What is a^b and (a&b)<<1?












8















I was doing this question in leetcode.



Request:




Calculate the sum of two integers a and b, but you are not allowed to use the operator + and -.




I can't understand the solution it gave



Could someone explain how this getSum function works?



Here is answer's JS:




var getSum=function(a,b){

         const Sum=a^b;//I can't understand it.Please give me an example to understand it

         const carry=(a&b)<<1;//I can't understand it too

               if(!carry){

                    return Sum

               }

        return getSum(Sum,carry);

};

console.log(getSum(5,1));








javascript







share|improve this question



























    8















    I was doing this question in leetcode.



    Request:




    Calculate the sum of two integers a and b, but you are not allowed to use the operator + and -.




    I can't understand the solution it gave



    Could someone explain how this getSum function works?



    Here is answer's JS:




    var getSum=function(a,b){
    
         const Sum=a^b;//I can't understand it.Please give me an example to understand it
    
         const carry=(a&b)<<1;//I can't understand it too
    
               if(!carry){
    
                    return Sum
    
               }
    
        return getSum(Sum,carry);
    
};
    
console.log(getSum(5,1));








    javascript







    share|improve this question

























      8












      8








      8


      1






      I was doing this question in leetcode.



      Request:




      Calculate the sum of two integers a and b, but you are not allowed to use the operator + and -.




      I can't understand the solution it gave



      Could someone explain how this getSum function works?



      Here is answer's JS:




      var getSum=function(a,b){
      
         const Sum=a^b;//I can't understand it.Please give me an example to understand it
      
         const carry=(a&b)<<1;//I can't understand it too
      
               if(!carry){
      
                    return Sum
      
               }
      
        return getSum(Sum,carry);
      
};
      
console.log(getSum(5,1));








      javascript







      share|improve this question














      I was doing this question in leetcode.



      Request:




      Calculate the sum of two integers a and b, but you are not allowed to use the operator + and -.




      I can't understand the solution it gave



      Could someone explain how this getSum function works?



      Here is answer's JS:




      var getSum=function(a,b){
      
         const Sum=a^b;//I can't understand it.Please give me an example to understand it
      
         const carry=(a&b)<<1;//I can't understand it too
      
               if(!carry){
      
                    return Sum
      
               }
      
        return getSum(Sum,carry);
      
};
      
console.log(getSum(5,1));








      var getSum=function(a,b){
      
         const Sum=a^b;//I can't understand it.Please give me an example to understand it
      
         const carry=(a&b)<<1;//I can't understand it too
      
               if(!carry){
      
                    return Sum
      
               }
      
        return getSum(Sum,carry);
      
};
      
console.log(getSum(5,1));





      var getSum=function(a,b){
      
         const Sum=a^b;//I can't understand it.Please give me an example to understand it
      
         const carry=(a&b)<<1;//I can't understand it too
      
               if(!carry){
      
                    return Sum
      
               }
      
        return getSum(Sum,carry);
      
};
      
console.log(getSum(5,1));





      javascript




      javascript






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 46 mins ago









      JackyJacky

      1758




      1758
























          3 Answers
          3






          active

          oldest

          votes


















          8














          Let's imagine that a = 3 and b = 5



          In binary notation they are a = 0011 and b = 0101



          XOR:
          a^b is XOR operator. When compare two bits it returns 0 if they are same and 1 if they are different. 01^10 => 11



          So when we're doing a^b result will be 0110 (6 in decimal)



          AND + SHIFT



          a&b performs logical AND operation. It returns 1 only when a = b = 1.



          In our case the result is 0001



          << shifts it(adds 0 on the right side) and result became 0010 which sets carry variable true. (only 0000 will be false).



          Next iterations:



          Everything repeats but now a = 0110 and b = 0010 (Sum and carry from last execution)



          Now a^b = 0100 and (a&b)<<1 = 0100



          Repeating again.



          Now a^b = 0000 and (a&b)<<1 = 1000



          And again.



          Now a^b = 1000 and (a&b)<<1 = 0000. Now carry is finally false. And we're returning 1000 which is decimal 8.



          Everything worked fine since 3+5=8






          share|improve this answer





















          • 2





            Thanks for your explanation!

            – Jacky
            8 mins ago











          • You're welcome!

