every other loyal general. Up to ths pont the problem was presented as a number of generals that need to reach an agreement and we found that the condtons for t to occur can be phrased n terms of how a sngle general sends hs value to the others. We therefore wll relate to the problem as a commandng general sendng an order to hs leutenants and reachng the followng problem: Byzantne Generals Problem. A commandng general must send an order to hs n 1leutenant generals such that: IC1. All loyal leutenants obey the same order. IC2. If the commandng general s loyal, then every loyal leutenant obeys the order he sends. The above condtons are called nteractve consstency condtons. The frst was mentoned several tmes n dfferent varatons, but the second s not trval. If the commander s loyal, then hs decson should be accepted by all the leutenants and the frst rule s derved from the second. If he s a trator then he wll probably send conflctng nformaton to hs leutenants, forcng them to use a non-trval algorthm to reach an agreement. 2. Impossblty results In our problem the generals can send only oral messages and later we wll extend our dscusson to wrtten messages. Ths model has two types of messages, oral and wrtten. Oral messages are completely under the control of the sender, whch means that a trator can send any message he desres, n contrary to wrtten messages whch cannot be forged. The use of oral messages makes the problem harder then t may seem. We wll later show that no soluton exsts to the problem unless more than two-thrds of the generals are loyal. The mpossblty result s true only for oral messages, n Secton 4 we wll dscuss the case of wrtten and sgned messages for whch ths result s not true. Our mpossblty result s based on a very smple case where there are three generals (a commander and two leutenants) and one of them s a trator. In order to smplfy the analyss we wll assume two possble messages: "attack" and "retreat". We wll show two dfferent scenarous where one leutenant s loyal n

3 both, we wll call hm L1. In the frst scenaro the commander s loyal but the second leutenant s a trator. In the second scenaro the commander s the trator. In both cases L1 wll receve conflctng nformaton, makng t mpossble for hm to make the rght decson and not volatng any of the nteractve consstency condtons. We wll present each scenaro n an ndependent fgure: Fgure 1: Leutenant 2 s a trator Commander attack" Leutenant 1 Leutenant 2 Commander sad retreat Fgure 1 shows us that L1 has receved conflctng messages. In order to satsfy IC2, L1 must obey the order to attack. We wll now examne Fgure 2, n whch the commander s a trator. Fgure 2: The commander s a trator Commander attack" retreat" Leutenant 1 Leutenant 2 Commander sad retreat In Fgure 2 we see that the commander s a trator and he s sendng conflctng nformaton to the leutenants. L1 receved the same messages as n Fgure 1 and he cannot tell what message the commander sent to L2 so ths scenaro wll appear exactly the same for L1. If he wll obey the order to attack then IC1 wll be volated n Fgure 2 and f he wll decde to retreat, then IC2 not be satsfed n Fgure 1. L1 has no way to dstngush between the dfferent scenaros f the trator les consstently. We therefore conclude that no soluton exsts for three generals that works n the presence of a sngle trator. Ths soluton wll be used as the bass of proofng that no soluton can cope wth m trators whle there are fewer than 3m+1 generals. We frst assume the exstence of such soluton, then we wll show a reducton from 3m or fewer generals to three generals, fnally we wll be able to show the exstence of a soluton for three generals n the presence of a sngle trator, whch we know s mpossble. In order to dstngush between the two dfferent groups of generals

4 we wll refer to the larger group (3m or fewer) as Albanan generals and the smaller group (three generals) as Byzantne generals. Let us assume the exstence of a soluton for the Albanan generals. Now we wll smulate wth each of the Byzantne generals approxmately one-thrd of the Albanan generals, so that each Byzantne general s smulatng at most m Albanan generals. The Byzantne commander wll smulate the Albanan commander and at most m-1 Albanan leutenants. There are m Albanan and one Byzantne trators so we wll choose a sngle Byzantne general to smulate all of the Albanan trators. Snce there s a soluton that satsfes IC1 and IC2 for the Albanan generals problem we receved a soluton for the Byzantne generals problem. IC1 holds for the Albanan generals whle two Byzantne generals are loyal and smulatng only loyal Albanan generals who obey the same order, thus, these two Byzantne generals obey the same order and therefore IC1 holds for the Byzantne generals problem. IC2 holds for the Albanan generals so f the Albanan commander s loyal and all loyal Albanan leutenants wll obey the order he sends, then the Byzantne commander and general whose smulatng loyal Albanan generals wll also obey the same order, therefor IC2 holds. Up to now we have dealt wth the problem of Byzantne generals wth the requrement to try and reach an exact agreement, but we wll now show that even reachng an approxmate agreement has the same dffculty. Frst we assume that the generals are tryng to agree on the approxmate tme to attack, nstead of a precse battle plan. The commander sends an order wth desred tme of attack and usng the nteractve consstency condtons we wll defne new condtons to match the case of reachng an approxmate agreement. These condtons wll be called approxmate nteractve conssteny condtons, let us present them: APIC1. All loyal leutenants attack wthn 10 mnutes of one another. APIC2. If the commandng general s loyal, then every loyal leutenant attacks wthn 10 mnutes of the tme gven n the commander`s orders. note: there s no mportance to the tme at whch the orders were sent and receved and t s rrelevant, only the attack tme gven n the order matters. We wll prove that ths problem s unsolvable n exactly the same condtons as the Byzantne Generals Problem f not less than one-thrd of the generals are trators. The proof wll be by contradcton, frst we assume the exstence of a soluton to the Byzantne Generals Problem wth the presence of a sngle trator. Then we construct a soluton to the orgnal Byzantne Generals Problem and by contradcton the assumed soluton does not exst. Let us assume a the exstence of a soluton to the above problem wth three generals and a sngle trator. In order to smplfy the analyss we assume the commander only wshes to send "attack" or "retreat" orders. He does so by sendng a message wth an attack tme and assgnng specfc tme for both orders. When the commander wll send a message wth attack tme 1:00 t means he orders to attack, when he wll send a message wth attack tme 2:00 t means he wshes to retreat. Each leutenant s executng the followng procedure to obtan hs order: 1. After recevng the attack tme from the commander, a leutenant does one of the followng: a. If the tme s 1:10 or earler, then attack.

