Birch Algorithm With Solved Example
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BIRCH (balanced iterative reducing and clustering using hierarchies) is an unsupervised data-mining algorithm used to perform hierarchical clustering over, particularly large data sets.
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Birch Algorithm With Solved Example
- 1. Solved Example Birch Algorithms Balanced Iterative Reducing And Clustering Using Hierarchies Dr. Kailash Shaw & Dr. Sashikala Mishra Symbiosis International University.
- 2. Introduction BIRCH (balanced iterative reducing and clustering using hierarchies) is an unsupervised data-mining algorithm used to perform hierarchical-clustering over particularly large data-sets. • The BIRCH algorithm takes as input a set of N data points, represented as real-valued vectors, and a desired number of clusters K. It operates in four phases, the second of which is optional. tree, while removing outliers and grouping crowded subclusters into larger ones. • Phase 1: Load data into memory Scan DB and load data into memory by building a CF tree. If memory is exhausted rebuild the tree from the leaf node. • Phase 2: Condense data Resize the data set by building a smaller CF tree Remove more outliers Condensing is optional • Phase 3: Global clustering Use existing clustering algorithm (e.g. KMEANS, HC) on CF entries • Phase 4: Cluster refining Refining is optional Fixes the problem with CF trees where same valued data points may be assigned to different leaf entries.
- 3. Example Let Have Following Data X1=(3,4), x2= (2,6), x3=(4,5), x4=(4,7), x5=(3,8), x6=(6,2), x7=(7,2), x8=(7,4), x9=(8,4), x10=(7,9) Cluster the Above Data Using BIRCH Algorithm, , considering T<1.5, and Max Branch = 2 For each Data Point we need to evaluate Radius and Cluster Feature: ->Consider Data Pint (3,4): As it is alone in the Feature map, Hence 1. Radius = 0 2. Cluster Feature CF1 <N, LS, SS> N = 1 as there is now one data point under consideration. LS = Sum of Data Point under consideration = (3,4) SS = Square Sum of Data Point Under Consideration = (32, 42)=(9,16) 3. Now construct the Leaf with Data Point X1 and Branch as CF1. CF1 <1, (3,4), (9,16)> Leaf X1 = (3, 4)
- 4. Example Let Have Following Data X1=(3,4), x2= (2,6), x3=(4,5), x4=(4,7), x5=(3,8), x6=(6,2), x7=(7,2), x8=(7,4), x9=(8,4), x10=(7,9) Cluster the Above Data Using BIRCH Algorithm, considering T<1.5, and Max Branch = 2 For each Data Point we need to evaluate Radius and Cluster Feature: ->Consider Data Pint x2 = (2,6): 1. Linear Sum LS = (3,4) + (2,6) = (5,10) 2. Square Sum SS = (32+22 , 42+62) =(13, 52) Now Evaluate Radius considering N=2 𝑅 = 𝑆𝑆−𝐿𝑆2/𝑁 𝑁 = (13,52)−(5,10)2/2 2 = (13,52)−(25,100)/2 2 = (13,52)−(12.5,50) 2 = 6.5,26 − (6.25,25) = (0.25,1) =(0.5, 1)<T As (0.25,1) < (T, T), hence X2 will cluster with Leaf X1. 2. Cluster Feature CF1 <N, LS, SS> = <2,(5,10),(13,52)> N = 2 as there is now two data point under CF1. LS = (3,4) + (2,6) = (5,10) SS = (32+22 , 42+62) =(13, 52) CF1 <1, (5,10), (13,52)> Leaf X1 = (3, 4), X2 = (2,6)
