Understanding the Pumping Lemma to Prove Irregularity
Utilizing the Pumping Lemma to demonstrate the irregularity of languages where the condition L>= {aibj, i>j} is not satisfied. The process involves fixing a pumping length, choosing a suitable string from L, and exploring possible splittings of the string to show irregularity.
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Pumping Lemma Examples
L>= {aibj: i > j} L>is not regular. We prove it using the Pumping Lemma.
L>= {aibj: i > j} L>is not regular. Fix an arbitrary pumping length n>0.
L>= {aibj: i > j} L>is not regular. Fix an arbitrary pumping length n>0. Choose a proper string s in L>.
L>= {aibj: i > j} L>is not regular. Fix an arbitrary pumping length n>0. Choose a proper string s in L>. |s| n
L> = {aibj : i > j} L> is not regular. Fix an arbitrary pumping length n>0. Choose a proper string s in L>. s = an+1bn L>. aaa aabb b n n+1
L> = {aibj : i > j} L> is not regular. Fix an arbitrary pumping length n>0. Choose a proper string s in L>. s = an+1bn L>. Consider all possible splittings of s in x,y,z with the desired properties. aaa aabb b n n+1
L> = {aibj : i > j} L> is not regular. Fix an arbitrary pumping length n>0. Choose a proper string s in L>. s = an+1bn L>. Consider all possible splittings of s in x,y,z with the desired properties. |xy| n |y| 1 aaa aabb b n n+1
L> = {aibj : i > j} L> is not regular. Fix an arbitrary pumping length n>0. Choose a proper string s in L>. s = an+1bn L>. Consider all possible splittings of s in x,y,z with the desired properties. |xy| n |y| 1 aaa aabb b n n+1
L> = {aibj : i > j} L> is not regular. Fix an arbitrary pumping length n>0. Choose a proper string s in L>. s = an+1bn L>. Consider all possible splittings of s in x,y,z with the desired properties. Y |xy| n |y| 1 aaa aabb b n+1 n
L> = {aibj : i > j} L> is not regular. Fix an arbitrary pumping length n>0. Choose a proper string s in L>. s = an+1bn L>. Consider all possible splittings of s in x,y,z with the desired properties: y = am, 1 m n. Y aaa aabb b n+1 n
L> = {aibj : i > j} L> is not regular. Fix an arbitrary pumping length n>0. Choose a proper string s in L>. s = an+1bn L>. Consider all possible splittings of s in x,y,z with the desired properties: y = am, 1 m n. xz =an+1-mbn L>. aaabb b n+1-m n n
L> = {aibj : i > j} L> is not regular. Fix an arbitrary pumping length n>0. Choose a proper string s in L>. s = an+1bn L>. Consider all possible splittings of s in x,y,z with the desired properties: y = am, 1 m n. xz =an+1-mbn L>. So L> is not regular!
L={ww : w in {a,b}*} First, figure out what this language is.
L={ww : w in {a,b}*} First, figure out what this language is. A string in the language?
L={ww : w in {a,b}*} First, figure out what this language is. A string in the language? aabaab
L={ww : w in {a,b}*} First, figure out what this language is. A string in the language? aabaab Another string in the language?
L={ww : w in {a,b}*} First, figure out what this language is. A string in the language? aabaab Another string in the language? aaaaaa
L={ww : w in {a,b}*} First, figure out what this language is. A string in the language? aabaab Another string in the language? aaaaaa A string not in the language?
L={ww : w in {a,b}*} First, figure out what this language is. A string in the language? aabaab Another string in the language? aaaaaa A string not in the language? abbb
L={ww : w in {a,b}*} First, figure out what this language is. A string in the language? aabaab Another string in the language? aaaaaa A string not in the language? abbb Is in the language?
L={ww : w in {a,b}*} First, figure out what this language is. A string in the language? aabaab Another string in the language? aaaaaa A string not in the language? abbb Is in the language? YES! ( = )
L={ww : w in {a,b}*} First, figure out what this language is. A string in the language? aabaab Another string in the language? aaaaaa A string not in the language? abbb Is in the language? YES! ( = ) Is aa in the language?
L={ww : w in {a,b}*} First, figure out what this language is. A string in the language? aabaab Another string in the language? aaaaaa A string not in the language? abbb Is in the language? YES! ( = ) Is aa in the language? YES!
L={ww : w in {a,b}*} First, figure out what this language is. A string in the language? aabaab Another string in the language? aaaaaa A string not in the language? abbb Is in the language? YES! ( = ) Is aa in the language? YES! Is a in the language?
L={ww : w in {a,b}*} First, figure out what this language is. A string in the language? aabaab Another string in the language? aaaaaa A string not in the language? abbb Is in the language? YES! ( = ) Is aa in the language? YES! Is a in the language? NO!
L={ww : w in {a,b}*} First, figure out what this language is. L = { , aa, bb, aaaa, abab, baba, bbbb, aaaaaa } abaabba|abaabba
L={ww : w in {a,b}*} We prove that L is not regular by using the pumping lemma.