5 b. If the tme s 1:50 or later, then retreat. c. Otherwse, contnue to step Ask the other leutenant what decson he reached n step 1. a. If the other leutenant reached a decson, then make the same decson he dd. b. Otherwse, retreat. Usng APIC condtons and the assumed soluton we wll show the presence of IC condtons n the case of three generals and a sngle trator. If the commander s loyal, and by usng APIC2, we receve that every loyal leutenant wll obtan the correct order, so IC2 s satsfed. In ths case IC1 follows from IC2, so the only thng left s to prove IC1 n case the commander s not loyal. If the commander s not loyal then both leutenants are loyal and from APIC1 we know they both wll reach the same decson. Hence, f one leutenant wll decde to attack at step 1 of the procedure the other cannot decde to retreat n the same step. So ether they both decde the same n step 1 or at least one wll reach step 2, n whch they can decde n sub-step a. or b.. If one leutenant wll make hs decson n step 2.a. then hs decson wll be the same as the second's leutenant. The other opton s that one wll reach step 2.b. and decde to retreat, the other wll ether reach the same step and decde the same or he wll use other's leutenant decson. We then receve that n any case both leutenants wll have the same decson so IC1 holds. Usng the assumpton regardng the exstence of a soluton that sustans APIC condtons and the above reducton to the orgnal Byzantne Generals Problem we now receve a soluton to the orgnal Byzantne generals problem wth three generals n the presence of one trator whch s mpossble. Therefore, no soluton exst that mantans APIC condtons n smlar condtons. It s now possble to use the same method as before to smulate m generals wth a sngle one n order to show that no soluton can cope wth exactly or more than one-thrd of the generals beng trators. The proof s the same as wth the demand of reachng an exact agreement. We have found that n the presence of 3m+1 generals no soluton can cope wth m trators, no matter f we want to reach exact or approxmate agreement.

6 3. A soluton wth oral messages Up to now we dscussed only the condtons and results that relate to mpossblty of soluton, now we wll show a soluton n form of an algorthm executed by the loyal generals and the condtons for t to work. A soluton to the Byzantne Generals Problem exst n the presence of at most m trators when there are at least 3m+1 generals. The algorthm wll nvolve sendng message from one general to the others (all or some), hence, we must explan exactly how the messages system work. The messages wll be "oral messages" whch we ntroduced n the begnnng of Secton 2. The followng assumptons wll defne the messages system that wll be used: A1. Every message that s sent s delvered correctly. A2. The recever of the message knows who sent t. A3. The absence of a message can be detected. These assumptons are meant to lmt the amount of harm a trator can make. The frst assumpton wll prevent hm from dsruptng messages of other generals. The second prevents hm from sendng spurous messages n the behalf of other generals. The thrd restrcts hm from sabotagng the army by not sendng any nformaton and thus preventng loyal generals from reachng a decson. Although we dscuss only generals, our man nterest s computer systems and practcal mplementaton of the algorthm, whch wll be explaned n Secton 6 n further detal. The algorthms n ths secton and n the followng one wll requre drect communcaton lnk between every two generals. The algorthms n Secton 5 wll not have ths requrement, meanng not every two generals must have drect connecton but we wll present cummuncaton paths condtons n order for the algorthms to work correctly. As stated before, generals can detect the absence of message, and snce every general must obtan some order sent by the others, even n ths case, we defne the message "retreat" as the default value. The soluton for the Byzantne generals problem wll be n the form of an nductve algorthms called Oral Message algorthm, OM(m), that wll solve the problem n the presence of at most m trators when there at least 3m+1 generals. Ths algorthm s defned for all nonnegatve ntegers m and wll be descrbed n terms of a leutenant "obtanng a value" rather than "obeyng an order" for convenence reasons. We assume the presence of a set of majorty functons, such that f a majorty of the values v equals v, then majorty( v 1, v 2,..., vn 1) equals v, for every nonnegatve nteger n. Ths property of the majorty functons s the only one nessecary for the correctness of the algorthm. Let us descrbe the OM(m) algorthm: Algorthm OM(0): 1) The commander sends hs value to every leutenant. 2) Each leutenant uses the value he receves from the commander. Algorthm OM(m) ( m> 0): 1) The commander sends hs value to every leutenant.

