Is a bitcoin address collision possible if generating 90 million addresses every 4 hours?












1















I am running a test to see if I can obtain a successful bitcoin address collision after generating billions of addresses. I am not entirely sure how I would check them yet. Basically I have an extra 10TB hard drive and am running supervanitygen on my 32-core, 128gb ram linux work computer. It seems to be generating about 90,000,000 (90 million) addresses (address + private key pair) every 4 hours.



Do you think it's possible or likely to run into an address collision this way? Or perhaps if everyone in the world did the same thing? Or what if I waited 10-20 years, and then checked the addresses?



This is all just a test to see if the bitcoin system is secure enough for big investment decision.










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    1















    I am running a test to see if I can obtain a successful bitcoin address collision after generating billions of addresses. I am not entirely sure how I would check them yet. Basically I have an extra 10TB hard drive and am running supervanitygen on my 32-core, 128gb ram linux work computer. It seems to be generating about 90,000,000 (90 million) addresses (address + private key pair) every 4 hours.



    Do you think it's possible or likely to run into an address collision this way? Or perhaps if everyone in the world did the same thing? Or what if I waited 10-20 years, and then checked the addresses?



    This is all just a test to see if the bitcoin system is secure enough for big investment decision.










    share|improve this question







    New contributor




    Anonymous User is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.























      1












      1








      1


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      I am running a test to see if I can obtain a successful bitcoin address collision after generating billions of addresses. I am not entirely sure how I would check them yet. Basically I have an extra 10TB hard drive and am running supervanitygen on my 32-core, 128gb ram linux work computer. It seems to be generating about 90,000,000 (90 million) addresses (address + private key pair) every 4 hours.



      Do you think it's possible or likely to run into an address collision this way? Or perhaps if everyone in the world did the same thing? Or what if I waited 10-20 years, and then checked the addresses?



      This is all just a test to see if the bitcoin system is secure enough for big investment decision.










      share|improve this question







      New contributor




      Anonymous User is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.












      I am running a test to see if I can obtain a successful bitcoin address collision after generating billions of addresses. I am not entirely sure how I would check them yet. Basically I have an extra 10TB hard drive and am running supervanitygen on my 32-core, 128gb ram linux work computer. It seems to be generating about 90,000,000 (90 million) addresses (address + private key pair) every 4 hours.



      Do you think it's possible or likely to run into an address collision this way? Or perhaps if everyone in the world did the same thing? Or what if I waited 10-20 years, and then checked the addresses?



      This is all just a test to see if the bitcoin system is secure enough for big investment decision.







      bitcoin-core address-generation cryptography vanitygen






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      Anonymous User is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      Anonymous UserAnonymous User

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          2 Answers
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          5














          Because of the Birthday paradox, you only need 280 addresses (despite there existing 2160 different address combinations) before a collision becomes probable.



          Thankfully, that is still an enormous number. At 90 million addresses per 4 hours, it will take about 445 times the age of the universe to reach that number.



          It's also irrelevant. Even if anyone - or everyone - generates 90 million addresses per 4 hours, there aren't that many addresses receiving funds. Thus only actually used addresses matter.






          share|improve this answer





















          • 3





            Also, in that case the collision would almost certainly be between two addresses you yourself generated... which isn't usually very useful for an attacker.

            – G. Maxwell
            4 hours ago











          • I'm confused at how you arrived at 2^80 for the birthday paradox. That's the square root of 2^160. Is that the calculation for all such schemes where the birthday paradox applies? In the wikipedia page, 23 marks the 50% line (is that what you mean be probable?), but it is 4 more than square root 366, a full 1% of the total more.

            – fredsbend
            9 mins ago











          • @fredsbend, the actual formula for the birthday paradox is complicated, but as the numbers get bigger, taking the square root becomes an increasingly better approximation to it. When talking about cryptography, the numbers are so huge that the difference isn't worth worrying about.

            – Mark
            1 min ago











          • @fredsbend Technically, the square root of the number of possibilities gives you a 39.3% probability for collisions (1-exp(-1/2)), that's about where you see the probability go from negligable to negligably not. 1/2 sqrt(n) gives you 11.75%. 2 sqrt(n) gives you 86.5%. In cryptography the exact factor is usually ignored, and we just say that you need approximately sqrt(n) elements to see a probable collision.

            – Pieter Wuille
            1 min ago



















          3














          A very relevant answer can be found here: Is Each Bitcoin Address Unique?