L={ww : w in {a,b}*} We prove that L is not regular by using the pumping lemma. First fix an arbitrary number n>0 to be the pumping length.
L={ww : w in {a,b}*} We prove that L is not regular by using the pumping lemma. Pumping length: n Choose a proper string in the language
L={ww : w in {a,b}*} We prove that L is not regular by using the pumping lemma. Pumping length: n Choose a proper string in the language. Choose wisely!!!
L={ww : w in {a,b}*} We prove that L is not regular by using the pumping lemma. Pumping length: n Choose a proper string in the language. Example: For s = a2n aaa aaa|aaa aaa n n
L={ww : w in {a,b}*} We prove that L is not regular by using the pumping lemma. Pumping length: n Choose a proper string in the language. Example: For s = a2n For x = , y = a2, z = a2n-2 z y aaa aaa|aaa aaa n n
L={ww : w in {a,b}*} We prove that L is not regular by using the pumping lemma. Pumping length: n Choose a proper string in the language. Example: For s = a2n For x = , y = a2, z = a2n-2 y y z aaaaa aa|aaaa aaa n+1 L n+1
L={ww : w in {a,b}*} We prove that L is not regular by using the pumping lemma. Pumping length: n Choose a proper string in the language. Example: For s = a2n For x = , y = a2, z = a2n-2 y y y z aaaaaaa a|aaaaa aaa n+2 L n+2
L={ww : w in {a,b}*} We prove that L is not regular by using the pumping lemma. Pumping length: n Choose a proper string in the language. Example: For s = a2n For x = , y = a2, z = a2n-2 z a aaaa|aa aaa n-1 L n-1
L={ww : w in {a,b}*} We prove that L is not regular by using the pumping lemma. Pumping length: n Choose a proper string in the language. Example: For s = a2n For x = , y = a2, z = a2n-2, there is no i: xyiz L!
L={ww : w in {a,b}*} We prove that L is not regular by using the pumping lemma. Pumping length: n Choose a proper string in the language. Example: For s = a2n For x = , y = a2, z = a2n-2, there is no i: xyiz L! s = a2ndoesn t work!!!
L={ww : w in {a,b}*} We prove that L is not regular by using the pumping lemma. Pumping length: n Choose a proper string in the language. Example: For s = (ab)2n abab abab|abab abab n n
L={ww : w in {a,b}*} We prove that L is not regular by using the pumping lemma. Pumping length: n Choose a proper string in the language. Example: For s = (ab)2n For x = , y = abab, z = (ab)2n-2 y z abab abab|abab abab n n
L={ww : w in {a,b}*} We prove that L is not regular by using the pumping lemma. Pumping length: n Choose a proper string in the language. Example: For s = (ab)2n For x = , y = abab, z = (ab)2n-2 y y z abababab ab|ababab abab n+1 L n+1
L={ww : w in {a,b}*} We prove that L is not regular by using the pumping lemma. Pumping length: n Choose a proper string in the language. Example: For s = (ab)2n For x = , y = abab, z = (ab)2n-2 For any i, xyiz = (ab)2i(ab)2n-2 = (ab)2(i-n-2) L!
L={ww : w in {a,b}*} We prove that L is not regular by using the pumping lemma. Pumping length: n Choose a proper string in the language. Example: For s = (ab)2n For x = , y = abab, z = (ab)2n-2 For any i, xyiz = (ab)2i(ab)2n-2 = (ab)2(i-n-2) L! s = (ab)2ndoesn t work!
L={ww : w in {a,b}*} We prove that L is not regular by using the pumping lemma. Pumping length: n Choose a proper string in the language. Use s = anbanb aaaa aab|aaaa...aab n n
L={ww : w in {a,b}*} We prove that L is not regular by using the pumping lemma. Pumping length: n Choose a proper string in the language. Use s = anbanb For any splitting of s in x,y,z with the desired properties: aaaa aab|aaaa...aab n n
L={ww : w in {a,b}*} We prove that L is not regular by using the pumping lemma. Pumping length: n Choose a proper string in the language Use s = anbanb For any splitting of s in x,y,z with the desired properties: y = amwith 1 m n.
L={ww : w in {a,b}*} We prove that L is not regular by using the pumping lemma. Pumping length: n Choose a proper string in the language Use s = anbanb For any splitting of s in x,y,z with the desired properties: y = amwith 1 m n. Observe that xy2z = am+nbanb is not in L QED
L = {w1w2 : w1,w2 {a,b}*,|w1|=|w2|} Is it regular?
L = {w1w2 : w1,w2 {a,b}*,|w1|=|w2|} Is it regular? A first attempt to design a FA a,b a,b a,b a,b ... q10 q11 q12 q13 q1n a,b a,b a,b a,b ... q2n-1 q2n-2 q2n-3 q2n q20
L = {w1w2 : w1,w2 {a,b}*,|w1|=|w2|} Is it regular? A first attempt to design a FA fails! a,b a,b a,b a,b ... q10 q11 q12 q13 q1n a,b a,b a,b a,b ... q2n-1 q2n-2 q2n-3 q2n q20