7 2) Every leutenant acts as the commander n OM(m-1) to dssemnate the value he obtaned from the commander to each of the n-2 leutenants. f no value was receved he uses "retreat". At the end of ths step each leutenant has a vector V contanng: a. The value he obtaned from the commander. b. The value dssemnated by every other leutenant. 3) Every leutenant uses the value majorty(v ). We wll now explan how the algorthm works, frst n general terms and later usng examples wth specfc parameters. In the frst step of the algorthm the commander sends hs order, f he s loyal, or orders (conflctng) f he s a trator, to every leutenant. In the next step of the algorthm each leutenant wll send the value he receved from the commander to the other leutenants, actng hmeslf as the commander n ths xt step. In ths way the leutenants wll share the values they obtaned wth the other leutenants for m steps. Fgure 3: Executon of OM(1) when Leutenant 1 s a trator Commander attack" attack" attack" retreat Leutenant 1 Leutenant 2 attack" attack attack" Leutenant 3 retreat attack" Let us examne an example of the algorthm's executon n the case m= 1, n= 4. Fgure 3 llustrates the stuaton graphcally when Leutenant 1 s the trator. In phase OM(1) the commander sends "attack" to every leutenant. In the next step each leutenant wll send the value he obtaned to the other leutenants, L2 and L3 wll send the value "attack" to the other leutenants. Snce L1 s a trator, he sends the value "retreat" to L2 and L3. Now L2 and L3 has obtaned the values: "retreat", "attack" and "attack" whch result n obtanng the value majorty("retreat", "attack", "attack")="attack". The commander s loyal and both loyal leutenants wll obey hs order, whch means that condton IC1 holds and thus IC2. Now we wll examne a bt more complcated example n the case m= 2, n= 7. In ths case the number of messages sent s substantaly larger than n the prevous case. Fgure 4 shows the values sent by the commander to hs leutenants n OM(2). Snce there are more than two steps of the algorthms

8 executon, there must be a way to dstngush among the dfferent messages from the dfferent steps. The problem s solved f every leutenant prefxes hs ndex number,, to the value he sends n step 2. Wth ths method every leutenant knows whch of the leutenants have passed ths message forward so he wll send t only to those who dd not receve t yet. Now let us revew the algorthm's executon n the case of Fgure 4. Fgure 5 presents the the path each message Leutenant 1 had receved by every other leutenant n the three steps of the algorthm's executon. The values sent n every message are summarzed n Fgure 6. Fgure 4: Executon of OM(2) when the commander and Leutenant 6 are trators. Commander retreat retreat x" (d.c.) attack" attack" attack" Leutenant 1 Leutenant 2 Leutenant 3 Leutenant 4 Leutenant 5 Leutenant 6 In Fgure 5, every message s orgnated at the commander as he begns the executon. In OM(2) the commander sends hs order drectly to L1, but as the

9 algorthm's steps progress we receve an ncreasng number of ntermedates for every message. The messages sent n OM(0) has two ntermedates, whle the ndex closest to L1 s the ndex of the leutenant who actually passed the message to L1. Fgure 6 wll contan the values sent n every message, but t wll have a dfferent structure. The messages wll be organzed n rows, when every row matches the messages sent to L1 by the last Leutenant ndex n Fgure 5.

10 Fgure 6: The messages Leutenant 1 has receved n the example presented n Fgure 4. ( a represents attack and r represents retreat ) OM(2): a OM(1): 2r, 3a, 4r, 5a, 6a OM(0): 2{ 3a, 4r, 5a, 6r} 3{ 2r, 4r, 5a, 6a} 4{ 2r, 3a, 5a, 6r} 5{ 2r, 3a, 4r, 5a, 6a} 6{ 2a, 3r, 4a, 5r } Fgure 7: All loyal leutenants' decsons after obtanng the values n step OM(1). L1: majorty(a, r, a, r, a, a) = attack L2: majorty(r, r, a, r, a, r) = retreat L3: majorty(a, r, a, r, a, a) = attack L4: majorty(r, r, a, r, a, r) = retreat L5: majorty(a, r, a, r, a, a) = attack All loyal leutenants do not choose the same acton Fgure 8: The decson Leutenant 1 has made after obtanng the values n step OM(0) and the decson made by every loyal Leutenant. From L2: majorty(r, r, r, a) = retreat From L3: majorty(a, a, a, r) = attack From L4: majorty(r, r, r, a) = retreat From L5: majorty(a, a, a, r) = attack From L6: majorty(a, r, a, r) = - (do nothng) All Leutenants` fnal decson: majorty(a, r, a, r, a, -) = attack Note: Leutenants 2-5 has obtaned the value attack as the value of L1. Let us now look how the algorthm works n ths complcated and extreme example. The frst message each leutenant has receved from the commander was n step OM(2), for Leutenant 1 t s shown n the matchng row presented n Fgure 6. If the commander s loyal then L1 must obey the order, but he does not know whether or not t s true. In step OM(1) L1 receves the orders from the rest of the leutenants, thus, by knowng there s a maxmal number of two trators and seeng the conflctng values he receved he can assume the commander s a trator (although ths s not a part from the algorthm ts purpose s not to fnd the trators).