          This is a question of the birthday attack on the hashes. Bitcoin addresses (assuming the "normal" style starting with a 1) encode 160 bit hashes, so the output space has a possible 2^160 hashes. Because its a hash function, we assume all outputs have equal probability of being output.



          At the rate you gave in the OP, you would generate 1.971 * 10^11 addresses in 1 year. Using the approximate birthday formula n^2 / 2H, the probability of a collision in 1 year is thus roughly on the order of 10^-26. That is, basically 0. In 20 years that probability becomes around the order of 10^-24. Still basically 0. If all 8 billion people in the world generated addresses at that rate for 20 years, in total 3.1536 * 10^22 addresses would be generated. Even then, the probability of a collision is only around 0.0003. And remember that a collision like this is completely different from a targeted attack to find the private key of a specific address. If there was a collision, chances are overwhelming that there would be no funds ever sent to the address anyway, making the attack useless.






          share|improve this answer
























          • "A collision can't be a collision if there were no funds ever sent to the address" in regards to the last sentence of your answer. Other than that, great answer! Thank you!

            – Anonymous User
            1 hour ago













          • @AnonymousUser You say you're able to generate 90 million addresses per 4 hours. MeshCollider is pointing out that that's actually irrelevant, as there aren't 90 million addresses being paid at that rate.

            – Pieter Wuille
            1 hour ago











          • Oh gotcha, you may want to be more clear in that. Thanks again.

            – Anonymous User
            1 hour ago











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          2 Answers
          2






          active

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          2 Answers
          2






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          5














          Because of the Birthday paradox, you only need 280 addresses (despite there existing 2160 different address combinations) before a collision becomes probable.



          Thankfully, that is still an enormous number. At 90 million addresses per 4 hours, it will take about 445 times the age of the universe to reach that number.



          It's also irrelevant. Even if anyone - or everyone - generates 90 million addresses per 4 hours, there aren't that many addresses receiving funds. Thus only actually used addresses matter.






          share|improve this answer





















          • 3





            Also, in that case the collision would almost certainly be between two addresses you yourself generated... which isn't usually very useful for an attacker.

            – G. Maxwell
            4 hours ago











          • I'm confused at how you arrived at 2^80 for the birthday paradox. That's the square root of 2^160. Is that the calculation for all such schemes where the birthday paradox applies? In the wikipedia page, 23 marks the 50% line (is that what you mean be probable?), but it is 4 more than square root 366, a full 1% of the total more.

            – fredsbend
            9 mins ago











          • @fredsbend, the actual formula for the birthday paradox is complicated, but as the numbers get bigger, taking the square root becomes an increasingly better approximation to it. When talking about cryptography, the numbers are so huge that the difference isn't worth worrying about.

            – Mark
            1 min ago











          • @fredsbend Technically, the square root of the number of possibilities gives you a 39.3% probability for collisions (1-exp(-1/2)), that's about where you see the probability go from negligable to negligably not. 1/2 sqrt(n) gives you 11.75%. 2 sqrt(n) gives you 86.5%. In cryptography the exact factor is usually ignored, and we just say that you need approximately sqrt(n) elements to see a probable collision.

            – Pieter Wuille
            1 min ago
















          5














          Because of the Birthday paradox, you only need 280 addresses (despite there existing 2160 different address combinations) before a collision becomes probable.



          Thankfully, that is still an enormous number. At 90 million addresses per 4 hours, it will take about 445 times the age of the universe to reach that number.



          It's also irrelevant. Even if anyone - or everyone - generates 90 million addresses per 4 hours, there aren't that many addresses receiving funds. Thus only actually used addresses matter.






          share|improve this answer





















          • 3





            Also, in that case the collision would almost certainly be between two addresses you yourself generated... which isn't usually very useful for an attacker.

            – G. Maxwell
            4 hours ago











          • I'm confused at how you arrived at 2^80 for the birthday paradox. That's the square root of 2^160. Is that the calculation for all such schemes where the birthday paradox applies? In the wikipedia page, 23 marks the 50% line (is that what you mean be probable?), but it is 4 more than square root 366, a full 1% of the total more.

            – fredsbend
            9 mins ago











          • @fredsbend, the actual formula for the birthday paradox is complicated, but as the numbers get bigger, taking the square root becomes an increasingly better approximation to it. When talking about cryptography, the numbers are so huge that the difference isn't worth worrying about.

