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Tuesday, August 12, 2008. Zhis iz a zesz. Ritchell Patrick Michael Linosoy. Sunday, August 3, 2008. 1) If E is a variable or constant then prefix(E)=E. 2) If E is E1 op E2 then prefix(E)=prefix(E1opE2)=op prefix(E1) prefix(E2). 3) If E is (E1) then prefix(E)=prefix(E1). 1) If E is a variable or constant then infix(E)=E. 2) If E is E1E2op then infix(E)=infix(E1E2op)=infix(E1) op infix(E2). 3) If E is (E1) then infix(E)=infix(E1). 1) If E is a variable or constant then prefix(E)=E. Thursday, July 31, 2008.

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Tuesday, August 12, 2008. Zhis iz a zesz. Ritchell Patrick Michael Linosoy. Sunday, August 3, 2008. 1) If E is a variable or constant then prefix(E)=E. 2) If E is E1 op E2 then prefix(E)=prefix(E1opE2)=op prefix(E1) prefix(E2). 3) If E is (E1) then prefix(E)=prefix(E1). 1) If E is a variable or constant then infix(E)=E. 2) If E is E1E2op then infix(E)=infix(E1E2op)=infix(E1) op infix(E2). 3) If E is (E1) then infix(E)=infix(E1). 1) If E is a variable or constant then prefix(E)=E. Thursday, July 31, 2008.

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JRJsolutions | jrjsolutions.blogspot.com Reviews

https://jrjsolutions.blogspot.com

Tuesday, August 12, 2008. Zhis iz a zesz. Ritchell Patrick Michael Linosoy. Sunday, August 3, 2008. 1) If E is a variable or constant then prefix(E)=E. 2) If E is E1 op E2 then prefix(E)=prefix(E1opE2)=op prefix(E1) prefix(E2). 3) If E is (E1) then prefix(E)=prefix(E1). 1) If E is a variable or constant then infix(E)=E. 2) If E is E1E2op then infix(E)=infix(E1E2op)=infix(E1) op infix(E2). 3) If E is (E1) then infix(E)=infix(E1). 1) If E is a variable or constant then prefix(E)=E. Thursday, July 31, 2008.

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jrjsolutions.blogspot.com

1

JRJsolutions: August 2008

http://jrjsolutions.blogspot.com/2008_08_01_archive.html

Tuesday, August 12, 2008. Zhis iz a zesz. Ritchell Patrick Michael Linosoy. Sunday, August 3, 2008. 1) If E is a variable or constant then prefix(E)=E. 2) If E is E1 op E2 then prefix(E)=prefix(E1opE2)=op prefix(E1) prefix(E2). 3) If E is (E1) then prefix(E)=prefix(E1). 1) If E is a variable or constant then infix(E)=E. 2) If E is E1E2op then infix(E)=infix(E1E2op)=infix(E1) op infix(E2). 3) If E is (E1) then infix(E)=infix(E1). 1) If E is a variable or constant then prefix(E)=E. View my complete profile.

2

JRJsolutions: Introduction

http://jrjsolutions.blogspot.com/2008/06/introduction.html

Tuesday, June 24, 2008. Members of the group:. Lim, Ritchell Royce M. Lleno, John Patrick R. Sasoy, Michael Jay G. Our group name is JRJsolutions; the first three characters taken from the first letters of first names, added to the first three characters is the word solutions. This blog was made in partial fulfillment to our requirement in our ICS40 class: Compiler Design and Principles. Ritchell Patrick Michael Linosoy. Subscribe to: Post Comments (Atom). Phases of a Compiler. View my complete profile.

3

JRJsolutions: activity 3

http://jrjsolutions.blogspot.com/2008/08/i.html

Sunday, August 3, 2008. 1) If E is a variable or constant then prefix(E)=E. 2) If E is E1 op E2 then prefix(E)=prefix(E1opE2)=op prefix(E1) prefix(E2). 3) If E is (E1) then prefix(E)=prefix(E1). 1) If E is a variable or constant then infix(E)=E. 2) If E is E1E2op then infix(E)=infix(E1E2op)=infix(E1) op infix(E2). 3) If E is (E1) then infix(E)=infix(E1). 1) If E is a variable or constant then prefix(E)=E. 2) If E is E1E2op then prefix(E)=prefix(E1E2op)=op prefix(E1) prefix(E2). View my complete profile.

4

JRJsolutions

http://jrjsolutions.blogspot.com/2008/07/s-ss-ss-a.html

Thursday, July 31, 2008. S  SS SS* a. A) show how a string aa a* can be generated by the grammar. B) construct a parse tree for this string. C) what language is generated by this language. A language that accepts a string with a minimum value of “a” and may also contain an infinite amount of a’s, ’s and *’s. A) S  0S1 01. A language that accepts a string which has equal numbers of 0s and 1s where the 0s precede the 1s, with the minimum value of 0011. B) S SS -SS a. C) SS(S)S Є. D) SaSbS bSaS Є.

5

JRJsolutions: July 2008

http://jrjsolutions.blogspot.com/2008_07_01_archive.html

Thursday, July 31, 2008. S  SS SS* a. A) show how a string aa a* can be generated by the grammar. B) construct a parse tree for this string. C) what language is generated by this language. A language that accepts a string with a minimum value of “a” and may also contain an infinite amount of a’s, ’s and *’s. A) S  0S1 01. A language that accepts a string which has equal numbers of 0s and 1s where the 0s precede the 1s, with the minimum value of 0011. B) S SS -SS a. C) SS(S)S Є. D) SaSbS bSaS Є.

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Tuesday, August 12, 2008. Zhis iz a zesz. Ritchell Patrick Michael Linosoy. Sunday, August 3, 2008. 1) If E is a variable or constant then prefix(E)=E. 2) If E is E1 op E2 then prefix(E)=prefix(E1opE2)=op prefix(E1) prefix(E2). 3) If E is (E1) then prefix(E)=prefix(E1). 1) If E is a variable or constant then infix(E)=E. 2) If E is E1E2op then infix(E)=infix(E1E2op)=infix(E1) op infix(E2). 3) If E is (E1) then infix(E)=infix(E1). 1) If E is a variable or constant then prefix(E)=E. Thursday, July 31, 2008.

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