11 Fgure 7 summarzes the decson every loyal leutenant has made after obtanng the nformaton from all leutenants n step OM(1). As we can see, all loyal leutenants dd not make the same decson n ths step. Step OM(0) wll gve L1 all the nformaton needed for hm to make the rght decson to correlate wth the rest of the loyal Leutenants n ther decson. After obtanng the values from all leutenants n step OM(0) L1 can calculate the rght decson that needs to be made. Fgure 8 shows the value L1 has used for every leutenant from the values obtaned n OM(0). Every Leutenant has made the same as L1 n ths step and has the rght value for every leutenant. Every leutenant's fnal decson s also presented n Fgure 8, as every value used for the other leutenants has been obtaned n a smlar manner as Leutenant 1 dd n Fgure 8. In ths phase we know that all leutenants use the same values and the same majorty functon, resultng n the same fnal decson. Now a lttle about the algorthm`s message complexty. Step OM(m) s the frst and executed only by the commander so t s executed only once and sends n-1 messages. Step OM(m-1) wll be executed by every leutenant so t wll have n 1 n 2 messages. Every n-1 executons, whch wll result n sendng ( )( ) executon of step OM(m-1) wll nvoke n-2 executons of OM(m-2) and so on whch n 1 n 2 n 3 messages to be sent. The method n whch each wll lead to ( )( )( ) leutenant s prefxng hs ndex for every message he sends, results n the fact that algorthm's step OM(m-k) wll nvoke n-k-1 executons of OM(m-k-1). Therefore, we receve that algorthm OM(m) wll nvoke the send of = 1 m ( n 1 )... ( n m) = ( n ) = O( n ) s exponental n n. messages. For n=3m+1, the algorthm = m Let us now prove the correctness of the algorthm. In order to show correctness for every m we frst prove the followng lemma: LEMMA 1: for any m and k, Algorthm OM(m) satsfes IC2 f there are more than 2k+m generals and at most k trators. PROOF: Condton IC2 specfes that f the commandng general s loyal then all loyal Leutenants have to obey the same order, so we know the commander s loyal. The proof wll be by nducton on m. Snce we know every message sent s delvered correctly we can see that OM(0) satsfes the lemma, so t s true for m=0. We wll now assume correctness for m-1, m>0, and prove t for m. In the begnnng of the algorthm (step OM(m)) the commander sends hs value v to hs n-1 leutenants. The next step s that every leutenant apples OM(m-1) wth n-1 generals. We know that there are more than 2k+m generals, so n> 2k+ m n 1> 2k+ m. By the nducton hypothess, the algorthm s correct for m-1 so we can conclude that every loyal leutenant uses the value v = v for each leutenant j. Snce n 1 k ( m 1) > k there s a majorty of j loyal leutenants and we receve that the majorty of all the values every loyal leutenant has for the other leutenants s v. Ths value wll be calculated n step (3) of the algorthm, provng IC2 and the lemma. The followng theorem wll prove that algorthm OM(m) solves the Byzantne Generals Problem.

12 THEOREM 1: For any m, Algorthm OM(m) satsfes condtons IC1 and IC2 f there are more than 3m genenrals and at most m trators. PROOF: By nducton on m. If m=0 then there are no trators, all generals has obtaned the same value from the commander and they followng t. Now we wll assume correctness of OM(m-1) and show that the theorem s true for OM(m), m>0. IC1 splts the prove nto two, The frst n whch the commander s loyal and the second where he s a trator. In the case that the commander s loyal, we choose k from Lemma 1 to sustan k=m and receve that OM(m) satsfes IC2, whch leads to the correctness of IC1 n ths case. Now the commander s a trator among the m trators, meanng there are at most m-1 tratorous leutenants. In step (2), each loyal Leutenant j apples OM(m-1) wth n-2 other leutenants actng as hs leutenatns. Snce there are more than 3m leutenants, there are at least 3m-1 leutenants and we receve that n> 3m n 1> 3m 1> 3 m 1. Therefore, we can apply the nducton ( ) hypothess to conclude that OM(m-1) satsfes condtons IC1 and IC2. By IC2, f one of the two leutenants s Leutenant j, and otherwse from IC1. Hence, for each j, any two loyal leutenants get the same value for v j n step (3). Thus, every two loyal leutenants use the same values calculated n OM(m-1) for all other leutenants and therefore obtan the same majorty value n step (3), provng IC1. 4. A soluton wth sgned messages The fact that trators can le and manpulate the data they send s what makes the Byzantne Generals Problem harder than s may seem, but f we somehow manage to restrct ther ablty to le the problem wll be easer. One way to do ths s to use unforgeable sgned messages as the communcaton method among the generals. Let us phrase the assumpton formaly and add t to our prevous assumptons, A1- A3: A4. (a) A loyal general's sgnature cannot be forged, and any alteraton of the contents of hs sgned message can be detected. (b) Anyone can verfy the authentcty of a general's sgnature. By nspectng the new assumpton we see that t refers only to a loyal leutenant's sgnature, whch means that trator's sgnature can be forged. Ths fact wll not nfluence on the loyal leutenants because trators wll be able to send fake data among themselves, and ther data wll stll be gnored by the algorthm when makng the fnal decson. The use of sgned messages has cancelled our prevos proof regardng the nonexstence of a soluton to the Byzantne Generals Problem wth three generals n the presence of a sngle trator, and we wll show an algorthm that solves the problem for m trators and any number of generals. Note that the problem s meanngless f there are less than m+2 generals a sngle loyal general can randomly choose any order to follow and stll satsfyng condtons IC1 and IC2. The proposed algorthm wll be smlar to OM, but t wll use assumpton A4 to adjust OM to our new case. The new algorthm wll be used for sgned messages and wll be called SM, t wll start lke OM when the commander sends hs order to hs leutenants but hs order s now sgned. Every leutenant receves