            – vicodin
            8 mins ago






          • 1





            Great explanation! I always find the bitwise operations hard to understand

            – Francisco Hanna
            2 mins ago



















          1














           int result = p ^ q; // XOR Operator, + without carry 0+0=0, 0+1=1+0=1, 1+1=0
          
int carry = (p & q) << 1; // Left Shift, 1+1=2
          
if (carry != 0) {
          
    return getSum(result, carry);
          
}
          
return result;


          Start By p=5,q=6. Then the XOR would be,



          0101
          
0110
          
------
          
0011


          So, XORing results in (0011) which is actually 3 in decimal. Then ANDing p and q we get,



          0101
          
0110
          
-------
          
0100


          We get 4 (100 in binary) by ANDing 5 & 6, now if we left shift this value by 1, we get



           0100<<1=1000


          So we get 8 (1000 in binary) after first recursion.As the result (carry variable) isnt zero, lets recursion again by xor value and carry value.



          getSum(3, 8);


          So, doing the first XORing we get,



          0011
          
1000
          
-------
          
1011


          The XORing this time yielded in 11 (1011 binary),so we perform the AND now,



          0011
          
1000
          
-------
          
0000


          We get all ZERO for ANDing 3 and 8, so this time the left shift operator also results in ZERO, as we have no 1 here which may give us a value by left shifing zeroes.
          As the carry variable is now Zero, we come to the end of recursion and the XORed value will be the Sum, which is 11 (1011 in Binary).



          Hope you get the working of the procedure. You can learn more by learning bitwise operation, as its the way the machine do the arithmatic operations.






          share|improve this answer

































            -1














            These are bitwise operations. They're close to hardware language.






            share|improve this answer



















            • 7





              Too short.You didn't explain why this works and how

              – Jacky
              38 mins ago






            • 1





              Link-only answers are discouraged here. Please add the relevant content in the answer itself.

              – Ian McLaird
              31 mins ago






            • 3





              It’s binary math. WofWca would have to give you a very very long explanation on binary to explain it. You should really read the W3 Schools article or watch some videos on Bitwise calculations. It’s the kind of thing that would be a few days of class work in college. The comment about link-only answers is fair, but I doubt you’ll grasp the concept from a StackOverflow post alone though.

              – Nate
              30 mins ago













            • @Nate Sorry,But I am not the computer science's student

              – Jacky
              27 mins ago











            Your Answer






            StackExchange.ifUsing("editor", function () {
            StackExchange.using("externalEditor", function () {
            StackExchange.using("snippets", function () {
            StackExchange.snippets.init();
            });
            });
            }, "code-snippets");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "1"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f55193135%2fwhat-is-ab-and-ab1%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            8














            Let's imagine that a = 3 and b = 5



            In binary notation they are a = 0011 and b = 0101



            XOR:
            a^b is XOR operator. When compare two bits it returns 0 if they are same and 1 if they are different. 01^10 => 11



            So when we're doing a^b result will be 0110 (6 in decimal)



            AND + SHIFT



            a&b performs logical AND operation. It returns 1 only when a = b = 1.



            In our case the result is 0001



            << shifts it(adds 0 on the right side) and result became 0010 which sets carry variable true. (only 0000 will be false).



            Next iterations:



            Everything repeats but now a = 0110 and b = 0010 (Sum and carry from last execution)



            Now a^b = 0100 and (a&b)<<1 = 0100



            Repeating again.



            Now a^b = 0000 and (a&b)<<1 = 1000



            And again.



            Now a^b = 1000 and (a&b)<<1 = 0000. Now carry is finally false. And we're returning 1000 which is decimal 8.



            Everything worked fine since 3+5=8






            share|improve this answer





















            • 2





              Thanks for your explanation!

              – Jacky
              8 mins ago











            • You're welcome!

              – vicodin
              8 mins ago






            • 1





              Great explanation! I always find the bitwise operations hard to understand

              – Francisco Hanna
              2 mins ago
















            8














            Let's imagine that a = 3 and b = 5



            In binary notation they are a = 0011 and b = 0101



            XOR:
            a^b is XOR operator. When compare two bits it returns 0 if they are same and 1 if they are different. 01^10 => 11



            So when we're doing a^b result will be 0110 (6 in decimal)



            AND + SHIFT



            a&b performs logical AND operation. It returns 1 only when a = b = 1.



            In our case the result is 0001



            << shifts it(adds 0 on the right side) and result became 0010 which sets carry variable true. (only 0000 will be false).



            Next iterations:



            Everything repeats but now a = 0110 and b = 0010 (Sum and carry from last execution)



            Now a^b = 0100 and (a&b)<<1 = 0100



            Repeating again.



            Now a^b = 0000 and (a&b)<<1 = 1000



            And again.