- 5. Example Let Have Following Data X1=(3,4), x2= (2,6), x3=(4,5), x4=(4,7), x5=(3,8), x6=(6,2), x7=(7,2), x8=(7,4), x9=(8,4), x10=(7,9) Cluster the Above Data Using BIRCH Algorithm, considering T<1.5, and Max Branch = 2 For each Data Point we need to evaluate Radius and Cluster Feature: ->Consider Data Pint x3 = (4,5) on CF1: 1. Linear Sum LS = (4,5) + (5,10) = (9,15) 2. Square Sum SS = (42+13 , 52 + 52) =(29, 77) Now Evaluate Radius considering N=3 𝑅 = 𝑆𝑆−𝐿𝑆2/𝑁 𝑁 = (29,77)−(9,15)2/3 3 =(0.47, 0.4714)<T As (0.47, 0.471) < (T, T), hence X3 will cluster with Leaf (X1, x2). 2. Cluster Feature CF1 <N, LS, SS> = <3,(9,15),(29,77)> N = 3 as there is now Three data point under CF1. LS = (4,5) + (5,10) = (9,15) SS = (42+13 , 52 + 52) =(29, 77) CF1 <1, (9,15), (29,77)> Leaf X1 = (3, 4), X2 = (2,6), X3 = (4,5)
- 6. Example Let Have Following Data X1=(3,4), x2= (2,6), x3=(4,5), x4=(4,7), x5=(3,8), x6=(6,2), x7=(7,2), x8=(7,4), x9=(8,4), x10=(7,9) Cluster the Above Data Using BIRCH Algorithm, considering T<1.5, and Max Branch = 2 For each Data Point we need to evaluate Radius and Cluster Feature: ->Consider Data Pint x4 = (4,7) on CF1: 1. Linear Sum LS = (4,7) + (9,15) = (13,22) 2. Square Sum SS = (42+29 , 72 + 77) =(45, 126) Now Evaluate Radius considering N=4 𝑅 = 𝑆𝑆−𝐿𝑆2/𝑁 𝑁 = (45,126)−(13,22)2/4 4 =(0.41, 0.55) As (0.41, 0.55) < (T, T), hence X4 will cluster with Leaf (X1, x2, x3). 2. Cluster Feature CF1 <N, LS, SS> = <4,(13,22),(45,126)> N = 4 as there is now four data point under CF1. LS = (4,7) + (9,15) = (13,22) SS = (42+29 , 72 + 77) =(45, 126) CF1 <1, (13,22), (45,126)> Leaf X1 = (3, 4), X2 = (2,6), X3 = (4,5), X4 = (4,7)
- 7. Example Let Have Following Data X1=(3,4), x2= (2,6), x3=(4,5), x4=(4,7), x5=(3,8), x6=(6,2), x7=(7,2), x8=(7,4), x9=(8,4), x10=(7,9) Cluster the Above Data Using BIRCH Algorithm, considering T<1.5, and Max Branch = 2 For each Data Point we need to evaluate Radius and Cluster Feature: ->Consider Data Pint x5 = (3,8) on CF1: 1. Linear Sum LS = (3,8) + (13,22) = (16,30) 2. Square Sum SS = (32+45 , 82 + 126) =(54, 190) Now Evaluate Radius considering N=5 𝑅 = 𝑆𝑆−𝐿𝑆2/𝑁 𝑁 = (54,190)−(16,30)2/5 5 =(0.33, 0.63) As (0.33, 0.63) < (T, T), hence X5 will cluster with Leaf (X1, x2, x3, x4). 2. Cluster Feature CF1 <N, LS, SS> = <5,(16,30),(54,190)> N = 5 as there is now four data point under CF1. CF1 <5,(16,30),(54,190)> Leaf X1 = (3, 4), X2 = (2,6), X3 = (4,5), X4 = (4,7) X5 = (3,8)