13 the order, adds hs sgnature and passes t forward to the rest of the leutenants. In general, when a leutenant receves a sgned order he adds hs sgnature and sends t to those whose sgnature s not ncluded n that order. One may see technchal dffcultes n recevng a sngle unforgeable order and sendng t to multple recpents, but t s not a matter of our concern snce we can assume that the commander sends a stack of hs sgned orders to every leutenant and the leutenant smply adds hs sgnature and passes them forward. We can also assume that each leutenant can copy sgned orders (does not matter how), add hs sgnature and dstrbute them. Algorthm SM assumes the exstence of functon chose whch s appled to a set of orders to obtan a sngle order. The only requrements we make for ths functon are: 1. If the set V conssts of the sngle element v, then chose(v) = v. 2. chose( ) = RETREAT, where s the empty set. The value RETREAT s an arbtrary default order and can be replaced by any other order. In the followng algorthm, a message contans the value x sgned by General wll be denoted by x:. We let the value x: sgned by General j be denoted by x::j. The commander wll be General 0, and the rest wll be hs leutenants. In ths algorthm, every leutenant wll have a set contanng the properly sgned orders he has receved so far. Ths set wll be denoted V for Leutenant. Ths set wll contan only the orders that the leutenant has receved and not the messages he has receved, snce there may be many dfferent messages wth the same order. In case the commander s loyal, the set V should not contan more than a sngle element. Let us now descrbe algorthm SM(m) explctly: we ntalze V =. 1) The commander sgns and sends hs value to every leutenant. 2) For each : a. If Leutenant receves a message of the form v:0 from the commander and he has not yet receved any order, then:. He lets V equal {v}.. He sends the message v:0: to every other leutenant. b. If leutenant receves a message of the form v :0: j1 :...: j and v k s not n the set V then:. He adds v to V.. If k<m, then he sends the message v :0: j1 :...: jk : to every leutenant other than j 1,..., j. k 3) For each : When Leutenant wll receve no more messages, he obeys the order chose( V ). Let us now clear a few problematc ssues regardng the algorthm. Frst, we see that every leutenant gnores any message contanng an order, v, that exsts n V n step 2).Second, n step 3) each leutenant decdes whch order to obey only

14 after he wll receve no more messages. The way each leutenant s able to know when he wll no longer receve messages can be shown by nducton on k, snce for each sequence of leutenants j,..., 1 j k when k<m, a leutenant can receve one message of the form v : 0 : j1 :...: j n step 2). The requrement can be that f k Leutenant j k wll not send such a message he wll send a message reportng t. In ths case we can easly decde whether all messages has been receved or not. Assumpton A3 protects us from the case of a trator not playng along wth the above requrement and decdes not to send any message. A dfferent approach s to use tme-out for the absence of messages. Ths wll be further dscussed n Secton 6. Thrd, every leutenant treats only messages of the proper form and throws any message that has llegal structure (vald structure s v :0: j1 :...: j ). If we use k the method of sendng enough copes of the same message so that the recpent wll not have to copy t by hmself, then every leutenant who throws a message wth llegal structure actually throws all of ts copes. The number of copes from a sngle message that should exst f t was sgned by k leutenants s: n k 2 n k 3... n m 2. ( )( ) ( ) Fgure 9 llustrates algorthm SM(1) n the case of three generals at whch the commander s a trator. The upper half shows the messages sent n step a. of the algorthm, and the lower half the messages sent n step b. The sets V obtaned by each leutenant are: V1 = { a, r}, V2 = { r, a}. Snce V1 = V and the fact 2 that they use the same chose functon, Leutenants 1 and 2 wll decde to follow the same order (chose(a, r)). In ths case, unlke n the examples of OM algorthm, both loyal leutenants can know who s the trator snce by assumpton A4 no message can be forged, and Leuteants 1 and 2 receve dfferent orders wth the same sgnature.

