Learning to Automatically Solve Algebra Word Problems Nate Kushman Yoav Artzi, Luke Zettlemoyer, Regina Barzilay 1
Task Automatically Solve Algebra Word Problems An amusement park sells 2 kinds of tickets. Tickets for children cost $1.50. Adult tickets cost $4. On a certain day, 278 people entered the park. On that same day the admission fees collected totaled $792. How many children were admitted on that day? How many adults were admitted? Two Training Scenarios: Goal: Generate Numerical Answers X + Y = 278 Full Equations 128 1.5*X + 4*Y = 729 150 128 Numerical Answers 150 2
Wide Variety of Problems j Interest Traveling Apart Math Problems Ticket Purchase An investor will invest a total of 15000 Two airplanes left the same airport An amusement park sells 2 kinds of dollars in 2 accounts , one paying 4 % traveling in opposite directions. If one A math test is worth 100 points and has 30 tickets. Tickets for children cost $1.50. annual simple interest and the other 3 %. airplane averages 400 miles per hour and problems. Each problem is worth either 3 Adult tickets cost $4. On a certain day, 278 If he wants to earn 550 dollars annual the other 250 miles per hour , how many points or 4 points. How many 4 point people entered the park. On that same interest , how much should he invest at 4 hours will it take for the distance between problems are there? day the admission fees collected totaled %? How much at 3 %? them be 1625 miles? $792. How many children were admitted … X + Y = 278 3.0*0.01*X+4.0*0.01*Y=550.0 X + Y = 30 (250.0*X)+(400.0*X)=1625.0 X+Y =15000 3*X + 4*Y = 100 1.5*X + 4*Y = 729 Ratio Fixed+Variable Coffee Beans Height Compare Colombian coffee beans cost 5.50 dollars A writing workshop enrolls novelists and Sunshine Car Rentals rents a basic car at a A physician 's assistant measures a child per pound, while Peruvian coffee beans poets in a ratio of 5 to 3. There are 24 daily rate of 17.99 dollars plus 0.18 per and finds that his height is 41.5 inches. At cost 4.25 dollars per pound. We want to people at the workshop. How many mile. City Rentals rents a basic car at his last visit to the doctor's office , the mix the beans together so as to produce a novelists are there? How many poets are 18.95 dollars plus 0.16 per mile. For what child was 38.5 inches tall. How much did 40-pound bag , costing 4.60 dollars per there ? mileage is the cost the same? the child grow , in inches? pound. How many pounds of Columbian … 24 = X+Y (5.5*X)+(4.25*Y)=40.0*4.6 17.99 + 0.18*X = 18.95 + 0.16*X X=41.5-38.5 3.0*X=5.0*Y X+Y=40.0 Value of Coins Mixture Row Upstream Animals Arianne is mixing a solution for Chemistry class. She has a 25 % copper solution and It takes a boat 4 hours to travel 24 miles There are 11 animals in a barnyard. Some Jill has 3.50 dollars in nickels and dimes. If a 50 % copper solution. How many down a river and 6 hours to return are chickens and some are cows. There she has 50 coins, how many nickels does milliliters of the 25 % solution and 50 % upstream to its starting point. What is the are 38 legs in all. How many chickens and she have? How many dimes? solution should she mix to make 10 rate of the current in the river? cows are in the barnyard? milliliters of a 45 % solution? X+Y=50.0 10 = X + Y (X+Y)*4.0=24.0 (2.0*X)+(4.0*Y)=38 0.05*X+0.1*Y=3.5 25.0*.01*X+ 50.0*0.01*Y=45.0*.01*10 (X-Y)*6.0=24.0 X+Y=11.0
Eventually Solve More Difficult Problems j Finance Problems Physics Problems You decide that you want to save 1,528,717 A block of mass m is pushed across a rough dollars for retirement. Assuming that you are surface by an applied force, 𝐺 , directed at an 25 years old today, will retire at the age of 65, angle 𝜄 relative to the horizontal. The block experiences a friction force, 𝑔 , in the opposite and can earn a 6 percent annual interest rate on your deposits, how much must you deposit direction. What is the coefficient of friction each year to meet your retirement goal? between the block and the surface? 1528717 𝑌 = 𝑔 X = 𝑍 𝑍 Z = 65-25 −𝑎 − 𝑛 + 𝑍 = 0 (1+0.01∗6) 𝑎 −1 Y = 𝑎 = 𝐺 ∗ sin 𝜄 0.01∗6
Challenge 1: Complexity of Semantic Inference An amusement park sells 2 kinds of tickets. Tickets for children cost $1.50. Adult tickets cost $4. On a certain day, 278 people entered the park. On that same day the admission fees collected totaled $792. How many children were admitted on that day? How many adults were admitted? Infer: part_of(people, children) part_of(people,adults) size(y)=sum parts(y) size(people) = size(children)+size(adults) B/g: 1 ticket per person size(s:chld-tk) = size(children) cost(s:chld-tk)=size(s:chld-tk)*cost(chld-tk) cost(s:x)=size(s:x)*cost(x) Infer: part_of($792, cost(s:adult-tk) part_of($792, cost(s:chld-tk)) B/g: size(y)=sum parts(y) $792 = cost(s:child-tk) + cost(s:adult-tk)
