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The I-Test Manish Gupta - 05305006 Santosh Kumar - 05305003 Soumitra Pal - 05305015 Group - 6

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Agenda Introduction Definitions Theorems Algorithm Conclusion

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Introduction a 1 I 1 + a 2 I 2 + … + a n I n = a 0 M k ≤ I k ≤ N k, 1 ≤ k ≤ n M k & N k can be * or integers

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Introduction (2) GCD-Test – Considers integrality – Ignores limits Banerjee-Test – Considers limits – Ignores integrality, checks for real solution – Not applicable if some limits are unknown How about considering both?

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Definition - Solvability a 1 I 1 + a 2 I 2 + … + a n I n = a 0 is (M 1,N 1 ; M 2,N 2 ; …; M n,N n ) integer solvable n > 0 If there exists j 1, j 2.. j n s.t. a 1 j 1 + a 2 j 2 + … + a n j n = a 0 For each k, 1 ≤ k ≤ n: if M k and N k are integers then M k ≤ j k ≤ N k If M k is an integer and N k = * then M k ≤ j k If N k is an integer and M k = * then j k ≤ N k n = 0 if a 0 =0

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Definition - Interval Equation a 1 I 1 + a 2 I 2 + … + a n I n = [L,U] --------- (1) L,U integers. M k ≤ I k ≤ N k, 1 ≤ k ≤ n It is a set of these equations a 1 I 1 + a 2 I 2 + … + a n I n = L a 1 I 1 + a 2 I 2 + … + a n I n = L+1 …. a 1 I 1 + a 2 I 2 + … + a n I n = U

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Solvability of Interval Equation Eq(1) is (M 1,N 1 ; M 2,N 2 ; …; M n,N n ) integer solvable – n>0, if at least 1 from the above set is integer solvable. – n=0, if L ≤ 0 ≤ U a 1 I 1 + a 2 I 2 + … + a n I n = a 0 can be written as an interval equation as follows: a 1 I 1 + a 2 I 2 + … + a n I n = [a 0, a 0 ]

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Theorem 1 GCD test for Interval Equation a 1 I 1 + a 2 I 2 + … + a n I n = [ L, U ] Has integer solution if and only if L ≤ d* ceil(L/d) ≤ U Where d = gcd(a 1, a 2,.., a n ) e.g. 4x + 6y = [9,12] 9≤10≤12 So, solution exists

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Lemma 2a Shift 1 variable to right If a n ≤ U-L+1, and a n >0 a 1 I 1 + a 2 I 2 + … + a n I n = [ L, U ] is (M 1,N 1 ; M 2,N 2 ; …; M n,N n ) integer solvable iff a 1 I 1 + a 2 I 2 + … + a n-1 I n-1 = [ L - a n N n, U - a n M n ] is (M 1,N 1 ; M 2,N 2 ; …; M n-1,N n-1 ) integer solvable.

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Lemma 2b Shift 1 variable to right If -a n ≤ U-L+1, and a n < 0 a 1 I 1 + a 2 I 2 + … + a n I n = [ L, U ] is (M 1,N 1 ; M 2,N 2 ; …; M n,N n ) integer solvable iff a 1 I 1 + a 2 I 2 + … + a n-1 I n-1 = [ L - a n M n, U - a n N n ] is (M 1,N 1 ; M 2,N 2 ; …; M n-1,N n-1 ) integer solvable.

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Theorem 2 Shift 1 variable to right If |a n | ≤ U-L+1 a 1 I 1 + a 2 I 2 + … + a n I n = [ L, U ] is (M 1,N 1 ; M 2,N 2 ; …; M n,N n ) integer solvable iff a 1 I 1 + a 2 I 2 + … + a n-1 I n-1 = [ L - a n + N n + a n - M n, U - a n + M n + a n - N n ] is (M 1,N 1 ; M 2,N 2 ; …; M n-1,N n-1 ) integer solvable.

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Example Usage of Theorem 2 gcd(1, -3, 7) = 1 GCD test says “solution may exist” Interval equation: x -3y +7z = [ 8,8] By result in the previous slide a n + = 1, a n - = 0, M n = 1, N n =3 -3y + 7z = [ 8 - 1.3 + 0.1, 8 - 1.1 + 0.3 ] x – 3y + 7z = 8 1≤x≤3; 1≤y≤2; 1≤z ≤ 4 Limits for Banerjee 2≤8≤28 Banerjee test says “solution may exist” And then 7z = [5+3, 7+6] So, 7z = [8, 13] Which is impossible So, I – test says no solution.

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Theorem 3 If |a n | ≤ U-L+1 and the IE a 1 I 1 + a 2 I 2 + … + a n I n = [ L, U ] is integer solvable but a 1 I 1 + a 2 I 2 + … + a n-1 I n-1 = [ L - a n + N n + a n - M n, U - a n + M n + a n - N n ] is not integer solvable, then solution of I n is outside [M n, N n ]

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Motivation for Theorem 4 To apply Theorem 2, |a n | ≤ U-L+1 Can we do something if the above does not hold?

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Theorem 4Reduce both sides a 1 I 1 + a 2 I 2 + … + a n I n = [ L, U ] is (M 1,N 1 ; M 2,N 2 ; …; M n,N n ) integer solvable, iff (a 1 /d) I 1 + (a 2 /d) I 2 + … + (a n /d) I n = [ceil(L/d), floor(U/d) ] is (M 1,N 1 ; M 2,N 2 ; …; M n,N n ) integer solvable Note that a 1 /d, a 2 /d, …, a n /d are integers.