            Now a^b = 1000 and (a&b)<<1 = 0000. Now carry is finally false. And we're returning 1000 which is decimal 8.



            Everything worked fine since 3+5=8






            share|improve this answer





















            • 2





              Thanks for your explanation!

              – Jacky
              8 mins ago











            • You're welcome!

              – vicodin
              8 mins ago






            • 1





              Great explanation! I always find the bitwise operations hard to understand

              – Francisco Hanna
              2 mins ago














            8












            8








            8







            Let's imagine that a = 3 and b = 5



            In binary notation they are a = 0011 and b = 0101



            XOR:
            a^b is XOR operator. When compare two bits it returns 0 if they are same and 1 if they are different. 01^10 => 11



            So when we're doing a^b result will be 0110 (6 in decimal)



            AND + SHIFT



            a&b performs logical AND operation. It returns 1 only when a = b = 1.



            In our case the result is 0001



            << shifts it(adds 0 on the right side) and result became 0010 which sets carry variable true. (only 0000 will be false).



            Next iterations:



            Everything repeats but now a = 0110 and b = 0010 (Sum and carry from last execution)



            Now a^b = 0100 and (a&b)<<1 = 0100



            Repeating again.



            Now a^b = 0000 and (a&b)<<1 = 1000



            And again.



            Now a^b = 1000 and (a&b)<<1 = 0000. Now carry is finally false. And we're returning 1000 which is decimal 8.



            Everything worked fine since 3+5=8






            share|improve this answer















            Let's imagine that a = 3 and b = 5



            In binary notation they are a = 0011 and b = 0101



            XOR:
            a^b is XOR operator. When compare two bits it returns 0 if they are same and 1 if they are different. 01^10 => 11



            So when we're doing a^b result will be 0110 (6 in decimal)



            AND + SHIFT



            a&b performs logical AND operation. It returns 1 only when a = b = 1.



            In our case the result is 0001



            << shifts it(adds 0 on the right side) and result became 0010 which sets carry variable true. (only 0000 will be false).



            Next iterations:



            Everything repeats but now a = 0110 and b = 0010 (Sum and carry from last execution)



            Now a^b = 0100 and (a&b)<<1 = 0100



            Repeating again.



            Now a^b = 0000 and (a&b)<<1 = 1000



            And again.



            Now a^b = 1000 and (a&b)<<1 = 0000. Now carry is finally false. And we're returning 1000 which is decimal 8.



            Everything worked fine since 3+5=8







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 8 mins ago

























            answered 26 mins ago









            vicodinvicodin

            1,088624




            1,088624








            • 2





              Thanks for your explanation!

              – Jacky
              8 mins ago











            • You're welcome!

              – vicodin
              8 mins ago






            • 1





              Great explanation! I always find the bitwise operations hard to understand

              – Francisco Hanna
              2 mins ago














            • 2





              Thanks for your explanation!

              – Jacky
              8 mins ago











            • You're welcome!

              – vicodin
              8 mins ago






            • 1





              Great explanation! I always find the bitwise operations hard to understand

              – Francisco Hanna
              2 mins ago








            2




            2





            Thanks for your explanation!

            – Jacky
            8 mins ago





            Thanks for your explanation!

            – Jacky
            8 mins ago













            You're welcome!

            – vicodin
            8 mins ago





            You're welcome!

            – vicodin
            8 mins ago




            1




            1





            Great explanation! I always find the bitwise operations hard to understand

            – Francisco Hanna
            2 mins ago





            Great explanation! I always find the bitwise operations hard to understand

            – Francisco Hanna
            2 mins ago













            1














             int result = p ^ q; // XOR Operator, + without carry 0+0=0, 0+1=1+0=1, 1+1=0
            
int carry = (p & q) << 1; // Left Shift, 1+1=2
            
if (carry != 0) {
            
    return getSum(result, carry);
            
}
            
return result;


            Start By p=5,q=6. Then the XOR would be,



            0101
            
0110
            
------
            
0011


            So, XORing results in (0011) which is actually 3 in decimal. Then ANDing p and q we get,



            0101
            
0110
            
-------
            
0100


            We get 4 (100 in binary) by ANDing 5 & 6, now if we left shift this value by 1, we get



             0100<<1=1000


            So we get 8 (1000 in binary) after first recursion.As the result (carry variable) isnt zero, lets recursion again by xor value and carry value.



            getSum(3, 8);


            So, doing the first XORing we get,



            0011
            
1000
            
-------
            
1011


            The XORing this time yielded in 11 (1011 binary),so we perform the AND now,



            0011
            
1000
            
-------
            
0000


            We get all ZERO for ANDing 3 and 8, so this time the left shift operator also results in ZERO, as we have no 1 here which may give us a value by left shifing zeroes.
            As the carry variable is now Zero, we come to the end of recursion and the XORed value will be the Sum, which is 11 (1011 in Binary).