            – Mark
            1 min ago











          • @fredsbend Technically, the square root of the number of possibilities gives you a 39.3% probability for collisions (1-exp(-1/2)), that's about where you see the probability go from negligable to negligably not. 1/2 sqrt(n) gives you 11.75%. 2 sqrt(n) gives you 86.5%. In cryptography the exact factor is usually ignored, and we just say that you need approximately sqrt(n) elements to see a probable collision.

            – Pieter Wuille
            1 min ago














          5












          5








          5







          Because of the Birthday paradox, you only need 280 addresses (despite there existing 2160 different address combinations) before a collision becomes probable.



          Thankfully, that is still an enormous number. At 90 million addresses per 4 hours, it will take about 445 times the age of the universe to reach that number.



          It's also irrelevant. Even if anyone - or everyone - generates 90 million addresses per 4 hours, there aren't that many addresses receiving funds. Thus only actually used addresses matter.






          share|improve this answer















          Because of the Birthday paradox, you only need 280 addresses (despite there existing 2160 different address combinations) before a collision becomes probable.



          Thankfully, that is still an enormous number. At 90 million addresses per 4 hours, it will take about 445 times the age of the universe to reach that number.



          It's also irrelevant. Even if anyone - or everyone - generates 90 million addresses per 4 hours, there aren't that many addresses receiving funds. Thus only actually used addresses matter.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 1 hour ago

























          answered 4 hours ago









          Pieter WuillePieter Wuille

          45.8k394154




          45.8k394154








          • 3





            Also, in that case the collision would almost certainly be between two addresses you yourself generated... which isn't usually very useful for an attacker.

            – G. Maxwell
            4 hours ago











          • I'm confused at how you arrived at 2^80 for the birthday paradox. That's the square root of 2^160. Is that the calculation for all such schemes where the birthday paradox applies? In the wikipedia page, 23 marks the 50% line (is that what you mean be probable?), but it is 4 more than square root 366, a full 1% of the total more.

            – fredsbend
            9 mins ago











          • @fredsbend, the actual formula for the birthday paradox is complicated, but as the numbers get bigger, taking the square root becomes an increasingly better approximation to it. When talking about cryptography, the numbers are so huge that the difference isn't worth worrying about.

            – Mark
            1 min ago











          • @fredsbend Technically, the square root of the number of possibilities gives you a 39.3% probability for collisions (1-exp(-1/2)), that's about where you see the probability go from negligable to negligably not. 1/2 sqrt(n) gives you 11.75%. 2 sqrt(n) gives you 86.5%. In cryptography the exact factor is usually ignored, and we just say that you need approximately sqrt(n) elements to see a probable collision.

            – Pieter Wuille
            1 min ago














          • 3





            Also, in that case the collision would almost certainly be between two addresses you yourself generated... which isn't usually very useful for an attacker.

            – G. Maxwell
            4 hours ago











          • I'm confused at how you arrived at 2^80 for the birthday paradox. That's the square root of 2^160. Is that the calculation for all such schemes where the birthday paradox applies? In the wikipedia page, 23 marks the 50% line (is that what you mean be probable?), but it is 4 more than square root 366, a full 1% of the total more.

            – fredsbend
            9 mins ago











          • @fredsbend, the actual formula for the birthday paradox is complicated, but as the numbers get bigger, taking the square root becomes an increasingly better approximation to it. When talking about cryptography, the numbers are so huge that the difference isn't worth worrying about.

            – Mark
            1 min ago











          • @fredsbend Technically, the square root of the number of possibilities gives you a 39.3% probability for collisions (1-exp(-1/2)), that's about where you see the probability go from negligable to negligably not. 1/2 sqrt(n) gives you 11.75%. 2 sqrt(n) gives you 86.5%. In cryptography the exact factor is usually ignored, and we just say that you need approximately sqrt(n) elements to see a probable collision.

            – Pieter Wuille
            1 min ago








          3




          3





          Also, in that case the collision would almost certainly be between two addresses you yourself generated... which isn't usually very useful for an attacker.

          – G. Maxwell
          4 hours ago





          Also, in that case the collision would almost certainly be between two addresses you yourself generated... which isn't usually very useful for an attacker.

          – G. Maxwell
          4 hours ago













          I'm confused at how you arrived at 2^80 for the birthday paradox. That's the square root of 2^160. Is that the calculation for all such schemes where the birthday paradox applies? In the wikipedia page, 23 marks the 50% line (is that what you mean be probable?), but it is 4 more than square root 366, a full 1% of the total more.