15

16 Fgure 10 llustrates SM(2) n case the commander and Leutenant 3 are trators. The upper half shows the messages sent by the commander, and although he s a trator he sent the same values to the loyal leutenants.the value he sent to the trator makes no dfference. In the lower half we see the next step of the algorthm, whch results n the followng values obtaned by both loyal leutenants: V = V = a r. Leadng to the same decson by both loyal leutenants, snce they { } 1 2, wll use chose(a, r). In algorthm SM(m), the sgnature added by each leutenant to every message he sends has the purpose of acknowledgng the recept of the message. In the specal case when the leutenant s the m-th to add hs sgnature to the order he relays, t s unnecessary to add hs sgnature snce the recpent of the message wll not pass t forward and he knows who sent the message even wthout the sgnature. Hence, the leutenants do not need to add ther sgnature n the case of SM(1). Let us now prove the correctness of the algorthm: THEOREM 2: For any m, Algorthm SM(m) solves the Byzantne Generals Problem f there are at most m trators. PROOF: As before, we need to show the fulfllment of IC1 and IC2. Let us start wth IC2. Snce the commander s loyal he sends the same value, v, wth hs sgnature, v:0, to every leutenant. All loyal leutenants wll obtan the same order, v,, and snce trators cannot forge the commander's sgnature (or anyone's sgnature for the matter) the leutenants wll not receve other message wth a dfferent order, u, "from the commander" (wth hs forged sgnature). Hence, every Lleutenant wll obtan the same set V, whch conssts only from the sngle order v, resultng them to make of the same decson usng chose functon. Snce all loyal leutenants wll obey the same order we have proved IC2. We wll now prove IC1 when the commander s a trator (IC2 covers the case he s loyal). In order to prove IC1 we must show that every two loyal leutenants obtan the same set of orders, V. Let us consder two loyal leutenants, and j. They wll follow the same order n step 3) f they have obtaned the orders sets, meanng V = V j. Therefore, n order to prove IC1 t suffces to prove that f puts an order v nto V n step 2), then j must put the same order v nto V j. Leutenant j wll put order v nto V j f he wll receve a properly sgned message wth the value v. If Leutenant receved the order v n step 2)a. then he sends t to j n step 2)a.. and by A1, j wll receve t. If adds the order to V n step 2)b., v : 0: j :...: j k. If j's sgnature then he must receve a frst message of the form 1 appears n the message t means he has receved t v earler. If not, we consder two cases: 1. k<m: In ths case, sends the message v :0: j1 :...: jk : to j, whch means j has receved the order v. 2. k=m: There are m-1 trators among the leutenants (the commander s the m-th trator), whch means that at least one of Leutenants j,..., 1 j m s loyal. Snce he s loyal he must have sent the value v to Leutenant j when he frst receved t, and by A1, j has receved the value.

17 We see that n every case Leutenants j and has receved v, whch shows the fulfllment of IC1 and IC2 by algorthm SM, resultng n the proof of ts correctness. 5. MISSING COMMUNICATION PATHS In prevous sectons we have assumed a clque topology, meanng there s a communcton path between each two generals. Ths secton wll dscuss the case of mssng paths. If every general wll be represented by node and a bdrectonal communcaton path by an edge, we wll receve a smple, undrectonal graph descrbng the network. Followng ths small preface of the secton we wll ntroduce algorthms OM(m) and SM(m) to more general graphs usng the assumpton that the network graph s completely connected. Before presentng the more general OM algorthm we frst defne the followng defnton n whch two generals wll be called neghbours f there s a communcaton path between them: DEFINITION 1: a) A set of nodes { 1,..., p } s sad to be a regular set of neghbours of a node f:. each j s a neghbour of, and. for any general k dfferent from, there exst paths γ j, k from to k not passng through such that any two dfferent paths j γ have no node n common other than k. j, k b) The graph G s sad to be p-regular f every node has a regular set of neghbours consstng of p dstnct nodes. We wll futher explan the term 'regular set'. In ths type of set every path from the neghbours of to k have no common nodes. Fgure 11 presents a graph wth a node that has a regular set of sze 4. In case we remove a node from a p-regular

18 graph there are two possble scenaros. The frst s when the removed node, n, s a neghbour of (one of the red nodes n Fgure 11), n the worst case of ths scenaro n s one of the nodes n the regular set of, after ts removal there reman p-1 nodes n the regular set of. Whch wll leave the graph (p-1)-regular. The second scenaro, n s not one of the neghbours of, and n the worst case of ths scenaro n s one of the nodes on a path from a node n the regular set of r, after ts removal there reman p-1 nodes n the regular set of, resultng n (p-1)-regular. We have found that removng a node from a p-regular graph wll leave t (p-1)-regualar Fgure 12: 3-regular graph Fgure 13: A graph that s not 3-regular Fgure 11 shows an example of a 3-regular graph, every node has a regular set of sze three. Fgure 12 shows an example of a graph that s not 3-regular, snce the central node has no regular set of neghbours contanng three nodes. Note that a 3-regular graph has at least four nodes, and n general, p-regular graph has at least p+1 nodes. We now extend OM(m) so that t wll solve the Byzantne Generals Problem when there are mssng communcaton paths. The condton from the generals network graph G s to be 3m-regular, whch, as prevously stated, contans at least 3m+1 nodes. For all postve ntegers m and p, we defne the algorthm OM(m, p) as follows when the graph G of generals s p-regular. In case G s not p- regular OM(m, p) s not defned. The defnton wll be smlar to OM(m), by nducton on m: Algorthm OM(m, p): 0) Choose a regular set N of neghbours of the commander consstng of p leutenants. 1) The commander sends hs value to every leutenant n N. 2) For each n N, let v be the value Leutenant receves from the commander, or else RETREAT f he receves no value. Leutenant sends v to every other leutenant k as follows: a. If m=1, then by sendng the value along the path j, k γ whose exstence s guaranteed by part a). of Defnton 1. b. If m>1, then by actng as the commander n the algorthm OM(m-1, p-1), wth the graph of generals obtaned by removng the orgnal commander from G.