            Hope you get the working of the procedure. You can learn more by learning bitwise operation, as its the way the machine do the arithmatic operations.






            share|improve this answer






























              1














               int result = p ^ q; // XOR Operator, + without carry 0+0=0, 0+1=1+0=1, 1+1=0
              
int carry = (p & q) << 1; // Left Shift, 1+1=2
              
if (carry != 0) {
              
    return getSum(result, carry);
              
}
              
return result;


              Start By p=5,q=6. Then the XOR would be,



              0101
              
0110
              
------
              
0011


              So, XORing results in (0011) which is actually 3 in decimal. Then ANDing p and q we get,



              0101
              
0110
              
-------
              
0100


              We get 4 (100 in binary) by ANDing 5 & 6, now if we left shift this value by 1, we get



               0100<<1=1000


              So we get 8 (1000 in binary) after first recursion.As the result (carry variable) isnt zero, lets recursion again by xor value and carry value.



              getSum(3, 8);


              So, doing the first XORing we get,



              0011
              
1000
              
-------
              
1011


              The XORing this time yielded in 11 (1011 binary),so we perform the AND now,



              0011
              
1000
              
-------
              
0000


              We get all ZERO for ANDing 3 and 8, so this time the left shift operator also results in ZERO, as we have no 1 here which may give us a value by left shifing zeroes.
              As the carry variable is now Zero, we come to the end of recursion and the XORed value will be the Sum, which is 11 (1011 in Binary).



              Hope you get the working of the procedure. You can learn more by learning bitwise operation, as its the way the machine do the arithmatic operations.






              share|improve this answer




























                1












                1








                1







                 int result = p ^ q; // XOR Operator, + without carry 0+0=0, 0+1=1+0=1, 1+1=0
                
int carry = (p & q) << 1; // Left Shift, 1+1=2
                
if (carry != 0) {
                
    return getSum(result, carry);
                
}
                
return result;


                Start By p=5,q=6. Then the XOR would be,



                0101
                
0110
                
------
                
0011


                So, XORing results in (0011) which is actually 3 in decimal. Then ANDing p and q we get,



                0101
                
0110
                
-------
                
0100


                We get 4 (100 in binary) by ANDing 5 & 6, now if we left shift this value by 1, we get



                 0100<<1=1000


                So we get 8 (1000 in binary) after first recursion.As the result (carry variable) isnt zero, lets recursion again by xor value and carry value.



                getSum(3, 8);


                So, doing the first XORing we get,



                0011
                
1000
                
-------
                
1011


                The XORing this time yielded in 11 (1011 binary),so we perform the AND now,



                0011
                
1000
                
-------
                
0000


                We get all ZERO for ANDing 3 and 8, so this time the left shift operator also results in ZERO, as we have no 1 here which may give us a value by left shifing zeroes.
                As the carry variable is now Zero, we come to the end of recursion and the XORed value will be the Sum, which is 11 (1011 in Binary).



                Hope you get the working of the procedure. You can learn more by learning bitwise operation, as its the way the machine do the arithmatic operations.






                share|improve this answer















                 int result = p ^ q; // XOR Operator, + without carry 0+0=0, 0+1=1+0=1, 1+1=0
                
int carry = (p & q) << 1; // Left Shift, 1+1=2
                
if (carry != 0) {
                
    return getSum(result, carry);
                
}
                
return result;


                Start By p=5,q=6. Then the XOR would be,



                0101
                
0110
                
------
                
0011


                So, XORing results in (0011) which is actually 3 in decimal. Then ANDing p and q we get,



                0101
                
0110
                
-------
                
0100


                We get 4 (100 in binary) by ANDing 5 & 6, now if we left shift this value by 1, we get



                 0100<<1=1000


                So we get 8 (1000 in binary) after first recursion.As the result (carry variable) isnt zero, lets recursion again by xor value and carry value.



                getSum(3, 8);


                So, doing the first XORing we get,



                0011
                
1000
                
-------
                
1011


                The XORing this time yielded in 11 (1011 binary),so we perform the AND now,



                0011
                
1000
                
-------
                
0000


                We get all ZERO for ANDing 3 and 8, so this time the left shift operator also results in ZERO, as we have no 1 here which may give us a value by left shifing zeroes.
                As the carry variable is now Zero, we come to the end of recursion and the XORed value will be the Sum, which is 11 (1011 in Binary).