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Example Usage of Theorem 4 2x -6y +14z =16 1≤x≤3;1≤y≤2;1≤z≤4 Both GCD and Banerjee say “maybe” Consider IE, 2x -6y +14z = [16,16] Theorem 2 is inapplicable because coefficients are greater than (U-L+1)=1 Using Theorem 4, x-3y+7z = [8,8] This we have seen before.

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Theorem 5 Analogous to Theorem 3 Combined with result of Theorem 4

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Theorem 6 Handle unknown limits when gcd = 1 If M i =N i =* for 1≤i≤k and gcd(a 1,a 2,..a k ) = 1 a 1 I 1 + a 2 I 2 +… + a k+1 I k+1 + … + a n I n = [L,U] is ( * (1),* (1) ; * (2),* (2) ; … ; * (k),* (k) ; M k+1,N k+1 ; …; M n,N n ) integer solvable for any M k+1,N k+1,..,M n,N n. This shows if both the limits are unknown for >=1 variables and if gcd of the coefficients of those variables is 1 then no need of Theorem 2 or 4, a 1 I 1 + a 2 I 2 +… + a k+1 I k+1 + … + a n I n = [L,U] is definitely integer solvable

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Theorem 7 Handle unknown limits when gcd > 1 If M i =N i =* for 1≤i≤k and gcd(a 1,a 2,..a k ) = d a 1 I 1 + a 2 I 2 + … + a k+1 I k+1 + … + a n I n = [L,U] is (* (1),* (1) ; * (2),* (2) ; … ; * (k),* (k) ; M k+1,N k+1 ; …; M n,N n ) integer solvable iff dI + a k+1 I k+1 + … + a n I n = [L,U] is (*,*; M k+1,N k+1 ; …; M n,N n ) is integer solvable. This shows that the set of variables with both limits unknown can be compressed into 1 variable, to reduce bookkeeping when theorem 2 and 4 are applied later.

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I – Test algorithm Input: a0,a1,a2,..,an,M1,N1,…Mn,Nn Output: no,yes,maybe Algorithm: 1. If there are some unknown bounds d = gcd of coefficients of variables with unknown bounds If d is 1, apply theorem 6.theorem 6 Else apply theorem 7.theorem 7

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I – Test algorithm (2) 2. While (1) While theorem 2 is applicable apply theorem 2theorem 2 if there are no more coefficients on LHS if L <= 0 <= UL <= 0 <= U return yes else return no Find gcd of the coefficients in the LHS list Perform the interval-equation GCD test.interval-equation GCD test If GCD test says “no” then return no If gcd != 1 apply theorem 4theorem 4 else return maybe.

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Example for Algo - 1 3I 1 + 5I 2 + a 3 I 3 + a 4 I 4 = a 0 1 ≤ a 3,a 4 ≤ 5 gcd (3, 5) = 1 algo returns ‘yes’ using theorem 6

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Example for Algo - 2 120I 1 + 180I 2 + 2I 3 + 4I 4 + 6I 5 = 8 1 ≤ a 3,a 4,a 5 ≤ 5 IE: 120I 1 + 180I 2 + 2I 3 + 4I 4 + 6I 5 = [8, 8] Bounds for I 1 & I 2 are unknown gcd(120,180) = 60 ≠ 1, apply theorem 7 60I + 2I 3 + 4I 4 + 6I 5 = [8, 8] => L=U=8 U-L+1 = 1, all coeffs > 1, can’t apply thm 2 apply theorem 4 30I + I 3 + 2I 4 + 3I 5 = [4, 4] => L=U=4 can apply theorem 2 for I 3

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Example for Algo – 2 (2) a 3 + =1, a 3 - =0, [U,L] = [4, 4] theorem 2 theorem 2 new [U,L] = [4 - 1.5 - 0.1, 4 - 1.1 + 0.5] = [-1, 3] 30I + 2I 4 + 3I 5 = [-1, 3] again theorem 2 applicable for I 4, 2 ≤ 3 - (-1) + 1 30I + 3I 5 = [-1-2.5+0.1, 3-2.1+0.5] = [-11, 1] Apply theorem 2 for I 5 that gives, 30I = [-11-3.5+0.1, 1-3.1+0.5] = [-26, -2] Now control comes out of the inner while loop Interval GCD test applied, says no. Algo returns no.

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Discussion on the algorithm Accurate when returns ‘yes’ or ‘no’ Returns ‘maybe’ when the equation has a solution which satisfies limits – Not only on all variables that I-test could move to RHS – But also may be on the rest of the variables. In case of ‘maybe’ (coefficients too big to apply Theorem 2 and none of the remaining limits are *) – then apply complete “step by step Banerjee test”. – [L ban, U ban ] ∩ [L,U] = ø

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Conclusion : Cost & Benefits Banerjee bound computation component is same as the Banerjee test (or less if it finishes before all terms are moved to the RHS). In worst case, I-test needs n GCD tests but practically it uses no more than 1 GCD test. Extends the range of applicability of Banerjee test Distinguishes “definite” yes from “tentative” Banerjee yes

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Thank you!

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