19 3) For each k, and each n N wth k, let v be the value Leutenant k receved from Leutenant n step 2), or RETREAT f he receved no value. Leutenant k uses the value majorty( v,..., v 1 ), where p N={,..., 1 p}. As we have prevously seen, removng a sngle node from a p-regular graph wll leave the graph (p-1)-regular, whch means we can apply the algorthm OM(m-1, p-1) n step 2)b. on a lawful graph. Let us now prove that OM(m, 3m) solves the Byzantne Generals Problem f there are at most m trators. The proof s smlar to the proof of the algorthm OM(m) and wll not be fully presented. It begns wth the followng extenson of Lemma 1: LEMMA 2: For any m>0 and any p 2k+ m, Algorthm OM(m, p) satsfes IC2 f there are at most k trators. PROOF: We frst examne the case of m=1. Each leutenant obtans the value majorty( v 1,..., v p ), whle every two values v, v j he receved was sent to hm along dfferent paths. Snce there are only k trators and we know that p 2k+ 1, more than half of the paths connect the leutenant drectly to other loyal leutenants. The commander s loyal (f not then the proof s trval), meanng more v v wll be equal to the than half of the values each leutenant receved n 1,..., p value sent by the commander, provng IC2. Now assume correctness for m-1, m>1 and prove t for m. Snce the commander s loyal, all p leutenants n N obtan the correct value. p 2k+ m> 2k whch means that the majorty of these leutenants are loyal. By nducton hypothess, each of the loyal leutenants sends the correct value to every loyal leutenant. Hence, each loyal leutenant has obtaned majorty of correct values, resultng n the correct order n step 3). Next we wll show a theorem from whch the correctness of algorthm OM(m, 3m) s mmedate. THEOREM 3: For any m>0 and any p 3m, Algorthm OM(m, p) solves the Byzantne Generals Problem f there are at most m trators. PROOF: Usng Lemma 2 and choosng k=m we receve that OM(m, p) satsfes IC2 and IC1 n case the commander s loyal. It s left to prove IC1 n the case that the commander s a trator. In order to prove IC1 n ths case we need to show that each loyal leutenant receves the same set of values v n step 3). If m=1, the commander s the only trator and all the rest leutenants are loyal, whch means they wll all get same value v from N. Eventually all loyal leutenants wll compute the same majorty and obtan the same order. If m>1, we know that p 3m p 1 3 m 1. Let us assume that OM(m-1, p-1) satsfes Theorem ( ) 3, and prove correctness for OM(m, p). We know that every leutenant gets the value each leutenant n N sent. Therefore, every loyal leutenant wll have the same nput for the same decson functon resultng n every leutenant arrvng the same decson.

20 We requre 3m-regularty from graph G whch s a strong connectvty hypothess. In fact, f there are 3m+1 generals the graph G s a complete graph and Algorthm OM(m, 3m) s reduced to Algorthm OM(m). The weakest connectvty hypothess for whch the Byzantne Generals Problem s solvable s that the sub-graph formed by the loyal generals s connected, snce a trator servng as the only ntermedate between two sub-graphs of G can easly block the messages and smply dsjont the sub-graphs. Under the above assumpton we wll show that SM(n-2) s a soluton to the Byzantne Generals Problem where n s the number of generals, regardless the number of trators. A few modfcatons must be made to match the new problem's restrctons, n whch the generals cannot send messages drectly every other general. We wll defne that the commander sends hs sgned order only to hs neghbourng leutenants, and each leutenant, n step 2)b., wll send hs message to every neghborng leutenant that s not among the j r. In order to contnue wth the presenton of the more general soluton we frst defne the dameter of graph, whch s the smallest number d, such that any two nodes are connected by a path contanng at most d arcs. If the graph s not connected, the dameter s. THEOREM 4: For any m and d, f there are at most m trators and the subgraph of loyal generals has dameter d, then Algorthm SM(m+d-1) (wth the above modfcaton) solves the Byzantne Generals Problem. PROOF: We frst consder the case of loyal commander. In ths case, when the commander sends hs message t wll go thorugh at most d-1 loyal leutenants and wll fnally reach every loyal leutenant. Snce all the letenants n the message's path were loyal t wll be relayed correctly n every step, and by relyng on assumpton A4 we know that the trator cannot forge a dfferent order. Ths proves IC2. In order to prove IC1 we need to prove the case of tratorous commander. We need to show that any order receved by Leutenant s also receved by Leutenant j (both loyal). Suppose receves an order v : 0: j1 :...: j, not sgend by j. k If k<m then wll send the message to every neghbour that has not yet receved the order and t wll be relayed to j wthn d-1 steps. If k m then one of the frst m sgners must be loyal (there are only m-1 tratorous leutenants) and send t to all of hs neghbours, whch wll eventually relay t to j wthn d-1 steps at most. COROLLARY: If the graph of loyal generals s connected, then SM(n-2) (as modfed above) solves the Byzantne Generals Problem for n generals. PROOF: Let d be the dameter of the graph of loyal generals, when d V. We also know that there are at most d loyal generals so there are fewer than n-d trators. If we choose m to be m= n d 1 we can derve the followng result m= n d 1 n 2= m+ d 1 and accordng Theorem 4 we have proved the corollary. We have assumed that the loyal generals graph s connected, but usng Theorem 4 we can show that even f the graph s not connected and there are at most m trators, then SM(m+d-1) has the followng propertes: 1) Any two loyal leutenants connected by a path of length at most d passng through only loyal leutenants wll obey the same order.