                Hope you get the working of the procedure. You can learn more by learning bitwise operation, as its the way the machine do the arithmatic operations.







                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited 11 mins ago

























                answered 16 mins ago









                Ayan_84Ayan_84

                520513




                520513























                    -1














                    These are bitwise operations. They're close to hardware language.






                    share|improve this answer



















                    • 7





                      Too short.You didn't explain why this works and how

                      – Jacky
                      38 mins ago






                    • 1





                      Link-only answers are discouraged here. Please add the relevant content in the answer itself.

                      – Ian McLaird
                      31 mins ago






                    • 3





                      It’s binary math. WofWca would have to give you a very very long explanation on binary to explain it. You should really read the W3 Schools article or watch some videos on Bitwise calculations. It’s the kind of thing that would be a few days of class work in college. The comment about link-only answers is fair, but I doubt you’ll grasp the concept from a StackOverflow post alone though.

                      – Nate
                      30 mins ago













                    • @Nate Sorry,But I am not the computer science's student

                      – Jacky
                      27 mins ago
















                    -1














                    These are bitwise operations. They're close to hardware language.






                    share|improve this answer



















                    • 7





                      Too short.You didn't explain why this works and how

                      – Jacky
                      38 mins ago






                    • 1





                      Link-only answers are discouraged here. Please add the relevant content in the answer itself.

                      – Ian McLaird
                      31 mins ago






                    • 3





                      It’s binary math. WofWca would have to give you a very very long explanation on binary to explain it. You should really read the W3 Schools article or watch some videos on Bitwise calculations. It’s the kind of thing that would be a few days of class work in college. The comment about link-only answers is fair, but I doubt you’ll grasp the concept from a StackOverflow post alone though.

                      – Nate
                      30 mins ago













                    • @Nate Sorry,But I am not the computer science's student

                      – Jacky
                      27 mins ago














                    -1












                    -1








                    -1







                    These are bitwise operations. They're close to hardware language.






                    share|improve this answer













                    These are bitwise operations. They're close to hardware language.







                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered 42 mins ago









                    WofWcaWofWca

                    40819




                    40819








                    • 7





                      Too short.You didn't explain why this works and how

                      – Jacky
                      38 mins ago






                    • 1





                      Link-only answers are discouraged here. Please add the relevant content in the answer itself.

                      – Ian McLaird
                      31 mins ago






                    • 3





                      It’s binary math. WofWca would have to give you a very very long explanation on binary to explain it. You should really read the W3 Schools article or watch some videos on Bitwise calculations. It’s the kind of thing that would be a few days of class work in college. The comment about link-only answers is fair, but I doubt you’ll grasp the concept from a StackOverflow post alone though.

                      – Nate
                      30 mins ago













                    • @Nate Sorry,But I am not the computer science's student

                      – Jacky
                      27 mins ago














                    • 7





                      Too short.You didn't explain why this works and how

                      – Jacky
                      38 mins ago






                    • 1





                      Link-only answers are discouraged here. Please add the relevant content in the answer itself.

                      – Ian McLaird
                      31 mins ago






                    • 3





                      It’s binary math. WofWca would have to give you a very very long explanation on binary to explain it. You should really read the W3 Schools article or watch some videos on Bitwise calculations. It’s the kind of thing that would be a few days of class work in college. The comment about link-only answers is fair, but I doubt you’ll grasp the concept from a StackOverflow post alone though.

                      – Nate
                      30 mins ago













                    • @Nate Sorry,But I am not the computer science's student

                      – Jacky
                      27 mins ago








                    7




                    7





                    Too short.You didn't explain why this works and how

                    – Jacky
                    38 mins ago





                    Too short.You didn't explain why this works and how

                    – Jacky
                    38 mins ago




                    1




                    1





                    Link-only answers are discouraged here. Please add the relevant content in the answer itself.

                    – Ian McLaird
                    31 mins ago





                    Link-only answers are discouraged here. Please add the relevant content in the answer itself.

                    – Ian McLaird
                    31 mins ago




                    3




                    3





                    It’s binary math. WofWca would have to give you a very very long explanation on binary to explain it. You should really read the W3 Schools article or watch some videos on Bitwise calculations. It’s the kind of thing that would be a few days of class work in college. The comment about link-only answers is fair, but I doubt you’ll grasp the concept from a StackOverflow post alone though.