          – fredsbend
          9 mins ago





          I'm confused at how you arrived at 2^80 for the birthday paradox. That's the square root of 2^160. Is that the calculation for all such schemes where the birthday paradox applies? In the wikipedia page, 23 marks the 50% line (is that what you mean be probable?), but it is 4 more than square root 366, a full 1% of the total more.

          – fredsbend
          9 mins ago













          @fredsbend, the actual formula for the birthday paradox is complicated, but as the numbers get bigger, taking the square root becomes an increasingly better approximation to it. When talking about cryptography, the numbers are so huge that the difference isn't worth worrying about.

          – Mark
          1 min ago





          @fredsbend, the actual formula for the birthday paradox is complicated, but as the numbers get bigger, taking the square root becomes an increasingly better approximation to it. When talking about cryptography, the numbers are so huge that the difference isn't worth worrying about.

          – Mark
          1 min ago













          @fredsbend Technically, the square root of the number of possibilities gives you a 39.3% probability for collisions (1-exp(-1/2)), that's about where you see the probability go from negligable to negligably not. 1/2 sqrt(n) gives you 11.75%. 2 sqrt(n) gives you 86.5%. In cryptography the exact factor is usually ignored, and we just say that you need approximately sqrt(n) elements to see a probable collision.

          – Pieter Wuille
          1 min ago





          @fredsbend Technically, the square root of the number of possibilities gives you a 39.3% probability for collisions (1-exp(-1/2)), that's about where you see the probability go from negligable to negligably not. 1/2 sqrt(n) gives you 11.75%. 2 sqrt(n) gives you 86.5%. In cryptography the exact factor is usually ignored, and we just say that you need approximately sqrt(n) elements to see a probable collision.

          – Pieter Wuille
          1 min ago











          3














          A very relevant answer can be found here: Is Each Bitcoin Address Unique?



          This is a question of the birthday attack on the hashes. Bitcoin addresses (assuming the "normal" style starting with a 1) encode 160 bit hashes, so the output space has a possible 2^160 hashes. Because its a hash function, we assume all outputs have equal probability of being output.



          At the rate you gave in the OP, you would generate 1.971 * 10^11 addresses in 1 year. Using the approximate birthday formula n^2 / 2H, the probability of a collision in 1 year is thus roughly on the order of 10^-26. That is, basically 0. In 20 years that probability becomes around the order of 10^-24. Still basically 0. If all 8 billion people in the world generated addresses at that rate for 20 years, in total 3.1536 * 10^22 addresses would be generated. Even then, the probability of a collision is only around 0.0003. And remember that a collision like this is completely different from a targeted attack to find the private key of a specific address. If there was a collision, chances are overwhelming that there would be no funds ever sent to the address anyway, making the attack useless.






          share|improve this answer
























          • "A collision can't be a collision if there were no funds ever sent to the address" in regards to the last sentence of your answer. Other than that, great answer! Thank you!

            – Anonymous User
            1 hour ago













          • @AnonymousUser You say you're able to generate 90 million addresses per 4 hours. MeshCollider is pointing out that that's actually irrelevant, as there aren't 90 million addresses being paid at that rate.

            – Pieter Wuille
            1 hour ago











          • Oh gotcha, you may want to be more clear in that. Thanks again.

            – Anonymous User
            1 hour ago
















          3














          A very relevant answer can be found here: Is Each Bitcoin Address Unique?



          This is a question of the birthday attack on the hashes. Bitcoin addresses (assuming the "normal" style starting with a 1) encode 160 bit hashes, so the output space has a possible 2^160 hashes. Because its a hash function, we assume all outputs have equal probability of being output.



          At the rate you gave in the OP, you would generate 1.971 * 10^11 addresses in 1 year. Using the approximate birthday formula n^2 / 2H, the probability of a collision in 1 year is thus roughly on the order of 10^-26. That is, basically 0. In 20 years that probability becomes around the order of 10^-24. Still basically 0. If all 8 billion people in the world generated addresses at that rate for 20 years, in total 3.1536 * 10^22 addresses would be generated. Even then, the probability of a collision is only around 0.0003. And remember that a collision like this is completely different from a targeted attack to find the private key of a specific address. If there was a collision, chances are overwhelming that there would be no funds ever sent to the address anyway, making the attack useless.






          share|improve this answer
























          • "A collision can't be a collision if there were no funds ever sent to the address" in regards to the last sentence of your answer. Other than that, great answer! Thank you!

            – Anonymous User
            1 hour ago













          • @AnonymousUser You say you're able to generate 90 million addresses per 4 hours. MeshCollider is pointing out that that's actually irrelevant, as there aren't 90 million addresses being paid at that rate.