Challenge 1: Complexity of Semantic Inference An amusement park sells 2 kinds of tickets. Tickets for children cost $1.50. Adult tickets cost $4. On a certain day, 278 people entered the park. On that same day the admission fees collected totaled $792. How many children were admitted on that day? How many adults were admitted? Infer: part_of(people, children) part_of(people,adults) size(y)=sum parts(y) size(people) = size(children)+size(adults) Solution: B/g: 1 ticket per person size(s:chld-tk) = size(children) Abstract to a restricted semantic representation – equations: cost(s:chld-tk)=size(s:chld-tk)*cost(chld-tk) cost(s:x)=size(s:x)*cost(x) X + Y = 278 1.5*X + 4*Y = 792 Infer: part_of($792, cost(s:adult-tk) part_of($792, cost(s:chld-tk)) Space of relations defined by equations seen in training data B/g: size(y)=sum parts(y) $792 = cost(s:child-tk) + cost(s:adult-tk)
Challenge 2: Complex Cross Sentence Relationships An amusement park sells 2 kinds of tickets. Tickets for children cost $1.50. Adult tickets cost $4. On a certain day, 278 people entered the park. On that same day the admission fees collected totaled $792. How many children were admitted on that day? How many adults were admitted? = ∗ + Adult tickets ∗ $4 $792 Tickets for children $1.50 Solution: Explore a very general space of alignments between the variables in an equation and the natural language 7
Challenge 3: Significant Domain Variation Math Problems Ticket Sales An amusement park sells 2 kinds of tickets. Tickets for children cost $1.50. Adult tickets cost $4. On a A math test is worth 100 points certain day, 278 people entered and has 30 problems. Each the park. On that same day the problem is worth either 3 points admission fees collected totaled or 4 points. How many 4 point $792. How many children were problems are there? admitted on that day? How many adults were admitted? X + Y = 278 X + Y = 30 1.5*X + 4*Y = 792 3*X + 4*Y = 100 8
Challenge 3: Significant Domain Variation Math Problems Ticket Sales An amusement park sells 2 kinds of tickets. Tickets for children cost $1.50. Adult tickets cost $4. On a A math test is worth 100 points certain day, 278 people entered and has 30 problems. Each the park. On that same day the problem is worth either 3 points admission fees collected totaled or 4 points. How many 4 point $792. How many children were problems are there? admitted on that day? How many adults were admitted? Solution: X + Y = 278 X + Y = 30 Move beyond lexicalized properties, e.g. syntax, discourse 1.5*X + 4*Y = 792 3*X + 4*Y = 100 9
Overview: Representation Space of possible Equation Types defined by generalizing labeled equations u 1 + u 2 = n 1 X + Y = 278 n 3 *u 1 + n 4 *u 2 = n 5 1.5*X + 4*Y = 792 For each word problem choose: System of u 1 + u 2 = n 1 n 3 *u 1 + n 4 *u 2 = n 5 equation types An amusement park sells 2 kinds of tickets. Tickets for children cost $1.50. Adult tickets cost $4. On a certain Alignment of equation day, 278 people entered the park. On that same day variables to text the admission fees collected totaled $792. How many children were admitted on that day? How many adults were admitted? 10
Overview: Representation Space of possible Equation Types defined by generalizing labeled equations u 1 + u 2 = n 1 X + Y = 278 n 3 *u 1 + n 4 *u 2 = n 5 1.5*X + 4*Y = 792 For each word problem choose: System of u 1 + u 2 = n 1 n 3 *u 1 + n 4 *u 2 = n 5 equation types An amusement park sells 2 kinds of tickets. Tickets for children cost $1.50. Adult tickets cost $4. On a certain Alignment of equation day, 278 people entered the park. On that same day variables to text the admission fees collected totaled $792. How many children were admitted on that day? How many adults were admitted? Solve resulting equations to get final answer 11
Overview: Model System of equation types Informed by Highly Highly availability of Ambiguous Ambiguous good alignment Alignment of equation variables to text 12
Overview: Model System of equation types Alignment of equation variables to text Joint Log-Linear Model 13
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