                    – Nate
                    30 mins ago







                    It’s binary math. WofWca would have to give you a very very long explanation on binary to explain it. You should really read the W3 Schools article or watch some videos on Bitwise calculations. It’s the kind of thing that would be a few days of class work in college. The comment about link-only answers is fair, but I doubt you’ll grasp the concept from a StackOverflow post alone though.

                    – Nate
                    30 mins ago















                    @Nate Sorry,But I am not the computer science's student

                    – Jacky
                    27 mins ago





                    @Nate Sorry,But I am not the computer science's student

                    – Jacky
                    27 mins ago


















                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Stack Overflow!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f55193135%2fwhat-is-ab-and-ab1%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    Ponta tanko

                    Image
                    Usona ponta tanko M60A1 Ponta tanko estas, simile kiel kirasita savveturilo, inter specialaj kirasitaj transportiloj. Kiel pontaj tankoj eblas konsideri ĉiujn memmovigaj rimedoj ebligantaj superi obstaklojn ĝis larĝeco 24 metroj rekte en spaco de bataloperacoj. Ties ĉefa tasko estas ebligi al la propraj taĉmentoj daŭrigi en rapida antaŭenirado sen senbezona reteniĝo. El tio pli-malpli rezultas taktikaj-teknikaj postuloj, kiujn ili devas plenumi. Precipe temas pri kapablo operacii komune kun unue-sinsekvaj taĉmentoj kaj transpontigi obstaklojn dum kelke da minutoj. This page is only for reference, If you need detailed information, please check here
                    Read more

                    Tantalo (mitologio)

                    Image
                    Ĉi tiu artikolo temas pri mitologio. Por aliaj signifoj vidu la paĝon Tantalo (apartigilo) . Tantalo , laŭ Gioacchino Assereto. Laŭ la helena mitologio, Tantalo (el la greka Τάνταλος , ŝancelanta aŭ tre malfeliĉa ) estis filo de Zeŭso kaj la oceanidino Pluto, laŭ la plej divastigata versio. Li estis reĝo de Frigio [1] aŭ de la monto Sipilo, en Lidio (Malgrandazio) [2] kaj post sia morto li suferis eternan torturadon ĉe la Tartaro, la plej profunda parto de Hadeso, rezervita por la plej perversaj homoj. Tamen, kelkaj rakontoj asertas ke li estis reĝo de Argo aŭ de Korinto [3] , aŭ ke li migris norden el la monto Sipilo por akiri la tronon de Paflagonio, el kie Ilo forpelis lin pro tio, ke Tantalo forkaptis kaj seksperfortis lian fraton Ganimedon [4] , kaj tiel ofendis la diojn. Verŝajne li ankaŭ regis urbon en okcidenta Anatolio, kiun oni nomis Tantalis [5] pro li, aŭ Sipilo , pro la monto sur kies flanko ankoraŭ staras ĝiaj ruinaĵoj [6] . Laŭ la geografiisto Strabo,...
                    Read more

                    Erzsébet Schaár

                    Image
                    Erzsébet Schaár Memportreto de Erzsébet Schaár en Pécs Personaj informoj Naskiĝo 27-an de julio 1908 ( 1908-07-27 ) en Budafok Morto 29-an de aŭgusto 1975 ( 1975-08-29 ) (67-jara) en Budapeŝto Ŝtataneco Hungario Profesio skulptisto Familianoj Edz(in)o Tibor Vilt v   •   d   •   r Erzsébet Schaár [erĵEbet ŝAr], laŭ hungarlingve kutima sinsekvo Schaár Erzsébet estis hungara skulptisto. Ŝia edzo estis Tibor Vilt. Déryné Széppataki Róza en Miskolc fare de Erzsébet Schaár Erzsébet Schaár [1] naskiĝis la 27-an de julio 1908 en Budafok, ŝi mortis la 29-an de aŭgusto 1975 en Budapeŝto. Enhavo 1 Biografio 2 Elektitaj solaj ekspozicioj 3 Elektitaj kolektivaj ekspozicioj 4 Elektitaj skulptaĵoj 5 Elektitaj disĉiploj 6 Fontoj 7 Referencoj Biografio | Erzsébet Schaár studis en Industriarta Lernejo, poste en Reĝa Belarta Altlernejo (1924-1926). Ŝia ĉefinstruisto estis Zsig...
                    Read more