            – Pieter Wuille
            1 hour ago











          • Oh gotcha, you may want to be more clear in that. Thanks again.

            – Anonymous User
            1 hour ago














          3












          3








          3







          A very relevant answer can be found here: Is Each Bitcoin Address Unique?



          This is a question of the birthday attack on the hashes. Bitcoin addresses (assuming the "normal" style starting with a 1) encode 160 bit hashes, so the output space has a possible 2^160 hashes. Because its a hash function, we assume all outputs have equal probability of being output.



          At the rate you gave in the OP, you would generate 1.971 * 10^11 addresses in 1 year. Using the approximate birthday formula n^2 / 2H, the probability of a collision in 1 year is thus roughly on the order of 10^-26. That is, basically 0. In 20 years that probability becomes around the order of 10^-24. Still basically 0. If all 8 billion people in the world generated addresses at that rate for 20 years, in total 3.1536 * 10^22 addresses would be generated. Even then, the probability of a collision is only around 0.0003. And remember that a collision like this is completely different from a targeted attack to find the private key of a specific address. If there was a collision, chances are overwhelming that there would be no funds ever sent to the address anyway, making the attack useless.






          share|improve this answer













          A very relevant answer can be found here: Is Each Bitcoin Address Unique?



          This is a question of the birthday attack on the hashes. Bitcoin addresses (assuming the "normal" style starting with a 1) encode 160 bit hashes, so the output space has a possible 2^160 hashes. Because its a hash function, we assume all outputs have equal probability of being output.



          At the rate you gave in the OP, you would generate 1.971 * 10^11 addresses in 1 year. Using the approximate birthday formula n^2 / 2H, the probability of a collision in 1 year is thus roughly on the order of 10^-26. That is, basically 0. In 20 years that probability becomes around the order of 10^-24. Still basically 0. If all 8 billion people in the world generated addresses at that rate for 20 years, in total 3.1536 * 10^22 addresses would be generated. Even then, the probability of a collision is only around 0.0003. And remember that a collision like this is completely different from a targeted attack to find the private key of a specific address. If there was a collision, chances are overwhelming that there would be no funds ever sent to the address anyway, making the attack useless.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 4 hours ago









          MeshColliderMeshCollider

          8,24131239




          8,24131239













          • "A collision can't be a collision if there were no funds ever sent to the address" in regards to the last sentence of your answer. Other than that, great answer! Thank you!

            – Anonymous User
            1 hour ago













          • @AnonymousUser You say you're able to generate 90 million addresses per 4 hours. MeshCollider is pointing out that that's actually irrelevant, as there aren't 90 million addresses being paid at that rate.

            – Pieter Wuille
            1 hour ago











          • Oh gotcha, you may want to be more clear in that. Thanks again.

            – Anonymous User
            1 hour ago



















          • "A collision can't be a collision if there were no funds ever sent to the address" in regards to the last sentence of your answer. Other than that, great answer! Thank you!

            – Anonymous User
            1 hour ago













          • @AnonymousUser You say you're able to generate 90 million addresses per 4 hours. MeshCollider is pointing out that that's actually irrelevant, as there aren't 90 million addresses being paid at that rate.

            – Pieter Wuille
            1 hour ago











          • Oh gotcha, you may want to be more clear in that. Thanks again.

            – Anonymous User
            1 hour ago

















          "A collision can't be a collision if there were no funds ever sent to the address" in regards to the last sentence of your answer. Other than that, great answer! Thank you!

          – Anonymous User
          1 hour ago







          "A collision can't be a collision if there were no funds ever sent to the address" in regards to the last sentence of your answer. Other than that, great answer! Thank you!

          – Anonymous User
          1 hour ago















          @AnonymousUser You say you're able to generate 90 million addresses per 4 hours. MeshCollider is pointing out that that's actually irrelevant, as there aren't 90 million addresses being paid at that rate.

          – Pieter Wuille
          1 hour ago





          @AnonymousUser You say you're able to generate 90 million addresses per 4 hours. MeshCollider is pointing out that that's actually irrelevant, as there aren't 90 million addresses being paid at that rate.

          – Pieter Wuille
          1 hour ago













          Oh gotcha, you may want to be more clear in that. Thanks again.

          – Anonymous User
          1 hour ago





          Oh gotcha, you may want to be more clear in that. Thanks again.

          – Anonymous User
          1 hour ago










          Anonymous User is a new contributor. Be nice, and check out our Code of Conduct.










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