21 2) If the commander s loyal, then any loyal leutenant connected to hm by path of length m+d passng only through loyal leutenants wll obey the same order. 6. RELIABLE SYSTEMS Up to ths pont we have mostly dscussed the theoretcal aspects of the Byzantne Generals Problem, n ths secton we wll explan the practcal mplementons of the algorthms presented n prevous sectons. The man purpose of ths paper s to promote the noton of relable systems. Nowadays relablty n computer systems s manly acheved wth redundancy. By usng a few dfferent replcas of the system's processor to compute the same output and takng ther majorty vote we are able to obtan a sngle value. Our reference to the system's processor s not necessarlly to a sngle chp, t may very well be a whole data-base system wth multple computers and processng unts. The way we choose to calculate the output does not matter, t can be calculated whtn the system or by the recpent of the raw data hmself. Majorty's vote among the dfferent processors assums that every nonfaulty processor wll produce the same output, whch s true as long as they use the same nput. However, snce nput may come from dfferent components t s enough that a fracture of them, or even a sngle one, wll have a malfuncton and the nput wll not be the same for every processor, resultng n dfferent outputs. Surely t s not the only way to get conflctng nput. In case of lack of synchronzaton, when processors read a value whle t s changng, can lead to dfferent nputs snce the reads were not n the exact same tme. Ths problem can be solved smply by synchronzaton, wth the mechansm of 'mutual excluson' (mutex, semaphore etc.). We receve that n order for majorty votng to yeld n a relable system, the followng two condtons must be satsfed: 1) All nonfaulty processors must use the same nput value (so they produce the same output). 2) If the nput unt s nonfaulty, then all nonfaulty processors use the value t provdes as nput (so they produce the correct output). Snce the "leutenants" are processors, the "commander" s the nput unt and "loyal" means nonfaulty we can see that the above condtons are smply the 'Interactve consstency' condtons. Here rses a dfferent queston, but smlar to the one we have dscussed so far n ths secton: "How can we ensure that all processors receve the same nput?". We can try to connect all processors to the same wre, but a faulty nput unt can send border-lne nput value that can be nterpreted as '0' by some processors and as '1' by others. Snce the nput unt s also a system of some type, the subject of relablty comes once agan. If the nput unt s faulty then ts output wll be meanngless and all that s left for the Byzantne Generals soluton s to ensure that all processors wll use the same nput. In case the nput s mportant and mght be faulty, t can be replcated usng redundancy. Although redundancy and relablty are connected, t s stll not enough to use many copes of the same data n order to obtan relablty. We stll need to nsure that the nonfaulty processors wll use the redundant data to produce the same output. Let us now handle the case of a nonfaulty nput unt, but the processors obtan dfferent nput values snce they sample them at dfferent tmes and the value s constantly changng. Despte the ncoherency and lack of synchronzaton n obtang the nput we stll want all processors to use the same nput values. In case

22 the functons majorty or chose are taken to be medan functons, we wll receve that the nput value wll belong to the doman of values obtaned from the nput unt. As long as the values produced by the nput unt wll be n reasonable range then all nonfaulty processors wll get reasonable values. All of the above solutons were mostly n terms of Byzantne generals rather than n terms of computng systems. We wll now examne the applcaton of these solutons to relable computer systems. The problem s not the mplementaton of "general's" algorthm wth a processor, t s the mplementaton of a message passng system that meets assumptons A1-A3 for the Oral Messages algorthm and A1-A4 for Sgned Messages algorthm. We wll refer to every assumpton n order: A1. Ths assumpton states that every message sent by a nonfaulty processor s delvered correctly. In real systems, communcaton lnes can fal. The oral messages algorthms wth parameter m wll work correctly n the presence of at most m faulty processors. Therefore, f we assume that the falure of a communcaton lne from a sngle processor s the falure of the processor (meanng he s a "trator") then the algorthms wll work correctly and the problems are equvalent. Now we examne the case for the sgned messages algorthms. If we assume that a faulty processor cannot forge a sgned message (whch s bascally assumpton A4), whch we wll see that s very reasonable, then a communcaton lne falure wll result only n reducng the connectvty of the graph of generals and algorthm SM(m) wll stll work correctly and Theorem 4 remans vald. Ths s true snce a nonfaulty processor wll smply gnore messages that are llegal due to communcaton lne falure, whch s equvalent to a mssng communcaton path n the Byzantne Generals Problem. A2. Accordng to ths assumpton each recever of a message knows who sent t. But most mportantly s that a faulty processor wll not dsguse hmself as other nonfaulty processor. The consequence of ths demand s that all nterprocessor communcaton wll be done by fxed lnes rather than a message swtchng network, manly because the use of swtchng system network wll reopen the Byzantne Generals Problem as we must consder faulty communcaton nodes. In the case of sgned messages, assumpton A2 s not necessary snce A4 prevents a processor from successfully forgng dfferent processor's message and thus mpersonatng ths processor. A3. The requrement that the absence of a message wll be detected s covered wth ths assumpton. Snce the absence of a message can only be detected f t s faled to arrve whthn some fxed length of tme, we must use tme-out conventon to answer the demand n A3. The use of tme-out to satsfy A3 requres two assumptons: 1. There s a fxed maxmum tme needed for the generaton and transmsson of a message. 2. The sender and recever have clocks that are synchronzed to whtn some maxmum error. The frst assumpton s qute basc snce t s a rephrasng of the prevous concluson that each dedcated recpent of a message must know the maxmal watng tme. However, the second assumpton s not trval, but t s crucal n order to solve the Byzantne Generals Problem (ths or an equvalent one). Let us assume an algorthm n whch the generals take acton only n the followng crcumstances: At some fxed ntal tme (the same for all generals). Upon the recept of a message.