- 8. Example Let Have Following Data X1=(3,4), x2= (2,6), x3=(4,5), x4=(4,7), x5=(3,8), x6=(6,2), x7=(7,2), x8=(7,4), x9=(8,4), x10=(7,9) Cluster the Above Data Using BIRCH Algorithm, considering T<1.5, and Max Branch = 2 For each Data Point we need to evaluate Radius and Cluster Feature: ->Consider Data Pint x6 = (6,2) on CF1: 1. Linear Sum LS = (6,2) + (16,30) = (22,32) 2. Square Sum SS = (62+54 , 22 + 190) =(90, 194) Now Evaluate Radius considering N=6 𝑅 = 𝑆𝑆−𝐿𝑆2/𝑁 𝑁 = (90,194)−(22,32)2/6 6 =(1.24, 1.97) As (1.24, 1.97) < (T, T), False. hence X6 will Not form cluster with CF1. CF1 will remain as it was in previous step. And New CF2 with leaf x6 will be created. 2. Cluster Feature CF2 <N, LS, SS> = <1,(6,2),(36,4)> N = 1 as there is now one data point under CF2. LS = (6,2) SS = (62, 22)= (36,4) CF1 <5,(16,30),(54,190)> CF2 <1,(6,2),(36,4)> Leaf X1 = (3, 4), X2 = (2,6), X3 = (4,5), X4 = (4,7) X5 = (3,8) Leaf X6 = (6, 2),
- 9. Example Let Have Following Data X1=(3,4), x2= (2,6), x3=(4,5), x4=(4,7), x5=(3,8), x6=(6,2), x7=(7,2), x8=(7,4), x9=(8,4), x10=(7,9) Cluster the Above Data Using BIRCH Algorithm, considering T<1.5, and Max Branch = 2 For each Data Point we need to evaluate Radius and Cluster Feature: ->Consider Data Pint x7 = (7,2). As There are Two Branch CF1 and CF2 hence we need to find with which branch X7 is nearer, then with that leaf radius will be evaluated. With CF1 = LS/N= (16,30)/5=(8,6) As there are N=5 Data Point With CF2 = LS/N= (6,2)/1=(6,2) As there is N=1 Data Point Now x7 is closer to (6,2) then (8,6). Hence X7 will calculate radius with CF2. 1. Linear Sum LS = (7,2) + (6,2) = (13,4) 2. Square Sum SS = (72+36 , 22 + 4) =(85, 8) Now Evaluate Radius considering N=2 𝑅 = 𝑆𝑆−𝐿𝑆2/𝑁 𝑁 = (85,8)−(13,4)2/2 2 =(0.5, 0) As (0.5, 0) < (T, T), True. hence X7 will form cluster with CF2 2. Cluster Feature CF2 <N, LS, SS> = <2,(13,4),(85,8)> N = 2 as there is now two data point under CF2. LS = (7,2) + (6,2) = (13,4) SS = (72+36 , 22 + 4) =(85, 8) CF1 <5,(16,30),(54,190)> CF2 <2,(13,4),(85,8)> Leaf X1 = (3, 4), X2 = (2,6), X3 = (4,5), X4 = (4,7) X5 = (3,8) Leaf X6 = (6, 2), X7=(7,2)
- 10. Example Let Have Following Data X1=(3,4), x2= (2,6), x3=(4,5), x4=(4,7), x5=(3,8), x6=(6,2), x7=(7,2), x8=(7,4), x9=(8,4), x10=(7,9) Cluster the Above Data Using BIRCH Algorithm, considering T<1.5, and Max Branch = 2 ->Consider Data Pint x8 = (7,4). As There are Two Branch CF1 and CF2 hence we need to find with which branch X8 is nearer, then with that leaf, radius will be evaluated. With CF1 = LS/N= (16,30)/5=(8,6) As there are N=5 Data Point With CF2 = LS/N= (13,4)/2=(6.5,2) As there is N=2 Data Point Now x8 is closer to (6.5,2) then (8,6). Hence X8 will calculate radius with CF2. 1. Linear Sum LS = (7,4) + (13,4) = (20,8) 2. Square Sum SS = (72+85 , 42 + 8) =(134, 24) Now Evaluate Radius considering N=3 𝑅 = 𝑆𝑆−𝐿𝑆2/𝑁 𝑁 = (134,24)−(20,8)2/3 3 =(0.47, 0.94) As (0.47, 94) < (T, T), True. hence X8 will form cluster with CF2 2. Cluster Feature CF2 <N, LS, SS> = <3,(20,8),(134,24)> N = 3 as there is now two data point under CF2. LS (7,4) + (13,4) = (20,8) SS = (134,24) CF1 <5,(16,30),(54,190)> CF2 <3,(20,8),(134,24)> Leaf X1 = (3, 4), X2 = (2,6), X3 = (4,5), X4 = (4,7) X5 = (3,8) Leaf X6 = (6, 2), X7=(7,2) , X8 = (7,4)
- 11. Example Let Have Following Data X1=(3,4), x2= (2,6), x3=(4,5), x4=(4,7), x5=(3,8), x6=(6,2), x7=(7,2), x8=(7,4), x9=(8,4) , x10=(7,9) Cluster the Above Data Using BIRCH Algorithm, considering T<1.5, and Max Branch = 2 ->Consider Data Pint x9 = (8,4). As There are Two Branch CF1 and CF2 hence we need to find with which branch X9 is nearer, then with that leaf, radius will be evaluated. With CF1 = LS/N= (16,30)/5=(8,6) As there are N=5 Data Point With CF2 = LS/N= (20,8)/3=(6.6,2.6) As there is N=3 Data Point Now x9 is closer to (6.6,2.6) then (8,6). Hence X8 will calculate radius with CF2. 1. Linear Sum LS = (8,4) + (20,8) = (28,12) 2. Square Sum SS = (82+134 , 42 + 24) =(198, 40) Now Evaluate Radius considering N=4 𝑅 = 𝑆𝑆−𝐿𝑆2/𝑁 𝑁 = (198,40)−(28,12)2/4 4 =(0.70, 1) As (0.7, 1) < (T, T), True. hence X9 will form cluster with CF2 2. Cluster Feature CF2 <N, LS, SS> = <4,(28,12),(198,40)> N = 4 as there is now four data point under CF2. LS = (28,12) SS = (198,40) CF1 <5,(16,30),(54,190)> CF2 <4,(28,12),(198,40)> Leaf X1 = (3, 4), X2 = (2,6), X3 = (4,5), X4 = (4,7) X5 = (3,8) Leaf X6 = (6, 2), X7=(7,2) , X8 = (7,4), X9 = (8,4)
- 12. Example Let Have Following Data X1=(3,4), x2= (2,6), x3=(4,5), x4=(4,7), x5=(3,8), x6=(6,2), x7=(7,2), x8=(7,4), x9=(8,4) , x10=(7,9) Cluster the Above Data Using BIRCH Algorithm, considering T<1.5, and Max Branch = 2 ->Consider Data Pint x10 = (7,9). As There are Two Branch CF1 and CF2 hence we need to find with which branch X9 is nearer, then with that leaf, radius will be evaluated. With CF1 = LS/N= (16,30)/5=(8,6) As there are N=5 Data Point With CF2 = LS/N= (28,12)/4=(7,3) As there is N=4 Data Point Now x10 is closer to (8,6) then (7,3). Hence X10 will calculate radius with CF1. 1. Linear Sum LS = (7,9) + (16,30) = (23,39) 2. Square Sum SS = (72+54 , 92 + 190) =(103, 271) Now Evaluate Radius considering N=6 𝑅 = 𝑆𝑆−𝐿𝑆2/𝑁 𝑁 = (103,271)−(23,39)2/6 6 =(1.57, 1.70) As (1.57, 1.70) < (T, T), False. hence X10 will become new leaf and Create new cluster feature CF3. But in a Branch only two CF is allowed hence Branch will Split. 2. Cluster Feature CF3 <N, LS, SS> = <1,(7,9),(49,81)> CF1 <5,(16,30),(54,190)> CF2 <4,(28,12),(198,40)> Leaf X1 = (3, 4), X2 = (2,6), X3 = (4,5), X4 = (4,7) X5 = (3,8) Leaf X6 = (6, 2), X7=(7,2) , X8 = (7,4), X9 = (8,4) CF12 <9,(44,42),(252,230)> CF3 <1,(7,9),(49,81)> CF3 <1,(7,9),(49,81)> Leaf X10= (7,9)
- 13. Thank You